This book is an indepth and modern presentation of important classical results in complex analysis and is suitable for
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English Pages 421 [422] Year 2019
Table of contents :
Contents
Preface
1. Complex numbers
2. Holomorphic functions
3. The complex integration
4. Sequences and series of holomorphic functions
5. Residue theory
6. Conformal representations
7. Solutions to the chapterwise exercises
Bibliography
Index
Teodor Bulboacǎ, Santosh B. Joshi, and Pranay Goswami Complex Analysis
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Teodor Bulboacǎ, Santosh B. Joshi, and Pranay Goswami
Complex Analysis 
Theory and Applications
Mathematics Subject Classification 2010 30XX Authors Prof. Dr. Habil. Teodor Bulboacǎ BabeşBolyai University Faculty of Mathematics and Computer Science Str. Mihail Kogălniceanu Nr. 1 400084 ClujNapoca Romania [email protected] Prof. Dr. Santosh B. Joshi Walchand College of Engineering Department of Mathematics 416415 Maharastra India [email protected] Prof. Dr. Pranay Goswami Ambedkar University Delhi Kashmere Gate Campus Lothian Road, Kashmere Gate 110006 Delhi India [email protected]
ISBN 9783110657821 eISBN (PDF) 9783110657869 eISBN (EPUB) 9783110658033 Library of Congress Control Number: 2019938423 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2019 Walter de Gruyter GmbH, Berlin/Boston Cover image: Lokal_Profil (Grafic »conformal_grid«) – Michael Bernhart (surface) Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com
Contents Preface  XI 1 1.1 1.2 1.3 1.3.1 1.4 1.5 1.5.1 1.5.2 1.6
Complex numbers  1 The field of the complex numbers  1 The complex plane  3 The topological and metric structure of the complex plane  6 Basic definitions and notation  6 Complex function, limits, continuity  9 The compactified complex plane  10 The geometric interpretation of the φ function  12 ̂  12 The topological structure of ℂ Exercises  14
2 2.1 2.2 2.3 2.3.1 2.4 2.5 2.5.1 2.5.2 2.5.3 2.5.4 2.6 2.6.1 2.7 2.8 2.8.1 2.8.2 2.8.3 2.9 2.9.1 2.9.2 2.9.3 2.9.4
Holomorphic functions  17 The derivative of the real valued complex functions  17 The differentiability of a complex function  19 The derivative of a complex function  23 The properties of the derivative  25 The geometric interpretation of the derivative  29 Entire functions  33 The polynomial function  33 The exponential function  33 Complex trigonometric functions  34 Complex hyperbolic functions  35 Bilinear transforms  35 Decompositions in elementary functions  36 The Möbiustype groups  39 Multivalued functions  43 The logarithmic function  43 Inverse trigonometric functions  45 The power function  46 Exercises  47 Real variable complex functions  47 The derivative of a complex function  47 Entire functions  49 Bilinear transforms  49
3
The complex integration  51
VI  Contents 3.1 3.1.1 3.1.2 3.2 3.2.1 3.3 3.3.1 3.3.2 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.10.1 3.10.2 3.10.3 3.10.4 3.10.5
The homotopic theory of the paths  51 Simply connected domains  55 Functions of bounded variation and paths  58 The complex integral  60 The Riemann–Stieltjes integral for complex valued functions  60 The Cauchy theorem  66 The connection between the integral and the primitive function  66 The Cauchy theorem  74 The Cauchy formula for the disc  79 The analytical branches of multivalued functions  82 The index of a path (curve) with respect to a point  84 Cauchy formula for closed curves  87 Some consequences of Cauchy formula  88 Schwarz and Poisson formulas  91 Exercises  94 The complex integral  94 The Cauchy theorem  96 The Cauchy formula for the disc  97 Some consequences of Cauchy formula  101 Multivalued functions analytical branches  102
4 Sequences and series of holomorphic functions  105 4.1 Sequences of holomorphic functions  105 4.2 Series of functions  107 4.3 Power series  108 4.4 The analyticity of holomorphic functions  111 4.5 The zeros of holomorphic functions  114 4.6 The maximum principle of the holomorphic functions  117 4.7 Laurent series  122 4.8 Isolated singular points  127 4.9 Meromorphic functions  131 4.10 Exercises  134 4.10.1 Power series  134 4.10.2 Taylor and Laurent series  135 4.10.3 Isolated singular points  138 4.10.4 The module maximum of the holomorphic functions  139 5 5.1 5.2 5.3
Residue theory  141 Residue theorem  141 Applications of the residue theorem to the calculation of the integrals  144 The study of meromorphic functions with the residue theorem  158
Contents  VII
5.4 5.4.1 5.4.2 5.4.3 5.4.4
Exercises  165 Residue theorem  165 Applications of the residue theorem to the calculation of the trigonometric integrals  168 Applications of the residue theorem to the calculation of the improper integrals  169 The study of meromorphic functions using the residue theorem  169
6 6.1 6.2 6.3 6.4 6.5
Conformal representations  171 Special classes of holomorphic functions  171 Univalent functions  175 The problem of conformal representation  179 The Riemann mapping theorem  181 Exercises  187
7 7.1 7.2 7.3 7.4 7.5 7.6
Solutions to the chapterwise exercises  191 Solutions to the exercises of Chapter 1  191 Solutions to the exercises of Chapter 2  199 Solutions to the exercises of Chapter 3  218 Solutions to the exercises of Chapter 4  277 Solutions to the exercises of Chapter 5  322 Solutions to the exercises of Chapter 6  380
Bibliography  405 Index  407

Dedicated to the memory of Professor Petru T. Mocanu (1931–2016)
Preface Complex analysis, as we know it now, is the culmination of over 500 years of mathematical development that has had tremendous influence in mathematics, physics and engineering. The numbers we now know as “complex” (a most unfortunate name due to C. F. Gauss) made their first appearance through the use of square roots of negative numbers in methods for the solution of cubic and quartic equations1 in the works of N. F. Tartaglia and G. Cardano and proved their value in “predicting” the correct values of roots. R. Descartes, who also coined the concept of a real number, named such square roots of negative numbers imaginary, since they could “only” be imagined. Leonhard Euler made extensive use of complex numbers and also introduced these2 in his textbooks. However, Euler also had only a vague geometric notion of complex numbers and the complex plane. The geometrization of the complex numbers had to wait until the end of the eighteenth century by the separate insights of JR. Argand, C. Wessel and Gauss. Much of the development of classical complex analysis, the topic of this book, occurred in the nineteenth century in the hands of AL. Cauchy and B. Riemann followed by the even more geometrical work of F. Klein, H. Poincare and many others. This book is an indepth and modern presentation of important classical results in complex analysis and is suitable for a first course in this topic. The organization of the material is as one would find it in a typical syllabus for an undergraduate course in complex analysis. This book is an outgrowth of, and has been tested during the courses we taught at BabeşBolyai University, Romania, Walchand College of Engineering, India and Ambedkar University Delhi, India. The level of difficulty of the material increases gradually from chapter to chapter. Each chapter contains exercises with solutions and applications of the results. We have strived for diversity in solution methods. Chapter 1 introduces the concept of complex numbers, and their arithmetic and geometric properties. We use stereographic projection to introduce the onepoint compactification of the complex plane, the Riemann sphere. Chapter 2 studies complex valued function, and various notions of differentiability of such functions and culminates with the concept of a holomorphic function. Our presentation of their basic properties ends with the Cauchy–Riemann equations. This chapter ends with a thorough overview of elementary entire functions and Möbius transformations that will be needed in the remaining chapters. Chapter 3 starts with the definitions of paths and complex integrals and is followed by Cauchy’s theorem and its consequences: the fundamental theorem of alge1 See, for example, Nahin, Paul J. (1998), An Imaginary Tale: The Story of √−1, Princeton University Press. 2 and the notation i for √−1 to reduce confusion in the applications of rules such as √ab = √a√b. https://doi.org/10.1515/9783110657869201
XII  Preface bra, the Cauchy integral formula for holomorphic functions defined on the disc and Morera’s theorem establishing sufficient conditions for holomorphy. The chapter ends with many applications including the theory of multivalent functions. Chapter 4 characterizes holomorphic function by their local analytic properties via power series expansions. We present important theorems on the zeroes of holomorphic functions, the uniqueness of holomorphic functions, the maximum modulus principle, the Schwarz lemma, Laurent series expansions, isolated singular points and some basic results on meromorphic functions. Chapter 5 develops the theory of residues and its principal applications: the computation of a variety of trigonometric and improper integrals. We also apply the theory of residues to the study of zeros and poles of meromorphic functions, the principle of the argument and Rouché’s theorem. This chapter ends with the open mapping theorem for nonconstant holomorphic functions and its topological consequences. Chapter 6 starts with the fundamental theorems of Montel, Vitali and Hurwitz and is then devoted to the topic of conformal mappings, univalent functions and the Riemann mapping theorem. Chapter 7 contains the solutions of all exercises that appeared at the end of the previous chapters. This book evolved from lectures delivered at universities in Romania and India, hence we would especially thank to students since it was their reaction for the course of “Complex Analysis” that made us to write this book. We would like to thank to our family members for their encouragement, timely help, and patience. We consider that this book could be used by the students that need a basic (and not only) preparation in complex analysis, and for all that are interested in this field, and any remarks and comments for improving the content of the book are welcome.
1 Complex numbers 1.1 The field of the complex numbers Let ℝ denote the set of all real numbers, where we defined the usual addition and multiplication, such that this set together with these two operations will be a field. Definition 1.1.1. 1. Let ℝ2 = ℝ × ℝ denote the set of all real ordered couples, i. e., (x, y) ∈ ℝ2 ⇔ x, y ∈ ℝ. 2. Define in the ℝ2 set the addition, and respectively, multiplication binary operations as follows: + : ℝ2 × ℝ2 → ℝ2 ,
(x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 );
⋅:ℝ ×ℝ →ℝ ,
(x1 , y1 ) ⋅ (x2 , y2 ) = (x1 x2 − y1 y2 , x1 y2 + y1 x2 ).
2
3.
2
2
The above defined algebraic structure (ℝ2 , +, ⋅) is called the set of complex numbers, briefly denoted by the symbol ℂ. We will use the notation ℂ∗ = ℂ \ {(0, 0)}.
Theorem 1.1.1. 1. (ℂ, +) is an Abelian group; the zero element is (0, 0). 2. (ℂ∗ , ⋅) is an Abelian group; the unit element is (1, 0). 3. The multiplication “⋅” is distributive with respect to the addition “+”. Proof. The points 1 and 3 can be proved directly. 2. Using simple computations, we may check that the operation “⋅” is associative and commutative. Supposing that there exists a unitary element, it follows easily that it will be (1, 0). We will prove that every nonzero element has an inverse. Let (x1 , y1 ) ∈ ℂ∗ be an arbitrary. Then (x1 , y1 ) ≠ (0, 0) ⇔ x12 + y12 ≠ 0, hence ∃(x, y) ∈ ℂ∗
such that (x1 , y1 ) ⋅ (x, y) = (1, 0)
⇔ ∃(x, y) ∈ ℂ∗
such that (x1 x − y1 y, x1 y + y1 x) = (1, 0) ⇔ {
x1 x − y1 y = 1 y1 x + x1 y = 0.
Since det A = x12 + y12 ≠ 0, because (x1 , y1 ) ≠ (0, 0), the above system is compatible (where x and y are unknown) and has a unique solution. −y1 1 This will be (x, y) = ( x2x+y 2 , x 2 +y 2 ), hence for all elements of the form (x1 , y1 ) ≠ (0, 0) 1
1
1
1
there exists an inverse given by the above relation, so ℂ is a commutative field.
The proof of the following result is straightforward, so it is left to the reader. https://doi.org/10.1515/9783110657869001
2  1 Complex numbers Theorem 1.1.2. The subset ℝ × {0} = {(x, 0) : x ∈ ℝ} ⊂ ℂ is a subfield of the ℂ field with respect to the above operations. The function φ : ℝ → ℝ × {0} ⊂ ℂ defined by φ(x) = (x, 0) is an isomorphism between the above field. Conventions. 1. According to the Theorem 1.1.2, we will denote the subfield ℝ × {0} of the ℂ field by ℝ, so we will simplify our notation. 2. We will introduce the notation i = (0, 1) ∈ ℂ. 3. Using the above convention, we get ℂ = ℝ + iℝ. Theorem 1.1.3. Any arbitrary complex number z = (x, y) may be written in the form x + iy, where i2 = −1. Proof. A simple calculation shows that (0, 1) ⋅ (0, 1) = (−1, 0) ⇒ i2 = −1, and hence z = (x, y) = (x, 0) + (0, 1) ⋅ (y, 0) = x + iy. Definition 1.1.2. 1. The form z = x + iy of the complex number z = (x, y), is called the algebraic form of the given complex number. 2. The numbers x, y, x − iy, and √x2 + y2 are called the real part, the imaginary part, the conjugate and the module, respectively, of the number z ∈ ℂ. 3. If z ∈ ℂ∗ , then the inverse of it, with respect to the multiplication, is denoted by z1 . Notation. x = Re z, y = Im z, x − iy = z, √x2 + y2 = z. Theorem 1.1.4. For any arbitrary numbers z, z1 , z2 ∈ ℂ, the next properties hold: 1. Re z = 21 (z + z), Im z = 2i1 (z − z); 2. 3.
z1 + z2 = z1 + z2 , z1 z2 = z1 z2 , z = z;
−z ≤ Re z ≤ z, −z ≤ Im z ≤ z, z = z;
4. z2 = zz, 5.
1 z
=
z z2
if z ≠ 0;
z = 0 ⇔ z = 0, z1 z2  = z1 z2 , z1 + z2  ≤ z1  + z2 .
Proof. The proof of the first four points is easy and straightforward, and we will provide the proof of the last point only. We can see that z1 + z2 2 = (z1 + z2 )(z1 + z2 ) = (z1 + z2 )(z1 + z2 ) = z1 z1 + z1 z2 + z2 z1 + z2 z2 = z1 2 + z2 2 + 2 Re(z1 z2 ).
1.2 The complex plane 
3
Since 2 Re(z1 z2 ) ≤ 2z1 z2  = 2z1 z2  = 2z1 z2 
⇒ z1 + z2 2 ≤ z1 2 + z2 2 + 2z1 z2  = (z1  + z2 )2 ⇒ z1 + z2  ≤ z1  + z2 .
1.2 The complex plane Let xOy an orthogonal axes system on a plane. We will identify the set ℝ2 with the plane xOy. Since ℝ2 and ℂ are identical, it follows that we could identify the complex set ℂ with the plane xy, such that x = ℝ and y = iℝ. Thus, the axis x will be called the real axis, and y will be called the imaginary axis. To every point from xy, it corresponds one and only one complex number of ℂ, and vice versa. Thus, there exists a bijective correspondence between the sets ℝ2 and ℂ; i. e., (x, y) ←→ x + iy. Letting z ∈ ℂ \ {0} = ℂ∗ , z = x + iy, from the relation y x z +i = z √x2 + y2 √x2 + y2
combined with elementary knowledge of trigonometry, it follows that the system cos θ = 2x 2 { √x +y { { { sin θ = y √x2 +y2 { is compatible, with respect to the θ unknown. This fact allows us to give the following definition. Definition 1.2.1. 1. Let z ∈ ℂ∗ . Every solution θ of the equation cos θ + i sin θ = 2.
z z
is called the argument of the complex number z. The equation has a unique solution in the interval (−π, π], and it is called the principal argument of the complex number z, and is denoted by arg z. If we denote by Arg z the set of all the arguments of the number z, then Arg z = {arg z + 2kπ : k ∈ ℤ}.
3.
The correspondence z → Arg z is a multivalued function from ℂ∗ to ℝ.
Theorem 1.2.1 (Properties of the argument). For any arbitrary numbers z1 , z2 ∈ ℂ∗ , we have:
4  1 Complex numbers 1. 2.
Arg (z1 z2 ) = Arg z1 + Arg z2 ; Arg zz1 = Arg z1 − Arg z2 . 2
Proof. Letting θj ∈ Arg zj , j = 1, 2, then cos θj + i sin θj =
zj zj 
,
j = 1, 2.
Using simple trigonometric properties, we obtain that z1 z2 = (cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 ) = cos(θ1 + θ2 ) + i sin(θ1 + θ2 ) z1  z2  and z1 z2 zz = 1 2 ⇒ θ1 + θ2 ∈ Arg z1 z2 , z1  z2  z1 z2  hence Arg z1 + Arg z2 ⊂ Arg (z1 z2 ).
(1.1)
Similarly, we deduce that z1 z2  zz1  2
=
z1 z1  z2 z2 
=
cos θ1 + i sin θ1 (cos θ1 + i sin θ1 )(cos θ2 − i sin θ2 ) = cos θ2 + i sin θ2 cos2 θ2 + sin2 θ2
= cos(θ1 − θ2 ) + i sin(θ1 − θ2 ),
hence θ1 − θ2 ∈ Arg (
z1 ), z2
or Arg z1 − Arg z2 ⊂ Arg (
z1 ). z2
(1.2)
Let θ ∈ Arg (z1 z2 ) be arbitrary, and consider an arbitrary angle θ1 ∈ Arg z1 . From the relation (1.2), we get θ − θ1 ∈ Arg (
z1 z2 ) = Arg z2 , z1
and thus θ ∈ Arg z2 + θ1 ⊂ Arg z1 + Arg z2 .
(1.3)
1.2 The complex plane 
5
In conclusion, from the relations (1.1) and (1.3) it follows the equality of the first point. Since Arg z1 = Arg ( we have ∀θ ∈ Arg hence
z1 z (1.1) z ) = Arg 1 + Arg z2 , z2 2 z2
z1 , ∃θ1 ∈ Arg z1 , ∃θ2 ∈ Arg z2 z2
such that θ1 = θ + θ2 .
θ = θ1 − θ2 ∈ Arg z1 − Arg z2 , and thus Arg
z1 ⊂ Arg z1 − Arg z2 . z2
(1.4)
From the relations (1.2) and (1.4), it follows the equality of the second point. Definition 1.2.2. If z ∈ ℂ∗ , then z = z(cos θ + i sin θ),
where θ ∈ Arg z
(1.5)
is called the trigonometric form of the complex number z. Remarks 1.2.1. 1. The set {z ∈ ℂ : z = Re z} is the real axis. 2. The set {z ∈ ℂ : z = i Im z} is the imaginary axis. 3. The set of those points z ∈ ℂ, such that Re z > 0, Re z < 0, Im z > 0, Im z < 0 is called the right, the left, the upper, and respectively the lower open halfplane. 4. The equation of a line of the complex plane is z = a + bt, 5.
where a ∈ ℂ, b ∈ ℂ∗ and t ∈ ℝ (t runs all over the ℝ).
This line contains the point a and it is parallel with the vector b. The above equation of the line can be written in the form z−a Im = 0. b
The set {z ∈ ℂ : Im z−a > 0} is called the right halfplane bounded by the above b line. 7. The set {z ∈ ℂ : Im z−a < 0} is called the left halfplane bounded by the above b line. 8. The angle of the lines z = a1 + b1 t and z = a2 + b2 t is given by Arg bb2 . It depends 1 on the considered order of the lines, and has a countable number of elements. 9. The set Sz0 (α1 , α2 ) = {z ∈ ℂ : α1 < arg(z − z0 ) < α2 } is an angular sector with the apex in z0 .
6.
6  1 Complex numbers
1.3 The topological and metric structure of the complex plane Let us define the function d : ℂ × ℂ → ℝ by the following formula: d(z1 , z2 ) = z1 − z2 . Theorem 1.3.1. The pair (ℂ, d) is a metric space, i. e., (m1 ) d(z1 , z2 ) ≥ 0, ∀z1 , z2 ∈ ℂ and d(z1 , z2 ) = 0 ⇔ z1 = z2 ; (m2 ) d(z1 , z2 ) = d(z2 , z1 ), ∀z1 , z2 ∈ ℂ; (m3 ) d(z1 , z3 ) ≤ d(z1 , z2 ) + d(z2 , z3 ), ∀z1 , z2 , z3 ∈ ℂ. Proof. The (m1 ) and (m2 ) properties are obvious, and it follows from the fact that z1 − z2  = 0 ⇔ z1 − z2 = 0. For (m3 ), it is well known that ∀w1 , w2 ∈ ℂ, we have w1 + w2  ≤ w1  + w2 .
(1.6)
Let w1 = z1 − z2 , w2 = z2 − z3 . Hence w1 + w2 = z1 − z3 , and replacing these numbers in the inequality (1.6), we get z1 − z3  ≤ z1 − z2  + z2 − z3 , which is in fact (m3 ). Remark 1.3.1. If we replace the complex numbers that appeared in the definition of the metric d by the corresponding couples of real numbers, we see that this metric represents the Euclidian distance of the Euclidian plane. Hence, the (ℂ, d) metric space can be identified with the couple consisting of the plane together with the Euclidian distance. 1.3.1 Basic definitions and notation 1.
If z0 ∈ ℂ and r > 0, then U(z0 ; r) = {z ∈ ℂ : z − z0  < r}
2. 3.
is the (open) disc with the center z0 and radius r. ̇ 0 ; r) = U(z0 ; r) \ {z0 }. We denote by U(z If z0 ∈ ℂ and 0 ≤ r1 < r2 , then U(z0 ; r1 , r2 ) = {z ∈ ℂ : r1 < z − z0  < r2 }
is the (open) ring with the center z0 and radius r1 and r2 . 4. The set G ⊂ ℂ is called an open set, if G = 0 or if ∀z ∈ G, ∃rz > 0 such that U(z; rz ) ⊂ G.
1.3 The topological and metric structure of the complex plane
5. 6.
 7
The set F ⊂ ℂ is called a closed set, if ℂ \ F is an open set. If z1 , z2 ∈ ℂ, then [z1 , z2 ] = {z ∈ ℂ : z = z1 + (z2 − z1 )t, t ∈ [0, 1]}
is called the closed segment that connects z1 and z2 . 7. For all A ⊂ ℂ, A ≠ 0, the couple (A, d) is a metric space. 8. In the (A, d) metric space, the neighborhood system of a point b is given by 𝒱 (b) = {V ⊂ A : ∃G ⊂ A, G is an open set in A, b ∈ G ⊂ V}.
9.
The set B ⊂ A is said to be a closed set in A (in the (A, d) metric space), if ∀(bn )n∈ℕ ⊂ B and ∃ lim bn = b ∈ A ⇒ b ∈ B. n→∞
10. The set A is closed and open in the (A, d) metric space. 11. The set A is called to be connected, if in the (A, d) metric space the set A is the only one set, excepted the empty set, that is closed and open. Example 1.3.1. The set A = {z1 , z2 }, with z1 ≠ z2 is not connected; also, the set A = U(2i; 1) ∪ ℝ, etc. The set A = U(i; 1) ∪ ℝ is connected. 12. The subset M ⊂ ℂ is said to be a domain, if M is an open and connected subset of ℂ. 13. The subset D ⊂ ℂ is said to be a starlike set with respect to the point z0 ∈ D, if [z0 , z] ⊂ D, ∀z ∈ D. 14. If the subset D ⊂ ℂ is starlike with respect to any arbitrary point z ∈ D, then D is called a convex set. 15. Let A ⊂ ℂ. We denote by A− the closure set of A, defined by A− = {z ∈ ℂ : U(z; r) ∩ A ≠ 0, ∀r > 0}. 16. Let A ⊂ ℂ. We denote by A the set of accumulation points of A, defined by ̇ r) ∩ A ≠ 0, ∀r > 0}. A = {z ∈ ℂ : U(z; It is easy to prove that A− = A ∪ A . 17. We denote by 𝜕A the boundary of the set A, defined by 𝜕A = A− ∩ (ℂ \ A)− . Exercise 1.3.1. 1. The subset A ⊂ ℂ is closed ⇔ A = A− . 2. If G ⊂ ℂ, then 𝜕(G− ) ⊂ 𝜕G. Give an example of an open set G ⊂ ℂ, such that 𝜕(G− ) ≠ 𝜕G.
8  1 Complex numbers Definition 1.3.1. Let A, B ⊂ ℂ be two nonempty sets (A, B ≠ 0). Then the distance between the sets A and B is given by d(A, B) = inf{d(a, b) : a ∈ A, b ∈ B}. Lemma 1.3.1. 1. z ∈ A− ⇔ d(z, A) = 0. 2. The subset A ⊂ ℂ is closed, if and only if d(z, A) = 0 ⇔ z ∈ A. Definition 1.3.2. The subset B ⊂ ℂ is said compact set, if such that ∃z ∗ = lim znj and z ∗ ∈ B.
∀(zn )n∈ℕ ⊂ B, ∃(znj )j∈ℕ ⊂ (zn )n∈ℕ
j→∞
Lemma 1.3.2. Every compact set is a closed set. Example 1.3.2. Let us consider the set A = ℝ ⊂ ℂ,
A = {(x, 0) : x ∈ ℝ}
and B = {(x, ex ) : x ∈ ℝ} = {x + iex : x ∈ ℝ} ⊂ ℂ. Thus, A ∩ B = 0, A and B are closed sets, but d(A, B) = 0. Exercise 1.3.2. If G ⊂ ℂ is an open set, then 𝜕G = G− \ G. Theorem 1.3.2. If A ⊂ ℂ is a compact set, B ⊂ ℂ is a closed set, and A ∩ B = 0, then d(A, B) > 0. Proof. Using the above notation, suppose that d(A, B) = 0. Then ∀n ∈ ℕ∗ , ∃an ∈ A, ∃bn ∈ B,
such that d(an , bn )
0 −
⇒ z ∉ (ℂ \ G) ⇒ z ∉ 𝜕G,
such that U(z; r) ⊂ G ⇒ U(z; r) ∩ (ℂ \ G) = 0
which is a contradiction.
Since K ⊂ G and 𝜕G ∩ G = 0 ⇒ K ∩ 𝜕G = 0, where K is a compact and 𝜕G is closed. According to Theorem 1.3.2, we conclude that d(K, 𝜕G) > 0. Corollary 1.3.2. If K ⊂ G and K is a compact set, K ≠ 0, and G ⊂ ℂ is an open set, then ∃r > 0 (where r depends only of K and G), such that ∀z ∈ K, U(z; r) ⊂ G.
1.4 Complex function, limits, continuity First, we will consider real variable complex functions. If [a, b] ⊂ ℝ, then a function f : [a, b] → ℂ has the form f (t) = α(t) + iβ(t) where α, β : [a, b] → ℝ are real valued functions. The limits and the continuity of the function f reduces to the limits and respectively to the continuity of the realvalued functions α and β. If A ⊂ ℂ, then the function f : A → ℂ may be written in the form f (z) = u(z) + iv(z),
where z → u(z), z → v(z)
are real valued functions defined on the complex subset A. Using the notation z = x + iy ∈ ℂ, with x, y ∈ ℝ, then w = f (z) = f (x + iy) = u(x, y) + iv(x, y), where u and v are two variables real functions. Definition 1.4.1. The function f : A → ℂ has the limit l ∈ ℂ at the point a ∈ A (where A represents the set of all the accumulation points of A), written limz→a f (z) = l, if ∀ε > 0, ∃η = η(ε) > 0
such that ∀z ∈ A and 0 < z − a < η ⇒ f (z) − l < ε.
10  1 Complex numbers We may easily prove that limz→a f (z) = l is equivalent to each of the following relations: lim f (z) = l;
(i)
z→a
(ii)
z→a
lim Re f (z) = Re l
and
lim Im f (z) = Im l.
z→a
Let f : A → ℂ and z0 ∈ A ∩ A . Then, the function f is called to be continuous at the point z0 , if limz→z0 f (z) = f (z0 ), i. e., if ∀ε > 0, ∃η = η(ε) > 0 ∀z ∈ A and
such that
0 < z − z0  < η ⇒ f (z) − f (z0 ) < ε.
If the function f is continuous at every point of the set A, then we say that the function f is continuous on the set A. From the triangle inequality, f (z) − f (z0 ) ≤ f (z) − f (z0 ) it follows, that if f is continuous at the point z0 , then f  is also continuous at the point z0 , where f (z) = f (z). The reverse implication is not true!
1.5 The compactified complex plane As we will see in the following chapters, it is useful to introduce the notion of the ∞ point, which will “complete” the complex plane ℂ with a unique “point” denoted by ∞, that is situated to an “infinite large distance” from the origin (and from any other point of ℂ), and satisfies the properties: ̂ = ℂ ∪ {∞}; 1. ℂ 2. a + ∞ = ∞ + a = ∞, ∀a ∈ ℂ; ̂ \ {0}, (hence − ∞ = ∞); 3. a ⋅ ∞ = ∞ ⋅ a = ∞, ∀a ∈ ℂ a ̂ 4. 0 = ∞, ∀a ∈ ℂ \ {0}; 5.
a ∞
= 0, ∀a ∈ ℂ.
The next forms have no sense (are not welldefined): ∞ − ∞, 0 ⋅ ∞, 00 , Also, we have ∞ = +∞ (the real +∞).
∞ . ∞
̂ set, together with the addition operation “+” Definition 1.5.1. The above defined ℂ and the multiplication operation “⋅” is called the compactified complex plane. Conventions. ̂ (i. e., lines of ℂ together with ∞). 1. The ∞ is an element of every line of ℂ 2. The ∞ is not element of any circle.
1.5 The compactified complex plane
 11
Let us denote by S2 ⊂ ℝ3 , S2 = {(x, y, u) ∈ ℝ3 : x2 + y2 + u2 = 1} the two dimensional sphere (ball) (Figure 1.1). ̂ by Theorem 1.5.1. Let us define the function Φ : S2 → ℂ z = Φ(x, y, u) = {
x+iy , 1−u
∞,
if (x, y, u) ≠ (0, 0, 1); if (x, y, u) = (0, 0, 1).
Let N = (0, 0, 1) ∈ S2 , and denote by φ the restriction of the function Φ to S2 \ {N}, i. e., φ = ΦS2 \{N} . Then the function φ is a continuous bijection between S2 \ {N} and ℂ, and φ−1 is also continuous. Proof. 1. The function φ is a bijection. Thus, x + iy x + iy x − iy x2 + y2 1 − u2 1+u ⇒ z2 = zz = = . = = 2 1−u 1−u 1−u (1 − u) (1 − u)2 1 − u
z=
From the above relation, we get u=
z2 − 1 z2 + 1
and
1−u=
2 . z2 + 1
z−z =
2iy 1−u
Since z+z =
2x , 1−u
it follows that φ−1 (z) = (x, y, u) = ( 2. The function φ(x, y, u) = 3. The inverse function
z−z z2 − 1 z+z , , ), 1 + z2 i(1 + z2 ) z2 + 1
x+iy 1−u
∀z ∈ ℂ.
is continuous on the set S2 \ {N}.
φ−1 (z) = (x, y, u) = (
z2 − 1 z−z z+z , 2 ) , 2 2 1 + z i(1 + z ) z + 1
is continuous on ℂ. ̂ by Theorem 1.5.2. Let N = (0, 0, −1) ∈ S2 , and define the function Φ1 : S2 → ℂ z = Φ1 (x, y, u) = {
x+iy , 1+u
∞,
if (x, y, u) ≠ (0, 0, −1); if (x, y, u) = (0, 0, −1).
Let us denote by φ1 the restriction of the function Φ1 to S2 \ {N }, i. e., φ1 = Φ1 S2 \{N } . Then the function a φ1 is a continuous bijection between S2 \ {N } and ℂ, and φ−1 1 is also continuous.
12  1 Complex numbers
Figure 1.1: The two dimensional sphere (ball).
1.5.1 The geometric interpretation of the φ function If NP(x, y, u − 1), let (NP) :
X x
=
Y y
=
U−1 u−1
(Figure 1.1).
Let it intersects the line (NP) with the plane U = 0, and let (x1 , y1 , 0) be the coordinate of this intersection. Then x1 y1 −1 x y = = ⇒ x1 = , y1 = . x y u−1 1−u 1−u Identifying the plane U = 0 with the set of the complex numbers ℂ, we get z = x1 + iy1 =
x + iy = φ(x, y, u). 1−u
It follows that the function φ maps each point P ∈ S2 \ {N} into the point z = = φ(x, y, u) of the plane U = 0, which is called the stereographic x1 + iy1 = x+iy 1−u projection of the point P to the complex plane ℂ. ̂ 1.5.2 The topological structure of ℂ The neighborhood system of the ∞ point is given by all the sets V that satisfy the next property: ∃r > 0
such that Wr = {∞} ∪ {z ∈ ℂ : z > r} ⊂ V.
We will denote it by ℬ(∞) = {Wr : r > 0} the base of the neighborhood system of the ∞. The images of the Wr sets by the function Φ−1 are open spherical caps with the apexes in the “North pole” N. ̂ and Φ and Φ−1 are conTheorem 1.5.3. The function Φ is a bijection between S2 and ℂ, tinuous functions. Proof. The function ΦS2 \{N} = φ is a bijection between S2 \ {N} and ℂ; the functions φ and φ−1 are continuous. From Theorem 1.5.1, the function Φ is a bijection.
1.5 The compactified complex plane
 13
Now we will prove that the function Φ is continuous at the point N, while the function Φ−1 is continuous at the point ∞. Since
it follows that
x + iy 2 1 + u r2 − 1 , > r2 ⇔ u > 2 Φ(x, y, u) > r ⇔ = 1 − u 1−u r +1 Φ(x, y, u) → ∞ ⇔ u → 1 ⇔ (x, y, u) → N.
Using these inequalities, we have ̂ \ U(0; r)) = {(x, y, u) ∈ S2 : u > Φ−1 (ℂ
r2 − 1 }, r2 + 1
̂ \ U(0; r)) is a neighborhood of the point (0, 0, 1) in S2 . hence Φ−1 (ℂ Using the fact that lim Φ−1 (z) = lim (
z→∞
z→∞
z2 − 1 z−z z+z , , ) = (0, 0, 1) = Φ−1 (∞), 1 + z2 i(1 + z2 ) 1 + z2
the function Φ−1 is continuous at the point z = ∞. Remarks 1.5.1. The twodimensional sphere (ball) S2 together with the functions Φ and Φ−1 is called the Riemann sphere (ball). Remarks 1.5.2. 1. It is not difficult to prove that the stereographic projection maps each circle of S2 , that does not contain the N point, into a circle of ℂ, and vice versa. 2. Moreover, the stereographic projection maps each circle of S2 that contains the N point, into a line of ℂ. ̂ we could determine the distance between the points z and Remark 1.5.3. In the set ℂ, z by using the Euclidian distance of ℝ3 between their images Φ−1 (z) and Φ−1 (z ). This new distance will be denoted by dc (z, z ). Since Φ−1 (z), Φ−1 (z ) ∈ S2 , then ̂×ℂ ̂ → [0, 2], dc : ℂ 2
2
2
dc (z, z ) = √(x − x ) + (y − y ) + (u − u ) = √2 − 2(xx + yy + uu ), where z = x1 + iy1 → P(x, y, u) ∈ S2 , and z = x1 + iy1 → P (x , y , u ) ∈ S2 . ̂ is called the Definition 1.5.2. The distance dc (z, z ) between the points z, z ∈ ℂ chordal distance, where dc (z, z ) =
2z − z  √(1 +
z2 )(1
+
z 2 )
and
dc (z, ∞) =
2
√1 + z2
.
14  1 Complex numbers ̂ dc ) is a metric space, and the topology of this metric Theorem 1.5.4. The couple (ℂ, ̂ space coincides with the already defined topology of ℂ. ̂ dc ) metric space is compact. Corollary 1.5.1. The (ℂ, Proof. The set S2 ⊂ ℝ3 is compact (since it is bounded and closed in ℝ3 ). Since Φ : ̂ is a continuous bijection, it follows that ℂ ̂ is compact. S2 → ℂ Remarks 1.5.4. 1. dc ( z1 , z1 ) = dc (z, z ) and dc ( z1 , ∞) = 2.
2z √1+z2
, ∀z, z ∈ ℂ.̂
If A ⊂ ℂ is a bounded set, i. e., ∃R > 0 such that A ⊂ U(0; R), then 2z − z  ≤ dc (z, z ) ≤ 2z − z , 1 + R2
3.
∀z, z ∈ A.
The behavior of the function f in a neighborhood of the ∞ point can be easily studied by considering the new function g(z) = f ( z1 ), and studying the behavior of the function g in a neighborhood of the point z = 0.
1.6 Exercises Exercise 1.6.1. Represent in the complex plane the following sets: 1. {z ∈ ℂ : Im(z + i) = 0}; 2. 3.
{z ∈ ℂ : arg(z − 2i) = π4 };
{z ∈ ℂ : z − α = z − β}, where α = −i and β = 4 + i;
4. {z ∈ ℂ : z + i = 2}; 5. 6. 7.
{z ∈ ℂ : z − 2i + z + 4i = 10}; z−1 = π3 }; {z ∈ ℂ : arg z+1
{z ∈ ℂ : − π4 < arg(z − i) < π6 };
8. {z ∈ ℂ : z < 1 − 9.
{z ∈ ℂ :
10. {z ∈ ℂ :
1 (z 2i
z  < 1};  z+3i 1−z  1+z  > 3};
− z)};
11. {z ∈ ℂ : z − 4 − z + 6 − 9 > 0}. Exercise 1.6.2. If a ∈ ℂ, prove that:  < 1 ⇔ Im a Im z > 0; 1. z ≠ a,  z−a z−a 2. 3.
 > 1 ⇔ Im a Im z < 0; z ≠ a,  z−a z−a  = 1 ⇔ Im a Im z = 0. z ≠ a,  z−a z−a
1.6 Exercises  15
Exercise 1.6.3. If a ∈ ℂ, prove that: 1. z ≠ −a,  z−a  < 1 ⇔ Re a Re z > 0; z+a
2. 3.
 > 1 ⇔ Re a Re z < 0; z ≠ −a,  z−a z+a z ≠ −a,  z−a  = 1 ⇔ Re a Re z = 0. z+a
Exercise 1.6.4. Prove that the following functions are bijections: 1. f : {z ∈ ℂ : Im z > 0} → U(0; 1), f (z) = z−a , where Im a > 0; z−a
2.
f : {z ∈ ℂ : Re z > 0} → U(0; 1), f (z) =
z−a , z+a
where Re a > 0.
Exercise 1.6.5. Let z1 , z2 ∈ ℂ, z1 ≠ z2 be two distinct complex numbers, and let A1 (z1 ), A2 (z2 ) be their corresponding points in the complex plane, and A is the line segment between points A1 and A2 . Prove that A(z) ∈ A1 A2 ⇔
z − z1 z − z1 = . z2 − z1 z2 − z1
Exercise 1.6.6. Let z1 , z2 , z3 ∈ ℂ be three distinct complex numbers, and let Ak (zk ), k ∈ {1, 2, 3} be their corresponding points in the complex plane. Prove that A1 , A2 and A3
are collinear ⇔
z3 − z1 ∈ ℝ. z2 − z1
Exercise 1.6.7. Let z1 , z2 , z3 ∈ ℂ be three distinct complex numbers, and let Ak (zk ), k ∈ {1, 2, 3} be their corresponding points in the complex plane. Prove that A3 ∈ [A1 A2 ] ⇔
z3 − z1 ∈ [0, 1] ⇔ ∃t ∈ [0, 1] : z3 = (1 − t)z1 + tz2 , z2 − z1
where [A1 A2 ] denotes the closed segment A1 A2 . Exercise 1.6.8. Let Ak (zk ), k ∈ {1, 2, 3, 4} be the corresponding images in the complex plane of the four distinct complex numbers z1 , z2 , z3 , z4 ∈ ℂ. Prove that: −z2 ∈ iℝ; 1. A1 A2 ⊥A3 A4 ⇔ zz1 −z 2.
A1 A2 ‖A3 A4 ⇔
3
4
z1 −z2 z3 −z4
∈ ℝ.
Exercise 1.6.9. Denote by C(a; r) = 𝜕U(a; r) the circle with the center in a ∈ ℂ, and the radius r > 0. Prove that A(z) ∈ C(a; r) ⇔ zz − az − az + aa = r 2 . Exercise 1.6.10. Let z1 , z2 , z3 ∈ ℂ be three complex numbers, such that their corresponding images in the complex plane Ak (zk ), k ∈ {1, 2, 3}, are not collinear points. Prove that zz z z 1 z z 1 1 z1 z 1 1 A(z) ∈ C(z1 , z2 , z3 ) ⇔ = 0, z2 z 2 z2 z 2 1 z3 z 3 z3 z 3 1
16  1 Complex numbers where C(z1 , z2 , z3 ) represents the circle that contains the points A1 (z1 ), A2 (z2 ) and A3 (z3 ). Exercise 1.6.11. Let z1 , z2 , z3 , z4 ∈ ℂ be four complex numbers, and let Ak (zk ), k ∈ {1, 2, 3, 4} be their corresponding images in the complex plane. Prove that A1 , A2 , A3 and A4
belong to the same circle ⇔
z2 − z1 z2 − z4 : ∈ ℝ. z3 − z1 z3 − z4
Exercise 1.6.12. Suppose that the continuous function f : ℂ → ℂ satisfies the conditions: 1. f (z1 + z2 ) = f (z1 ) + f (z2 ) and 2. f (z1 z2 ) = f (z1 )f (z2 ), ∀z1 , z2 ∈ ℂ. Prove that the function f has one of the following form: f (z) = 0, ∀z ∈ ℂ, or f (z) = z, ∀z ∈ ℂ, or f (z) = z, ∀z ∈ ℂ.
2 Holomorphic functions 2.1 The derivative of the real valued complex functions Definition 2.1.1. The function f : [a, b] → ℂ is said to be differentiable at the point t0 ∈ [a, b], if there exist the limit lim
t→t0
f (t) − f (t0 ) = f (t0 ) ∈ ℂ. t − t0
Using the fact that f (t) − f (t0 ) α(t) − α(t0 ) β(t) − β(t0 ) = +i , t − t0 t − t0 t − t0 the function f is differentiable at t0 ∈ [a, b] if and only if there exists the limits lim
t→t0
α(t) − α(t0 ) = α (t0 ) and t − t0
lim
t→t0
β(t) − β(t0 ) = β (t0 ), t − t0
i. e., the real valued functions α and β are differentiable at the point t0 . In such a case, we have f (t0 ) = α (t0 ) + iβ (t0 ). Theorem 2.1.1. If the function f : [a, b] → ℂ is differentiable on the interval [a, b] if f (t) = 0, ∀t ∈ [a, b], then f is constant on the interval [a, b]. Proof. We have f (t) = 0 ⇒ α (t) = β (t) = 0. But α and β are real valued real functions, and according to the Lagrange meanvalue theorem we deduce that α(t) = α, β(t) = β, ∀t ∈ [a, b], i. e., f (t) = α + iβ, ∀t ∈ [a, b] is a constant function on [a, b]. In the proof of the above theorem, we used the Lagrange’s meanvalue theorem1 for the real valued real functions. But this theorem does not hold for the complex valued real functions, as we may see in the following example. Example 2.1.1. Let f (t) = t 2 + it 3 , t ∈ [a, b], a < b, and suppose that there exists c ∈ (a, b), such that f (b) − f (a) = f (c)(b − a),
i. e.,
b2 + ib3 − a2 − ia3 = (2c + 3ic2 )(b − a),
1 Lagrange meanvalue theorem states that a function is continuous in the closed interval [a, b] and differentiable in an open interval (a, b). Then there is at least one point c ∈ (a, b) such that f (c) = f (b)−f (a) . b−a https://doi.org/10.1515/9783110657869002
18  2 Holomorphic functions or b3 − a3 = 3c2 (b − a) b2 + ab + a2 = 3c2 ⇔{ 2 2 b − a = 2c(b − a) b + a = 2c
{
⇒ b2 + ab + a2 = 3(
2
a+b ) 2
⇒ 4b2 + 4ba + 4a2 = 3a2 + 6ab + 3b2 ⇔ b2 − 2ab + a2 = 0 ⇔ (b − a)2 = 0 ⇔ a = b, which contradicts the assumption. The next weaker version of the Lagrange theorem holds for these functions. Theorem 2.1.2 (The Lagrange theorem for complex valued real functions). If the function f : [a, b] → ℂ is continuous on [a, b], and it is differentiable on (a, b), then there exist the numbers c ∈ (a, b) and σ ∈ ℂ, with σ ≤ 1, such that: 1. f (b) − f (a) ≤ f (c)(b − a); 2. f (b) − f (a) = σf (c)(b − a). Proof. Letting f (t) = α(t) + iβ(t), then f (t) = α (t) + iβ (t), t ∈ (a, b). We will define the function φ : [a, b] → ℝ,
φ(t) = (α(b) − α(a))α(t) + (β(b) − β(a))β(t).
The function φ is continuous on [a, b], and it is differentiable on (a, b), then according to the Lagrange theorem for the real valued real functions, we deduce that ∃c ∈ (a, b) such that φ(b) − φ(a) = φ (c)(b − a). On the other hand, 2 2 2 φ(b) − φ(a) = (α(b) − α(a)) + (β(b) − β(a)) = f (b) − f (a)
and φ (c) = (α(b) − α(a))α (c) + (β(b) − β(a))β (c). From the Cauchy inequality, we have 2 2 2 2 √ φ (c) ≤ (α(b) − α(a)) + (β(b) − β(a)) √α (c) + β (c) = f (b) − f (a)f (c),
and combining with the above relations it follows that 2 f (b) − f (a) = φ(b) − φ(a) = φ (c)(b − a) ≤ (b − a)f (b) − f (a)f (c).
2.2 The differentiability of a complex function  19
If f (b) ≠ f (a), the previous inequality implies that f (b) − f (a) ≤ f (c)(b − a). It is obvious that this inequality holds for the case when f (b) = f (a), which completes the proof of the first point. Supposing that f (b) = f (a), then we can choose σ = 0, and for this value the second conclusion holds. Supposing that f (b) ≠ f (a), from the inequality of the first point it follows that f (c) ≠ 0, and thus f (c)(b − a) ≠ 0. Hence we may denote f (b) − f (a) = σ ∈ ℂ, f (c)(b − a) and from the first point we have σ ≤ 1. Multiplying with f (c)(b − a), we obtain the second relation of the theorem.
2.2 The differentiability of a complex function From the basic knowledge of mathematical analysis, it is well known that if G ⊂ ℝ2 is an open set, the function f : G → ℝ2 is said to be (Fréchet) differentiable at the point (x0 , y0 ) ∈ G, if there exists a linear operator A : ℝ2 → ℝ2 , such that ‖f (x, y) − f (x0 , y0 ) − A(x − x0 , y − y0 )‖ √(x − x0 )2 + (y − y0 )2
→ 0,
if (x, y) → (x0 , y0 ).
(2.1)
In the ℝ2 set, with an orthogonal coordinates axis, the operator A can be associa a ated with a matrix ( a2111 a2212 ), and with this notation we have A(x − x0 , y − y0 ) = (a11 (x − x0 ) + a12 (y − y0 ), a21 (x − x0 ) + a22 (y − y0 )). Replacing this form of the operator A in the limit formula (2.1), and denoting by u and v the components of the vector function f , it follows that 1 √(x − x0
)2
+ (y − y0
)2
(u(x, y) − u(x0 , y0 ), v(x, y) − v(x0 , y0 ))
−(a11 (x − x0 ) + a12 (y − y0 ), a21 (x − x0 ) + a22 (y − y0 )) → 0, if (x, y) → (x0 , y0 ).
(2.2)
20  2 Holomorphic functions From here, taking y = y0 , and letting x → x0 we get a11 =
𝜕u(x0 , y0 ) , 𝜕x
a21 =
𝜕u(x0 , y0 ) , 𝜕y
a22 =
𝜕v(x0 , y0 ) . 𝜕x
Similarly, taking x = x0 , and letting y → y0 we have a12 =
𝜕v(x0 , y0 ) , 𝜕y
hence we deduce that A=(
𝜕u(x0 ,y0 ) 𝜕x 𝜕v(x0 ,y0 ) 𝜕x
𝜕u(x0 ,y0 ) 𝜕y 𝜕v(x0 ,y0 ) 𝜕y
).
If we identify ℝ2 with ℂ, then the functions f and A will become complex functions of the forms: f (z) = f (x, y) = u(x, y) + iv(x, y);
f (z0 ) = f (x0 , y0 ) = u(x0 , y0 ) + iv(x0 , y0 ),
where
z = x + iy, z0 = x0 + iy0 ;
A(x − x0 , y − y0 ) = (a11 + ia21 )(x − x0 ) + (a12 + ia22 )(y − y0 ). as
Letting A1 = a11 +ia21 and A2 = a12 +ia22 , then the limit formula (2.2) can be written f (z) − f (z0 ) − A1 (x − x0 ) − A2 (y − y0 ) → 0, z − z0 
if z → z0 .
(2.3)
Definition 2.2.1. Let G ⊂ ℂ be an open set, and let f : G → ℂ. If there exist the numbers A1 , A2 ∈ ℂ such that the limit formula (2.3) holds, then the function f is said to be differentiable at the point z0 ∈ G. Remarks 2.2.1. 1. Writing the complex function in this vectorial form, we see that the concept of Differentiability is identical with the concept of the (Fréchet) differentiability. 2. The next equivalent form is the following: The function f : G → ℂ (G ⊂ ℂ is an open set) is differentiable at the point z0 ∈ G, if and only if there exist the complex numbers A1 , A2 ∈ ℂ, and there exists the function g : G \ {z0 } → ℂ with limz→z0 g(z) = 0, such that f (z) = f (z0 ) + A1 (x − x0 ) + A2 (y − y0 ) + g(z)z − z0 ,
∀z ∈ G \ {z0 }.
In fact, if (2.3) holds, letting g(z) =
f (z) − f (z0 ) − A1 (x − x0 ) − A2 (y − y0 ) , z − z0 
then g satisfies the required condition.
(2.4)
2.2 The differentiability of a complex function  21
Conversely, if there exists the function g with the above property, then the fraction that appeared in the formula (2.3) is g(z), and thus limz→z0 g(z) = 0. If we let f = u + iv, and we suppose that the function f is differentiable at z0 = x0 + iy0 ∈ G, we obtain 𝜕u 𝜕v 𝜕f +i = , 𝜕x 𝜕x 𝜕x 𝜕u 𝜕v 𝜕f A2 = a12 + ia22 = +i = . 𝜕y 𝜕y 𝜕y
A1 = a11 + ia21 =
If we rewrite the relation (2.4) using the fact that x − x0 =
z − z0 + (z − z0 ) , 2
y − y0 =
z − z0 − (z − z0 ) , 2i
then we deduce that 𝜕f z − z0 + (z − z0 ) 𝜕f z − z0 − (z − z0 ) + + g(z)z − z0  𝜕x 2 𝜕y 2i 𝜕f (z0 ) 1 𝜕f (z0 ) = f (z0 ) + ( −i )(z − z0 ) 2 𝜕x 𝜕y 𝜕f (z0 ) 1 𝜕f (z0 ) + ( +i )(z − z0 ) + g(z)z − z0 . 2 𝜕x 𝜕y
f (z) = f (z0 ) +
Remarks 2.2.2. 1. We will use the following important notation: 𝜕f (z0 ) 𝜕f (z0 ) 1 𝜕f (z0 ) = ( −i ), 𝜕z 2 𝜕x 𝜕y 2.
With the above notation, the relation (2.4) becomes f (z) = f (z0 ) +
3.
𝜕f (z0 ) 1 𝜕f (z0 ) 𝜕f (z0 ) = ( +i ). 𝜕z 2 𝜕x 𝜕y
𝜕f (z0 ) 𝜕f (z0 ) (z − z0 ) + g(z)z − z0 , (z − z0 ) + 𝜕z 𝜕z
∀z ∈ G \ {z0 }. (2.5)
From the mathematical analysis, it is well known that the function f : ℝ2 → ℝ2 is differentiable, if and only if its coordinates u and v are differentiable real valued functions. It is also known that a real valued function is differentiable, if it has continuous firstorder partial derivatives.
Definition 2.2.2. Let G ⊂ ℂ be an open set. We say that the function f : G → ℂ has firstorder continuous partial derivatives on G (denoted by f ∈ C 1 (G)), if both real and imaginary parts of the function f have firstorder continuous partial derivatives on G. Theorem 2.2.1. If f ∈ C 1 (G), where G ⊂ ℂ is an open set, then the function f is differentiable at all the points of G.
22  2 Holomorphic functions Proof. Suppose that the function f : G → ℂ is differentiable at the point z0 ∈ G. Let θ ∈ (−π, π] be an arbitrary angle, and let us compute the directional derivative of the function f in the point z0 . In fact, this will be the derivative of a complex valued real variable function. For this, suppose that z = z0 + h, where arg h = θ and h → 0. For the computation of the directional derivative denoted by fθ (z0 ), we will use the formula (2.5), i. e., f (z0 + h) − f (z0 ) =
𝜕f (z0 ) 𝜕f (z0 ) h + g(z0 + h)h. h+ 𝜕z 𝜕z
(2.6)
We conclude that fθ (z0 ) = lim h→0
arg h=θ
f (z0 + h) − f (z0 ) , h
and the existence of this last limit follows from the relation (2.6), because hh is constant whenever arg h = θ. Theorem 2.2.2 (Kasner theorem). Let G ⊂ ℂ be an open set, let z0 ∈ G, and suppose that the function f is Differentiable at the point z0 . If the angle θ runs once the whole (−π, π] interval, then the directional derivative fθ (z0 ) runs twice the whole circle with 𝜕f (z ) 𝜕f (z ) the center 𝜕z0 and radius  𝜕z0 . Proof. Using the notation cos θ + i sin θ = eiθ , then h = heiθ , h = he−iθ , and hence h = e−2iθ . If we divide by h the equality (2.6), we get h f (z0 + h) − f (z0 ) 𝜕f (z0 ) 𝜕f (z0 ) −2iθ h = + e + g(z0 + h) , h 𝜕z 𝜕z h hence fθ (z0 ) = lim h→0
arg h=θ
f (z0 + h) − f (z0 ) 𝜕f (z0 ) 𝜕f (z0 ) −2iθ = + e , h 𝜕z 𝜕z
(2.7)
i. e., 𝜕f (z0 ) 𝜕f (z0 ) = . fθ (z0 ) − 𝜕z 𝜕z The proof follows immediate from these last two relations. Remarks 2.2.3. 1. The circle that appeared in Theorem 2.2.2 is called the Kasner circle. 2. The relation (2.7) shows that if the angle θ runs once in positive orientation the complete circle, then the directional derivative fθ (z0 ) runs twice the Kasner circle in negative direction.
2.3 The derivative of a complex function  23
2.3 The derivative of a complex function Definition 2.3.1. Let G ⊂ ℂ be an open set, and let z0 ∈ G. We say that the function f : G → ℂ is differentiable at the point z0 , if ∃ lim
z→z0
f (z) − f (z0 ) = f (z0 ) ∈ ℂ. z − z0
The above limit is denoted by f (z0 ), and it is called the derivative of the function f at the point z0 . Definition 2.3.2. Let G ⊂ ℂ be an open set. The function f : G → ℂ is said to be differentiable at the point z0 ∈ G, if there exists the complex number α ∈ ℂ, and there exists the function g : G \ {z0 } → ℂ with limz→z0 g(z) = 0, such that f (z) = f (z0 ) + α(z − z0 ) + g(z)(z − z0 ),
z ∈ G \ {z0 }.
(2.8)
Theorem 2.3.1 (The Cauchy–Riemann theorem). The function f = u + iv, f : G → ℂ (where G ⊂ ℂ is an open set) is differentiable at the point z0 ∈ G, if and only if it is differentiable at the point z0 , and the partial derivatives of u and v satisfy the relations 𝜕u(z0 ) 𝜕v(z0 ) = 𝜕x 𝜕y
and
𝜕u(z0 ) 𝜕v(z0 ) =− 𝜕y 𝜕x
(called Cauchy–Riemann conditions). Proof. Suppose that ∃f (z0 ). From the previous theorem ∃α ∈ ℂ, and ∃ g : G\{z0 } → ℂ, with limz→z0 g(z) = 0, such that f (z) = f (z0 ) + α(z − z0 ) + g(z)(z − z0 ). Letting A1 = α and A2 = iα, the above relation becomes f (z) = f (z0 ) + A1 (x − x0 ) + A2 (y − y0 ) + g(z)
z − z0 z − z0 . z − z0 
This represents the differentiability of the function f , where the condition (2.4) is writz−z ten in another equivalent form, because g(z) z−z0  → 0, if z → z0 . 0
Since f is differentiable at z0 , and since there exists f (z0 ) it follows that fθ (z0 ) = f (z0 ), ∀θ ∈ (−π, π]. This last relation holds if and only if the coefficient of e−2iθ from the righthand side of (2.7) vanishes (it is zero), i. e., 𝜕f (z0 ) 𝜕f (z0 ) 1 𝜕f (z0 ) = ( +i ) = 0. 𝜕z 2 𝜕x 𝜕y
From this equality, we have 𝜕u(z0 ) 𝜕v(z0 ) 𝜕u(z0 ) 𝜕v(z0 ) +i +i − = 0, 𝜕x 𝜕x 𝜕y 𝜕y
24  2 Holomorphic functions hence 𝜕u(z0 ) 𝜕v(z0 ) = 𝜕x 𝜕y
and
𝜕u(z0 ) 𝜕v(z0 ) =− , 𝜕y 𝜕x
which represents the Cauchy–Riemann conditions. Suppose now that the function f is differentiable at the point z0 , and the functions u and v satisfy the Cauchy–Riemann conditions. From the differentiation of the function f at the point z0 , it follows that u and v are two differentiable real functions at z0 = x0 + iy0 . Then 𝜕u(z0 ) 𝜕u(z0 ) (x − x0 ) + (y − y0 ) + u1 (z)z − z0 , 𝜕x 𝜕y 𝜕v(z0 ) 𝜕v(z0 ) (x − x0 ) + (y − y0 ) + v1 (z)z − z0 , v(z) = v(z0 ) + 𝜕x 𝜕y
u(z) = u(z0 ) +
(2.9) (2.10)
where u1 (z) → 0 and v1 (z) → 0, if z → z0 . We will multiply the both sides of (2.10) by i, and add equation (2.9) to it. Using the Cauchy–Riemann conditions, we deduce that f (z) = u(z) + iv(z) = f (z0 ) + (
𝜕u(z0 ) 𝜕v(z0 ) +i )(z − z0 ) + g(z)(z − z0 ), 𝜕x 𝜕x
(2.11)
where g(z) =
u1 (z) + iv1 (z) z − z0  → 0, z − z0
if z → z0 .
From the relation (2.11), we conclude that the function f is differentiable at the point z0 , and using Theorem 2.2.1 it follows the existence of the derivative f (z0 ). Remarks 2.3.1. 1. Let G ⊂ ℂ be an open set. The function f : G → ℂ is derivable at the point z0 ∈ G, 𝜕f (z ) if and only if it is differentiable at this point, and 𝜕z0 = 0. 2. The relation (2.11) shows the next very important formula: f (z0 ) =
𝜕u(z0 ) 𝜕v(z0 ) +i . 𝜕x 𝜕x
Using the Cauchy–Riemann conditions, the derivative of the function f at the point z0 ∈ G can be written in the following forms: f (z0 ) =
𝜕u(z0 ) 𝜕v(z0 ) 𝜕v(z0 ) 𝜕v(z0 ) 𝜕u(z0 ) 𝜕u(z0 ) −i = +i = −i , 𝜕x 𝜕y 𝜕y 𝜕x 𝜕y 𝜕y
hence f (z0 ) =
𝜕f (z0 ) 𝜕f (z0 ) = −i . 𝜕x 𝜕y
2.3 The derivative of a complex function  25
3.
It is easy to see that 2
2
2
2
2
2
2
2
𝜕u(z0 ) 𝜕v(z0 ) 𝜕u(z0 ) 𝜕u(z0 ) 2 ) +( ) =( ) +( ) f (z0 ) = ( 𝜕x 𝜕x 𝜕x 𝜕y =( =
𝜕u(z0 ) 𝜕v(z0 ) 𝜕v(z0 ) 𝜕v(z0 ) ) +( ) =( ) +( ) 𝜕y 𝜕y 𝜕y 𝜕x
𝜕u(z0 ) 𝜕v(z0 ) 𝜕u(z0 ) 𝜕v(z0 ) − . 𝜕x 𝜕y 𝜕y 𝜕x
Consequently, 2 f (z0 ) =
𝜕u(z0 ) 𝜕x 𝜕v(z0 ) 𝜕x
𝜕u(z0 ) 𝜕y 𝜕v(z0 ) 𝜕y
D(u, v)(z0 ) . = J(u, v)(z0 ) = D(x, y)
2.3.1 The properties of the derivative Property 2.3.1. 1. Let G ⊂ ℂ be an open set. Every differentiable function at the point z0 ∈ G is continuous at that point. 2. (f (z) + g(z)) = f (z) + g (z); (f (z)g(z)) = f (z)g(z) + f (z)g (z); 3.
f (z) ( g(z) ) =
f (z)g(z)−f (z)g (z) . g 2 (z)
Let G1 , G2 ⊂ ℂ be two open sets, and let f1 : G1 → G2 , f2 : G2 → ℂ, z1 ∈ G1 , z2 ∈ f1 (G1 ). If the functions fk are differentiable in Gk , k ∈ {1, 2}, then f2 ∘ f1 : G1 → ℂ is differentiable at z1 ∈ G1 , and (f2 ∘ f1 ) (z1 ) = f2 (f1 (z1 ))f1 (z1 ).
4. If the function h : (a, b) → ℂ is differentiable on (a, b), h((a, b)) ⊂ G, where G ⊂ ℂ is an open set, and the function f : G → ℂ is differentiable at the point h(t0 ), then (f ∘ h) (t0 ) = f (h(t0 ))h (t0 ). Proof. 3. Let z ∈ G1 . Since the function f1 is differentiable at z1 , we have f1 (z) = f1 (z1 ) + f1 (z1 )(z − z1 ) + g1 (z)(z − z1 ), and since f2 is differentiable at z2 ∈ G2 we have that for all z ∈ G2 , f2 (z) = f2 (z2 ) + f2 (z2 )(z − z2 ) + g2 (z)(z − z2 ),
26  2 Holomorphic functions where gj : Gj \ {zj } → ℂ with limz→zj gj (z) = 0, j = 1, 2. Replacing in the last relation z by f1 (z), and z2 = f1 (z1 ), we obtain (f2 ∘ f1 )(z) = f2 (f1 (z1 ))) + f2 (f1 (z1 ))(f1 (z) − f1 (z1 )) + g2 (f1 (z))(f1 (z) − f1 (z1 ))
= f2 (f1 (z1 )) + f2 (f1 (z1 ))(f1 (z1 )(z − z1 ) + g1 (z)(z − z1 )) + g2 (f1 (z))(f1 (z) − f1 (z1 )) = f2 (f1 (z1 )) + f2 (f1 (z1 ))f1 (z1 )(z − z1 ) + g(z)(z − z1 ),
where g(z) = f2 (f1 (z1 ))g1 (z) + g2 (f1 (z))
f1 (z) − f1 (z1 ) z − z1
and
lim g(z) = 0,
z→z1
because limz→z1 g1 (z) = 0, limz→z1 f1 (z) = z2 , hence limz→z1 g2 (f1 (z)) = 0, and because there exists the limit limz→z1 Consequently,
f1 (z)−f1 (z1 ) z−z1
= f1 (z1 ) ∈ ℂ.
(f2 ∘ f1 ) (z1 ) = f2 (f1 (z1 ))f1 (z1 ). Definition 2.3.3. 1. Let G ⊂ ℂ be an open set, G ≠ 0, and let f : G → ℂ. The function f is said to be a holomorphic function on G, if it is differentiable in all the points of G. 2. The set of all the holomorphic functions on the open set G ⊂ ℂ is denoted by H(G); the set H(G) is a complex vectorial space. 3. Those functions that are holomorphic on ℂ are called entire functions. Theorem 2.3.2. Let f = u + iv a complex valued function, defined on the open disc U(z0 ; r), with r > 0. If the functions u and v have partial derivatives in a neighborhood of the point z0 , that are continuous at z0 , and satisfy the Cauchy–Riemann conditions in z0 , then f is differentiable at z0 . Proof. From the assumption, the functions u and v are differentiable at the point z0 . Hence the function f is differentiable at z0 . Since the Cauchy–Riemann conditions are satisfied, according to the Cauchy–Riemann theorem, we get that there exists the derivative f (z0 ). Theorem 2.3.3. If G ⊂ ℂ is an open set, f = u + iv ∈ C 1 (G), and the functions u and v satisfy the Cauchy–Riemann conditions in every point of the set G, then f ∈ H(G). Let denote by R(G) the set of all the differentiable functions on the open set G, and let denote by F(G) the set of all the complex valued functions defined on G. Denoting by 𝜕f 1 𝜕f 𝜕f = ( + i ), 𝜕z 2 𝜕x 𝜕y
2.3 The derivative of a complex function  27
then 𝜕 : R(G) → F(G). 𝜕z
From the Cauchy–Riemann theorem, we have H(G) ⊂ R(G), and from the point 1 of Remark 2.3.1 we deduce that H(G) = ker
𝜕 𝜕 ⇔ ≡ 0. 𝜕z 𝜕z
Remarks 2.3.2. 1. There are “many” functions f ∈ C 1 (ℂ), that are not differentiable (derivable) in any point. The most simple example is given by the function f (z) = z, that is not differentiable in any point, since it does not satisfy the Cauchy–Riemann conditions in any point of ℂ. 2. The Cauchy–Riemann conditions are necessary conditions for the differentiability, but they are not sufficient. For z = x + iy, let define the function
Thus, u = Re f =
xy √x2 +y2
xy xy { √x2 +y2 + i √x2 +y2 , f (z) = { { 0,
if z ≠ 0 if z = 0.
= Im f = v. A simple computation shows that 𝜕u(0) 𝜕v(0) 𝜕u(0) 𝜕v(0) = = = = 0. 𝜕x 𝜕x 𝜕y 𝜕y
Hence the functions u and v have partial derivatives at the point z = 0, and satisfy the Cauchy–Riemann conditions. We will prove that the function f is continuous at z = 0. Denoting x = r cos θ, y = r sin θ, then f (z) = {
r(sin θ cos θ + i sin θ cos θ), 0,
r ≠ 0 r=0
sin 2θ √2 ⇒ f (z) ≤ r sin θ cos θ1 + i ≤ r < r ⇒ lim f (z) = 0. 1 + i ≤ r 2 z→0 2
On the other hand
f (z) − f (0) r cos θ sin θ(1 + i) sin θ(1 + i) = = . z−0 r(cos θ + i sin θ) 1 + i tan θ
Taking z ∈ ℝ+ , z → 0, then θ = 0, sin θ = 0, and thus Taking θ = arg z =
√2 (1+i) 2
1+i
=
√2 . 2
π , 4
z → 0, then sin θ =
√2 , 2
f (z)−f (0) z−0
→ 0.
tan θ = 1, and thus
From the above two limits, it follows that ∄ limz→0
f (z)−f (0) . z−0
f (z)−f (0) z−0
→
28  2 Holomorphic functions
Figure 2.1: Proof of Theorem 2.3.4.
Theorem 2.3.4. Let D ⊂ ℂ be a domain, and let f be a holomorphic function on D, i. e., f ∈ H(D). Then the function f is constant on D if and only if f (z) = 0, ∀z ∈ D. Proof. If f is a constant function on D, it follows immediately that f ∈ H(D) and f (z) = 0, ∀z ∈ D. Suppose that f (z) = 0, ∀z ∈ D. Let z0 ∈ D be an arbitrary point. Since D is an open set, there exists r > 0, such that U(z0 ; r) ⊂ D. For all z ∈ U(z0 ; r), we have f (z) = 𝜕u(z) + i 𝜕v(z) = 0. 𝜕x 𝜕x Let z1 ∈ U(z0 ; r) be a fixed point for the moment (Figure 2.1). The real valued function φ : [x0 , x1 ] → ℝ,
φ(x) = u(x, y0 )
𝜕u(x,y )
is derivable, and φ (x) = 𝜕x 0 = 0. Hence φ is a constant function, thus u(x, y0 ) = u(x0 , y0 ) = u(x1 , y0 ), ∀x ∈ [x0 , x1 ]. From the similar reasons for the function v, we deduce that v(x, y0 ) = v(x0 , y0 ) = v(x1 , y0 ). Since f (z) = 𝜕v(z) − i 𝜕u(z) = 0, differentiating the function ψ(y) = u(x1 , y) we get 𝜕y 𝜕y ψ (y) =
𝜕u(x1 , y) = 0, 𝜕y
∀y ∈ [y0 , y1 ] ⇒ u(x1 , y0 ) = u(x1 , y1 ).
Similarly, we have v(x1 , y0 ) = v(x1 , y1 ). Combining all of the above results, we deduce that f (z1 ) = f (z0 ). Since z1 ∈ U(z0 ; r) was an arbitrary point, we obtain that f (z) = f (z0 ), ∀z ∈ U(z0 ; r). Let A = {z ∈ D : f (z) = f (z0 )}. Using the same proof as to the above, it is easy to show that for all arbitrary z1 ∈ A, there exists r1 > 0, such that U(z1 ; r1 ) ⊂ D and f (z) = f (z1 )(= f (z0 )), ∀z ∈ U(z1 ; r1 ). Thus, the set A is an open set in D. Let zn ∈ A, ∀n ∈ ℕ∗ , with limn→∞ zn = z ∈ D. Since f is a continuous function on D, we have limn→∞ f (zn ) = f (z). But f (zn ) = f (z0 ), ∀n ∈ ℕ∗ , because of the fact that zn ∈ A, ∀n ∈ ℕ∗ . Thus f (z) = f (z0 ), so the set A is closed in D. Since D is a connected set, from the above two results we conclude that D = A.
2.4 The geometric interpretation of the derivative  29
Theorem 2.3.5. Let D ⊂ ℂ be a domain, and let f ∈ H(D). If at least one of the following real valued functions f , Re f , Im f , arg f is constant on the domain D, then the function f is constant on D. Proof. For example, let, u(x, y) = Re f (z) = c, ∀z ∈ D. Then from the Cauchy–Riemann conditions we have
𝜕v 𝜕x
=
𝜕v 𝜕y
𝜕u 𝜕x
=
𝜕u 𝜕y
= 0 on D, and
= 0 on the domain D. Then
f (z) = 0, ∀z ∈ D, and according to the Theorem 2.3.4, the function f is constant on the domain D. The proof is similar to the case that Im f is constant on D. If f (z)2 = u2 (x, y) + v2 (x, y) = c, ∀z ∈ D, then
{
𝜕v u 𝜕u + v 𝜕x =0 𝜕x
𝜕v u 𝜕u + v 𝜕y = 0. 𝜕y
From the Cauchy–Riemann conditions, this system is equivalent to {
𝜕v + v 𝜕x =0 u 𝜕u 𝜕x
𝜕v v 𝜕u − u 𝜕x = 0, 𝜕x
𝜕v and 𝜕x , so the determinant of the system will be −(u2 + v2 ). with the unknown 𝜕u 𝜕x 2 2 If u + v = c = 0, then f (z) = 0, ∀z ∈ D. If u2 + v2 = c ≠ 0, then the system has in every point z ∈ D only the trivial solution, 𝜕v i. e., 𝜕u = 𝜕x = 0, ∀z ∈ D, and thus f (z) = 0, ∀z ∈ D. According to Theorem 2.3.4, the 𝜕x function f will be constant on the domain D. We have arg f = arctan uv = k0 ⇒ uv = k, ∀z ∈ D. If u = 0 ⇒ u = Re f is constant ⇒ f is constant. If u ≠ 0 ⇒ v − ku = 0 ⇒ Re(−k − i)f = Re(−k − i)(u + iv) = Re(−ku + v + i(−u − kv)) = −ku + v = 0. Hence the function (−k − i)f is constant, then f is constant on the domain D.
2.4 The geometric interpretation of the derivative Definition 2.4.1. The real variable continuous function γ : [0, 1] → ℂ,
γ(t) = α(t) + iβ(t)
is called a path of curve in ℂ. Then the system {
x = α(t) y = β(t),
t ∈ [0, 1]
represents the parametric equation of a continuous curve in ℝ2 .
30  2 Holomorphic functions The point γ(0) is called the starting point, while the point γ(1) is called the end point of the path γ. Definition 2.4.2. 1. The set {γ} = γ([0, 1]) is called the image of the path γ. 2. The paths γ1 and γ2 are called equivalent paths, denoted by γ1 ∼ γ2 , if ∃h : [0, 1] → [0, 1] a continuous bijection, with h−1 continuous, and h(0) = 0, h(1) = 1, such that γ1 = γ2 ∘ h. 3. If γ ∈ C 1 [0, 1], with γ (t) ≠ 0, ∀t ∈ [0, 1], then γ is called a smooth path. Remarks 2.4.1. 1. If γ ∈ C 1 [0, 1], with γ (t0 ) = α (t0 ) + iβ (t0 ) ≠ 0, then α (t0 ) and β (t0 ) are called the directional parameters of the tangent to the path {γ} into the point γ(t0 ). 2. If {γ} is a smooth path, then there exists a tangent to it in every point on {γ}. If θ is the angle between the Ox axis and this tangent, then θ = arg γ (t0 ) = arctan 3.
β (t0 ) . α (t0 )
If γ1 and γ2 are two smooth paths, with γ1 (t0 ) = γ2 (t0 ) = z0 , then the angle between γ1 and γ2 at the point z0 is the angle between the tangents to both of these paths into the point z0 .
Definition 2.4.3. 1. Let G ⊂ ℂ be an open set and let f ∈ C 1 (G). The function f is said to be a firstorder conformal map at the point z0 ∈ G, if the angle at z0 ∈ G between any arbitrary smooth paths γ1 and γ2 , that start from z0 , is the same with the angle at f (z0 ) between their images f ∘ γ1 and f ∘ γ2 . 2. The function f is called direct conformal mapping or inverse conformal mapping, if the directions of the angles is preserved, or respectively is changed. 3. The function f is said to be a secondorder conformal map at the point z0 ∈ G, if the length of any rectifiable smooth path that starts from z0 = γ(0) is modified “in the same way” by the function f , i. e., (f ∘ γ) (0) = k γ (0),
∀γ smooth path, γ(0) = z0 ,
and the constant k is independent of γ and it is called the linear deformation (distortion) coefficient of the function f into the point z0 . Theorem 2.4.1. If the function f : G → ℂ is differentiable at the point z0 ∈ G, where G ⊂ ℂ is an open set, and f (z0 ) ≠ 0, then: 1. The function f is first order direct conformal mapping at the point z0 . 2. The function f is second order conformal mapping at the point z0 .
2.4 The geometric interpretation of the derivative  31
Figure 2.2: Proof of Theorem 2.4.1.
Proof. Let γ be an arbitrary smooth path that starts from z0 , i. e., γ(0) = z0 , and let γ1 = f ∘ γ (Figure 2.2). Since (f ∘ γ) (0) = f (γ(0))γ (0), we have γ1 (0) = f (z0 )γ (0), so it follows that arg γ1 (0) = arg f (z0 ) + arg γ (0). It means that the tangent to the image path γ1 into the point f (z0 ) is obtained by rotating the tangent to the path γ into the point z0 with the angle arg f (z0 ) (which is independent of γ), i. e., arg γ1 (0) − arg γ (0) = arg f (z0 ). This implies that the angle and the direction of the angle between any two arbitrary smooth paths that starts from z0 will be preserved by the function f , which proves the first part of the theorem. For any arbitrary smooth path that starts from z0 , we have (f ∘ γ) (0) = f (z0 )γ (0), hence f is secondorder conformal mapping at the point z0 , and the linear distortion coefficient of the function f in z0 will be f (z0 ). Remarks 2.4.2. 1. If the function f has derivative at the point z0 ∈ G, where G ⊂ ℂ is an open set, and if f (z0 ) ≠ 0, then arg f (z0 ) is the rotational angle of the tangent in z0 to any smooth path that starts from z0 . 2. The condition f (z0 ) ≠ 0 is necessary. For example, the function f (z) = z 2 doubles all the angles that starts from z0 = 0. Theorem 2.4.2. Let f ∈ C 1 (G), where G ⊂ ℂ is an open set, and let z0 ∈ G. 1. If the function f is firstorder direct conformal mapping at the point z0 , then f has derivative at this point.
32  2 Holomorphic functions 2.
If the function f is secondorder conformal mapping at the point z0 , then f or f have derivatives at this point.
Proof. 1. Let γ(t) = (1 − t)z0 + tz1 , where arg(z1 − z0 ) = θ, and let γ1 (t) = f (γ(t)), t ∈ [0, 1]. Then γ1 (0) = fθ (γ(0))(z1 − z0 ) = fθ (z0 )γ (0), hence arg γ1 (0) − arg γ (0) = arg fθ (z0 ).
(2.12)
For the lefthand side to be independent of θ, it is necessary that arg fθ (z0 ) is independent of θ. During the proof of Kasner theorem, we obtained that fθ (z0 ) = Using this relation, it follows that if arg(fθ (z0 ) − Since
𝜕f (z0 ) 𝜕z
and
𝜕f (z0 ) 𝜕z
𝜕f (z0 ) 𝜕f (z0 ) −2iθ + e . 𝜕z 𝜕z 𝜕f (z0 ) 𝜕z
≠ 0, then
𝜕f (z0 ) 𝜕f (z0 ) ) = arg − 2θ. 𝜕z 𝜕z
are constants, then arg(fθ (z0 ) − 𝜕f (z0 ) 𝜕z
𝜕f (z0 ) ) 𝜕z
cannot be constant.
Hence, the vector with the starting point in and the end point in fθ (z0 ) runs an angle of 4π whenever θ runs an interval of the length of 2π. Thus, the righthand side 𝜕f (z ) of the equality (2.12) is constant only if 𝜕z0 = 0. Since this last equality is equivalent to the Cauchy–Riemann conditions, it follows that the function f has derivative at the point z0 . 2. Let γ be the same path as to the point 1 of the proof. From the assumption (f ∘ γ) (0) = k γ (0), where the constant k is independent of θ. On the other hand, (f ∘ γ) (0) = fθ (z0 )γ (0), hence fθ (z0 ) = k, ∀θ ∈ (−π, π]. According to the Kasner theorem, are possible the next two situations: 𝜕f (z0 ) = 0, and then the function f has derivative at the point z0 ; (i) 𝜕z (ii)
or
𝜕f (z0 ) 𝜕z
= 0, and this represents the Cauchy–Riemann conditions for the function f . In this case, we have that the function f has derivative at the point z0 .
2.5 Entire functions  33
2.5 Entire functions 2.5.1 The polynomial function Definition 2.5.1. The function p : ℂ → ℂ, p(z) = c0 + c1 z + ⋅ ⋅ ⋅ + cn z n , where cj ∈ ℂ, j ∈ {0, 1, . . . , n} and cn ≠ 0, is called nth order polynomial function. Remarks 2.5.1. 1. Any nonzero constant polynomial is a 0th order polynomial. 2. The zero constant polynomial is a −∞th order polynomial. 3. The function f (z) = z n is an entire function, and its derivative is f (z) = nz n−1 . 4. From the previous remark, any polynomial function in an entire function, as a sum of entire functions. 2.5.2 The exponential function Let z = x + iy → ez = ex (cos y + i sin y). Since Re e = e cos y and Im ez = ex sin y have continuous firstorder partial derivatives on ℂ, and satisfy the Cauchy–Riemann conditions on ℂ, it follows that the function ez is holomorphic on the complex plane ℂ. The function, denoted by ez , exp z or exp(z), is called the exponential function. From z
x
(ez ) =
𝜕 z e = ez , 𝜕x
we have (ez )(n) = ez , ∀n ∈ ℕ. If x = 0, then eiy = cos y + i sin y,
e−iy = cos y − i sin y ⇒ cos y =
eiy + e−iy , 2
sin y =
eiy − e−iy . 2i
The above relations are called Euler formulas. 1. ez+2πi = ez , ∀z ∈ ℂ, hence the exponential function is a periodic function, with the period equal with 2πi, so it is not an injective function on ℂ. 2. Re ez = ex cos y, Im ez = ex sin y, ez  = ex , where z = x + iy. 3. ez = ez , ∀z ∈ ℂ, and arg ez = y (mod 2π), where z = x + iy. 4. ez ≠ 0, ∀z ∈ ℂ, because ez  = ex ≠ 0, ∀x ∈ ℝ. 5. z = reiθ , where r = z, θ = arg z, ∀z ∈ ℂ∗ .
34  2 Holomorphic functions Remark 2.5.2. Let us define the following Bk strips, parallel with the Ox axis, by Bk = {z = x + iy ∈ ℂ : 2kπ ≤ y < 2(k + 1)π},
k ∈ ℤ.
Then ℂ = ⋃k∈ℤ Bk , and the exponential function is injective on the every Bk strip, k ∈ ℤ. Since the exponential is holomorphic and injective in each Bk strip, we say that it is a univalent function on Bk . 2.5.3 Complex trigonometric functions Starting with the Euler formulas, we will define the complex valued trigonometric functions as follows: 1 cos z = (eiz + e−iz ); 2
sin z =
1 iz (e − e−iz ). 2i
Thus, cos ix = cosh x,
sin ix = i sinh x.
It is easy to prove the next formulas: cos(z + w) = cos z cos w − sin z sin w, sin(z + w) = sin z cos w + cos z sin w, which imply cos z = cos(x + iy) = cos x cos iy − sin x sin iy = cos x cosh y − i sin x sinh y. Exercise 2.5.1. 1. Prove that  sinh y ≤  cos z ≤ cosh y, if z = x + iy. 2. Prove that  cos z2 = 21 (cosh 2y + cos 2x), if z = x + iy. 3. The function sin and cos are unbounded functions, because lim
z=x0 +iy→∞
sin z = ∞
and
lim
z=x0 +iy→∞
cos z = ∞.
Solution. It is well known that cosh2 y − sinh2 y = 1,
(2.13)
 cos z2 = cos2 x cosh2 y + sin2 x sinh2 y = cos2 x(1 + sinh2 y) + sin2 x sinh2 y = cos2 x + (cos2 x + sin2 x) sinh2 y = cos2 x + sinh2 y ⇒ sinh2 y ≤  cos z2 ,
(2.14)
2.6 Bilinear transforms  35
 cos z2 = cos2 x cosh2 y + sin2 x sinh2 y = cos2 x cosh2 y + sin2 x(cosh2 y − 1) = cosh2 y − sin2 x ≤ cosh2 y ⇒  cos z2 ≤ cosh2 y.
(2.15)
From the relations (2.13) and (2.15), we obtain  cos z2 = cosh2 y − sin2 x,
1 cosh2 y = (cosh 2y + 1) and 2
1 sin2 x = (1 − cos 2x), 2
respectively. Hence 1 1 1  cos z2 = (cosh 2y + 1) − (1 − cos 2x) = (cosh 2y + cos 2x). 2 2 2 Since
and
1 lim cos z = ∞,  cos z2 ≥ (cosh 2y − 1) ⇒ z=x0 +iy→∞ 2 1 1  sin z2 = (cosh 2y − cos 2x) ⇒  sin z2 ≥ (cosh 2y − 1) 2 2 ⇒ lim sin z = ∞. z=x0 +iy→∞
The above defined functions cos z and sin z, and the function tan z = called (complex) trigonometric functions.
sin z , cos z
are
2.5.4 Complex hyperbolic functions Similar to the definition of the complex trigonometric functions, we will define the complex hyperbolic functions as follows: 1 cosh z = (ez + e−z ); 2
1 sinh z = (ez − e−z ). 2
Thus, the functions cosh and sinh are entire periodic functions, with the period equal with 2πi.
2.6 Bilinear transforms Let the function T be defined as z → Tz =
az + b , cz + d
a, b, c, d ∈ ℂ.
Then T is a constant function, if and only if ad − bc = 0. (For example, if ac ≠ 0, then ad − bc = 0 ⇔
b a
= dc , hence Tz =
b a z+ a c z+ d c
= ac .)
36  2 Holomorphic functions Definition 2.6.1. The function T is called bilinear transform, if ad − bc ≠ 0 and ̂ → ℂ, ̂ T:ℂ
az+b , if z ∈ ℂ \ {− dc } { cz+d { { Tz = { ∞, if z = − dc { { a if z = ∞, { c,
if c ≠ 0
and Tz = {
a z d
+ db ,
∞,
if z ∈ ℂ if z = ∞,
if c = 0,
where a, b, c, d ∈ ℂ. Theorem 2.6.1. A bilinear transform is an injective singlevalued continuous function of ̂ onto ℂ. ̂ ℂ Proof. When c = 0, Tz = da z + db , where da ≠ 0. Hence ∀w ∈ ℂ, ∃!z ∈ ℂ: Tz = w. Further, T and T −1 are continuous in ℂ, limz→∞ Tz = ∞ and limw→∞ T −1 w = ∞, so it follows that T and T −1 are continuous at the ∞ point. If c ≠ 0 and z0 ≠ − dc , then az + b and cz + d are continuous in a neighborhood of z0 and cz0 + d ≠ 0, hence T is continuous at the z0 point. If z → − dc ⇒ cz + d → 0, az + b → a(− dc ) + b ≠ 0 ⇒ Tz → ∞ ⇒ T is continuous at the − dc point. If z ≠ − dc , then w=
−dw + b az + b ⇒z= ⇒ Tℂ\{− d } c cz + d cw − a
is injective.
The value of the function T cannot be ac , because
a az + b = ⇔ ad = bc, c cz + d
hence it is impossible. Hence, the function T is a bijection between ℂ \ {− dc } and ℂ \ { ac }, T(− dc ) = ∞ and T(∞) = ac . Tz → ∞, if z → − dc and T −1 w → ∞, if w → ac , so it follows that T is continuous at the − dc point, and T −1 is continuous at the ac point. 2.6.1 Decompositions in elementary functions If c ≠ 0, then Tz =
az + b
c(z + dc )
=
az + a dc − a dc + b c(z + dc )
=
d a −a c + b a bc − ad = + . + c c(z + d ) c c2 (z + d ) c c
2.6 Bilinear transforms  37
If c ≠ 0, then Tz =
a c
+
It is well known that:
bc−ad . c2 (z+ dc )
(1)
z1 = z +
(2)
z2 =
(3)
z3 = kz2
1 z1
d c
translation symmetryinversion
a c
(4) w = z3 +
(k =
bc−ad ) c2
complex scaling translation.
If c = 0, then Tz = da z + db is a first degree polynomial. Hence, in the case c ≠ 0, the function T can be written by composing (1), (2), (3) and (4) transformations: (1) and (4) translations, and (3) complex homotheticity: z3 = k
k z, k 2
where k (i) z3 = k z2 rotation (i. e., the product between z2 and a complex number with unitary module) (ii) z3 = kz3 real homotheticity (i. e., dilatation or shrinking). The (2) number transform is called symmetryinversion and z2 =
1 ⇒ z1 z2  = 1. z1
If ζ = z1  ei arg z1 , then ζ and z1 are inverse points with respect to the unit circle, because 1 (i) ζ and z1 belongs on a halfline that starts from the origin O, (ii) ζ z1  = 1. On the other hand, ei arg z1 =
z1 , z1 
ζ =
1 1 −i arg z1 1 ⇒ z2 = ζ . e = = z1  z1 z1 ei arg z1
Hence, the point z2 t can be obtained from z1 as follows: (i) First, we find the inverse of z1 with respect to the unit circle, which will be ζ . (ii) Second, we find the symmetric of ζ with respect to the real axis Ox, which will be ζ = z2 . ̂ onto ℂ’s ̂ circles. (The lines Theorem 2.6.2. A bilinear transform maps the circles from ℂ are also circles which contain the ∞ point.) Proof. The general equation of a circle is A(x2 + y2 ) + Bx + Cy + D = 0, where A + B + C ≠ 0.
38  2 Holomorphic functions If A = 0, then the above equation represents a line. If A ≠ 0, we may suppose that A > 0. The circle is a really circle, if B2 + C 2 − 4AD ≥ 0.
(2.16)
We will obtain the complex equation of the circle, if we change the variables (parameters) 1 x = (z + z), 2
x2 + y2 = zz,
y=
1 i i (z − z) = − z + z, 2i 2 2
i. e., 1 i Azz + B (z + z) + C (z − z) + D = 0, 2 2
or
Azz + αz + αz + D = 0
(2.17)
where α = 21 (B − iC), α = 21 (B + iC). The condition (2.16), for the equation (2.17) becomes αα − AD ≥ 0. Let us consider the z =
1 w
(2.18)
transform. Then the relation (2.17) becomes A + αw + αw + Dww = 0,
(2.19)
that also represents the equation of a circle. Theorem 2.6.3. The cross ratio is invariant by a bilinear transform T, i. e., (z1 , z2 , z3 , z4 ) = (Tz1 , Tz2 , Tz3 , Tz4 ), where (z1 , z2 , z3 , z4 ) =
z1 − z2 z3 − z2 : . z1 − z4 z3 − z4
Proof. It is evidently that Tz − Tz =
az + b az + b (ad − bc)(z − z ) − = . cz + d cz + d (cz + d)(cz + d)
Hence, we have Tz1 − Tz2 Tz3 − Tz2 : Tz1 − Tz4 Tz3 − Tz4 (ad − bc)(z1 − z2 ) (cz1 + d)(cz4 + d) (ad − bc)(z3 − z2 ) (cz3 + d)(cz4 + d) ⋅ : ⋅ = (cz1 + d)(cz2 + d) (ad − bc)(z1 − z4 ) (cz3 + d)(cz2 + d) (ad − bc)(z3 − z4 ) z − z2 z3 − z2 = 1 : . z1 − z4 z3 − z4
2.7 The Möbiustype groups  39
Theorem 2.6.4. Every arbitrary T bilinear transform is conformal in every z ∈ ℂ \ {− dc } point. Proof. We have Tz − Tz ad − bc , = z−z (cz + d)(cz + d) and considering z → z in this relation, we obtain that T (z) =
ad−bc (cz+d)2
≠ 0.
Theorem 2.6.5. If a, b, c, d ∈ ℝ and if ad − bc > 0, then T maps the upper halfplane into upper halfplane, and the lower halfplane into the lower halfplane. If a, b, c, d ∈ ℝ and if ad − bc < 0, then T maps the upper halfplane into lower halfplane, and the lower halfplane into the upper halfplane. Proof. Since Tz = ( then
az + b az + b )= = Tz, cz + d cz + d
Tz − Tz = Tz − Tz =
ad − bc (z − z). cz + d2
Hence, if z ∈ ℝ ⇒ z − z = 0 ⇒ Tz − Tz = 0 ⇒ Tz ∈ ℝ. If Im z > 0, (2.20)
z − z = 2i Im z ⇒ 2i Im Tz = Tz − Tz =
(2.20)
ad − bc 2i Im z. cz + d2
From here, it follows that Im Tz > 0 whenever ad − bc > 0, and Im Tz < 0 whenever ad − bc < 0. Remark 2.6.1. Each bilinear transform depends on three complex parameters. Hence, if we know the values of an unknown bilinear transform in three different points, then we may determine this transform.
2.7 The Möbiustype groups For two given bilinear transforms, T1 z =
a1 z + b1 c1 z + d1
and T2 z =
a2 z + b2 c2 z + d2
T1 ∘ T2 and T2 ∘ T1 are also bilinear transforms. To an arbitrary Tj bilinear transform, corresponds a matrix of the form (
aj cj
obtained with the aid of its coefficients.
bj ), dj
40  2 Holomorphic functions Denoting T3 = T1 ∘ T2 , the bilinear transform T3 has the corresponding matrix given by (
a1 c1
b1 a )⋅( 2 d1 c2
b2 a a + b1 c2 )=( 1 2 d2 a2 c1 + c2 d1
a1 b2 + b1 d2 ). b2 c1 + d1 d2
It follows that the matrix of T3 may be obtained from as the product of the corresponding matrices of T1 and T2 . Theorem 2.7.1. The bilinear transforms set is a group with respect to the functions composition. This group is called Möbiustype group. a b
Proof. Tj is a bilinear transform ⇔ det( c j dj ) ≠ 0. The determinant of the T1 ∘ T2 correj j sponding matrix is det T1 ⋅ det T2 , hence the matrix of Tj belongs to the group GL2 (ℂ). From this remark, we deduce that the set of all bilinear transforms represent a group with respect to the composition, which is a subgroup GL2 (ℂ). We call that the bilinear transformation group as it is the holomorphic image of GL2 (ℂ). Remark 2.7.1. An arbitrary bilinear transform remains unchanged if we multiply all its coefficients by the same complex number k, k ≠ 0. That is, az + b = cz + d
a z k c z k
+ +
b k d k
,
d ∀z ∈ ℂ \ {− }. c
The constant k may be chosen such that ad − bc = 1. Theorem 2.7.2. Those bilinear transforms with a, b, c, d ∈ ℝ and ad − bc = 1, represent the subgroup of the Möbiustype group that maps the upper halfplane into the upper halfplane. Proof. According to Theorem 2.6.5, if a, b, , c, d ∈ ℝ and ad − bc > 0, the transform Tz = az+b maps the upper halfplane into the upper halfplane. Dividing all its coefficients cz+d by √ad − bc, and changing the notation, we obtain ad − bc = 1. Conversely, suppose that T maps the upper halfplane into the upper halfplane. Hence, we will deduce that Tz ∈ ℝ if z ∈ ℝ, so T can be written in such a form that a, b, c, d ∈ ℝ. Case I: cd = 0 b 1. If d = 0, then Tz = ac + cz ∈ ℝ. Further, z ∈ ℝ, 2.
z→∞⇒
b a ∈ ℝ and z = 1 ⇒ ∈ ℝ. c c
If c = 0, then Tz = da z + db . Further, z=0⇒
b ∈ℝ d
and z = 1 ⇒
a ∈ ℝ. d
2.7 The Möbiustype groups  41
Figure 2.3: Proof of Theorem 2.7.3.
Case II: cd ≠ 0. In this case, we obtain z ∈ ℝ,
z→∞⇒
b a ∈ ℝ and z = 0 ⇒ ∈ ℝ, c d
hence a b ad − bc − = ∈ ℝ. c d cd From here, we have Tz =
a + c
bc−ad cd c z+1 d
∈ ℝ,
if z ∈ ℝ ⇒
c z + 1 ∈ ℝ, d
if z ∈ ℝ ⇒
c ∈ ℝ. d
Hence, if T maps ℝ into ℝ, the coefficients a, b, c and d may be chosen to be real numbers. From here, according to Theorem 2.6.5 we obtain that ad − bc > 0. Definition 2.7.1. Let (C) be the circle in ℝ2 with the center in O, and with the radius r > 0. The points A, B ∈ ℝ2 are called inverse with respect to the (C) circle, if O, A and B belong to the same halfline with the origin in O, and OA ⋅ OB = r 2 . Theorem 2.7.3. The points A and B are inverse points with respect to the circle (C), if and only if every circle (C ) which contains both these points is perpendicular (or orthogonal) to (C). (Two circles are called orthogonal if their tangents in the intersection points are orthogonal.) Proof. The circles (C) and (C ) are orthogonal, if their tangents in the intersection points are orthogonal, which is equivalent to the fact that the radius corresponding to one of the intersection points are orthogonal. This last affirmation is true, if and only if the radius (corresponding to one of the intersection points) of one circle is tangent to the other circle. Suppose that (C )⊥(C), and let P ∈ (C) ∩ (C ) (Figure 2.3). Then OP is tangent to (C ) and r 2 = OP 2 = OA ⋅ OB (the power of point O with respect to the circle (C )).
42  2 Holomorphic functions Conversely, let OP the tangent from O to the circle (C ), P ∈ (C ). Then r 2 = OA ⋅ OB = OP 2 ⇒ OP = r ⇒ P ∈ (C) ∩ (C ), and since OP is tangent to (C ) then OP is orthogonal to the radius O P of the circle (C ), hence we obtain that (C)⊥(C ). Theorem 2.7.4. Let (C) ⊂ ℂ be a circle, and let T be a bilinear transform. Then T((C)) = (C ) is also a circle. The T bilinear transform maps every z1 and z2 two inverse points with respect to the (C) circle, into the points z1 = Tz1 and z2 = Tz2 that are inverse points with respect to the (C ) circle. Proof. Let (C ) be an arbitrary circle that contains the points z1 and z2 . From Theorem 2.7.3, this arbitrary circle is orthogonal to (C). Since T is a conformal mapping, it maps orthogonal circles onto orthogonal circles. Let (C1 ) and (C2 ) be two arbitrary circles that contain the points z1 and z2 . Then the images (Cj ) = T((Cj )), j = 1, 2 are distinct circles that are orthogonal to (C ), and their intersections are zj = Tzj , j = 1, 2. But (Cj )⊥(C ), j = 1, 2, and using again Theorem 2.7.3 it follows that z1 and z2 are inverse points with respect to the circle (C ). Theorem 2.7.5. The functions of the group of the bilinear transforms that maps the disc U(0; r) onto itself, has the form Tz = r 2 eiθ
z−a , az − r 2
θ ∈ ℝ, a < r.
Proof. Let a ∈ U(0; r) be such a point, so that Ta = 0, and let z1 the inverse of a with respect to the circle (C) = {z ∈ ℂ : z = r}. Then az1  = r 2 Since a and
r2 a
and
arg a = arg z1 ⇒ z1 =
r2 r 2 i arg a r 2 a r 2 i arg z1 e e = = . = a a a a a 2
are inverse points, then Ta = 0 and T( ra ) are also inverse points with 2
respect to the (C) circle (T maps the circle (C) onto itself). But 0 and T( ra ) could be 2
inverse points if and only if T( ra ) = ∞. Suppose that Tz = Hence,
a1 z+b1 . From aa1 +b1 c1 z+d1
a z+ Tz = 1 c1 z +
b1 a1 d1 c1
=
2
= 0 it follows that a = − ba1 and c1 ra +d1 = 0.
a1 z − a z−a , =k c1 z − r 2 az − r 2 a
1
where k =
a1 a . c1
If z = r, then Tz = r. We have z = r ⇒ r 2 = Tr ⋅ Tr = kk
r−a r−a r 2 − r(a + a) + aa 1 = k2 2 = kk 2 2 4 3 2 ra − r ra − r r − r (a + a) + r aa r
⇒ k2 = r 4 ⇒ k = r 2 eiθ , z−a Hence, Tz = r 2 eiθ az−r 2.
θ ∈ ℝ.
2.8 Multivalued functions  43
Remarks 2.7.2. 1. If we multiply the fraction by −1 = eiπ , and we replace θ + π by θ , then Tz = r 2 eiθ 2.
z−a , r 2 − az
θ ∈ ℝ, a < r.
A special case is that when r = 1. Those bilinear transforms that maps the U = U(0; 1) unit disc onto itself has the general form Tz = eiθ
z−a , 1 − az
θ ∈ ℝ, a < 1.
Example 2.7.1. Determine those bilinear transforms that map the upper halfplane onto the unit disc U = U(0; 1). Solution. Suppose that the bilinear transform T maps the upper halfplane onto the unit disc U. Let a ∈ {z ∈ ℂ : Im z > 0} that point, such that Ta = 0 ∈ U. The function T maps the line Im z = 0 onto the circle z = 1. The points a and a are inverse points with respect to the “circle” Im z = 0, since every arbitrary circle that contain these points are orthogonal to the line Im z = 0. Hence, Ta and Ta are inverse with respect to the circle z = 1, and since Ta = 0 it is necessary that Ta = ∞. Hence, Tz = k
z−a , z−a
where Im a > 0.
But 0 ∈ {z ∈ ℂ : Im z = 0} ⇒ T0 belongs to the boundary of U, hence 0 − a T0 = k = k = 1, 0 − a so it follows that k = eiθ , θ ∈ ℝ. We obtained in such a way that the required bilinear transforms have the general form: Tz = eiθ
z−a , z−a
θ ∈ ℝ, Im a > 0.
(2.21)
2.8 Multivalued functions 2.8.1 The logarithmic function The realvalued logarithmic function is the inverse of the exponential function. There exists this inverse because the realvalued exponential function is a bijection between ℝ and ℝ+ \ {0}.
44  2 Holomorphic functions We saw previously that the complexvalued exponential function ez is a periodical function, with the period equal to 2πi. Thus, the complexvalued exponential function could be injective at most in an open set included into a strip, parallel with the Ox axis, with the width equal to 2π. For any arbitrary z ∈ ℂ∗ , let us consider the equation ew = z, with the unknown w. Letting w = u + iv, then z = eu+iv = eu eiv = reiθ ,
where z = reiθ .
Hence eu = r, i. e., u = ln r = ln z and v = θ + 2kπ, k ∈ ℤ. We conclude that any complex number of the form wk = ln r + i(θ + 2kπ), k ∈ ℤ, satisfy the given equation. We will denote by Log z, the set of all the solutions of the equation ew = z, i. e., Log z = {wk : k ∈ ℤ} = {ln r + i(θ + 2kπ) : k ∈ ℤ}, or Log z = ln z + iArg z. Let D = ℂ \ ({z ∈ ℂ : Re z < 0, Im z = 0} ∪ {0}) = {z ∈ ℂ :  arg z < π}, where k ∈ ℤ is a given integer. Then we denote logk z := ln z + i(arg z + 2kπ),
 arg z < π,
and the function logk : D → ℂ is called a branch of the multivalued Log function. The log0 function is called the main branch of the multivalued Log function, and we will denote it by log, i. e., log z := log0 z; we have log 1 = 0 (Figure 2.4). If we let z = reiθ , θ ∈ (−π, π), then log z = ln r + iθ. Denoting log z = u(z) + iv(z), where z = x+iy, then u(x, y) = ln √x2 + y2 and v(x, y) = arctan xy +kπ, k ∈ ℤ. Since u, v ∈ C 1 (D) and these functions satisfy the Cauchy–Riemann conditions on D, it follows that the function log is differentiable on D and (log z) =
1 . z
Thus, computing the derivative of the elog z = z identity, we deduce that (elog z ) = elog z (log z) = 1 ⇒ (log z) =
1 1 = . elog z z
2.8 Multivalued functions  45
Figure 2.4: The log function.
The next differential equation system (Cauchytype problem) {
f (z) =
1 z
f (1) = 0,
z∈D
has the solution f = log, while the generalized differential equation system {
f (z) =
1 z
f (1) = 2kπi,
z ∈ D, k ∈ ℤ
has the solution f = logk . Since logk : D → ℂ,
logk z = ln r + i(θ + 2kπ),
where z = reiθ , θ ∈ (−π, π),
it follows that logk = log +2kπi. 2.8.2 Inverse trigonometric functions Let z ∈ ℂ be a fixed number, and let us consider the next equation, with the unknown w: iw −iw 1. cos w = z, where cos w = e +e . 2 eiw +e−iw 2iw Hence, from = z we get e − 2zeiw + 1 = 0, i. e., eiw = z ± √z 2 − 1. It follows that
2
iw = Log (z ± √z 2 − 1) and Arccos z := w = −iLog (z ± √z 2 − 1).
46  2 Holomorphic functions Similarly, we will consider the equation: iw −iw 2. sin w = z, where sin w = e −e . 2i
Hence, from follows that
eiw −e−iw 2i
= z, we get e2iw − 2izeiw − 1 = 0, i. e., eiw = iz ± √1 − z 2 . It iw = Log (iz ± √1 − z 2 )
and Arcsinz := w = −iLog (iz ± √1 − z 2 ). Like in the previous two cases, let us consider the equation: iw −iw sin w . 3. tan w = z, where tan w = cos = i(ee iw−e w +e−iw ) Hence, from
i. e., e2iw =
1+iz . 1−iz
eiw −e−iw eiw +e−iw
= iz we get e2iw − 1 = iz(e2iw + 1) = 0 ⇔ (1 − iz)e2iw = 1 + iz,
It follows that
2iw = Log
1 + iz 1 − iz
and 1 + iz i . Arctan z := w = − Log 2 1 − iz 2.8.3 The power function Letting α ∈ ℂ, then we will define the function z α by z α = eαLog z . It follows that the function F : ℂ∗ → 𝒫 (ℂ),
F(z) = z α
is a multivalued function. Considering the main branch log of the multivalued function Log , let us define the function f by f (z) = eα log z ,
f (1) = 1.
This function is called the main branch of the multivalued function z α , and moreover, f (z) =
α α log z αz α e = = αz α−1 . z z
2.9 Exercises  47
For the special case α = n ∈ ℕ, we deduce that n
z n = enLog z = en(ln r+i(θ+2kπ)) = en ln r eniθ e2knπi = r n eniθ = (reiθ ) = z n . Thus, the new definition of the power functions reduces for the special case α = n ∈ ℕ to the wellknown definition of the power functions with natural exponents. Similarly, for α = n1 , n ∈ ℕ, we have 1
1
1
1
z n = e n Log z = e n (ln r+i(θ+2kπ)) = e n ln r ei = √n r(cos
θ + 2kπ θ + 2kπ + i sin ), n n
θ+2kπ n
= √n rei
θ+2kπ n
k ∈ {0, 1, . . . , n − 1},
and we obtain the wellknown definitions of the nth order root functions.
2.9 Exercises 2.9.1 Real variable complex functions Exercise 2.9.1. Represent in the complex plane the graphics of the following functions: 1. f : ℝ → ℂ, f (t) = −3 + i + (1 + i)t;
2.
3.
f : [1, 3] → ℂ, f (t) = 2 − 3i + (1 − 2i)t;
f : ℝ → ℂ, f (t) =
4. f : ℝ → ℂ, f (t) =
16 ; 3−i+(1+2i)t −12+2i−(1−3i)t . 1+3i−(2−i)t
2.9.2 The derivative of a complex function Exercise 2.9.2. Prove that the function f : ℂ → ℂ, f (z) = z, is not differentiable in any point of ℂ. Exercise 2.9.3. Let us consider the function f : ℂ → ℂ, f (z) = z 2 + zz − z 2 + 2z − z. Determine those points from ℂ in which f is differentiable, and compute its derivative f in these points. Exercise 2.9.4. Determine those points of ℂ in which the function f (z) = z Re z is differentiable, and compute its derivative f in these points. Exercise 2.9.5. Determine those points of ℂ in which the function f (z) = 2z + z 2 is differentiable, and compute its derivative f in these points. Exercise 2.9.6. Let us consider the function defined by f (z) = ln √x2 + y2 + i arctan xy , where z = x + iy. Prove that f ∈ H(ℂ∗ ), and compute its derivative f .
48  2 Holomorphic functions Exercise 2.9.7. Prove that the function defined by f (z) = (ex − e−x ) cos y + i(ex + e−x ) sin y, where z = x + iy, is an entire function, and compute its derivative f . Exercise 2.9.8. Determine the values of a, b, c, d ∈ ℝ, such that the next functions will be differentiable in ℂ: 1. f : ℂ → ℂ, f (z) = x + ay + i(bx + cy); 2. f : ℂ → ℂ, f (z) = x2 + axy + by2 + i(cx 2 + dxy + y2 ); 3. f : ℂ → ℂ, f (z) = cos x(cosh y+a sinh y)+i sin x(cosh y+b sinh y), where z = x +iy. Exercise 2.9.9. Write the Cauchy–Riemann equation system, and compute the derivative f , in the next three cases: 1. If z = reiθ and f (z) = u(z) + iv(z), where the variables are r and θ; 2. If f (z) = R(z)eiΦ(z) , z = x + iy, where the variables are x and y; 3. If z = reiθ and f (z) = R(z)eiΦ(z) , where the variables are r and θ, supposing that the function f is holomorphic, and the functions R and Φ are real differentiable functions. Exercise 2.9.10. Prove that the function f (z) = z n is an entire function ∀n ∈ ℕ, and calculate its derivative. Exercise 2.9.11. Let f : D → ℂ a differentiable non constant function on the domain D ⊂ ℂ. Prove that the image f (D) cannot be a segment, i. e., there does not exist points a, b ∈ ℂ, such that f (D) = {z ∈ ℂ : z = (1 − t)a + tb, t ∈ [0, 1]}. Exercise 2.9.12. Let G be an open set, and let the function f : G → ℂ, f = u + iv. Prove that: f (ζ )−f (z) 1. If for all z ∈ G there exists limζ →z Re ζ −z , then there exist the partial derivatives
2.
𝜕u 𝜕v , 𝜕x 𝜕y
in G, and
𝜕u 𝜕v , 𝜕y 𝜕x
in G, and
𝜕u(z) 𝜕x
=
𝜕v(z) , 𝜕y
∀z ∈ G;
=
− 𝜕v(z) , 𝜕x
If for all z ∈ G there exists limζ →z Im
tives
𝜕u(z) 𝜕y
f (ζ )−f (z) , ζ −z
then there exists the partial deriva
∀z ∈ G.
Exercise 2.9.13. Let us define the function ez as follows: ez = lim (1 + n→+∞
n
z ) , n
where z = x + iy.
Prove that ez = ex+iy = ex (cos y + i sin y),
x, y ∈ ℝ.
2.9 Exercises  49
2.9.3 Entire functions Exercise 2.9.14. Solve the following equations: 1.
sin z − cos z = i; 2.
4. cos z = 7.
3+i ; 4
5.
ei3z = −1;
8.
cosh z − i sinh z = 1;
3.
sin z =
4i ; 3 5i ; 3
sinh z = 2i ;
6. tan z =
e
9. tanh z = 2.
1 z2
= 1;
Exercise 2.9.15. Find the algebraic form of the complex numbers: 1.
ei ;
5.
tan(1 − 2i); 6.
2.
sinh 2i;
3. cosh(2 + 3i);
log(−2i);
7.
4.
log(−3 + 4i); 8.
cos(1 − i); log √1−i . 3+i
Exercise 2.9.16. Determine the values of the next powers: 1. i1−i ;
2.
(1 + i√3)i ;
3. 1−i ;
4. e i . √
Exercise 2.9.17. Prove the following identities: 1. sin z = −i sinh(iz), sinh(iz) = i sin z;
2.
cosh z = cos(iz), cosh(iz) = cos z;
3.
sinh(z1 ± z2 ) = sinh z1 cosh z2 ± sinh z2 cosh z1 ;
5.
sin 2z = 2 sin z cos z;
7.
sinh2 z
4. cos(z1 ± z2 ) = cos z1 cos z2 ∓ sin z1 sin z2 ;
6.
2 tan z ; 1−tan2 z cosh 2z−1 = , 2
tan 2z =
cosh2 z =
cosh 2z+1 ; 2
8.  sin z = √sin2 x + sinh2 y = √ 21 (cosh 2y − cos 2x), where z = x + iy; 9.
 cos z = √cos2 x + sinh2 y = √ 21 (cosh 2y + cos 2x), where z = x + iy;
10. tan z = −i tanh iz, 2
tan iz = i tanh z;
11. cosh 2z = cosh z + sinh2 z =
1+tanh2 z . 1−tanh2 z
2.9.4 Bilinear transforms Exercise 2.9.18. Determine the bilinear transform that maps the points z1 = 1 + i, z2 = ∞ and z3 = 1 into the points w1 = i, w2 = −1 and w3 = ∞, respectively. Then prove that this bilinear transform conformally maps the disc {z = x + iy ∈ ℂ : (x − 1)2 + (y − 1)2 < 1} onto the halfplane {w = u + iv ∈ ℂ : u + v > 0}. Exercise 2.9.19. Prove that the function w = z−1 conformally maps the halfplane z+1 {z ∈ ℂ : Re z > 0} onto the disc {w ∈ ℂ : w < 1}.
50  2 Holomorphic functions Exercise 2.9.20. Determine the bilinear transform w = f (z) that conformally maps the halfplane {z ∈ ℂ : Re z < 1} onto the disc {w ∈ ℂ : w < R}, such that f (−1) = 0 and f (−1) = 1. Find also the value of the radius R. 1+(2β−1)z
Exercise 2.9.21. Suppose that β < 1, and let f (z) = 1+z . Prove that the function f conformally maps the unit disc {z ∈ ℂ : z < 1} onto the halfplane {w ∈ ℂ : Re w > β}. Exercise 2.9.22. Prove that the function f (z) = z + az 2 , a ∈ ℂ∗ , is injective on the disc 1 }. {z ∈ ℂ : z < 2a Exercise 2.9.23. Let us define the function w(z) = z 2 + 5z + 6. 1. Determine the image of the domain {z ∈ ℂ : 0 < Re z < 1, Im z > 0} by the w function; 2. Prove that the restriction of the function w to {z ∈ ℂ : Re z > − 52 } has an inverse, and determine this inverse; 3. Prove that the restriction of the function w to {z ∈ ℂ : Re z < − 52 } has an inverse, and determine this inverse. Exercise 2.9.24. Determine a function w = f (z) that conformally maps the strip {z ∈ ℂ : 0 < Re z < a}, onto the upper halfplane {w ∈ ℂ : Im w > 0}.
a > 0,
3 The complex integration 3.1 The homotopic theory of the paths The homotopic theory of the paths deals with the continuous deformations of the path, and it allows to define and to study the properties of the complex integral. Definition 3.1.1. 1. Let G be an open set of the complex plane ℂ. The continuous function γ : [0, 1] → G is called a path in G, where [0, 1] is the close real interval between 0 and 1, where the topology of ℝ is given by the euclidian topology. 2. The point z1 = γ(0) is called the starting point of the path γ, while the point z2 = γ(1) is called the end point of the path γ. 3. We denote by 𝒟G (z1 , z2 ) the set of all the paths in G with the starting point z1 and the end point z2 . 4. Whenever z1 = γ(0) = γ(1) = z2 , we say that the path γ is a closed path. 5. The set of all the path in G with the same starting and end point z0 , is called the set of all the closed paths in G starting from z0 , and it is denoted by 𝒟G (z0 ). 6. The set {γ} = γ([0, 1]) is called the image of the path γ. Definition 3.1.2. Let γ1 , γ2 ∈ 𝒟G (z1 , z2 ). We say that the path γ1 is homotopic with γ2 in G, if there exists a continuous function φ : S × T → G, where S = T = [0, 1], that satisfies the next conditions: φ(0, t) = γ1 (t),
and
φ(s, 0) = z1 ,
φ(1, t) = γ2 (t), φ(s, 1) = z2 ,
∀t ∈ [0, 1]
∀s ∈ [0, 1].
The function φ is called the homotopy between the paths γ1 and γ2 (in the set G), or the function that establishes the homotopy between the paths γ1 and γ2 (in the set G). Hence, for all s ∈ S we have φ(s, ⋅) ∈ 𝒟G (z1 , z2 ), which geometrically means that the function φ, with the variable s, is a continuous deformation of the path γ1 to the path γ2 , whenever the argument s runs on the whole interval [0, 1]. Notation. We denote by a γ1 ∼ γ2 the fact that the path γ1 is homotopic with the path γ2 in the set G.
G
Theorem 3.1.1. The relation “∼”1 is an equivalence relation in the set of the paths G
𝒟G (z1 , z2 ).
1 A relation ∼ is said to be an equivalence relation ⇔ it is reflexive, transitive and symmetric. G
https://doi.org/10.1515/9783110657869003
52  3 The complex integration Proof. If γ is an arbitrary path in G, then γ ∼ γ, because the function φ(s, t) = γ(t), G
φ : S × T → G is continuous and satisfies the conditions of the homotopy. Thus, the relation “∼” is reflexive, i. e., the path γ is homotopic with itself. G
Supposing that γ1 ∼ γ2 , we will prove that γ2 ∼ γ1 . Let φ : S × T → G be the G
G
homotopy between γ1 and γ2 , i. e.,
φ(0, t) = γ1 (t),
and
φ(s, 0) = z1 ,
φ(1, t) = γ2 (t), φ(s, 1) = z2 ,
∀t ∈ [0, 1]
∀s ∈ [0, 1].
If we let φ1 (s, t) = φ(1 − s, t), then φ1 : S × T → G is continuous, and φ1 (0, t) = φ(1, t) = γ2 (t), and
φ1 (s, 0) = φ(1 − s, 0) = z1 ,
φ1 (1, t) = φ(0, t) = γ1 (t),
∀t ∈ [0, 1]
φ1 (s, 1) = φ(1 − s, 1) = z2 ,
∀s ∈ [0, 1].
Hence γ2 ∼ γ1 , i. e., the relation “∼” is a symmetric relation. G
G
Suppose that γ1 ∼ γ2 and γ2 ∼ γ3 . Then there exist the continuous functions φ1 , φ2 : G
G
S × T → G, such that
φ1 (0, t) = γ1 (t),
φ1 (s, 0) = z1 ,
φ1 (1, t) = γ2 (t),
∀t ∈ [0, 1]
φ2 (1, t) = γ3 (t),
∀t ∈ [0, 1]
φ1 (s, 1) = z2 ,
∀s ∈ [0, 1]
and φ2 (0, t) = γ2 (t),
φ2 (s, 0) = z1 ,
φ2 (s, 1) = z2 ,
∀s ∈ [0, 1].
Let us define the function φ : S × T → G by φ(s, t) = {
φ1 (2s, t), φ2 (2s − 1, t),
s ∈ [0, 21 ]
s ∈ [ 21 , 1].
The function φ is welldefined, and since φ1 (1, t) = γ2 (t) = φ2 (t), ∀t ∈ [0, 1], the function φ is continuous. From the facts, φ(0, t) = γ1 (t), and
φ(s, 0) = z1 ,
φ(1, t) = γ3 (t), φ(s, 1) = z2 ,
∀t ∈ [0, 1]
∀s ∈ [0, 1],
it follows that φ is the function that establishes the homotopy γ1 ∼ γ3 , thus the relation “∼” is a transitive relation. G
G
3.1 The homotopic theory of the paths  53
Definition 3.1.3. 1. Let γ1 ∈ 𝒟G (z1 , z2 ), γ2 ∈ 𝒟G (z2 , z3 ). We denote by γ1 ∪ γ2 the path defined as follows: (γ1 ∪ γ2 )(t) = {
2.
γ1 (2t), γ2 (2t − 1),
if t ∈ [0, 21 ]
if t ∈ [ 21 , 1].
Then γ1 ∪ γ2 ∈ 𝒟G (z1 , z3 ), and the path γ1 ∪ γ2 is called the union of the paths γ1 and γ2 . We may similarly define the union of many paths. Thus, if γ1 ∈ 𝒟G (z1 , z2 ), γ2 ∈ 𝒟G (z2 , z3 ), . . . , γn ∈ 𝒟G (zn , zn+1 ), then the union of these paths is denoted by γ1 ∪ γ2 ∪ ⋅ ⋅ ⋅ ∪ γn , and is given by γ(t) = (γ1 ∪ γ2 ∪ ⋅ ⋅ ⋅ ∪ γn )(t) = γk+1 (nt − k), k ∈ {0, 1, . . . , n − 1}.
k k+1 ], t∈[ , n n
Definition 3.1.4. Let γ be a path, and let be Δ = (t0 , t1 , . . . , tn ) be a division of the interval [0, 1], i. e., 0 = t0 < t1 < ⋅ ⋅ ⋅ < tn = 1. The path system (γ1 , γ2 , . . . , γn ) is called the decomposition of the path γ by the division Δ, if γk = γ ∘ hk , where hk are the affine functions hk : [0, 1] → [tk−1 , tk ], defined by hk (t) = tk−1 + t(tk − tk−1 ), k ∈ {1, . . . , n}. We may see easily that {γk } = (γ ∘ hk )([0, 1]) = γ([tk−1 , tk ]). Theorem 3.1.2. If γ is a path in G, and f : [0, 1] → [0, 1] is continuous and surjective function, with f (0) = 0, f (1) = 1, then the path γ1 = γ ∘ f is homotopic with the path γ. Proof. Letting φ : S × T → G, where φ(s, t) = γ(t + s(f (t) − t)), then φ is a continuous function and φ(0, t) = γ(t),
φ(1, t) = (γ ∘ f )(t),
φ(s, 0) = γ(0) = z1 ,
∀t ∈ [0, 1],
φ(s, 1) = γ(1) = z2 ,
∀s ∈ [0, 1],
which prove the conclusion. It is obvious that {γ} = {γ ∘f }, that is the images of the both of these paths coincide. Theorem 3.1.3. If (γ1 , γ2 , . . . , γn ) is a decomposition of the path γ in G, then the path γ is homotopic with the path γ1 ∪ γ2 ∪ . . . γn in G. Proof. Let (γ1 , γ2 , . . . , γn ) the decomposition of the path γ by the division Δ = (t0 , t1 , . . . , tn ), i. e., 0 = t0 < t1 < ⋅ ⋅ ⋅ < tn = 1 and hk (t) = tk−1 + t(tk − tk−1 ), t ∈ [0, 1].
54  3 The complex integration Then γ1 ∪γ2 ∪⋅ ⋅ ⋅∪γn = γ∘(h1 ∪h2 ∪⋅ ⋅ ⋅∪hn ). Denote by h the function h : [0, 1] → [0, 1], with h = h1 ∪h2 ∪⋅ ⋅ ⋅∪hn . Thus, the function h is continuous with h(0) = h1 (0) = 0, h(1) = hn (1) = 1, and according to Theorem 3.1.2 we conclude γ ∼ γ ∘ h = γ1 ∪ γ2 ∪ . . . γn . G
Theorem 3.1.4. Let γ1 ∈ 𝒟G (z1 , z2 ), γ2 ∈ 𝒟G (z2 , z3 ), and γ3 ∈ 𝒟G (z3 , z4 ). Then (γ1 ∪ γ2 ) ∪ γ3 ∼ γ1 ∪ (γ2 ∪ γ3 ). G
Proof. Define the function h : [0, 1] → [0, 1] by 2t, { { { h(t) = { t + 41 , { { 1+t { 2 ,
t ∈ [0, 41 ],
t ∈ [ 41 , 21 ],
t ∈ [ 21 , 1].
Then (γ1 ∪ γ2 ) ∪ γ3 = [γ1 ∪ (γ2 ∪ γ3 )] ∘ h, and from Theorem 3.1.2 we deduce that (γ1 ∪ γ2 ) ∪ γ3 = [γ1 ∪ (γ2 ∪ γ3 )] ∘ h ∼ γ1 ∪ (γ2 ∪ γ3 ). G
Definition 3.1.5. If γ ∈ 𝒟G (z1 , z2 ), then the path γ − ∈ 𝒟G (z2 , z1 ) defined by γ − (t) = γ(1 − t), t ∈ [0, 1], is called the inverse of the path γ. Theorem 3.1.5. If γ1 , γ2 ∈ 𝒟G (z1 , z2 ) and γ1 ∼ γ2 , then γ1− ∼ γ2− . G
G
Proof. If φ : S × T → G is the function that establishes the homotopy γ1 ∼ γ2 , then the function φ− (s, t) = φ(s, 1 − t) will be the homotopy between γ1− and γ2− .
G
Theorem 3.1.6. Let γ1 , γ3 ∈ 𝒟G (z1 , z2 ) and γ2 , γ4 ∈ 𝒟G (z2 , z3 ). If γ1 ∼ γ3 and γ2 ∼ γ4 , then G G γ1 ∪ γ2 ∼ γ3 ∪ γ4 . G
Proof. Let φ1 and φ2 the functions that establish the homotopies γ1 ∼ γ3 , and γ2 ∼ γ4 G
respectively. Then the function
φ(s, t) = {
φ1 (s, 2t), φ2 (s, 2t − 1),
G
if t ∈ [0, 21 ], if t ∈ [ 21 , 1]
is the homotopy between γ1 ∪ γ2 and γ3 ∪ γ4 . Definition 3.1.6. If z ∈ G, then we will denote by ez the constant path defined by ez (t) = z, ∀t ∈ [0, 1]. We have that ez ∈ 𝒟G (z). The path γ ∈ 𝒟G (z) is said to be homotopic to a point in G, if γ ∼ ez . G
Theorem 3.1.7. For all γ ∈ 𝒟G (z1 , z2 ), we have ez1 ∪ γ ∼ γ and γ ∼ γ ∪ ez2 . G
Proof. Let us define the function f : [0, 1] → [0, 1] by f (t) = {
0, 2t − 1,
if t ∈ [0, 21 ],
if t ∈ [ 21 , 1].
G
3.1 The homotopic theory of the paths  55
Thus, ez1 ∪ γ = γ ∘ f , and from Theorem 3.1.2 we have γ ∘ f ∼ γ, hence ez1 ∪ γ ∼ γ. The G
G
second conclusion may be proved in a similar way.
Theorem 3.1.8. Let γ ∈ 𝒟G (z1 , z2 ). Then γ ∪ γ − ∼ ez1 , i. e., γ ∪ γ − is homotopic to a point G
in G.
Proof. Let φ(s, t) = {
γ(2st), γ(2s(1 − t)),
if t ∈ [0, 21 ], if t ∈ [ 21 , 1].
Thus, the function φ establishes the homotopy ez1 ∼ γ ∪ γ − . G
Theorem 3.1.9. Let γ1 , γ2 ∈ 𝒟G (z1 , z2 ) and γ1 ∼ γ2 . Then γ1 ∪ γ2− ∼ ez1 , that is γ1 ∪ γ2− is G
homotopic to a point in G.
G
Proof. Combining Theorem 3.1.5 with Theorem 3.1.6, we have γ1 ∪ γ2− ∼ γ1 ∪ γ1− . AccordG
ing to the Theorem 3.1.8, we get ez1 ∼ γ1 ∪ γ1− , and the conclusion follows by using the G
transitivity of the homotopy relation.
3.1.1 Simply connected domains Definition 3.1.7. 1. The subset A ⊂ ℂ is said to be arcconnected, if any arbitrary two points of A can be connected by a path in A (i. e., there exists a path in A, whose starting and end points are any two points in A). 2. If z1 , z2 ∈ ℂ, the linear path that connects the points z1 and z2 is the path λ, defined by λ(t) = z1 + t(z2 − z1 ),
t ∈ [0, 1].
3.
The path γ is called a broken line, if there exists such a decomposition of it which elements are linear path. 4. The subset A ⊂ ℂ is said to be connected by broken lines, if any arbitrary two points of A can be connected by a broken line in A. We see immediately that every broken line connected set is a connected by arcs set. The reverse is not true! Theorem 3.1.10. The set D ⊂ ℂ is a domain, if and only if D is open and a broken line connected set. Proof. “⇐” 1. First, we will prove that any open set D, which is connected by arcs will be a domain.
56  3 The complex integration If this is not true, there exists a nonempty subset A ⊂ D, such that A is open and closed in the D metric space. Then the set D \ A = B has the same property, i. e., it is open and closed in the D metric space. Let z1 ∈ A, z2 ∈ B and let γ be a path that connects these two points. Then A ∩ {γ} and B ∩ {γ} are nonempty, disjoint and closed subsets of {γ}, and hence the sets M = γ −1 (A ∩ {γ}) and N = γ −1 (B ∩ {γ}) are disjoint, nonempty and closed subsets of [0, 1], with M ∪ N = [0, 1]. Suppose that 1 ∉ M, and let tm the maximum (the biggest number) of the set M (this exists, because M is compact). Then tm ∉ N and (tm , 1] ⊂ N, hence there exists a sequence of points of N that tends to tm . This contradicts the fact that N is a closed set, and the above property is proved. 2. If D is an open set and it is broken line connected, then D is connected by arcs, and according to the first point, we conclude that it is a domain. “⇒” Let z1 , z2 be two arbitrary points of the domain D. Let define by A, A ⊂ D, the set of all the points of D that can be connected by a broken line with the point z1 . Let z ∈ A be an arbitrary point. Since D is open, there exists a number r > 0, such that U(z; r) ⊂ D. Any arbitrary point z ∈ U(z; r) can be connected by broken lines with z1 in D (the points z1 and z can be connected by a broken line, while the points z and z can be connected by a linear path). It follows that U(z; r) ⊂ A, hence A is open in D. Let z ∈ D be the limit of an arbitrary sequence of the set A. Since D is open, there exists a number r > 0, such that U(z; r) ⊂ D. Then the above mentioned sequence has an element z ∈ U(z; r). The points z1 and z can be connected by a broken line, while z and z can be connected by a linear path in D, hence the points z1 and z can be connected by a broken line in D. Thus, the set A is closed in the metric space D. Since D is a domain, it follows that A = D, hence z2 ∈ A, and thus z1 and z2 can be connected by a broken line in D. Definition 3.1.8. 1. Let γ be a path, and let Δ = (t0 , t1 , . . . , tn ) be a division of the interval [0, 1]. Then the sum n
V(γ, Δ) = ∑ γ(tk ) − γ(tk−1 ) k=1
2.
is called the variation of the path γ corresponding to the division Δ. If l(γ) = sup{V(γ, Δ) : Δ division} < +∞, then we say that the path γ has finite variation. The number l(γ) is called the total variation of the path γ, it represents the length of the path (curve) γ. In such a case, the path γ is said to be a rectifiable path.
3.1 The homotopic theory of the paths  57
Corollary 3.1.1. Since every broken line has finite variation, from Theorem 3.1.10 it follows immediately that any two points of a domain can be connected by a rectifiable path. Definition 3.1.9. The domain D ⊂ ℂ is called a simply connected domain, if any closed path in D is homotopic to a point in D. Theorem 3.1.11. The domain D ⊂ ℂ is simply connected, if and only if any arbitrary two paths in D, with the same start and the same end points, are homotopic in D. Proof. Suppose that the domain D is simply connected, and let γ1 , γ2 ∈ 𝒟D (z1 , z2 ). Then the path γ1 ∪ γ2− ∈ 𝒟D (z1 ) is closed, and from the above definition we have γ1 ∪ γ2− ∼ ez1 .
Using some of the properties prove in the previous section, we obtain that
D
γ1 ∼ γ1 ∪ ez2 ∼ γ1 ∪ (γ2− ∪ γ2 ) ∼ (γ1 ∪ γ2− ) ∪ γ2 ∼ ez1 ∪ γ2 ∼ γ2 , D
D
D
D
D
thus γ1 ∼ γ2 . D
Conversely, if any arbitrary two paths in D, with the same start and the same end points, are homotopic in D, then every path γ ∈ 𝒟D (z) satisfies γ ∼ ez , because γ and D
ez have the same start and end points. Hence, the domain D is simply connected.
Definition 3.1.10. 1. The domain D is said to be starlike with respect to the point z0 ∈ D, if ∀z ∈ D we have {λz } ⊂ D, where λz is the linear path that connects the points z0 and z. 2. The domain D is said to be convex, if it is starlike with respect to any arbitrary point of D. Remarks 3.1.1. 1. The above last condition is equivalent to the fact, that the linear path that connects any arbitrary point of D is included in D. 2. Any starlike set is a connected by arcs set because every point of the set can be connected by a broken line that consists in the union of two linear paths (each linear paths of the broken line connects the corresponding point with z0 ). Theorem 3.1.12. Any starlike domain is a simply connected domain. Proof. Suppose that the domain D is starlike with respect to the point z0 ∈ D, and let γ ∈ 𝒟D (z0 ). Then the function φ(s, t) = z0 +s(γ(t)−z0 ) establishes the ez0 ∼ γ homotopy. D
Consider any z ∈ D, and let γ ∈ 𝒟D (z) be an arbitrary path. If λ is the linear path that connects z0 and z (there exists, because D is starlike), then λ ∪ γ ∪ λ− ∈ 𝒟D (z0 ) and from some of the previous results λ ∪ γ ∪ λ− ∼ ez0 . Using the properties proved in the previous section, we get
D
γ ∼ λ− ∪ (λ ∪ γ ∪ λ− ) ∪ λ ∼ λ− ∪ ez0 ∪ λ ∼ ez . D
D
D
58  3 The complex integration Thus, we proved that any closed path in D is homotopic to a point, hence D is simply connected.
3.1.2 Functions of bounded variation and paths Definition 3.1.11. 1. Let [a, b] ⊂ ℝ, f : [a, b] → ℂ, and let Δ = (t0 , t1 , . . . , tn ) be a division of the interval [a, b]. Let denote by ‖Δ‖ = max{tk − tk−1 : k ∈ {1, 2, . . . , n}} the norm of the division Δ. The variation of the complex valued function f corresponding to the division Δ is the sum n
V(f , Δ) = ∑ f (tk ) − f (tk−1 ). k=1
2.
The number V(f ; [a, b]) = sup V(f , Δ) Δ
3.
is called the total variation of the function f on the interval [a, b]. The function f is said to be a function with bounded variation, if V(f ; [a, b]) < +∞. The set of all the functions with bounded variation defined on the interval [a, b], is denoted by BV[a, b].
Remarks 3.1.2. 1. If γ : [0, 1] → ℂ and Δ is a division of the interval [0, 1], then the variation of f corresponding to Δ, i. e., V(γ, Δ), represents the length of the broken line that connects the points γ(tk ), k ∈ {0, 1, . . . , n}, that belong to the image {γ}. 2. If f = u + iv, from the inequalities u(tk ) − u(tk−1 ) ≤ f (tk ) − f (tk−1 ) ≤ u(tk ) − u(tk−1 ) + v(tk ) − v(tk−1 ) and v(tk ) − v(tk−1 ) ≤ f (tk ) − f (tk−1 ) ≤ u(tk ) − u(tk−1 ) + v(tk ) − v(tk−1 ) it follows that the function f is a function with bounded variation, if and only if the real and the imaginary parts of f are functions with bounded variation.
3.1 The homotopic theory of the paths  59
3.
If f , g : [a, b] → ℂ are two functions with bounded variation, then αf + βg with α, β ∈ ℂ is a function with bounded variation. Let c ∈ (a, b). The function f is a function with bounded variation on the interval [a, b], if f is a function with bounded variation on the intervals [a, c] and [c, b]. If f is a function with bounded variation, then f  is also a function with bounded variation. 4. Every path γ : [0, 1] → ℂ can be uniformly approximated by a broken line consisting of the points γ(tk ), k ∈ {0, 1, . . . , n} (that belong to the image {γ}), i. e., ∀ε > 0, ∃γε a broken line consisting of the points γ(tk ), k ∈ {0, 1, . . . , n}, such that γ(t) − γε (t) < ε,
∀t ∈ [0, 1].
Remark 3.1.3. If (γn )n∈ℕ∗ is a sequence of paths of G that uniformly converges to the path γ of G, i. e., ∀ε > 0, ∃nε ∈ ℕ∗
such that γ(t) − γn (t) < ε, ∀n > nε , ∀t ∈ [0, 1],
and f : G → ℂ is a continuous function on the open set G, then the function sequence (f ∘ γn )n∈ℕ∗ uniformly converges to the path f ∘ γ. Proof. Let ε > 0 be an arbitrary number, and let η > 0 be a sufficiently enough small number, such that −
U − (γ; η) = ( ⋃ U(γ(t); η)) ⊂ G. t∈[0,1]
Since U − (γ; η) is bounded and closed, then U − (γ; η) is compact. According to Cantor theorem (every continuous function on a compact is uniformly continuous on this compact), we have ∃η1 > 0, η1 ≤ η such that f (z ) − f (z ) < ε, ∀z , z ∈ U − (γ; η), z − z < η1 . Let n0 ∈ ℕ∗ be an enough big number, such that ∀n > n0 ,
γ(t) − γn (t) < η1 , ∀t ∈ [0, 1].
Then f (γ(t)) − f (γn (t)) < ε,
∀t ∈ [0, 1] and ∀n > n0 .
thus the function sequence (f ∘ γn ) uniformly converges to the function f ∘ γ.
60  3 The complex integration
3.2 The complex integral 3.2.1 The Riemann–Stieltjes integral for complex valued functions Definition 3.2.1. Consider the closed interval [a, b] ⊂ ℝ, let u, U : [a, b] → ℝ be two realvalued functions, and define the sum n
σ(u, U, Δ) = ∑ u(τk )(U(tk ) − U(tk−1 )), k=1
where Δ = (t0 , t1 , . . . , tn ) is a division of the interval [a, b], and τk ∈ [tk−1 , tk ], k = 1, n are arbitrary points. If there exists a real number I ∈ ℝ, such that ∀ε > 0, ∃η > 0
such that ‖Δ‖ < η ⇒ I − σ(u, U, Δ) < ε,
we say that the function u is integrable with respect to the function U (in the Riemann–Stieltjes sense), and we denote b
I = ∫ udU. a
The number I is called the Riemann–Stieltjes integral of u on the interval [a, b] with respect U. Definition 3.2.2. Let f = u + iv and F = U + iV, f , F : [a, b] → ℂ be two given functions. We say that f is integrable on the interval [a, b] with respect to F (in the Riemann– Stieltjes sense), if u and v are integrable functions on [a, b] with respect to U, and respectively to V. According to the above definition, the integral of the function f with respect to the function F on the interval [a, b] will be b
b
b
b
b
∫ f dF = ∫ udU − ∫ vdV + i(∫ udV + ∫ vdU). a
a
a
a
a
Theorem 3.2.1. If f is integrable with respect to F on [a, b], and if for any division Δ = (t0 , t1 , . . . , tn ) of [a, b] we define the sum n
σ(f , F, Δ) = ∑ f (τk )(F(tk ) − F(tk−1 )), k=1
where τk ∈ [tk−1 , tk ], then for all ε > 0 there exists a number η > 0, such that ‖Δ‖ < η implies b ∫ f dF − σ(f , F, Δ) < ε. a
3.2 The complex integral  61
Proof. We have σ(f , F, Δ) = σ(u, U, Δ) − σ(v, V, Δ) + i(σ(u, V, Δ) + σ(u, U, Δ)). b
Using the relations ∫a udU, etc. together with the above identity, the proof follows immediately. Theorem 3.2.2. Let f = u + iv, F = U + iV, and fn , Fn : [a, b] → ℂ real variable complex valued functions, and let α1 , α2 ∈ ℂ. 1. If f is integrable on [a, b] with respect to F, then F is integrable on [a, b] with respect to f , and b
b
∫ f dF + ∫ Fdf = f (b)F(b) − f (a)F(a). a
2.
a
If f1 and f2 are integrable with respect to F, then α1 f1 + α2 f2 is integrable with respect to F, and b
b
b
∫(α1 f1 + α2 f2 )dF = α1 ∫ f1 dF + α2 ∫ f2 dF. a
a
a
3.
If f is continuous and F is a function with bounded variation on [a, b], then f is integrable with respect to F on [a, b]. 4. Let (fn )n∈ℕ∗ a sequence of functions that uniformly converges to f , and let (Fn )n∈ℕ∗ a sequence of functions consisting on functions with bounded variation, such that (Fn )n∈ℕ∗ converges pointwise to F and the set of the total variations V(Fn , [a, b]) is bounded. Then b
b
lim ∫ fn dFk = ∫ f dF. n→∞
k→∞ a
5.
a
If f is continuous, and F is continuously differentiable on [a, b], then f is integrable on [a, b] with respect to F, and b
b
∫ f dF = ∫ f (t)F (t)dt, a
a
where the righthand side integral is given by b
b
∫(u(t)U (t) − v(t)V (t))dt + i ∫(u(t)V (t) + v(t)U (t))dt.
a
a
62  3 The complex integration 6.
If c ∈ (a, b) and f is integrable with respect to F on the intervals [a, c] and [c, b], then f is integrable with respect to F on [a, b], and b
c
b
∫ f dF = ∫ f dF + ∫ f dF. a
7.
c
a
If f is integrable with respect to F on [a, b], and if h : [a , b ] → [a, b] is a homeomorphism with h(a ) = a and h(b ) = b, then f ∘ h is integrable with respect to F ∘ h on [a , b ], and b
b
∫ f dF = ∫(f ∘ h)d(F ∘ h). a
a
Proof. The points 1–6 follow from the corresponding properties of the Riemann– Stieltjes integral of the realvalued functions. We will prove the point 7 of this theorem. If f = u + iv, F = U + iV, then f ∘ h = u ∘ h + iv ∘ h and F ∘ h = U ∘ h + iV ∘ h. It is sufficient to prove, using the definition of the complex Riemann–Stieltjes integral, that the next formula holds: b
b
∫(u ∘ h)d(U ∘ h) = ∫ udU. a
a b
Let I = ∫a udU, and let ε > 0 be arbitrary. Then ∃η1 > 0
such that I − σ(u, U, Δ) < ε, whenever ‖Δ‖ < η1 .
Since h is a homeomorphism with h(a ) = a and h(b ) = b, then h is a strictly increasing continuous function, hence it is uniformly continuous. From the definition of the uniform continuity, it follows that ∃η > 0,
such that h(t ) − h(t ) < η1 , ∀t , t ∈ [a , b ], t − t < η.
Let Δ = (t0 , t1 , . . . , tn ) a division of the interval [a , b ], with ‖Δ ‖ < η, and let τk ∈ [tk−1 , tk ] be arbitrary points. Let denote tk = h(tk ) and τk = h(τk ), k = 1, n. Then Δ = (t0 , t1 , . . . , tn ) will be a division of the interval [a , b ], such that ‖Δ‖ < η1 , and n
σ(u ∘ h, U ∘ h, Δ ) = ∑ (u ∘ h)(τk )((U ∘ h)(tk ) − (U ∘ h)(tk−1 )) k=1 n
= ∑ u(τk )(U(tk ) − U(tk−1 )) = σ(u, U, Δ). k=1
3.2 The complex integral  63
Thus I − σ(u ∘ h, U ∘ h, Δ ) = I − σ(u, U, Δ) < ε, and consequently b
b
∫(u ∘ h)d(U ∘ h) = ∫ udU. a
a
Definition 3.2.3. Let γ : [0, 1] → ℂ be a path with finite variation (i. e., a rectifiable path), and let f : {γ} → ℂ be a continuous function. Then the function f ∘ γ : [0, 1] → ℂ is continuous, hence it is integrable on the interval [0, 1] with respect to the function γ. Denoting the corresponding integral by 1
∫(f ∘ γ)dγ = ∫ f (z)dz = ∫ f , 0
γ
γ
this integral will be called the complex integral (or the Cauchy integral) of the function f on the path γ. Theorem 3.2.3 (The properties of the complex integral). Let γ ∈ 𝒟(z1 , z2 ) and γ1 ∈ 𝒟(z2 , z3 ) be two rectifiable paths, let f , g : {γ} → ℂ be two continuous functions, and let α, β ∈ ℂ. Then: 1. ∫γ αf + βg = α ∫γ f + β ∫γ g. 2.
3.
∫γ− f = − ∫γ f .
If f is continuous also on {γ1 }, then ∫γ∪γ f = ∫γ f + ∫γ f . 1
1
4. If (γ1 , γ2 , . . . , γn ) is a division of the path γ, then ∫γ f = ∑nk=1 ∫γ f . 5.
6.
k
If f (γ(t)) ≤ M, ∀t ∈ [0, 1], then  ∫γ f  ≤ M V(γ), where V(γ) is the length of the path γ. For the linear path γ(t) = z1 + t(z2 − z1 ), t ∈ [0, 1], we have 1
∫ f = (z2 − z1 ) ∫ f (z1 + t(z2 − z1 ))dt. γ
7.
0
If G ⊂ ℂ is an open set, the function f : G → ℂ is a continuous, (γn )n∈ℕ∗ is a sequence of rectifiable paths in G that uniformly converges on [0, 1] to the path γ, where {γ} ⊂ G, and the real numbers sequence (V(γn ))n∈ℕ∗ is bounded, then the sequence (∫γ f )n∈ℕ∗ converges, and n
lim ∫ f = ∫ f .
n→∞
γn
γ
64  3 The complex integration 8. Let γ be a rectifiable path, and let fn : {γ} → ℂ, n ∈ ℕ∗ be a sequence of continuous functions that uniformly converges on the set {γ} to the function f . Then there exists the integral ∫γ f , and lim ∫ fn = ∫ f .
n→∞
9.
γ
γ
If γ1 and γ2 are equivalent paths in G, if the function f is continuous on {γ1 }, and γ1 is rectifiable, then γ2 is also rectifiable and ∫f = ∫f. γ1
γ2
Proof. The points 1, 6, 7, 8 and 9 follow immediately from the points 2, 4, 5 and 7 of the Theorem 3.2.2. 2. The integral sums on the path γ − are n
σ(f ∘ γ − , γ − , Δ) = ∑ f (γ(1 − τk ))(γ(1 − tk ) − γ(1 − tk−1 )). k=1
Since tk = 1 − tn−k , τk = 1 − τn−k+1 , where Δ = (t0 , t1 , . . . , tn ), then ‖Δ ‖ = ‖Δ‖ and τk ∈ [tk , tk−1 ]. If we let n − k + 1 = m, then n
)) ) − γ(tn−k+1 )(γ(tn−k σ(f ∘ γ − , γ − , Δ) = ∑ f (τn−k+1 k=1
n
= − ∑ f (γ(τm )) = −σ(f ∘ γ, γ, Δ ). ) − γ(tm−1 )(γ(tm m=1
Thus, if ‖Δ‖ = ‖Δ ‖ → 0 then we get ∫γ f = − ∫γ− f . 3. A simple computation shows that 1
∫ f = ∫(f ∘ (γ ∪ γ1 ))(t)d(γ ∪ γ1 )(t)
γ∪γ1
0 1 2
1
= ∫ f (γ(2t))dγ(2t) + ∫ f (γ1 (2t − 1))dγ1 (2t − 1). 0
1 2
According to the point 7 of the Theorem 3.2.2, if h : [a , b ] → [a, b] is an increasing homeomorphism of the interval [a , b ] onto the interval [a, b], then b
b
∫(f ∘ h)d(g ∘ h) = ∫ f dg. a
a
3.2 The complex integral  65
Letting h : [0, 21 ] → [0, 1], h(t) = 2t, then 1 2
1
∫ f (γ(2t))dγ(2t) = ∫ f (γ(t))dγ(t) = ∫ f , 0
γ
0
and similarly 1
∫ f (γ1 (2t − 1))dγ1 (2t − 1) = ∫ f . γ1
1 2
Thus we get ∫ f = ∫f + ∫f, γ1
γ
γ∪γ1
and the proof of the point 4 is similar to the above. Remarks 3.2.1. 1. The complex integral can be defined by using the second type curve integral. In fact, if we let f = u + iv and γ = α + iβ, then 1
1
1
∫ f = ∫(f ∘ γ)dγ = ∫ u(α(t), β(t))dα(t) − ∫ v(α(t), β(t))dβ(t) γ
0
1
0
0
1
+ i(∫ u(α(t), β(t))dβ(t) + ∫ v(α(t), β(t))dα(t)), 0
2.
0
where the integrals that appeared in last sums are second type real line integrals, written with the aid of the real line integrals. The connection between the complex integral and the first type curve integral is given by the next relation: 1
∫ f (ζ )dζ  = ∫ f (γ(t))γ (t)dt γ
0
1
2
2
= ∫ f (γ(t))√α (t) + β (t)dt = ∫ f ds, γ
0
where n
∫ f (ζ )dζ  = lim ∑ (f ∘ γ)(τk )γ(tk ) − γ(tk−1 ). γ
‖Δ‖→0
k=1
66  3 The complex integration 3.
Using the previous definition, we deduce that n σ(f ∘ γ, γ, Δ) = ∑ (f ∘ γ)(τk )[γ(tk ) − γ(tk−1 )] k=1 n
≤ ∑ (f ∘ γ)(τk )γ(tk ) − γ(tk−1 ), k=1
and thus ∫ f (ζ )dζ ≤ ∫ f (ζ )dζ . γ γ 4. The complex integral along a closed path does not depend on the starting point. In fact, let γ : [0, 1] → ℂ, with γ(0) = γ(1) = z0 , and let z1 ∈ {γ} be an arbitrary point. Then there exists t1 ∈ [0, 1] such that γ(t1 ) = z1 . Let (γ1 , γ2 ) be the decomposition of the path γ by the division Δ = (0, t1 , 1). Then ∫f = ∫ f = ∫f + ∫f = ∫f + ∫f = ∫ f. γ
γ1 ∪γ2
γ1
γ2
γ2
γ1
γ2 ∪γ1
Remark that the union γ2 ∪ γ1 is welldefined, because the end point of γ2 is z0 that it is also the starting point of γ1 , and γ2 ∪ γ1 ∈ 𝒟(z1 ).
3.3 The Cauchy theorem 3.3.1 The connection between the integral and the primitive function Definition 3.3.1. Let G ⊂ ℂ be an open set, and let f be a complex function, with f : G → ℂ. The function g ∈ H(G) is said to be a primitive function of the function f , if g (z) = f (z) for all z ∈ G. For the real valued functions, the connection between the (definite) integral and the primitive of a function is given by the Newton–Leibniz formula. If f is a real valued real variable continuous function f : [a, b] → ℝ, then the function x
F : [a, b] → ℝ,
F(x) := ∫ f (t)dt a
is a primitive function of f . For the complex valued complex variable functions, we will try to find a similar property. Let f ∈ C(D), where D ⊂ ℂ domain, and let γ ∈ 𝒟D (z0 , z) be an arbitrary rectifiable path. A natural question is the following:
3.3 The Cauchy theorem
 67
Which are the sufficient conditions, such that the function g defined by z
g(z) = ∫ f = ∫ f z0
γ
is a primitive function of f , where the rectifiable path γ connects the fixedpoint z0 with the variable point z? The question has sense, only if the integral ∫γ f is independent of the path γ, and depends on only to the end point of γ, which is z. Remark 3.3.1. The integral of the function f is independent of the integration path, if and only if ∫γ f = 0 for all arbitrary, closed and rectifiable path γ. In fact, if γ = γ1 ∪ γ2− is an arbitrary decomposition of the closed path γ, then 0 = ∫f = ∫ f = ∫f + ∫ f = ∫f − ∫f ⇔ ∫f = ∫f. γ
γ1
γ1 ∪γ2−
γ2−
γ1
γ2
γ1
γ2
Conversely, if the integral is independent of the path, then from the above reasons the integral will be equal to zero for all arbitrary, closed and rectifiable path γ. Theorem 3.3.1 (The connection between the primitive and the integral). Let D ⊂ ℂ be a domain (an open and connected set), and let f : D → ℂ be a continuous function. Then the following relations hold: 1. If ∫f = 0
(3.1)
γ
2.
for any arbitrary closed and rectifiable path γ, with {γ} ⊂ D, then the function f has a primitive g in the domain D. Any other primitive of the function f in the domain D is of the form g +c, where c ∈ ℂ. If the function f has a primitive function g in the domain D, then for all rectifiable curve γ, with {γ} ⊂ D, the following Newton–Leibniz formula holds: ∫ f = g(γ(1)) − g(γ(0)). γ
Thus, if γ a closed path, then ∫γ f = 0. Proof. 1. Let z0 ∈ D be a given point, and let z ∈ D be arbitrary. There exists a number r > 0, such that U(z; r) ⊂ D. If we consider the point h ∈ U(0; r), then z + h ∈ U(z; r). Let us denote by λ the linear path that connects z with z+h, and let γ be an arbitrary rectifiable path that connects z0 with z, and let γ1 be an arbitrary rectifiable path that
68  3 The complex integration connects z0 with z + h. Then γ ∪ λ ∪ γ1− is a closed path in the domain D, hence from our assumption we get ∫
f = 0.
γ∪λ∪γ1−
If we let g(z) = ∫γ f , according to the previous remark, it follows that the function g is
welldefined, and g(z + h) = ∫γ f . 1 Supposing that h ≠ 0, from the relation (3.1), we obtain that 0 = ∫ f + ∫ f + ∫ f = g(z) + ∫ f − g(z + h), γ
λ
λ
γ1−
and thus g(z + h) − g(z) 1 1 1 = ∫ f (ζ )dζ = ∫(f (ζ ) − f (z))dζ + f (z) ∫ dζ . h h h h λ
(3.2)
λ
λ
Using the fact that ∫λ dζ = h, because 1
1
∫ dζ = ∫(z + th)t dt = ∫ hdt = h, 0
λ
0
the relation (3.2) reduces to 1 g(z + h) − g(z) = f (z) + ∫(f (ζ ) − f (z))dζ . h h λ
Since the function f is continuous at the point z, then such that ζ − z ≤ h < δ ⇒ f (ζ ) − f (z) < ε.
∀ε > 0, ∃δ = δε > 0 In a such a case, we have
1 1 εh = ε. ∫(f (ζ ) − f (z))dζ ≤ h h λ
Since ε > 0 is arbitrary, from the above inequality we obtain that lim
h→0
1 ∫(f (ζ ) − f (z))dζ = 0. h λ
Thus, according to the relation (3.3), there exists the limit lim
h→0
1 g(z + h) − g(z) = lim ∫ f (ζ )dζ = f (z) h→0 h h γ
which is in fact the derivative g (z), and it is equal with f (z).
(3.3)
3.3 The Cauchy theorem
 69
If g1 is another primitive of the function f on the domain D, then g1 − g will be a holomorphic function with the derivative identical to zero. Since D is a domain, by using a wellknown result concerning the holomorphic functions in a domain with the derivative equal to zero, it follows that g1 − g = c, where c is a constant. 2. Let γ ∈ 𝒟D (z0 , z1 ) be a rectifiable path, let Δ = (t0 , t1 , . . . , tn ) be a division of the interval [0, 1], and let γn be the broken line that connects consequently the points γ(tk ), k ∈ {0, 1, . . . , n}. Then 1
tk
n
tk
n
∫ f = ∫(f ∘ γn )dγn (t) = ∑ ∫ (f ∘ γn )dγn = ∑ ∫ f (γn (t))γn (t)dt.
γn
k=1 t
0
k=1 t
k−1
(3.4)
k−1
From the assumption g (z) = f (z), we have (g(γn (t))) = g (γn (t))γn (t) = f (γn (t))γn (t).
Combining this relation with (3.4), we deduce that n
tk
n
∫ f = ∑ ∫ (g(γn (t))) dt = ∑ (g(γn (tk )) − g(γn (tk−1 )))
γn
k=1 t
k=1
k−1
= g(γ(1)) − g(γ(0)) = g(z1 ) − g(z0 ). If ‖Δ‖ → 0, then γn uniformly converges to γ, and thus ∫γ f → ∫γ f . Consequently, n
∫ f = g(z1 ) − g(z0 ). γ
For the special case when γ closed, i. e., z1 = z0 , from the above formula we get ∫γ f = 0. Remarks 3.3.2. Consider the function f (z) = 1.
If n ∈ ℤ \ {1}, then g(z) = hence
1 (1−n)(z−z0 )n−1
∫ γ
2.
1 , (z−z0 )n
n ∈ ℤ.
is a primitive function for f in D = ℂ \ {z0 },
1 dζ = 0, (ζ − z0 )n
where γ(t) = z0 + re2πit , t ∈ [0, 1], i. e., {γ} = 𝜕U(z0 ; r). 1 If n = 1, then the function f (z) = z−z has derivative on D = ℂ \ {z0 }, but it has not 0 a primitive function in D, because ∫f = ∫ γ
γ
1 dζ = 2πi, ζ − z0
where γ(t) = z0 + re2πit , t ∈ [0, 1], i. e., {γ} = 𝜕U(z0 ; r).
70  3 The complex integration
Figure 3.1: Proof of Theorem 3.3.2.
Theorem 3.3.2 (Cauchy theorem for triangles, or Goursat lemma). Let T be the interior of the triangle with the apex in the points a, b, c ∈ ℂ and let f ∈ H(T) ∩ C(T − ). Let γ = 𝜕T denote the path consisting of the boundary of the triangle T. Then ∫γ f = 0. Proof. Suppose that f ∈ H(G), where G ⊂ ℂ is an open set, such that and T − ⊂ G, and let I = ∫γ f .
Let denote by a , b and c the middle points of the segments bc, ac and, respectively, ab, and let connect these points by segments. We will denote by T (1) , T (2) , T (3) and T (4) the triangles that we obtained using the above method (Figure 3.1), and let γ (k) = 𝜕T (k) , k = 1, 4. With these notations, supposing that all the path γ (k) have direct orientation, it follows that 4
I = ∫f = ∑ ∫ f. γ
k=1
γ (k)
From this identity, we deduce that 4 I ≤ ∑ ∫ k=1 (k) γ
f .
Thus, there exists at least a triangle T (k) (denoted by T1 , and the boundary by γ1 = 𝜕T1 ), such that T1− ⊂ T − ,
γ1 = 𝜕T1
I and ∫ f ≥ . 4 γ 1
Denoting by l = l(γ), the length of the path γ, then l(γ1 ) = l1 = 2l .
3.3 The Cauchy theorem
 71
Now, we will apply the above method to the triangle T1 . It follows that there exists a triangle T2 , such that T2− ⊂ T1− ,
γ2 = 𝜕T2 ,
l(γ2 ) = l2 =
I ∫ f ≥ 2 . 4 γ
l1 l = , 2 22
2
If we continue this method, we deduce that there exists a sequence of triangles Tk , such that − ⊃ ⋅⋅⋅, G ⊃ T − ⊃ T1− ⊃ ⋅ ⋅ ⋅ ⊃ Tn− ⊃ Tn+1 I l ln = l(γn ) = n , ∫ f ≥ n . 4 2 γ
γn = 𝜕Tn ,
n
Using the fact that diam(Tn− ) → 0, whenever n → ∞, according to the Cantor theorem there exists a point z0 ∈ Tn− for all n ∈ ℕ∗ , and z0 ∈ T − ⊂ G. Since f ∈ H(G), we have that f (z) = f (z0 ) + f (z0 )(z − z0 ) + g(z)(z − z0 ), where limz→z0 g(z) = 0. Now, using this previous relation we have ∫ f = [f (z0 ) − f (z0 )z0 ] ∫ dζ + f (z0 ) ∫ ζ dζ + ∫ g(ζ )(ζ − z0 )dζ . γn
γn
γn
γn
Since the functions 1 and z have primitives, according to Theorem 3.3.1 ∫ dζ = ∫ ζ dζ = 0, γn
γn
hence ∫ f = ∫ g(ζ )(ζ − z0 )dζ .
γn
γn
Let ε > 0 be arbitrary. Since limz→z0 g(z) = 0, there exist a number δ = δε > 0 and an index n0 ∈ ℕ∗ , such that z ∈ G,
z − z0  < δ ⇒ g(z) < ε
and Tn− ⊂ U(z0 ; δ),
∀n > n0 .
Consequently, εl2 I ≤ ∫ f = ∫ g(ζ )(ζ − z0 )dζ ≤ εln ∫ dζ ≤ εln2 = n . n 4 4 γ γ γ n
n
n
Thus, I ≤ εl2 , and since ε > 0 was an arbitrary number it follows that I = 0.
72  3 The complex integration Let return now to the general case f ∈ H(T) ∩ C(T − ). Let us consider the sequences (an )n∈ℕ∗ ⊂ T, (bn )n∈ℕ∗ ⊂ T, (cn )n∈ℕ∗ ⊂ T, such that an → a, bn → b, and cn → c, whenever n → ∞. For the boundaries γn = δ(an , bn , cn ) of the triangle (an , bn , cn ), we have γn γ = δ(a, b, c), and thus 0 = ∫f → ∫f, γn
hence ∫ f = 0.
γ
γ
Theorem 3.3.3 (The connection between the holomorphic functions and the primitives). Let D be a starlike domain with respect to the point z0 ∈ D. Let us consider the lines d1 , d2 , . . . , dn that contain the point z0 , and let d = ⋃nk=1 dk . If f : D → ℂ is a continuous function on the domain D, and holomorphic on D \ d, i. e., f ∈ C(D) ∩ H(D \ d), then f has a primitive function on D. Proof. The set D \ d is open, hence for every point of D \ d there exists a neighborhood that does not intersect the set d. 1. Suppose that z ∈ D \ d. If we denote by λz the linear path that connects z0 with z, then the function g(z) = ∫λ f is welldefined. z Letting U(z; r) ⊂ D \ d and h ∈ U(0; r), then z + h ∈ U(z; r). Let us consider the − path γ = λz ∪ λ̃ ∪ λz+h , where λ̃ is the linear path that connects z with z + h. The path γ represents the boundary of the triangle with the apexes z0 , z, and z + h. On the interior of this triangle the function f is holomorphic, and it is continuous on the boundary, hence by Theorem 3.3.2 we have that ∫γ f = 0, i. e., ∫ f = ∫ f + ∫ f + ∫ f = g(z) + ∫ f − g(z + h) = 0. γ
λz
λ̃
− λz+h
λ̃
Thus, if h ≠ 0 we obtain g(z + h) − g(z) 1 = ∫ f → f (z), h h
h → 0.
λ̃
This result is obtained using the same technique like to the proof of the point 1 of the Theorem 3.3.1. Thus, there exists g (z), and g (z) = f (z), ∀z ∈ D \ d. 2. Suppose that z ∈ (d ∩ D) \ {z0 }, z ∈ dj . Then there exists a number r > 0, such that the disc U(z; r) intersects only the line dj . Letting h ∈ U(0; r), then the function f is holomorphic in the interior of the triangle with the apexes in z0 , z and z + h, and it is continuous on the boundary of this triangle. In this case, we may use the same method as to the proof of the point 1 of Theorem 3.3.2, hence it follows that the function g(z) = ∫λ f has derivative, and g (z) = f (z). z
3. Finally, we will prove that there exists g (z0 ) and g (z0 ) = f (z0 ). From the definition of g, we have g(z0 ) = 0, and 1
g(z) = ∫ f = (z − z0 ) ∫ f (z0 + t(z − z0 ))dt, λz
0
3.3 The Cauchy theorem
 73
hence, for z ≠ z0 we have 1
g(z) − g(z0 ) = ∫ f (z0 + t(z − z0 ))dt. z − z0 0
We will prove that 1
lim ∫ f (z0 + t(z − z0 ))dt = f (z0 ).
z→z0
0
Since the function f is continuous on D and z0 ∈ D, for all ε > 0 there exists a number δ > 0, such that if z ∈ D and z − z0  < δ, then f (z) − f (z0 ) < ε. If z − z0  < δ, then z0 + t(z − z0 ) − z0  = tz − z0  ≤ z − z0  < δ, ∀t ∈ [0, 1], hence f (z0 + t(z − z0 )) − f (z0 ) < ε for all t ∈ [0, 1]. From here, we deduce that 1 1 1 ∫ f (z0 + t(z − z0 ))dt − f (z0 ) = ∫ f (z0 + t(z − z0 ))dt − ∫ f (z0 )dt 0 0 0
1 = ∫[f (z0 + t(z − z0 )) − f (z0 )]dt ≤ max f (z0 + t(z − z0 )) − f (z0 ) < ε, t∈[0,1] 0 1
whenever z − z0  < δ, hence limz→z0 ∫0 f (z0 + t(z − z0 ))dt = f (z0 ). Thus there exists g (z0 ), and g (z0 ) = f (z0 ).
Consequently, the function g is holomorphic on the domain D, and g = f .
Remarks 3.3.3. 1. According to the result of the point 2 of the Theorem 3.3.1, if the assumptions of the Theorem 3.3.3 hold, then the integral of the function f is independent of the path. As a special case, the integral of f on any closed and rectifiable path is zero. 2. If the function f is continuous on the starlike domain D, and it has derivatives in all the points of D except at most a finite number of points, then the function f has a primitive in D. 3. It seems that the existence of the derivative it is a “too strong” condition for the existence of the primitive. We will see that this condition it is not only sufficient, but it is necessary for the existence of the primitive. Example 3.3.1. Define the function f (z) = {
sin(z−i)z , (z−i)z
1,
if z ∈ ℂ \ {0, i}, if z ∈ {0, i}.
Since f has derivatives on ℂ \ {0, i} and it is continuous on ℂ, using the previous theorem we deduce that there exists a primitive of the function f on the entire complex plane ℂ.
74  3 The complex integration 3.3.2 The Cauchy theorem Using the Goursat theorem (Theorem 3.3.2), we will prove the Cauchy theorem for any arbitrary closed and rectifiable path. Theorem 3.3.4 (Cauchy theorem). Let G ⊂ ℂ be an open set, and let f ∈ H(G). Then the next two equivalent affirmations hold: (i) If γ0 and γ1 are two rectifiable paths in G, with γ0 ∼ γ1 , then G
∫f = ∫f. γ0
γ1
(ii) For any arbitrary closed and rectifiable path in G, that is homotopic to a point in G, that is γ ∼ 0, we have G
∫ f = 0. γ
Proof. Using a similar proof with those of the Remark 3.3.1, the equivalence between (i) and (ii) follows immediately. Then it is sufficient to prove the part (i) of the above theorem. Since γ0 and γ1 are homotopic path in G, there exists a continuous function φ : S × T → G,
S = T = [0, 1],
such that φ(0, t) = γ0 (t), φ(s, 0) = z0 ,
φ(1, t) = γ1 (t), φ(s, 1) = z1 ,
∀t ∈ T,
∀s ∈ S,
where z0 and z1 are the starting and the end points of the paths γ0 and γ1 , respectively. Since S × T is a compact set and φ is a continuous function, the image K = φ(S × T) is a compact subset of G. Thus, d(K, 𝜕G) > 0. Let ρ > 0 be a number, such that 0 < ρ < d(K, 𝜕G). The function φ is continuous on the compact set S × T, then φ will be a uniformly continuous function on S × T. Thus, there exists a number δ = δρ > 0, such that ∀(s , t ), (s , t ) ∈ S × T,
√(s − s )2 + (t − t )2 < δ
⇒ φ(s , t ) − φ(s , t ) < ρ. Let us consider two divisions of T and S, denoted by Δ = (t0 , t1 , . . . , tn ), and respectively by Δ = (s0 , s1 , . . . , sn ), such that tk = sk for all k ∈ {0, 1, . . . , n}, and ‖Δ ‖ = ‖Δ ‖ < √δ2 .
3.3 The Cauchy theorem
 75
Let Rjk = [sj , sj+1 ] × [tk , tk+1 ], j, k = 0, n − 1, and zjk = φ(sj , tk ),
j, k ∈ {0, 1, . . . , n}.
Then z0k = γ0 (tk ),
zj0 = z0 ,
znk = γ1 (tk )
zjn = z1
(because s0 = 0, sn = 1) and
(because t0 = 0, tn = 1).
Consider Ujk = U(zjk ; ρ). Then Ujk ⊂ G and φ(Rjk ) ⊂ Ujk , because zjk ∈ K and d(K, 𝜕G) > ρ > 0, and respectively, ∀(s, t) ∈ Rjk
we have s − sj 
0 such that U − (z0 ; r) ⊂ G, and consider an arbitrary number ε > 0. Since g is a uniformly continuous function on the compact U − (z0 ; r) × K, there exists η>0
ε such that z − z0  < η, implies g(z, ζ ) − g(z0 , ζ ) < , V(γ)
where V(γ) the length of the path γ. Thus ε ∫ dζ  = ε, h(z) − h(z0 ) ≤ ∫ g(z, ζ ) − g(z0 , ζ )dζ  < V(γ) γ
hence the function h is continuous at the point z0 .
γ
80  3 The complex integration 2. If gz (z, ζ ) continuous, and z0 ∈ G is a given point, let define the function g1 by g(z,ζ )−g(z ,ζ )
0 { z−z0 g1 (z, ζ ) = { { gz (z0 , ζ ),
(z, ζ ) ∈ (G \ {z0 }) × K,
,
(z, ζ ) ∈ {z0 } × K.
Then g1 : G×K → ℂ is a continuous function, and according to the point 1. the function h1 (z) = ∫ g1 (z, ζ )dζ γ
will be continuous on G, hence h1 (z0 ) = limz→z0 h1 (z). On the other hand, h1 (z0 ) = ∫ gz (z0 , ζ )dζ γ
and h1 (z) = ∫ γ
g(z, ζ ) − g(z0 , ζ ) h(z) − h(z0 ) dζ = . z − z0 z − z0
It follows that there exists the limit lim
z→z0
h(z) − h(z0 ) = h1 (z0 ) = ∫ gz (z0 , ζ )dζ , z − z0 γ
which proves the existence of h (z0 ). Corollary 3.4.1. Let φ : K → ℂ be a continuous function, let γ be a rectifiable path in ℂ and let denote K = {γ} and G = ℂ \ K. If we define the function g : G × K → ℂ by φ(ζ ) g(z, ζ ) = ζ −z , then the function h defined by φ(ζ ) dζ ζ −z
h(z) = ∫ γ
is holomorphic on G, there exists the derivatives h(n) , ∀n ∈ ℕ∗ , and moreover, h(n) (z) = n! ∫ γ
φ(ζ ) dζ . (ζ − z)n+1
Theorem 3.4.2 (Cauchy formulas for the disc). Let f : U − (z0 ; r) → ℂ be a continuous function on the closed disc U − (z0 ; r), such that f is holomorphic on the open disc U(z0 ; r). Then, for any arbitrary k ∈ ℕ there exists the derivatives f (k) (z), ∀z ∈ U(z0 ; r) and f (k) (z) =
k! 2πi
∫ 𝜕U(z0 ;r)
f (ζ ) dζ . (ζ − z)k+1
Proof. 1. First, we will prove the above results for the special case k = 0. Let z ∈ U(z0 ; r) be a given point, and let the number rn > 0 such that z − z0  < rn < r. Let denote by γn the boundary of the disc U(z0 ; rn ), considering a direct (positive) orientation.
3.4 The Cauchy formula for the disc  81
It is obvious that I=
f (ζ ) f (ζ ) − f (z) dζ 1 1 f (z) dζ = dζ + . ∫ ∫ ∫ 2πi ζ − z 2πi ζ −z 2πi ζ − z γn
γn
(3.9)
γn
Now we will compute the integral ∫ γn
dζ . ζ −z
Let ρ > 0 be an enough small number, such that U − (z; ρ) ⊂ U(z0 ; rn ), and let 1 G = U(z0 ; rn ) \ {z}. Then the function ζ −z is holomorphic with respect to the variable ζ ∈ G. Since the path γ = 𝜕U(z; ρ) has the equation γ(t) = ρei2πt + z, t ∈ [0, 1], it follows that ∫ γ
1
1
0
0
dζ 2πiρei2πt =∫ dt = 2πi ∫ dt = 2πi. ζ −z ρei2πt
Because γ ∼ γn , from the Cauchy theorem we obtain that G
∫ γn
dζ = 2πi, ζ −z
hence the relation (3.9) reduces to I=
f (ζ ) − f (z) 1 dζ + f (z). ∫ 2πi ζ −z
(3.10)
γn
Since f is a holomorphic function on the disc U(z0 ; r), it follows that the function g defined by f (ζ )−f (z)
, { ζ −z g(ζ ) = { { f (z),
ζ ∈ U(z0 ; r) \ {z}, ζ =z
is holomorphic on U(z0 ; r) \ {z}, and it is continuous on U(z0 ; r). Using Theorem 3.3.3 for the function g, we deduce that there exists a primitive of g on the disc U(z0 ; r), and according to the point 2. of the Theorem 3.3.1 we get ∫γ g(ζ )dζ = 0. Thus, the integral n from righthand side of the equality (3.10) is zero, hence f (z) =
f (ζ ) 1 dζ . ∫ 2πi ζ − z γn
If we suppose that rn → r, then γn uniformly converges to 𝜕U(z0 ; r), and f (z) =
f (ζ ) 1 1 dζ → ∫ 2πi ζ − z 2πi γn
∫ 𝜕U(z0 ;r)
which proves the formula of the theorem for k = 0.
f (ζ ) dζ , ζ −z
82  3 The complex integration 2. Using the Corollary 3.4.1 for the integral f (z) =
1 2πi
∫ 𝜕U(z0 ;r)
f (ζ ) dζ , ζ −z
we deduce that ∃f (k) (z), ∀k ∈ ℕ, and f (k) (z) =
k! 2πi
∫ 𝜕U(z0 ;r)
f (ζ ) dζ . (ζ − z)k+1
Theorem 3.4.3 (Morera theorem). If a complex function has a primitive on the open set G ⊂ ℂ, then the function is holomorphic on this set. Proof. Let g be a primitive for the function f : G → ℂ. Then g ∈ H(G). Let z0 ∈ G be a given point, and let r > 0 be a number such that U − (z0 ; r) ⊂ G. Applying the Cauchy formula for the function g, we obtain that there exists the derivative g (z0 ). Since g (z) = f (z), z ∈ G, it follows that there exists f (z0 ) = g (z0 ). Thus, the function f is holomorphic on G. Corollary 3.4.2. The derivative of an holomorphic function is also holomorphic. Consequently, any holomorphic function has derivatives of any arbitrary order, and these derivatives are also holomorphic. Definition 3.4.1. Let G ⊂ ℂ. A family of lines is said to be locally finite in the set G, for every point of G there exists a neighborhood of it that intersects only a finite number of these lines. For example, the family of the lines x = n1 , n ∈ ℕ∗ , is locally finite in the disc U(1; 1), but it is not locally finite in the disc U(0; 1). Theorem 3.4.4. Let G ⊂ ℂ be an open set, and let d be a family of lines that are locally finite in the set G. If the function f is holomorphic on G \ d, and it is continuous on G, then f ∈ H(G). Proof. If we let z0 ∈ G, then there exists a number r > 0 such that U(z0 ; r) ⊂ G, and such that the disc U(z0 ; r) intersects only a finite number of these lines that contain the point z0 . According to the Theorem 3.3.3, the function f has a primitive in the disc U(z0 ; r), and then from Theorem 3.4.3 we deduce that f ∈ H(U(z0 ; r)). Since z0 is an arbitrary point of G, we conclude that f ∈ H(G).
3.5 The analytical branches of multivalued functions Theorem 3.5.1 (Existence theorem of the analytical branches of multivalued functions). Let D ⊂ ℂ a simply connected domain in ℂ, and let g ∈ H(D) a such a function, such that g(z) ≠ 0, ∀z ∈ D.
3.5 The analytical branches of multivalued functions  83
Let z1 ∈ D and let w1 , w2 , α ∈ ℂ be complex numbers, such that w1 ∈ Log g(z1 ), w2 ∈ (g(z1 ))α . Then there exists only the functions f1 , f2 ∈ H(D), such that: 1. f1 is the analytical branch of the multivalued function Log g, with f1 (z1 ) = w1 , 2. f2 is the analytical branch of the multivalued function g α , with f2 (z1 ) = w2 . (z) Proof. 1. Let us define h(z) = gg(z) . Then h ∈ H(D), since g(z) ≠ 0, ∀z ∈ D. Because D is a simply connected domain, every arbitrary closed path of D is homotopic with 0 in D, hence on every closed path of D the integral of h is zero. It follows that the function h has a primitive in D, and let denote it by l. Then
(
(l (z)g(z) − g (z))el(z) (g (z) − g (z))el(z) el(z) ) = = = 0, g(z) g 2 (z) g 2 (z)
hence the expression from the first bracket is constant, i. e., el(z) = cg(z),
c ∈ ℂ.
Since el(z) cannot be equal to 0, it follows that c ≠ 0. Let choose the constant k ∈ ℂ, such that the function f1 (z) = l(z) + k satisfies the condition f1 (z1 ) = w1 , i. e., k = w1 − l(z1 ). Then the function f1 ∈ H(D) is the analytical branch of Log g which we want to find, since ef1 (z) = el(z) ek = el(z) ew1 e−l(z1 ) =
cg(z) w1 e = g(z), cg(z1 )
because w1 ∈ Log g(z1 ), hence ew1 = g(z1 ). Obviously, we have f1 (z) ∈ Log g(z), ∀z ∈ D
and f1 (z) = l (z) =
g (z) , g(z)
since according to its definition, the function l is a primitive function for h = gg . If ̃f1 ∈ H(D) is another branch of Log g(z) that satisfies the condition ̃f1 (z1 ) = w1 , then
g(z) = ef1 (z) = ef1 (z) , ̃
hence ef1 (z)−f1 (z) = 1. ̃
If we differentiate the both sides of this relation, we get ̃ (̃f1 (z) − f1 (z))ef1 (z)−f1 (z) ≡ 0,
so ̃f1 (z) ≡ f1 (z). Since ̃f1 (z1 ) = f1 (z1 ), we deduce that ̃f1 = f1 . 2. Let be the number w2 ∈ ℂ, such that w2 ∈ (g(z1 ))α = eαLog g(z1 ) and let be w1 ∈ Log g(z1 ) a such a number that w2 = eαw1 . Using the first point of the proof, there exists the function f1 ∈ H(D), such that w1 = f1 (z1 ) and f1 (z) ∈ Log g(z), ∀z ∈ D.
84  3 The complex integration Now define the function f2 (z) = eαf1 (z) , z ∈ D. Then α
f2 (z) ∈ eαLog g(z) = (g(z)) ,
∀z ∈ D,
and f2 (z1 ) = eαf1 (z1 ) = eαw1 = w2 .
Hence there exists a such a function f2 ∈ H(D), with f2 (z) ∈ (g(z))α , ∀z ∈ D, and f2 (z1 ) = w2 . From here, we have f2 (z) = αf1 (z)eαf1 (z) = α α−1
= αg (z)(g(z))
g (z) αf1 (z) (g(z))α e ∈ αg (z) g(z) g(z)
,
∀z ∈ D.
Let ̃f2 ∈ H(D) another analytic branch of (g(z))α , with ̃f2 (z1 ) = w2 . Now define the function h(z) =
̃f (z) 2 . f2 (z)
Then h ∈ H(D) and α2πik(z) −2πk(z) Im α , = e h(z) = e
z ∈ D,
h(z) where k(z) ∈ ℤ, ∀z ∈ D. It follows that k(z) = − ln , z ∈ D. Since h ∈ H(D) and 2π Im α the function k : D → ℤ is continuous in D, from here we deduce that k is a constant function on D. Hence, according to the above formula, the function h will be constant on the domain D, so it follows that h is also constant on D. Since h(z1 ) = 1, h(z) = 1, ∀z ∈ D, we finally get that ̃f2 = f2 .
3.6 The index of a path (curve) with respect to a point Let us consider the following two paths, γ1 (t) = e2πit and γ2 (t) = e4πit . When t runs to the whole interval [0, 1], then the image γ1 (t) runs once, while the image γ2 (t) runs twice over the unit circle in positive direction. We see that dζ 1 = 1, ∫ 2πi ζ γ1
dζ 1 = 2. ∫ 2πi ζ γ2
It follows that the above integrals count however many times the images γ1 (t) and γ2 (t) run over the image curve; in this case, the unit circle, when the argument runs over the interval [0, 1]. Definition 3.6.1. Let γ be a rectifiable path (curve) in ℂ (not necessary closed), and let be the point z0 ∈ ℂ \ {γ}. The index of the curve γ with respect to the point z0 is given by the number n(γ, z0 ) =
dζ 1 . ∫ 2πi ζ − z0 γ
3.6 The index of a path (curve) with respect to a point  85
Definition 3.6.2. The function f : G → ℂ is said to be locally constant if f is constant in every connected component of G. If the subset G ⊂ ℂ is open, and f ∈ H(G), then the function f is locally constant on G if and only if f (z) = 0, ∀z ∈ G. Theorem 3.6.1 (The index theorem). Let γ be a rectifiable curve in ℂ. 1. If γ1 ∼ γ2 , then n(γ1 , z0 ) = n(γ2 , z0 ). 2. 3.
ℂ\{z0 }
If γ = γ1 ∪ γ2 , then n(γ, z0 ) = n(γ1 , z0 ) + n(γ2 , z0 ). n(γ − , z0 ) = −n(γ, z0 ). Let γ be a rectifiable closed curve in ℂ. Then: 4. If z0 ∈ ℂ \ {γ}, then n(γ, z0 ) ∈ ℤ (integer number). 5. The function j(z) = n(γ, z) is locally constant on ℂ \ {γ}. 6. If γ = 𝜕U(z0 ; r) and z ∈ U(z0 ; r), then n(γ, z) = 1. 7. The set ℂ \ {γ} has exactly one not bounded connected component. If z0 is an arbitrary point of this not bounded connected component, then n(γ, z0 ) = 0. 8. The set {z ∈ ℂ : n(γ, z) ≠ 0} is bounded.
Proof. Point 1 follows immediately from the Cauchy’s theorem. Points 2 and 3 are immediate consequences of the properties of the complex integrals. 1 4. Let z ∈ ℂ \ {γ}. The function f (ζ ) = ζ −z is holomorphic on the ℂ \ {z} set, but it has no primitive, since it is not simply connected set. But it has primitive on every simply connected subset of ℂ \ {z}. Next, we will divide such an open neighborhood of {γ} in the corresponding simply connected domains, such that every one of these components of the function f has a primitive. Let ρ = d(z; {γ}). Since {γ} is a closed set, and z ∉ {γ}, ρ > 0, the function γ is uniformly continuous on the closed interval [0, 1], hence ∃η > 0
such that, if t − t < η, then γ(t ) − γ(t ) < ρ.
Let us consider the division Δ = (t0 , t1 , . . . , tn ), 0 = t0 < t1 < ⋅ ⋅ ⋅ < tn = 1 of the interval [0, 1], such that ‖Δ‖ < η, and let zk = γ(tk ), k ∈ {0, 1, . . . , n}. Then γ([tk , tk+1 ]) ⊂ U(zk ; ρ), k = 0, n − 1. Let γ1 , γ2 , . . . , γn be the decomposition of the curve γ corresponding to the division Δ. Then {γk } = γ([tk−1 , tk ]) ⊂ U(zk−1 ; ρ), and n(γ, z) =
n dζ dζ 1 1 =∑ . ∫ ∫ 2πi ζ − z k=1 2πi ζ − z γ
γk
Now we will compute the integrals which appears in this sum. We know that z0 = γ(0), and let us consider w0 ∈ Log (z0 −z). Now we will use the existence theorem of the analytical branch for the multivalued function Log , for Log g(ζ ), where g(ζ ) = ζ − z, in the U(z0 ; ρ) disc (which is a simply connected domain). It follows that there exists
86  3 The complex integration
Figure 3.4: Proof of Theorem 3.6.1.
the function f0 ∈ H(U(z0 ; ρ)), such that f0 (ζ ) =
1 , ζ −z
f0 (ζ ) ∈ Log (ζ − z),
∀ζ ∈ U(z0 ; ρ) and f0 (z0 ) = w0 .
Hence f0 (z1 ) ∈ Log (z1 −z), because z1 ∈ U(z0 ; ρ) (Figure 3.4). In the disc U(z1 ; ρ), we will choose that analytic branch of Log (ζ − z), denoted by f1 , such that f1 (z1 ) = f0 (z1 ), and we will repeat this algorithm for all the points z2 , . . . , zn . 1 has the primitive Because of the fact that in the domain U(zk ; ρ), the function ζ −z fk (ζ ) ∈ Log (ζ − z), it follows that ∫ γk
dζ = fk (zk ) − fk (zk−1 ). ζ −z
So, we deduce that n(γ, z) =
dζ 1 n 1 n = ∑∫ ∑ (f (z ) − fk (zk−1 )) 2πi k=1 ζ − z 2πi k=1 k k γk
1 1 = (f (z ) − f0 (z0 )) = (f (z ) − f0 (z0 )), 2πi n n 2πi n 0 because fk (zk ) = fk−1 (zk ) and zn = z0 (γ is closed path). Since fn (z0 ) ∈ Log (z0 − z) and f0 (z0 ) ∈ Log (z0 − z), fn (z0 ) − f0 (z0 ) = 2kπi,
for a convenient k ∈ ℤ.
(The number k will be the same, for any arbitrary choose of the starting value w0 ∈ Log (z0 − z), because in its definition the integral is welldefined.) It follows that n(γ, z) =
1 ⋅ 2kπi = k ∈ ℤ. 2πi
3.7 Cauchy formula for closed curves  87
dζ
1 5. Let the function j(z) = n(γ, z) = 2πi ∫γ ζ −z defined on ℂ \ {γ}. It is well known that j is differentiable on the whole ℂ \ {γ} set. On the other hand, from the above point we know that j(z) is an integer number, so j needs to be constant in an enough small neighborhood of z, because it is continuous on z. So, j (z) = 0, and then the function j(z) = n(γ, z) is locally constant on the open set ℂ \ {γ}. Another proof of this point is the next ____. Let be the function j(z) = n(γ, z) = dζ 1 defined on ℂ \ {γ}. It is well known that j is differentiable on the set ℂ \ {γ}, ∫ 2πi γ ζ −z
and j (z) =
dζ 1 . ∫ 2πi γ (ζ −z)2
Since the function g(ζ ) =
1 (ζ −z)2
has a primitive on ℂ \ {γ}, and
γ is a closed rectifiable path, it follows that j (z) = 0, ∀z ∈ ℂ \ {γ}, hence j is locally constant on the open set ℂ \ {γ}. 6. If γ = 𝜕U(z0 ; r), then we may easily see that ∫ γ
dζ = 2πi, ζ −z
∀z ∈ U(z0 ; r),
hence n(γ, z) = 1. 7. Since {γ} = γ([0, 1]) is compact, close and bounded, there exists r > 0 such that {γ} ⊂ U(0; r), and thus ℂ \ U(0; r) ⊂ ℂ \ {γ}. Because ℂ \ U(0; r) is a connected domain, it is contained in a connected component of ℂ \ {γ}. Evidently, this is the unique component of ℂ \ {γ} that contains the ∞. Using the previous point of the theorem, the function n(γ, z) is constant on the set ℂ \ U(0; r). Let zn ∈ ℂ \ U(0; r), zn → ∞. Then dζ V(γ) 1 1 ⋅ → 0. 0 ≤ n(γ, zn ) = ≤ ∫ 2π ζ − zn 2π zn  − r γ Hence n(γ, z) = 0, ∀z ∈ ℂ\U(0; r), and the same is valid for all the points z that belongs to the component of ℂ \ {γ} that contains the ∞. 8. This point is a direct consequence of the point 7.
3.7 Cauchy formula for closed curves Theorem 3.7.1 (Cauchy formulas for closed curves). If G ⊂ ℂ, where G is an open set, and if f ∈ H(G), then the function f has derivatives of arbitrary order in the whole set G. For every rectifiable closed γ curve with γ ∼ 0, and ∀z ∈ G \ {γ} the next formula G
holds: n(γ, z) ⋅ f (k) (z) =
f (ζ ) k! dζ , ∫ 2πi (ζ − z)k+1
∀k ∈ ℕ.
(3.11)
γ
Proof. Since the differentiability is a local property, using the Cauchy theorem for the disc, the first part of the theorem is true, hence the function f has derivatives of arbitrary order in the whole G set.
88  3 The complex integration Next, we will prove the formula (3.11) for the case k = 0. Let z ∈ G \ {γ} be fixed. Then f (ζ ) f (ζ ) − f (z) dζ 1 1 f (z) dζ = dζ + . ∫ ∫ ∫ 2πi ζ − z 2πi ζ −z 2πi ζ − z γ
γ
γ
The value of the last integral is 2πi n(γ, z). The value of the first integral of the righthand side is zero, since the function g given by f (ζ )−f (z)
if ζ ∈ G \ {z},
, { ζ −z g(ζ ) = { { f (z),
if ζ = z
is analytic on G \ {z}, and continuous on G. Hence g has a primitive in a neighborhood of z, and according to the theorem of Morera, it is differentiable at z. So, we have that g ∈ H(G), γ ∼ 0, and according to Cauchy’s integral theorem we G
obtain ∫γ g = 0. We obtained that the formula (3.11) is true for the case k = 0. Since it is well known that we can differentiate under the integral sign, it follows that the relations (3.11) hold for all k ∈ ℕ.
3.8 Some consequences of Cauchy formula Definition 3.8.1. The real valued function u, defined on the open complex subset G is said to be harmonic, if it has secondorder continuous partial derivatives on G, which satisfy 𝜕2 u(z) 𝜕2 u(z) + = 0, 𝜕x 2 𝜕y2
∀z = x + iy ∈ G.
Theorem 3.8.1. Let G ⊂ ℂ an open set, and let f = u + iv be holomorphic on G. Then the real and the imaginary parts of f have arbitrary order partial derivatives on G, and both are harmonic functions on G, i. e., 𝜕2 u 𝜕2 u + = 0, 𝜕x 2 𝜕y2 Proof. If f = u + iv, then f =
𝜕u 𝜕x
𝜕v + i 𝜕x =
𝜕v 𝜕y
𝜕2 v 𝜕2 v + = 0. 𝜕x 2 𝜕y2 − i 𝜕u and since f is differentiable we have 𝜕y
that 𝜕u , 𝜕v , 𝜕u , 𝜕v ∈ C(G). Using the Cauchy–Riemann theorem, the functions u and v 𝜕x 𝜕x 𝜕y 𝜕y have secondorder continuous partial derivatives. 2 2 𝜕v 𝜕v 𝜕2 v 𝜕2 v = 𝜕y and 𝜕u = − 𝜕x , it follows that 𝜕𝜕xu2 = 𝜕y𝜕x and 𝜕𝜕yu2 = − 𝜕x𝜕y , and Since 𝜕u 𝜕x 𝜕y 2 𝜕2 u + 𝜕𝜕yu2 𝜕x 2 𝜕2 v 𝜕2 v + 𝜕y 2 = 0. 𝜕x 2
according to the Schwarz theorem we obtain Similarly, it is possible to prove that
= 0.
3.8 Some consequences of Cauchy formula  89
Remark 3.8.1. Since the function u and v are connected by the Cauchy–Riemann relations, they are called harmonically conjugate functions. Theorem 3.8.2. 1. If u : D → ℝ is a harmonic function on the simply connected domain D ⊂ ℂ, then there exists at least one function v : D → ℝ, such that f = u + iv ∈ H(D). 2. If v : D → ℝ is a harmonic function on the simply connected domain D ⊂ ℂ, then there exists at least one function u : D → ℝ, such that f = u + iv ∈ H(D). ̃ + iṽ, where u ̃= Proof. 1. Let ̃f : D → ℂ, ̃f = u have firstorder partial derivatives on D, and
𝜕u , 𝜕x
̃ and ṽ ṽ = − 𝜕u . Then the functions u 𝜕y
̃ 𝜕2 u 𝜕u 𝜕2 u 𝜕ṽ = 2 =− 2 = 𝜕x 𝜕x 𝜕y 𝜕y
̃ 𝜕u 𝜕2 u 𝜕2 u 𝜕ṽ = = =− . 𝜕y 𝜕x𝜕y 𝜕y𝜕x 𝜕x
According to the Cauchy–Riemann theorem, we have ̃f ∈ H(D), hence the function ̃f has primitives on the simply connected domain D ⊂ ℂ. 𝜕v 𝜕v If f = u + iv is a primitive for ̃f , hence 𝜕u = 𝜕y and 𝜕u = − 𝜕x . From the fact 𝜕x 𝜕y f=
𝜕v ̃ 𝜕u 𝜕u 𝜕u +i =f = −i , 𝜕x 𝜕x 𝜕x 𝜕y
it follows that 𝜕u = 𝜕u and 𝜕u = 𝜕u , so u = u + c, c ∈ ℝ, and we obtain f = f − c = 𝜕x 𝜕x 𝜕y 𝜕y u + iv ∈ H(D). 2. The proof is similar to the previous point. Remark 3.8.2. 1. According to the previous theorem, if u : D → ℝ is a harmonic function on the simply connected domain D ⊂ ℂ, then there exists at least one function v : D → ℝ, such that f = u + iv ∈ H(D). The function v can be found as follows. Since dv =
𝜕v 𝜕u 𝜕u 𝜕v dx + dy = − dx + dy, 𝜕x 𝜕y 𝜕y 𝜕x
which is a total differential on D, if we integrate it along the curve [MN]∪[NP] ⊂ D, where M(x0 , y0 ), N(x, y0 ) and P(x, y), then x
y
x0
y0
𝜕u(x, y0 ) 𝜕u(x, y) v(x, y) − v(x0 , y0 ) = − ∫ dx + ∫ dy. 𝜕y 𝜕x 2.
Similarly, we know that if v : D → ℝ is a harmonic function on the simply connected domain D ⊂ ℂ, then there exists at least one function u : D → ℝ, such that f = u + iv ∈ H(D).
90  3 The complex integration The function u can be obtained similarly: 𝜕u 𝜕u 𝜕v 𝜕v dx + dy = dx − dy, 𝜕x 𝜕y 𝜕y 𝜕x
du =
is a total differential on D, if we integrate it along the curve [MN]∪[NP] ⊂ D, where M(x0 , y0 ), N(x, y0 ) and P(x, y), then x
u(x, y) − u(x0 , y0 ) = ∫ x0
3.
y
𝜕v(x, y0 ) 𝜕v(x, y) dx − ∫ dy. 𝜕y 𝜕x y0
If f ∈ H(G), where G ⊂ ℂ is an open set, and z0 = x0 + iy0 ∈ G such that f (z0 ) ≠ 0, then 𝜕u(x0 ,y0 ) 𝜕(u, v)(x0 , y0 ) 𝜕x = 𝜕u(x0 ,y0 ) 𝜕(x, y) 𝜕y =
𝜕v(x0 ,y0 ) 𝜕x
𝜕v(x0 ,y0 ) 𝜕y
𝜕u(x0 , y0 ) 𝜕v(x0 , y0 ) 𝜕u(x0 , y0 ) 𝜕v(x0 , y0 ) − 𝜕x 𝜕y 𝜕y 𝜕x 2
=(
2
𝜕u(x0 , y0 ) 𝜕v(x0 , y0 ) 2 ) +( ) = f (z0 ) ≠ 0. 𝜕x 𝜕x
Since the Jacobian determinant is not zero in z0 ∈ G, and f has firstorder continuous partial derivatives on G, according to the theorem concerning the inverse function, there exist such open sets U(z0 ; r) ⊂ G and U(f (z0 ); ρ) ⊂ ℂ; for those, the function f : U(z0 ; r) → U(f (z0 ; ρ)) has an inverse f −1 : U(f (z0 ; ρ)) → U(z0 ; r), which is continuously differentiable and
(f −1 ) (w) =
1 , f (f −1 (w))
∀w ∈ U(f (z0 ; ρ)).
Theorem 3.8.3 (Cauchy inequalities). Let M be an upper bound for f  on the closed disc U − (z0 ; r) ⊂ G, where f is a holomorphic function on open set G ⊂ ℂ. Then, for every n ∈ ℕ we have M (n) f (z0 ) ≤ n! n . r Proof. Using Cauchy formula for the discs, we directly deduce that n! (n) ∫ f (z0 ) = 2π
𝜕U(z0 ;r)
≤
n! M 2π r n+1
n! f (ζ ) dζ ≤ n+1 2π (ζ − z0 ) ∫
𝜕U(z0 ;r)
dζ  =
∫ 𝜕U(z0 ;r)
f (ζ ) dζ  ζ − z0 n+1
n! n! M 2πr = n M. 2π r n+1 r
3.9 Schwarz and Poisson formulas  91
Theorem 3.8.4 (Liouville theorem). Every bounded entire function is constant. Proof. Suppose that f ∈ H(ℂ) and f (z) < M, ∀z ∈ ℂ. Let us consider z0 ∈ ℂ, and r > 0 be arbitrary. From Theorem 3.8.3, it follows that M → 0, f (z0 ) ≤ r
if r → ∞.
Hence f (z0 ) = 0, ∀z0 ∈ ℂ ⇒ f is a constant function. Corollary 3.8.1 (Fundamental theorem of the algebra). Every polynomial P(z) = a0 + a1 z + ⋅ ⋅ ⋅ + an z n ,
ak ∈ ℂ, an ≠ 0, n ≥ 1
has at least a zero (root) in ℂ. Proof. We are presenting the second proof of the fundamental theorem of algebra which is based on the Liouville theorem. Suppose that the conclusion is not true. Then 1 f (z) = P(z) is an entire function, and limz→∞ f (z) = 0, hence there exists an enough big number r > 0, such that f (z) ≤ 1 if z > r. Let M = sup{f (z) : z ∈ U(0; r)}. Sine f is holomorphic, then it is continuous on the compact disc U(0; r), hence f  has a finite upper bound M < +∞. Evidently, f (z) ≤ M + 1,
∀z ∈ ℂ,
hence f is a bounded entire function. According to the Liouville theorem, the function f is constant on ℂ, i. e., f (z) = c, ∀z ∈ ℂ. Hence, P(z) =
1 , c
∀z ∈ ℂ,
which contradicts the assumption n ≥ 1 and an ≠ 0.
3.9 Schwarz and Poisson formulas Let f a holomorphic function on the disc U(z0 ; r), such that f is continuous on U(z0 ; r). According to the Cauchy formula, all the values of the function f (and its derivatives) on the disc U(z0 ; r) are given by the values of f on the boundary of U(z0 ; r) disc. The corresponding result for the harmonic functions is given by the Poisson formula. The real part of u of a holomorphic function f on the disc U(z0 ; r) enable us to determine the function f , except to a constant, if we know its values on the boundary of U(z0 ; r) disc. Schwarz’ formula gives this last connection. To simplify the computations, we assume in both of the cases that z0 = 0.
92  3 The complex integration Theorem 3.9.1 (Poisson formula). Let u be a harmonic function on the open set G ⊂ ℂ, such that U(0; R) ⊂ G. Then 2π
1 R2 − r 2 dt, ∫ u(Reit ) 2 2π R − 2Rr cos(t − θ) + r 2
u(reiθ ) =
∀z = reiθ ∈ U(0; R).
0
Proof. Since the set U(0; R) ⊂ G is compact, and ℂ \ G is closed, there exists a number ε > 0 such that D = U(0; R + ε) ⊂ G. Also, there exists a function f ∈ H(D) such that Re f = u, because D is a simply connected domain and u is a harmonic function on D. Denoting γ = 𝜕U(0; R), using the Cauchy formula for the discs (for the special case k = 0), we have 1
f (z) =
f (ζ ) f (γ(τ)) 1 1 dζ = γ (τ)dτ. ∫ ∫ 2πi ζ − z 2πi γ(τ) − z γ
0
Denote γ(τ) = Re2πiτ . Using the changing of variables 2πτ = t, we get 2π
1 f (Reit )Reit f (z) = dt. ∫ 2π Reit − reiθ
(3.12)
0
Denote by z ∗ =
R2 z
=
R2 iθ e r
the inverse of z with respect to the circle 𝜕U(0; R), and let us
consider the function g(ζ ) = integral theorem we have
f (ζ ) . ζ −z ∗
Since z ∗  > R and g ∈ H(U(0; R)), from the Cauchy
f (ζ ) 1 dζ = 0, ∫ 2πi ζ − z ∗ γ
i. e., 2π
0=
f (Reit )reit 1 dt. ∫ it 2π re − Reiθ 0
Subtracting the equality (3.12) from the above relation, we deduce f (reiθ ) =
2π
1 r R − ]dt ∫ f (Reit )eit [ it 2π Re − reiθ reit − Reiθ 0
2π
R2 − r 2 1 dt, = ∫ f (Reit ) 2 2π R − 2Rr cos(t − θ) + r 2 0
and from this last formula immediately follows the required relation.
(3.13)
3.9 Schwarz and Poisson formulas  93
Remark 3.9.1. For the special case r = 0, we obtain 2π
u(0) =
1
1 ∫ u(Reit )dt = ∫ u(Re2πiτ )dτ. 2π 0
(3.14)
0
As a consequence, the value to the origin of a disc for any harmonic function, is the meanvalue of those from the boundary of the disc. Theorem 3.9.2 (Schwarz formula). Let f = u + iv ∈ H(G), where G ⊂ ℂ is an open set, and U(0; R) ⊂ G. Let γ = 𝜕U(0; R). Then f (z) =
ζ + z dζ 1 + iv(0), ∫ u(ζ ) 2πi ζ −z ζ
∀z ∈ U(0; R).
γ
Proof. Adding the relations (3.12) and (3.13), we obtain 2π
R r 1 + it ]dt f (re ) = ∫ f (Reit )eit [ it iθ 2π Re − re re − Reiθ iθ
0
2π
=
1 2Rr sin(θ − t) ]dt, ∫ f (Reit )[1 + i 2 2π R − 2Rr cos(θ − t) + r 2 0
and hence v(reiθ ) =
2π
2π
0
0
1 1 2Rr sin(θ − t) dt. ∫ v(Reit )dt + ∫ u(Reit ) 2 2π 2π R − 2Rr cos(θ − t) + r 2
Since v is a harmonic function, the first integral of the righthand side is equal to v(0), and applying Poisson formula we get f (reiθ ) = u(reiθ ) + iv(reiθ ) 2π
= iv(0) +
1 R2 + 2iRr sin(θ − t) − r 2 dt ∫ u(Reit ) 2 2π R − 2Rr cos(θ − t) + r 2 0
2π
R2 − r 2 + Rr[ei(θ−t) − e−i(θ−t) ] 1 dt = iv(0) + ∫ u(Reit ) 2 2 2π R − r − Rr[ei(θ−t) + e−i(θ−t) ] 0
2π
= iv(0) +
Reit + reiθ 1 dt. ∫ u(Reit ) it 2π Re − reiθ 0
Since z = reiθ and τ =
t , 2π
it follows that 1
1 Re2πiτ + z f (z) = iv(0) + 2πidτ, ∫ u(Re2πiτ ) 2πiτ 2πi Re −z 0
94  3 The complex integration and from γ(τ) = Re2πiτ , 1
γ(τ) + z γ (τ) 1 f (z) = iv(0) + dτ, ∫ u(γ(τ)) 2πi γ(τ) − z γ(τ) 0
we conclude ζ + z dζ 1 . ∫ u(ζ ) 2πi ζ −z ζ
f (z) = iv(0) +
γ
3.10 Exercises 3.10.1 The complex integral Exercise 3.10.1. Calculate the integral 1 ∫ dz z γ
in the following cases: ⌢
⌢
⌢
⌢
⌢
⌢
1.
{γ} = AB, where AB is the arc of the circle with the center in O that lies the points A(−i) and B(i), directly oriented;
2.
{γ} = AB, where AB is the arc of the circle with the center in O that lies the points A(−i) and B(i), inversely oriented;
3.
{γ} = CD, where CD is the arc of the circle with the center in O that lies the points C(1 − i) and D(1 + i), directly oriented; 4. {γ} = CD, where CD is linear path that connects the points C(1 − i) and D(1 + i).
Exercise 3.10.2. Calculate the integral ∫ γ
z dz, z
where γ = γ1 ∪ γ2 ∪ γ3 , with {γ1 } = [−3, −2], {γ3 } = [2, 3] are directly oriented segments, and {γ2 } is the arc of the circle with the center in O that lies the points A(−2) and B(2), inversely oriented. Exercise 3.10.3. Calculate the integral ∫ log zdz, γ π
where the path γ has the equation γ(t) = ei(2t−1) 2 , t ∈ [0, 1], and log z is the main branch of the multivalued function Log z, i. e., log 1 = 0.
3.10 Exercises  95
Exercise 3.10.4. Calculate the integral 1 dz, √z
∫ γ
πt
where the path γ has the equation γ(t) = ei 2 , t ∈ [0, 1], and √z is the main branch of 1 the multivalued function z 2 , i. e., √1 = 1. Exercise 3.10.5. Calculate the integral 1 dz z√z
∫ γ
in the next cases: 1. {γ} = 𝜕U(0; r 2 ) is a directly oriented path, and √z is the main branch of the multi1 valued function z 2 , i. e., √1 = 1; 1 2. γ(t) = r 2 e4πit , t ∈ [0, 1], and √z is the branch of the multivalued function z 2 with √1 = −1. Exercise 3.10.6. Calculate the integral ∫ zdz γ
in the next cases: 1. {γ} = [−1, 1] is a directly oriented segment; 2. the path γ has the equation γ(t) = eπit , t ∈ [0, 1]; 3. the path γ has the equation γ(t) = eπi(t−1) , t ∈ [0, 1]. Exercise 3.10.7. Calculate the integral ∫ zdz γ
in the next cases: 1. the path γ has the equation γ(t) = −i + 2it, t ∈ [0, 1]; π(2t−1) 2. the path γ has the equation γ(t) = ei 2 , t ∈ [0, 1]; 3. {γ} = 𝜕U(0; 1) is the directly oriented circle. Exercise 3.10.8. Calculate the integral ∫ Re zdz γ
in the next cases:
96  3 The complex integration 1. 2. 3.
γ = γ1 ∪ γ2 , where γ1 and γ2 are given by γ1 (t) = −1 + (i + 1)t, t ∈ [0, 1], and γ2 (t) = i + (1 − i)t, t ∈ [0, 1]; γ = γ3 , where γ3 (t) = ei γ = γ3 ∪ (γ1 ∪ γ2 )− .
π(t−1) 2
, t ∈ [0, 1];
3.10.2 The Cauchy theorem Exercise 3.10.9. Calculate the integral ∫ γ
1 dz z−α
in the following cases: 1. {γ} = 𝜕U(α; r) is a directly oriented path; 2. {γ} = 𝜕T, where T is the triangle determined by the points 1, i and −1, directly oriented, such that α ∈ ̸ 𝜕T. Exercise 3.10.10. Calculate the integral ∫(z − a)n dz γ
in the following cases: 1. γ is given by γ(t) = a + reπit , t ∈ [0, 1] and n ∈ ℤ; 2. {γ} = 𝜕U(a; r) is a directly oriented path, and n ∈ ℤ; 3. {γ} is an arbitrary smooth path, such that γ ∈ 𝒟(z0 , z1 ), and n ∈ ℤ \ {−1}; 4. γ is given by γ(t) = a + re2πit , t ∈ [0, 1] and n = 21 .
Exercise 3.10.11. Calculate the integral ∫ γ
ez dz, sin z + cos z
where {γ} = 𝜕U(0; 41 ) is a directly oriented path. Exercise 3.10.12. Calculate the integral ∫(z + ez sin z)dz, γ
where {γ} = 𝜕U(0; 2) is a directly oriented path.
3.10 Exercises  97
Exercise 3.10.13. Calculate the integral ∫ γ
ez dz, cos z cos iz
where {γ} ⊂ U(0; 1) is an arbitrary closed, rectifiable and directly oriented path. Exercise 3.10.14. Calculate the integral ∫ γ
1 dz, (z 2 + 4)(z 2 + 16)
where {γ} = 𝜕T, and T is the triangle determined by the points 1, i and −1, directly oriented. Exercise 3.10.15. Calculate the integral 1
e z−3 cos z dz, ∫ (z − 2)n γ
where {γ} = 𝜕U(0; 1) is a directly oriented path, and n ∈ ℤ. Exercise 3.10.16. Calculate the integral ∫ γ
z2
z dz, −1
where {γ} = 𝜕U(0; a) is a directly oriented path, and a ≠ 1.
3.10.3 The Cauchy formula for the disc Exercise 3.10.17. Calculate the following integral: ∫ γ
sin z dz, (z − π2 )(z 2 + 5)
where {γ} = 𝜕U(0; 2) is a directly oriented path. Exercise 3.10.18. Calculate the following integral: ∫ γ
z5 dz, ez (z + 1)6
where {γ} = 𝜕U(0; 2) is a directly oriented path.
98  3 The complex integration Exercise 3.10.19. Calculate the following integral: ∫ γ
zez dz, (z − 2)n
where {γ} = 𝜕U(0; 3) is a directly oriented path and n ∈ ℕ. Exercise 3.10.20. Calculate the following integral: ∫ γ
ez sin z dz, z−a
where {γ} = 𝜕U(0; r) is a directly oriented path, and r ≠ a. Exercise 3.10.21. Calculate the integral, ∫ γ
1 dz, z2 + 1
in the next cases: 1. {γ} = 𝜕U(i; 1) is a directly oriented path; 2. {γ} = 𝜕U(−i; 1) is a directly oriented path; 3. {γ} = 𝜕U(0; 2) is a directly oriented path. Exercise 3.10.22. Calculate the following integral: ∫ γ
ez cos z dz, z3 + 8
where {γ} = 𝜕U(0; 1) is a directly oriented path. Exercise 3.10.23. Calculate the following integral: ∫ γ
sin z dz, zk
k ∈ ℤ,
where {γ} = 𝜕U(0; 1) is a directly oriented path. Exercise 3.10.24. Calculate the integral, ∫ γ
ez dz, z−a
in the next cases: 1. {γ} = 𝜕U(a; a), a ∈ ℂ∗ , is a directly oriented path; 2. {γ} = 𝜕U(−a; a), a ∈ ℂ∗ , is a directly oriented path.
3.10 Exercises  99
Exercise 3.10.25. Calculate the following integral: ∫ γ
sinh z z−
πi 2
dz,
where {γ} = 𝜕U( πi2 ; 1) is a directly oriented path. Exercise 3.10.26. Calculate the following integral: ∫(z − α)n dz,
n ∈ ℤ,
γ
where {γ} = 𝜕U(0; r), α ≠ r, is a directly oriented path. Exercise 3.10.27. Calculate the following integral: ∫ γ
z(z 2
1 dz, + z − 2)
in the next cases: 1. {γ} = 𝜕U(0; 32 ) is a directly oriented path; 2. {γ} = 𝜕U(0; 3) is a directly oriented path. Exercise 3.10.28. Calculate the integral ∫ γ
f (z) dz, z(z 2 + 1)n
in the next cases: 1. f (z) = 1, n = 1 and {γ} = {z = x + iy : x2 + 4y2 − 1 = 0} is a directly oriented path; 2. f (z) = 1, n = 1 and {γ} = 𝜕U(0; 2) is a directly oriented path; 3. f (z) = eiz , n = 1 and {γ} = {z = x + iy : x2 + 4y2 − 1 = 0} is a directly oriented path; 4. f (z) = eiz , n = 1 and {γ} = 𝜕U(0; 2) is a directly oriented path; 5. f (z) = eiz , n = 2 and {γ} = {z = x + iy : x2 + 4y2 − 1 = 0} is a directly oriented path; 6. f (z) = eiz , n = 2 and {γ} = 𝜕U(0; 2) is a directly oriented path. Exercise 3.10.29. Calculate the integral ∫ γ
cos πz dz, z2 − 1
where {γ} = {z = x + iy : x2 + y2 − 2ax = 0}, a ∉ {− 21 , 0, 21 }, is a directly oriented path. Exercise 3.10.30. Calculate the following integral: ∫ γ
1 dz, (z 2 + 1)n
n ∈ ℤ,
where {γ} = {z = x + iy : x2 + y2 − ay = 0}, a ∈ ℝ∗ \ {1}, is a directly oriented path.
100  3 The complex integration Exercise 3.10.31. Calculate the integral iπ
∫ γ
ze 2 z dz, z−1
where {γ} = 𝜕U(0; a), a ≠ 1, is a directly oriented path. Exercise 3.10.32. Calculate the integral ∫ γ
zn dz, z−a
n ∈ ℕ,
where {γ} = 𝜕U(0; 1), a ≠ 1 is a directly oriented path. Exercise 3.10.33. Calculate the following integral: eπz dz, z(z − i)
∫ γ
where {γ} = 𝜕U(0; a), a ≠ 1 is a directly oriented path. Exercise 3.10.34. Calculate the integral ∫ γ
eiz dz, z2 − π2
where {γ} = 𝜕U(0; a), a ≠ π is a directly oriented path. Exercise 3.10.35. Calculate the following integral: ∫ γ
cos πz dz, (z 2 − 4)2
where {γ} = {z = x + iy : x2 + y2 − ax = 0}, a ∉ {−2, 0, 2} is a directly oriented path. Exercise 3.10.36. Calculate the following integral: ∫ γ
z 100 eiπz dz, z2 + 1
where {γ} = {z = x + iy : 4x 2 + y2 − 4 = 0} is a directly oriented path. Exercise 3.10.37. Calculate the following integral: ∫ γ
cosh πz 2 (z + i)4
dz,
where {γ} = {z ∈ ℂ : z + 2i = 2} is a directly oriented path.
3.10 Exercises  101
Exercise 3.10.38. Let f : {z ∈ ℂ : Re z ≥ 0} → ℂ be a continuous function, such that limz→∞ zf (z) = 0. Prove that, for all a ≤ 0, we have lim ∫ eaz f (z)dz = 0,
r→∞
where γr (t) = rei
π(2t−1) 2
γr
, t ∈ [0, 1].
Exercise 3.10.39. Let f : {z ∈ ℂ : Re z ≤ 0} → ℂ be a continuous function, such that limz→∞ zf (z) = 0. Prove that, for all a ≥ 0, we have lim ∫ eaz f (z)dz = 0,
r→∞
where γr (t) = rei
π(2t+1) 2
γr
, t ∈ [0, 1].
3.10.4 Some consequences of Cauchy formula Exercise 3.10.40. Determine the entire functions f of the form f (z) = u(x, y) + iv(x, y), z = x + iy, if we know that: 1. v(x, y) = ex sin y and f (0) = 1; 2. v(x, y) = x2 − y2 + xy and f (0) = 0; 3. v(x, y) = 2xy and f (0) = 0; 4. u(x, y) = ex (x cos y − y sin y) and f (0) = 0. Exercise 3.10.41. Prove that there exists a function of the form f (z) = u(x, y) + iv(x, y), z = x + iy, which is holomorphic on the domain D ⊂ ℂ, with 0 ∉ D and 1 ∈ D, such that y v(x, y) = x2 +y 2 and f (1) = 0. Exercise 3.10.42. Did there exists functions of the form f (z) = u(x, y) + iv(x, y), where z = x + iy, that are holomorphic on a domain D ⊂ ℂ, such that: 2 −y2 1. u(x, y) = (xx2 +y 2 )2 ;
2. 3.
v(x, y) = ln(x2 + y2 ) − x2 + y2 ; y u(x, y) = e x ?
Exercise 3.10.43. Determine the function f of the form f (z) = u(x, y)+iv(x, y), z = x+iy, which is holomorphic on the domain D ⊂ ℂ, in the following cases: y 1. D = ℂ \ {z ∈ ℂ : Re z ≥ 0, Im z = 0}, v(x, y) = x2 +y 2 and f (−2) = 0; 2.
D = ℂ \ {z ∈ ℂ : Im z ≤ 0, Re z = 0}, v(x, y) = 3 + x2 − y2 − 1 2
2
2
y 2(x2 +y2 )
3. D = ℂ \ {z ∈ ℂ : Re z ≤ 0, Im z = 0}, u(x, y) = ln(x + y ); 4. D = ℂ \ {z ∈ ℂ : Re z ≤ 0, Im z = 0}, v(x, y) = ln(x2 + y2 ) + x − 2y;
and f (2) = 0;
102  3 The complex integration 5. 6. 7.
D = ℂ, v(x, y) = 1 + xy; y D = ℂ \ {z ∈ ℂ : Im z ≤ 0}, v(x, y) = y − x2 +y 2; D = ℂ \ ({z ∈ ℂ : Re z ≤ 0, Im z = 1} ∪ {z ∈ ℂ : Re z ≥ 0, Im z = −1}), u(x, y) =
x x2 +(y+1)2
+
x x2 +(y−1)2
and f (1) = 1;
8. D = ℂ \ ({z ∈ ℂ : Im z ≤ 0, Re z = −1} ∪ {z ∈ ℂ : Im z ≥ 0, Re z = 1}), 9.
v(x, y) = − (x+1)y2 +y2 −
y (x−1)2 +y2
and f (0) = 0;
D = ℂ \ ({z ∈ ℂ : Im z ≥ 0, Re z = −1} ∪ {z ∈ ℂ : Im z ≤ 0, Re z = 1}), 2xy 1 v(x, y) = − (x2 −y2 −1) 2 +4x 2 y 2 and f (i) = − 2 .
3.10.5 Multivalued functions analytical branches Exercise 3.10.44. Determine the maximal domain D ⊂ ℂ, such that the following functions are holomorphic in D: z ; 1. w(z) = √3 2i−z
2. w(z) = √3 z + 1;
4. w(z) = Log
5.
z−1 ; z+1
w(z) =
Log z ; z
3.
w(z) = Log (z 2 + 1);
6.
w(z) = √z − i + √z + i.
Exercise 3.10.45. Did there exist holomorphic branches of the following functions in the corresponding domain D ⊂ ℂ? 1. w(z) = √3 z+1 , where D = {z ∈ ℂ : 1 < z < ∞}; z−1 2.
3.
w(z) = 2Log
z+1 , z−1
where D = {z ∈ ℂ : 1 < z < ∞};
w(z) = √(z 2 − 1)(z 2 − 4), where D = {z ∈ ℂ : Re z > 0, z − 3 > 52 }.
Exercise 3.10.46. Define the function f (z) = √3 z + 1. 1. If D = ℂ \ T, where T = {z ∈ ℂ : Im z = 0, Re z ≥ −1}, then determine the branch f0 of the function f , such that f0 ∈ H(D) and f0 (−2) = −1; 2. Compute the values f0 (i), f0 (−i), f0 (i) and f0 (−i); 3. Denote by ̃f the extension of f0 to D ∪ T. Determine the values of the function ̃f on the borders Ta and Tf of the halfline T. Exercise 3.10.47. Let us define the function f (z) = √z − i + √z + i. 1. If D = ℂ \ (T1 ∪ T2 ), where T1 = {z ∈ ℂ : Re z = 0, Im z ≤ −1}, and T2 = {z ∈ ℂ : Re z = 0, Im z ≥ 1}, then determine the branch f0 of the function f , such that f0 ∈ H(D) and f0 (0) = −i√2; 2. Compute the values f0 ( √13 ) and f0 (2); 3. Denote by ̃f the extension of f0 to D∪(T1 ∪T2 ). Determine the values of the function ̃f on both of the borders of T and T . 1
2
Exercise 3.10.48. Let us define the function f (z) = Log (z 2 + 1).
3.10 Exercises  103
1.
2. 3.
If D = ℂ \ (T1 ∪ T2 ), where T1 = {z ∈ ℂ : Re z = 0, Im z ≤ −1}, and T2 = {z ∈ ℂ : Re z = 0, Im z ≥ 1}, then determine the branch f0 of the function f , such that f0 (0) = 0; Compute the values f0 (2 + i), f0 (−2 − i) and f0 ( 2i ); Denote by ̃f the extension of f0 to D∪(T1 ∪T2 ). Determine the values of the function ̃f on the borders of T ∪ T . 1 2
Exercise 3.10.49. Let us define the function f (z) = √z 2 + 1. Determine the holomorphic branches of the function f in the corresponding domains D ⊂ ℂ: 1. D = ℂ \ (T1 ∪ T2 ), where T1 = {z ∈ ℂ : Re z = 0, Im z ≤ −1}, and T2 = {z ∈ ℂ : Re z = 0, Im z ≥ 1}; 2. D = ℂ \ T3 , where T3 = {z ∈ ℂ : Re z = 0, Im z ≥ −1}; 3. D = ℂ \ T4 , where T4 = {z ∈ ℂ : Re z = 0, −1 ≤ Im z ≤ 1}. Exercise 3.10.50. Let us define the function f (z) = Logz z . 1. If D = ℂ \ T, where T = {z ∈ ℂ : Re z ≤ 0, Im z = 0}, then determine the branch f0 of the function f , such that f (z) ∈ ℝ whenever z > 0; 2. Compute the values f0 (i) and f0 (−i). Exercise 3.10.51. Calculate the index n(γ, z) for z ∈ ℂ \ {γ}, if γ = γ1 ∪ γ2 , where γ1 and γ2 are given by γ1 (t) = (2t − 1)r, t ∈ [0, 1] and γ2 (t) = reπit , t ∈ [0, 1]. Exercise 3.10.52. Calculate the index n(γ, z) for z ∈ {z ∈ ℂ : r < z < ρ, Im z > 0}, if γ = γ1 ∪ γ2 ∪ γ3 ∪ γ4 , where γ1 , γ2 , γ3 and γ4 are given by γ1 (t) = −ρ(1 − t) − rt, t ∈ [0, 1], γ2 (t) = reπ(1−t)i , t ∈ [0, 1], γ3 (t) = r(1 − t) + ρt, t ∈ [0, 1] and γ4 (t) = ρeπit , t ∈ [0, 1]. Exercise 3.10.53. Let G ⊂ ℂ be an open set, and let f : G → ℂ such that f ∈ C(G), and f ∈ H(G \ [a, b]). Prove that f ∈ H(G). Exercise 3.10.54. Let f ∈ C(ℂ) a bounded function, such that f ∈ H(ℂ \ [a, b]). Prove that the function f is constant on ℂ. Exercise 3.10.55. Did there exist nonconstant entire functions f ∈ H(ℂ), such that f (ℂ) ⊂ {w ∈ ℂ : Im w > 0}? Exercise 3.10.56. Did there exist nonconstant entire functions f ∈ H(ℂ), such that f (ℂ) ⊂ {w ∈ ℂ : Re w > 0}? Exercise 3.10.57. Did there exist nonconstant entire functions f ∈ H(ℂ), such that f (ℂ) ⊂ ℂ \ U(0; 1)? Exercise 3.10.58. Let the function f ∈ H(ℂ), such that there exists limz→∞ f (z) = a ∈ ℂ. Prove that f is a constant function on ℂ.
104  3 The complex integration Exercise 3.10.59. Let f ∈ H(ℂ) such that Re f is a bounded function. Prove that f is a constant function on ℂ. Exercise 3.10.60. Let f ∈ C(ℂ) such that, for all r > 0 we have f (z) =
∫ 𝜕U(0;r)
Prove that f (z) = 0, ∀z ∈ ℂ.
f (ζ ) dζ , ζ −z
∀z ∈ ℂ : z > r.
4 Sequences and series of holomorphic functions 4.1 Sequences of holomorphic functions Definition 4.1.1. Let (fn )n∈ℕ∗ be a sequence of holomorphic functions defined on the open set G ⊂ ℂ. We say that the sequence (fn )n∈ℕ∗ uniformly converges on compact sets to the function f : G → ℂ (briefly, compact converges to f ), if for any arbitrary compact set K, with K ⊂ G, the function sequence (fn K )n∈ℕ∗ uniformly converges to f K . (Denoted by fn K f .) Remarks 4.1.1. 1. The compact convergence is stronger than the pointwise convergence on the whole set G, but it is weaker than the uniformly convergence on G. 2. The compact convergence can be written as follows: The sequence of the functions (fn )n∈ℕ∗ uniformly converges on compact sets to f , if and only if for any arbitrary number ε > 0, and any arbitrary compact set K ⊂ G, there exists an index n0 = n0 (ε, K) ∈ ℕ, such that for all n > n0 we have fn (z) − f (z) < ε, 3.
∀z ∈ K.
The next Cauchy uniformly convergence criteria holds: The sequence of functions (fn )n∈ℕ∗ uniformly converges on compact sets, if and only if for any arbitrary number ε > 0, and any arbitrary compact set K ⊂ G, there exists an index n0 = n0 (ε, K) ∈ ℕ, such that for all n, m > n0 we have fn (z) − fm (z) < ε,
∀z ∈ K.
4. Since every compact set K ⊂ ℂ can be covered by a finite number of closed discs of G, it follows that the sequence of functions (fn )n∈ℕ∗ uniformly converges on compact sets, if and only if uniformly converges on any arbitrary closed disc of G. Theorem 4.1.1. If the functions fn : G → ℂ, n ∈ ℕ∗ , are continuous on the open set G, and (fn )n∈ℕ∗ compact converges to f : G → ℂ, then f is a continuous function on G. Proof. Let z0 ∈ G be an arbitrary point, and let r > 0 such that K = U(z0 ; r) ⊂ G. Since K is compact and fn K f , it follows that f is continuous on K, thus f is continuous at the point z0 . Theorem 4.1.2 (Weierstrass lemma). If the functions (fn )n∈ℕ∗ are holomorphic on the disc D = U(z0 ; R), are continuous on D− , and the sequence of functions (fn )n∈ℕ∗ uniformly converges on the boundary 𝜕D, then: 1. (fn )n∈ℕ∗ is compact convergent on D; 2. the limit f of the sequence of functions (fn )n∈ℕ∗ is holomorphic in D, i. e., f ∈ H(D); https://doi.org/10.1515/9783110657869004
106  4 Sequences and series of holomorphic functions 3.
for any k ∈ ℕ, the sequence of the derivative functions (fn(k) )n∈ℕ∗ compact converges to f (k) on D.
Proof. Since (fn )n∈ℕ∗ ⊂ C(D) and (fn )n∈ℕ∗ uniformly converges on the set 𝜕D, the limit g of the sequence (fn 𝜕D )n∈ℕ∗ is continuous on 𝜕D. Let γ = 𝜕D = 𝜕U(z0 ; R). Then, according to the Theorem 3.4.1, the function f defined by f (z) =
g(ζ ) 1 dζ , ∫ 2πi ζ − z γ
z∈D
is holomorphic on D, and from Corollary 3.4.1 the function f has derivatives of any order in D. Moreover, f (k) (z) =
g(ζ ) k! dζ , ∫ 2πi (ζ − z)k+1 γ
z ∈ D.
We will prove that (fn(k) )n∈ℕ∗ compact converges to f (k) in D, for all natural numbers k. Using the Cauchy formula for the functions fn , we have fn(k) (z) =
fn (ζ ) k! dζ , ∫ 2πi (ζ − z)k+1 γ
z ∈ D.
Hence k! fn (ζ ) − g(ζ ) (k) (k) dζ fn (z) − f (z) = ∫ , 2π (ζ − z)k+1 γ
z ∈ D.
It is sufficient to check that (fn(k) )n∈ℕ∗ uniformly converges to f (k) on the closed disc U(z0 ; r), for any arbitrary r < R. Let d = R − r, and let ε > 0 be an arbitrary number. Since the sequence of functions (fn 𝜕D )n∈ℕ∗ uniformly converges to g, there exists an index n0 = n0 (ε) ∈ ℕ, such that ∀n > n0 we have k+1 εd , fn (ζ ) − g(ζ ) < Rk!
∀ζ ∈ 𝜕D.
It follows that ∀n > n0 , and ∀z ∈ U(z0 ; r) the next inequality holds: k! εdk+1 1 k! fn (ζ ) − g(ζ ) (k) (k) dζ  ≤ 2πR = ε, ∫ fn (z) − f (z) ≤ k+1 2π 2π Rk! dk+1 ζ − z γ
hence fn(k) U(z
0 ;r)
f (k) , ∀k ∈ ℕ.
Theorem 4.1.3 (Weierstrass theorem). Let G ⊂ ℂ be an open set, and let fn ∈ H(G), n ∈ ℕ∗ . If the sequence of functions (fn )n∈ℕ∗ compact converges to f : G → ℂ in G, then:
4.2 Series of functions  107
1. 2.
f ∈ H(G); the sequence of the derivative functions (fn(k) )n∈ℕ∗ compact converges to f (k) in G, ∀k ∈ ℕ.
Proof. According to the point 4 of the Remark 4.1.1, it is sufficient to prove the compact convergence on an arbitrary closed disc of G. But, using the Weierstrass lemma (Theorem 4.1.2), this last condition holds.
4.2 Series of functions Definition 4.2.1. 1. Let G ⊂ ℂ be an open set, and let fn : G → ℂ, ∀n ∈ ℕ. The sum ∞
∑ fn = ∑ fn = f0 + f1 + ⋅ ⋅ ⋅ + fn + ⋅ ⋅ ⋅ n=0
is called series of functions. The notation ∞
∑ fn (z) = f0 (z) + f1 (z) + ⋅ ⋅ ⋅ + fn (z) + ⋅ ⋅ ⋅
n=0
2.
is also used. The finite sum Sn = f0 + f1 + ⋅ ⋅ ⋅ + fn
is called the partial sum of the above series of function. The series of functions ∑ fn converges, uniformly converges or compact converges in G if the partial sum (Sn )n∈ℕ converges, uniformly converges, or compact converges in G, respectively. 4. The limit S = limn→∞ Sn is called the sum of the series of functions, and it is denoted by the symbol S = ∑ fn . The notation
3.
∞
S(z) = ∑ fn (z), n=0
z∈G
is also used. Remark 4.2.1. The Cauchy convergence criteria for the series of functions can be written as follows: The series of functions ∑ fn uniformly converges on the set E, if and only if ∀ε > 0 there exists an index n0 = n0 (ε) ∈ ℕ, such that ∀n > n0 and ∀p ∈ ℕ∗ we have fn+1 (z) + ⋅ ⋅ ⋅ + fn+p (z) < ε,
∀z ∈ E.
108  4 Sequences and series of holomorphic functions An immediate consequence of this criteria is the next theorem. Theorem 4.2.1 (Weierstrass uniformly convergence criteria). If there exists a convergent series of positive numbers ∑ un , and there exists an index n0 ∈ ℕ, such that for all n > n0 we have fn (z) ≤ un ,
∀z ∈ E,
then the series of functions ∑ fn uniformly converges in E. In fact, the series of functions ∑ fn absolutely converges, in the sense that the series ∑ fn  (uniformly) converges. Theorem 4.2.2 (Weierstrass theorem for holomorphic functions series). Let G ⊂ ℂ be an open set, and let fn ∈ H(G), n ∈ ℕ. If the series of functions ∑ fn compact converges in G, then: 1. the sum S = ∑ fn ∈ H(G) and 2. the series of the derivative functions ∑ fn(k) compact converges to S(k) in G, ∀k ∈ ℕ.
4.3 Power series Definition 4.3.1. A power series is a series of functions defined as follows: ∞
∑ an (z − z0 )n = a0 + a1 (z − z0 ) + ⋅ ⋅ ⋅ + an (z − z0 )n + ⋅ ⋅ ⋅ ,
n=0
where z0 is a fixed number, the variable is z and an ∈ ℂ. The above series of functions is called the power series with the coefficients an , about the point z0 . n Theorem 4.3.1 (Abel theorem). If the power series ∑∞ n=0 an (z − z0 ) converges in z1 ≠ z0 , then it absolutely converges in all the points z ∈ ℂ, with z − z0  < z1 − z0 , and the series of functions compact converges on the disc U(z0 ; z1 − z0 ).
Proof. Let r = z − z0 , and let denote r1 = z1 − z0 . Since the series of complex numbers ∑ an (z1 − z0 )n converges, the general term tends to zero. Hence, there exists an index n0 ∈ ℕ such that an r1n = an (z1 − z0 )n < 1,
∀n > n0 .
Considering the series of positive numbers ∑ an r n , for all n > n0 we have an r n < ( rr )n , because an  < r1n . 1
1
If r < r1 , then the geometric series ∑( rr )n is convergent, and using the Weierstrass 1 criteria (Theorem 4.2.1) we deduce that the series of functions ∑ an (z − z0 )n absolutely and uniformly converges on the closed disc U(z0 ; r). Since we can choose the number r to be arbitrary close to r1 , the above result is proved.
4.3 Power series  109
Theorem 4.3.2 (Cauchy–Hadamard theorem or the radius of convergence theorem). n Let ∑∞ n=0 an (z − z0 ) be a power series, and define the number R using the following Cauchy–Hadamard relations: 1 = lim sup √n an , R n→+∞
if lim sup √n an  ∈ ℝ∗ ;
R = 0,
if lim sup √n an  = +∞;
R = +∞,
if lim sup √n an  = 0.
n→+∞
n→+∞
n→+∞
The number R, called the radius of convergence of the power series, satisfies the following properties: 1. The power series is absolute and compact convergent on the disc U(z0 ; R); 2. the power series is divergent in all the points of ℂ \ U(z0 ; R); 3. the sum S of the power series is holomorphic on U(z0 ; R); 4. the power series can be derived term by term, and the obtained power series n−1 has the radius of convergence equal to R, and the sum equal to ∑∞ n=1 nan (z − z0 ) S (the derivative of the sum S), i. e., ∞
S (z) = ∑ nan (z − z0 )n−1 , n=1
5.
∀z ∈ U(z0 ; R).
the power series can be integrated term by term, and the obtained power series n ∑∞ n=1 an ∫C (z − z0 ) dz has the radius of convergence equal to R, and the sum equal to ∫C S where C is smooth piecewise curve which lie inside U(z0 , R).
n n n n Proof. Let L = lim supn→+∞ √a n , and let denote r = z − z0 . Since √an r = r √an , using the Cauchy root convergence criteria for the series of positive numbers ∑ an r n , we deduce that this series converges if rL < 1, and diverges if rL > 1. Suppose that L ≠ 0, and L ≠ +∞. Then, for z − z0  = r < R = L1 (i. e., if rL < 1), the power series ∑ an (z − z0 )n absolutely converges. We will prove that if z − z0  = r > R, then the power series diverges. In fact, if there exists a number z1 , with z1 − z0  > R, such that the series ∑ an (z1 − z0 )n converges, then for the number r with z1 − z0  > r > R we may use a similar proof like of the Abel theorem (Theorem 4.3.1), and we deduce that the series ∑ an r n is convergent. On the other hand, r > L1 , i. e., Lr > 1, which contradicts the Cauchy root convergence criteria. Hence, in this case, the power series absolutely converges in the all the points of the disc U(z0 ; R), and by the Abel theorem, it is compact convergent on this open disc. The power series diverges in all the points of the set ℂ \ U(z0 ; with R). If L = 0, then rL < 1 for all r > 0, thus the above results hold if R = +∞. If L = +∞, then rL > 1 for all r > 0, thus the power series diverges in all the points z ∈ ℂ \ {z0 }, so R = 0 and the convergence domain will be {z0 }.
110  4 Sequences and series of holomorphic functions Determine now the convergence radius R for the series of the derivative functions
n−1 . Since the nature of the power series remains unchanged if we mul∑∞ n=1 nan (z − z0 )
tiply it by (z − z0 ), from the above results it follows that the radius of convergence for the series of the derivative functions is the same with those of ∞
∑ nan (z − z0 )n .
n=1
Hence, 1 1 = lim sup √n nan  = lim √n n ⋅ lim sup √n an  = lim sup √n an  = , n→+∞ R R n→+∞ n→+∞ n→+∞ and thus R = R. This result could be also obtained by using the Weierstrass theorem for the series of functions (see the Theorem 4.2.2). From this previously mentioned theorem, we conclude that ∞
S (z) = ∑ nan (z − z0 )n−1 . n=1
Remarks 4.3.1. 1. If there exists the finite or the infinite limit limn→+∞ R = lim
n→+∞
2.
an+1  an 
= l ∈ ℝ, then l = L and
an  . an+1 
If we derive k times the given power series, and then we replace z = z0 , it follows that ak =
S(k) (z0 ) . k!
3.
The disc U(z0 ; R) does not represent the domain of convergence of the power series. Theorem 4.3.2 (the Cauchy–Hadamard theorem) does not give any information about the points of the boundary 𝜕U(z0 ; R) (where the convergence of the power series needs to be studied in every given point). 4. For two given power series, ∞
S(z) = ∑ an (z − z0 )n n=0
the sum of the series S and T will be ∞
∑ (an + bn )(z − z0 )n ,
n=0
∞
and T(z) = ∑ bn (z − z0 )n , n=0
4.4 The analyticity of holomorphic functions  111
while the product of the series S and T will be ∞
∑ cn (z − z0 )n ,
where cn = a0 bn + a1 bn−1 + ⋅ ⋅ ⋅ + an b0 .
n=0
5.
The convergence radius of both of these power series will be greater or equal to the minimum of the convergence radius of S and T. 1 n The power series ∑∞ n=0 z has the convergence radius R = 1, and the sum 1−z . The above power series converges if z < 1, and diverges if z ≥ 1.
Theorem 4.3.3 (The theorem of the identity of the power series coefficients). Let ∞
S(z) = ∑ an (z − z0 )n n=0
∞
T(z) = ∑ bn (z − z0 )n
and
n=0
be two power series with the convergence radius at least r > 0 (i. e., their sums exist in the disc U(z0 ; r)). If there exists a subset E ⊂ U(z0 ; r), such that z0 ∈ E and S(z) = T(z), ∀z ∈ E, then the two given power series are identical, i. e., an = bn , ∀n ∈ ℕ. n Proof. The difference of the power series ∑∞ n=0 (an − bn )(z − z0 ) will be S − T. If we suppose that k ∈ ℕ is the smallest index, such that ak ≠ bk , then ∞
S(z) − T(z) = (z − z0 )k (ak − bk + ∑ (ak+n − bk+n )(z − z0 )n ) = (z − z0 )k Q(z),
n=1
z ∈ U(z0 ; r).
Let (zm )m∈ℕ∗ ⊂ E \ {z0 } be a sequence, with limm→+∞ zm = z0 . Since S(zm ) − T(zm ) = 0, we have that Q(zm ) = 0, m ∈ ℕ∗ . Also, the function Q is continuous at z0 , because it is the sum of a power series about the point z0 . Hence Q(z0 ) = ak − bk = limm→+∞ Q(zm ) = 0, so we get that ak = bk , which represents a contradiction.
4.4 The analyticity of holomorphic functions Theorem 4.4.1 (Taylor series expansion theorem). Let D = U(z0 ; r), with r > 0, and let n f ∈ H(D). Then there exists a unique power series ∑∞ n=0 an (z − z0 ) , with the radius of convergence R ≥ r, such that ∞
f (z) = ∑ an (z − z0 )n , n=0
∀z ∈ D,
and the power series coefficients are given by an =
f (n) (z0 ) f (ζ ) 1 = dζ , ∫ n! 2πi (ζ − z0 )n+1
where γ = 𝜕U(z0 ; ρ), with 0 < ρ < r.
γ
112  4 Sequences and series of holomorphic functions Proof. We will use the Cauchy formula for an arbitrary disc with the radius ρ ∈ (0, r), and we obtain f (n) (z0 ) f (ζ ) 1 = dζ , ∫ n! 2πi (ζ − z0 )n+1
γ = 𝜕U(z0 ; ρ).
γ
Thus, the righthand side integrals are independent on the radius of the circle γ. Let z ∈ D, and let choose a radius ρ, such that r0 = z − z0  < ρ < r. From the Cauchy formula, we get f (z) = The fraction
1 ζ −z
f (ζ ) 1 dζ . ∫ 2πi ζ − z γ
can be written as follows: 1 1 1 1 . = = ζ − z ζ − z0 + z0 − z ζ − z0 1 − z−z0 ζ −z 0
z−z
r
If ζ ∈ {γ}, from the above choosing method of ρ we have  ζ −z0  = ρ0 < 1. Consequently, 0 from the Weierstrass criteria (Theorem 4.2.1) we deduce that the sum ∞
∑(
n=0
n
z − z0 ) ζ − z0
(where the variable is ζ ∈ {γ}) uniformly converges on the set {γ}, because n z − z n r 0 ) = ( 0 ) ( ρ ζ − z0
and the geometric series
∞
∑(
n=0
n
r0 ) ρ
converges. But n
∞
∑(
n=0
z − z0 1 ) = z−z , ζ − z0 1 − ζ −z0 0
1 and if we replace under the integral the fraction ζ −z , by using the above relations, then the series n
f (ζ ) ∞ z − z0 ) ∑( ζ − z0 n=0 ζ − z0
also uniformly converges on the set {γ}, and its sum will be f (z) =
f (ζ ) . ζ −z
Hence n
f (ζ ) f (ζ ) ∞ z − z0 1 1 ) dζ . dζ = ∑( ∫ ∫ 2πi ζ − z 2πi ζ − z0 n=0 ζ − z0 γ
γ
4.4 The analyticity of holomorphic functions  113
The uniformly convergence allows us to integrate term by term, and thus ∞
f (z) = ∑ ( n=0
f (ζ ) 1 dζ )(z − z0 )n . ∫ 2πi (ζ − z0 )n+1 γ
From here, by using the notation from the theorem we obtain ∞
f (z) = ∑ an (z − z0 )n . n=0
From the above proof, we conclude that the power series of the righthand side converges. Since z is an arbitrary point of the disc U(z0 ; r), it follows that the given power series convergence with radius R is greater or equal than r, i. e., R ≥ r. From Theorem 3.3.2, and respectively from Theorem 3.3.3, using the Theorem 4.4.1 we deduce the following results. Theorem 4.4.2. Let D = U(z0 ; r), and let denote by A the set of all the complex numbers n sequences (an )n∈ℕ ⊂ ℂ, such that the power series ∑∞ n=0 an (z − z0 ) converges, ∀z ∈ D. Then there exists a unique bijective and linear function L : H(D) → A, such that if n f ∈ H(D) and if L(f ) = (an ), then f (z) = ∑∞ n=0 an (z − z0 ) , ∀z ∈ D. The coefficients an , n ∈ ℕ, are given by the formulas an =
f (n) (z0 ) . n!
Definition 4.4.1. 1. Let G ⊂ ℂ open set, and let f : G → ℂ. We say that the function f may be expanded in Taylor series about the point z0 ∈ G, if there exists a number r > 0 n with U(z0 ; r) ⊂ G, and there exists a power series ∑∞ n=0 an (z − z0 ) which converges on the disc U(z0 ; r), such that ∞
f (z) = ∑ an (z − z0 )n , n=0
2.
∀z ∈ U(z0 ; r).
If the above property holds for any arbitrary point z0 ∈ G, then we say that the function f is analytic on G.
Theorem 4.4.3 (The analyticity of holomorphic functions). The complex valued function f defined on the open set G ⊂ ℂ is holomorphic on G, if and only if it is analytic on G. Proof. Let f ∈ H(G) and let z0 ∈ G. Then there exists a number r > 0, such that U(z0 ; r) ⊂ G, and the function f is holomorphic on this disc. Then, from the Theorem 4.4.1 the function f may be expanded in Taylor series on the disc U(z0 ; r), and thus f is analytic in z0 . Since z0 ∈ G is arbitrary, then function f will be analytic on G.
114  4 Sequences and series of holomorphic functions If the function f is analytic on G, then ∀z0 ∈ G, ∃r > 0, such that U(z0 ; r) ⊂ G and the function f may be expanded in Taylor series on the disc U(z0 ; r). Hence, the function f is the sum of the power series, and from the Cauchy–Hadamard theorem it will be holomorphic on this disc. So, there exists the derivative f (z0 ) ∈ ℂ, and since z0 ∈ G is arbitrary we conclude that f ∈ H(G). Examples 4.4.1. 1. Let us consider the geometric power series ∞ 1 = ∑ zn, 1 − z n=0
z ∈ U(0; 1).
Replacing in this formula the variable z by −z, we obtain that ∞ 1 = ∑ (−1)n z n , 1 + z n=0
z ∈ U(0; 1).
From the Weierstrass lemma, we can derive term by term the above relation, and we deduce that ∞ 1 = ∑ nz n−1 , 2 (1 − z) n=1
2.
z ∈ U(0; 1).
Since ez , cos z, and sin z are integer functions, from the Theorem 4.4.1 we get that zn , n! n=0 ∞
ez = ∑ 3.
∞ 1 = ∑ (−1)n−1 nz n−1 , 2 (1 + z) n=1
∞
cos z = ∑ (−1)n n=0
z 2n , (2n)!
∞
sin z = ∑ (−1)n n=0
z 2n+1 , (2n + 1)!
z ∈ ℂ.
Let f (z) = log(1 + z) be those branch of the multivalued function Log (1 + z), such that f (0) = 0. Since f ∈ H(U(0; 1)), from the Theorem 4.4.1 we obtain the following formula: ∞
log(1 + z) = ∑ (−1)n+1 n=1
zn , n
z ∈ U(0; 1).
4. Let α ∈ ℂ, and let f (z) = (1+z)α be those branch of the multivalued power function (1 + z)α , such that f (0) = 1. We deduce similarly that α(α − 1) ⋅ . . . ⋅ (α − n + 1) n z , n! n=1 ∞
(1 + z)α = 1 + ∑
z ∈ U(0; 1),
and this last series is called the binomial series.
4.5 The zeros of holomorphic functions Definition 4.5.1. Let G ⊂ ℂ an open set, and let f ∈ H(G). The number a ∈ G is said to be an nth order zero for the f function, if f (a) = f (a) = ⋅ ⋅ ⋅ = f (n−1) (a) = 0,
f (n) (a) ≠ 0.
4.5 The zeros of holomorphic functions  115
Theorem 4.5.1 (Characterization of the zeros of holomorphic functions). If G is an open subset of ℂ, then a ∈ G is an nth order zero for the holomorphic function f , if and only if there exists a function g ∈ H(G), such that f (z) = (z − a)n g(z),
∀z ∈ G and g(a) ≠ 0.
Proof. Since f is analytic on G, there exists the number r > 0 such that U(a; r) ⊂ G, and f (z) = f (a) +
f (a) f (n−1) (a) f (n) (a) (z − a) + ⋅ ⋅ ⋅ + (z − a)n−1 + (z − a)n + ⋅ ⋅ ⋅ 1! (n − 1)! n!
= (z − a)n (
f (n) (a) f (n+1) (a) + (z − a) + ⋅ ⋅ ⋅), n! (n + 1)!
∀z ∈ U(a; r).
(4.1)
Let us define the function { g(z) = { {
f (z) , (z−a)n f
(n)
(a) , n!
if z ∈ G \ {a}, if z = a.
From the relation (4.1), it follows that lim g(z) =
z→a
f (n) (a) = g(a) ≠ 0, n!
hence g is continuous at a. Since g is holomorphic on G \ {a}, we obtain evidently that g is differentiable also in a, so g ∈ H(G). Conversely, if f has the form of the conclusion of the theorem, then f may be written as in the above expansion, and then the number a is an nth order zero for f . Theorem 4.5.2 (The set of the zeros of holomorphic functions). Let f be an holomorphic function on the domain D ⊂ ℂ, and let introduce the following notation: A = A(f ) = {a ∈ D : f (a) = 0}
and
B = B(f ) = {a ∈ D : f (k) (a) = 0, ∀k ∈ ℕ}. Then the next statements are equivalent: (i) (ii) (iii)
f ≡ 0;
B ≠ 0;
A ∩ D ≠ 0
(where A represents the set of all accumulation points of A).
116  4 Sequences and series of holomorphic functions Proof. It is obvious that from (i) it follows (ii) and (iii). We will prove that (ii) ⇒ (i). For this purpose, we will show that B is a closed, and in the same time an open in D. We see that, if a ∈ B, from the analyticity of f it follows that there exists a number r > 0, such that ∞
f (z) = ∑ an (z − a)n , n=0
∀z ∈ U(a; r),
where an = f n!(a) = 0, ∀n ∈ ℕ, because a ∈ B. Then f (z) = 0, ∀z ∈ U(a; r), and consequently, f (n) (z) = 0, ∀n ∈ ℕ and ∀z ∈ U(a; r). Thus z ∈ B, ∀z ∈ U(a; r), that is the set B is open in D. Let (zn )n∈ℕ ⊂ B and a = limn→+∞ zn ∈ D. Since f (k) ∈ H(D), the function f (k) is continuous at a, ∀k ∈ ℕ. Hence, limn→+∞ f (k) (zn ) = f (k) (a). But f (k) (zn ) = 0, because zn ∈ B, and thus f (k) (a) = 0, ∀k ∈ ℕ, so a ∈ B. Hence, the set B is closed in D. From the fact that D is connected, we conclude that B = D, and then f ≡ 0. Now we will prove that (iii) ⇒ (ii). Let a ∈ A ∩ D. From the continuity of f , we have f (a) = 0. We will prove that a ∈ B. If this would not be true, then there exists the smallest m ∈ ℕ natural number, such that f (m) (a) ≠ 0. Then the number a ∈ D will be an mth order zero for f , and by Theorem 4.5.1 there exists a function g ∈ H(D), such that f (z) = (z − a)m g(z) and g(a) ≠ 0. Since a ∈ A , there exists a sequence (an )∈ℕ ⊂ A \ {a} such that limn→+∞ an = a. Since f (an ) = 0 and an − a ≠ 0, ∀n ∈ ℕ, g(an ) = 0, ∀n ∈ ℕ, from the continuity of g it follows that g(a) = limn→+∞ g(an ) = 0, which represents a contradiction. (n)
Corollary 4.5.1 (Properties of the zeros holomorphic functions). If D ⊂ ℂ is a domain, f ∈ H(D), f ≢ 0 and a ∈ D is a zero for the function f , then: ̇ r) = U(a; r) \ {a}; 1. ∃r > 0 such that f (z) ≠ 0, ∀z ∈ U(a; 2. the number a is a finite order zero for the function f ; 3. ∃n ∈ ℕ∗ and ∃g ∈ H(D), such that f (z) = (z − a)n g(z), ∀z ∈ D and g(a) ≠ 0. Theorem 4.5.3 (Uniqueness theorem of holomorphic functions). Let D ⊂ ℂ a domain, and let f , g ∈ H(D). Suppose that one of the next assumptions holds: 1. ∃E ⊂ D, such that E ∩ D ≠ 0 and f E = gE (that is {z ∈ D : f (z) = g(z)} ∩ D ≠ 0); 2. ∃a ∈ D, such that f (n) (a) = g (n) (a), ∀n ∈ ℕ. Then f ≡ g. Proof. We will apply Theorem 4.5.2 for the f − g function. Remarks 4.5.1. 1. From the Cauchy formula, it follows that the function f ∈ H(D) is well determined if we know its values on a closed curve (path).
4.6 The maximum principle of the holomorphic functions  117
2.
According to Theorem 4.5.3, the function f ∈ H(D) is well determined if we know its values on an arbitrary sequence that converges in D.
Corollary 4.5.2. Let D ⊂ ℂ be a domain. Then the set H(D) is a domain of integrity with respect to the addition and the multiplication of functions. Proof. Suppose that f , g ∈ H(D) are such functions, that f ⋅ g ≡ 0. If f ≢ 0, then the set A = A(f ) (the set of f function zeros) consists from isolated points. In order that f (z)g(z) = 0 holds ∀z ∈ D, it is necessary that g(z) = 0, ∀z ∈ D \ A. Since (D \ A) ∩ D ≠ 0, from Theorem 4.5.3, we deduce that g ≡ 0. Hence, the assumption f ⋅ g ≡ 0 implies that f ≡ 0 or g ≡ 0. Theorem 4.5.4 (The principle of the continuation of holomorphic functions). Let D ̃ ̃ ̃ ̃ and D two domains in ℂ, and D ⊂ D, D ≠ 0. If f ∈ H(D) and f ∈ H(D) are two functions, such that ̃f D = f , then the function ̃f is the unique continuation (extension) of f to the domain ̃ D. The function ̃f is called the holomorphic (analytic) continuation (extension) of f . Proof. In fact, if ̃f1 ∈ H(̃ D) and ̃f1 D = f , then ̃f1 D = ̃f D . Since D ∩ ̃ D ≠ 0, from Theõ ̃ rem 4.5.3 we have f1 = f .
4.6 The maximum principle of the holomorphic functions Theorem 4.6.1 (The maximum principle of the holomorphic functions). If D ⊂ ℂ is a domain, and f ∈ H(D) is a nonconstant function, then f  cannot take its maximum value in the domain D. Proof. Suppose that there exists z0 ∈ D, such that f (z0 ) ≥ f (z), ∀z ∈ D. Let r > 0 such that U(z0 ; r) ⊂ D, and let denote γ = 𝜕U(z0 ; r). Thus {γ} ⊂ D, and using Cauchy formula we get f (z0 ) =
f (ζ ) 1 dζ . ∫ 2πi ζ − z0 γ
We know that γ(t) = z0 + re2πit , t ∈ [0, 1], hence 1
f (z0 + re2πit ) 1 d(re2πit ) f (z0 ) = ∫ 2πi re2πit 0
=
1
1
0
0
f (γ(t)) 1 ∫ 2πit ⋅ r2πie2πit dt = ∫ f (γ(t))dt. 2πi re
118  4 Sequences and series of holomorphic functions It follows 1 1 f (z0 ) = ∫ f (γ(t))dt ≤ ∫ f (γ(t))dt, 0 0 or 1
1
0 ≤ ∫(f (γ(t)) − f (z0 ))dt ≤ ∫(f (z0 ) − f (z0 ))dt = 0. 0
0
This yields that f (γ(t)) = f (z0 ), ∀t ∈ [0, 1]. Hence, on the set {γ} we have f (z) = f (z0 ). But the radius of γ can be changed, to be arbitrary and less or equal to r. Thus we deduce that f (z) is constant for all z ∈ U(z0 ; r). From a wellknown previous result, we get that f is constant on the U(z0 ; r) disc. Then, according to the principle of the continuation of holomorphic functions, we conclude that f is constant on the whole domain D. Remark 4.6.1. The previous theorem has the next equivalent form: If D ⊂ ℂ is a domain, f ∈ H(D) and there exists z0 ∈ D such that f (z) ≤ f (z0 ), ∀z ∈ D, then f is constant on D. Corollary 4.6.1. If f ∈ H(D) is a nonconstant function, and if f (z) ≠ 0, ∀z ∈ D, where D is a domain, then f  cannot take its minimum value in the domain D. Proof. From the assumption, we have g = f1 ∈ H(D), and since g cannot take its maximum value in D, it follows that f  cannot take its minimum value in D. Corollary 4.6.2. Let D ⊂ ℂ be a bounded domain, and let f ∈ C(D) ∩ H(D). Then max f (z) = max f (z). z∈D
z∈𝜕D
Proof. Since f ∈ C(D), there exists z0 ∈ D such that f (z) ≤ f (z0 ), ∀z ∈ D. If the function f is nonconstant, from Theorem 4.6.1 we get that z0 ∉ D, hence z0 ∈ 𝜕D. The next result is one of the extensions of Liouville theorem. Theorem 4.6.2. If f ∈ H(ℂ), that is f is an entire function, and if 1 , f (z) ≤  Im z
∀z ∈ ℂ,
(4.2)
then f (z) = 0, ∀z ∈ ℂ. Proof. Evidently, the function f is bounded on the set ℂ \ Δε for all arbitrary ε > 0, where Δε = {z ∈ ℂ :  Im z < ε}, but it is not sure that it is bounded on Δε . Let r > 0, and let define the function g(z) = (z 2 − r 2 )f (z), z ∈ ℂ.
4.6 The maximum principle of the holomorphic functions  119
If z ∈ ℂ with z = r and Re z ≥ 0, then there exists θ(z) ∈ [0, π4 ] such that cos θ(z) = Then, according to relation (4.2), we get
 Im z . z−r
1 z − r = ≤ √2, (z − r)f (z) ≤  Im z cos θ(z)
∀z ∈ ℂ : z = r, Re z ≥ 0.
(4.3)
If z ∈ ℂ with z = r and Re z < 0, similarly we obtain 1  − z − r = ≤ √2, (z + r)f (z) ≤  Im(−z) cos θ(−z)
∀z ∈ ℂ : z = r, Re z < 0.
(4.4)
Now, from (4.3) and (4.4), we respectively deduce that g(z) = (z − r)f (z)z + r ≤ 2√2r, if z = r, Re z ≥ 0 g(z) = (z + r)f (z)z − r ≤ 2√2r, if z = r, Re z < 0, hence g(z) ≤ 2√2r, if z ∈ ℂ and z = r. From here, using Theorem 4.6.1, we conclude that g(z) ≤ 2√2r,
∀z ∈ ℂ : z ≤ r,
hence 2√2r f (z) ≤ 2 2 , z − r 
∀z ∈ ℂ : z < r,
and if in this last inequality we take r → +∞, it follows that f (z) = 0, ∀z ∈ ℂ. Theorem 4.6.3. Let Δ = {z ∈ ℂ : 0 < Re z < 1}, let f ∈ H(Δ) such that f ∈ C(Δ), and suppose that f is bounded on the compact set Δ. For θ ∈ [0, 1], let define the function Mθ (f ) = sup{f (θ + iy) : y ∈ ℝ}. Then Mθ (f ) ≤ [M0 (f )]
1−θ
θ
⋅ [M1 (f )] .
Proof. Without loss of the generality, suppose that M0 (f ) > 0 and M1 (f ) > 0. Moreover, if we replace the function f by f (z) ⋅ [M0 (f )]θ−1 ⋅ [M1 (f )]−θ , we may suppose that M0 (f ) = M1 (f ) = 1, and in this last case we need to prove that f (z) ≤ 1, ∀z ∈ Δ. Let M = sup{f (z) : z ∈ Δ} < +∞, and define the function fn (z) =
f (z)
1+
M z n
,
z ∈ Δ, n ∈ ℕ∗ .
If we denote by 𝜕Rn the boundary of the rectangle with corners in −in, 1 − in, 1 + in and in, i. e., Rn = {z ∈ ℂ : 0 < Re z < 1,  Im z < n}, n ∈ ℕ∗ ,
120  4 Sequences and series of holomorphic functions then we may easily prove that fn (z) ≤ 1,
∀z ∈ 𝜕Δ :  Im z ≤ n, or ∀z ∈ Δ :  Im z = n.
It follows that fn (z) ≤ 1, ∀z ∈ 𝜕Rn , ∀n ∈ ℕ∗ . From here, according to Theorem 4.6.1, we deduce that fn (z) ≤ 1,
∀z ∈ Rn , ∀n ∈ ℕ∗ ,
and since limn→+∞ fn (z) = f (z), ∀z ∈ Δ, we conclude that f (z) ≤ 1, ∀z ∈ Δ. Theorem 4.6.4 (Schwarz lemma). If f ∈ H(U(0; 1)) satisfies the conditions f (0) = 0 and f (z) < 1, ∀z ∈ U(0; 1), then: 1. f (z) ≤ z, ∀z ∈ U(0; 1); 2. f (0) ≤ 1. If there exists z0 ∈ U(0; 1) such that f (z0 ) = z0 , or such that f (0) = 1, then f has the form f (z) = cz, where c ∈ ℂ, c = 1. Proof. Define the function g by g(z) = {
f (z) , z
f (0),
if z ∈ U̇ = U(0; 1) \ {0}, if z = 0.
̇ and according to a known result we have g ∈ H(U). Let r be an Then g ∈ C(U) ∩ H(U), arbitrary number, such that 0 < r < 1. From Corollary 4.6.2, it follows max g(z) = max g(z). z∈𝜕U(0;r)
z∈U(0;r)
If z ∈ 𝜕U(0; r), from the assumption of the theorem we obtain f (z) 1 ≤ , g(z) = z r and according to the previous remarks 1 g(z) ≤ , r
∀z ∈ U(0; r).
g(z) ≤ 1,
∀z ∈ U(0; 1),
Letting r → 1, we get
and from the definition of g we have f (z) ≤ z,
∀z ∈ U(0; 1).
4.6 The maximum principle of the holomorphic functions  121
If z = 0, then g(0) ≤ 1,
that is f (0) ≤ 1.
Suppose that there exists z0 ∈ U such that f (z0 ) = z0 . Then g(z0 ) = 1 and since g(z) ≤ 1, ∀z ∈ U, according to the Theorem of the module maximum of the holomorphic functions it follows that g(z) = 1, ∀z ∈ U. So, the holomorphic function g has a constant module on the domain U, hence g is constant on U, i. e., g(z) = c,
∀z ∈ U,
where c = 1. From the definition of g, we now deduce that f (z) = cz,
∀z ∈ U.
We may will use the same proof if f (0) = 1, because g(0) = f (0). Remark 4.6.2. The geometric interpretation of Schwarz lemma is the next: Let f ∈ H(U(0; 1)) a such a function, with f (0) = 0 and f (U(0; 1)) ⊂ U(0; 1). Then the inclusion f (U(0; r)) ⊂ U(0; r) holds for every r ∈ (0, 1). There exists a number r∗ ∈ (0, 1) such that f (U(0; r∗ )) = U(0; r∗ ) if and only if the function f has the form f (z) = cz, where c = 1. The Schwarz lemma can be generalized as follows. Theorem 4.6.5 (Generalized Schwarz lemma). If the function f ∈ H(U(z0 ; r)) satisfies the condition f (z) − f (z0 ) < R, ∀z ∈ U(z0 ; r), then: 1. f (z) − f (z0 ) ≤ Rr z − z0 , ∀z ∈ U(z0 ; r); 2. f (z0 ) ≤ Rr . If there exists z∗ ∈ U(z0 ; r) such that f (z∗ ) − f (z0 ) = Rr z∗ − z0 , or such that f (z0 ) = Rr , then the function f has the form f (z) = f (z0 ) + c Rr (z − z0 ), where c ∈ ℂ, c = 1. Proof. Let us define the function g(ζ ) = R1 [f (z) − f (z0 )], where z = z0 + rζ and ζ ∈ U. It follows that g ∈ H(U), g(0) = 0 and g(ζ ) < 1, ∀ζ ∈ U. Using the Schwarz lemma, we get g(ζ ) ≤ ζ , ∀ζ ∈ U and g (0) ≤ 1, hence R f (z) − f (z0 ) ≤ z − z0 , r
R ∀z ∈ U(z0 ; r) and f (z0 ) ≤ . r
If there exists z∗ ∈ U(z0 ; r) such that f (z∗ ) − f (z0 ) = Rr z∗ − z0 , or such that f (z0 ) = Rr , then g(ζ∗ ) = ζ∗  or g (0) = 1, where z∗ = z0 + rζ∗ . Applying the Schwarz lemma, we deduce that g has the form g(ζ ) = cζ , where c ∈ ℂ and c = 1, i. e., f (z) = f (z0 ) + c Rr (z − z0 ), with c ∈ ℂ, c = 1.
122  4 Sequences and series of holomorphic functions Remark 4.6.3. The generalized Schwarz lemma has the following geometric interpretation: Let f ∈ H(U(z0 ; r)) a function that satisfies f (U(z0 ; r)) ⊂ U(f (z0 ); R). Then f (U(z0 ; r1 )) ⊂ U(f (z0 ); R1 ) for all r1 ∈ (0, r), with RR1 = rr1 . There exists a number
r1 ∈ (0, r) such that RR1 = rr1 and f (U(z0 ; r1 )) = U(f (z0 ); R1 ), if and only if the function f has the form f (z) = f (z0 ) + c Rr (z − z0 ), c = 1.
4.7 Laurent series Definition 4.7.1. 1. If z0 ∈ ℂ is a given point, then a Laurent series about z0 will be the sum of the form ∞
∑ an (z − z0 )n = ⋅ ⋅ ⋅ +
n=−∞
2. 3.
a a−n + ⋅ ⋅ ⋅ + −1 + a0 + ⋅ ⋅ ⋅ + an (z − z0 )n + ⋅ ⋅ ⋅ , n (z − z0 ) z − z0
where an ∈ ℂ, n ∈ ℤ, are called the Laurent series coefficients. If an = 0, ∀n < 0, then the Laurent series will be a power series. The sum ∞
P(z) = ∑ a−n (z − z0 )−n n=1
is called the main part of the Laurent series. 4. The sum ∞
T(z) = ∑ an (z − z0 )n n=0
5.
is called the Taylor part of the Laurent series. The Laurent series is said to be pointwise convergent (or uniformly convergent) on the set E ⊂ ℂ \ {z0 }, if the main and the Taylor parts are pointwise convergent (or respectively, uniformly convergent) on the set E.
Theorem 4.7.1 (Theorem of the annulus of convergence). Let us consider the Laurent series ∞
S(z) = ∑ an (z − z0 )n = P(z) + T(z), n=−∞
and let r = lim sup √n a−n , n→+∞
1 = lim sup √n an , R n→+∞
where for the values of R we used the conventions from the Cauchy–Hadamard theorem (Theorem 4.3.2).
4.7 Laurent series  123
1. 2. 3.
If r < R, then: on the circular ring U(z0 ; r, R) the Laurent series S is absolute and compact convergent; the Laurent series S diverges in all the points of the set ℂ \ U(z0 ; r, R); n the sum S(z) = ∑∞ n=−∞ an (z − z0 ) is holomorphic on the circular ring U(z0 ; r, R).
Proof. The Taylor part of the Laurent series S has a radius of convergence R, given by the Cauchy–Hadamard theorem. Also, the Taylor part T(z) compactly converges on the disc U(z0 ; R), and diverges in all the points of ℂ \ U(z0 ; R). Moreover, the sum T(z) is a holomorphic function in the disc U(z0 ; R). 1 In the main part P(z) of the Laurent series S(z), let use the substitution w = z−z . 0 The power series obtained in this way, i. e., ∞
∑ a−n wn ,
n=1
has the disc of convergence U(0; 1r ), where it is absolute and compact convergent, and the sum will be a holomorphic with respect to the variable w in U(0; 1r ). The inequality w < 1r is equivalent to the assumption z − z0  > r. It follows that the main part is absolute and compact convergent on the open set ℂ \ U(z0 ; r), and thus, the sum of the main part is holomorphic on ℂ \ U(z0 ; r). Also, the main part diverges in all the points that satisfy the inequality z − z0  < r. We conclude that the Laurent series S is absolute and compact convergent on the circular ring U(z0 ; r, toR), and its sum is holomorphic in this open set. In all the points of the set ℂ \ U(z0 ; r, R), at least one of the parts P(z) or T(z) are divergent, hence the Laurent series S diverges in these points. Theorem 4.7.2 (The identity of the Laurent series coefficients). If the next two Laurent ∞ n n series ∑∞ n=−∞ an (z − z0 ) and ∑n=−∞ bn (z − z0 ) have the same sums on the circular ring U(z0 ; r, R), with 0 ≤ r < R, then an = bn , ∀n ∈ ℤ. n Proof. Denoting cn = an − bn , then the sum of the Laurent series ∑∞ n=−∞ cn (z − z0 ) will −n−1 be identically zero on the circular ring U(z0 ; r, R). Multiplying this by (z − z0 ) , we deduce that the sum of this new Laurent series will be also identically zero, i. e., ∞
∑ ck (z − z0 )k−n−1 = 0,
k=−∞
∀z ∈ U(z0 ; r, R).
Letting γ = 𝜕U(z0 ; r+R ), since the above series uniformly converges on the compact 2 {γ}, it can be integrated term by term, and thus, ∞
∑ ck ∫(z − z0 )k−n−1 dz = 2πicn = 0,
k=−∞
γ
124  4 Sequences and series of holomorphic functions
Figure 4.1: Proof of Theorem 4.7.3.
i. e., cn = an − bn = 0. Since n ∈ ℤ was chosen arbitrary, we conclude that an = bn ,
∀n ∈ ℤ.
Theorem 4.7.3 (Laurent series expansion theorem). Let f be a holomorphic function on the circular ring D = U(z0 ; r, R) (0 ≤ r < R). Then the Laurent series ∞
∑ an (z − z0 )n ,
n=−∞
where an =
f (ζ ) 1 dζ ∫ 2πi (ζ − z0 )n+1 γ
(with γ = 𝜕U(z0 ; ρ), r < ρ < R) has a ring of convergence that contains the domain D, and the sum of the Laurent series in the domain D is equal to f , i. e., ∞
f (z) = ∑ an (z − z0 )n , n=−∞
∀z ∈ D.
Proof. We remark that according to the Cauchy theorem, the integrals that appeared in the definitions of the coefficients an are independent of ρ, hence the only assumption for ρ is r < ρ < R. Let z ∈ D be a given arbitrary point, let choose the real numbers r1 , r2 ∈ ℝ such that r < r2 < z − z0  < r1 < R, and let denote γ1 = 𝜕U(z0 ; r1 ), γ2 = 𝜕U(z0 ; r2 ). Let us consider the linear path λ that connects the paths γ2 with γ1 , such that z0 ∈ {γ}. Then the path γ̃ = λ ∪ γ1 ∪ λ− satisfies the condition γ̃ ∼ γ2 (Figure 4.1). D
4.7 Laurent series  125
Considering the function f (ζ )−f (z)
if ζ ∈ D \ {z},
, { ζ −z g(ζ ) = { { f (z),
if ζ = z,
then g ∈ H(D), and thus, ∫ g = ∫ g ⇒ ∫ g + ∫ g + ∫ g = ∫ g, γ̃
γ2
λ
γ1
λ−
γ2
i. e., ∫ g = ∫ g.
γ1
(4.5)
γ2
Further, ∫g = ∫ γ1
γ1
f (ζ ) − f (z) f (ζ ) dζ dζ = ∫ dζ − f (z) ∫ ζ −z ζ −z ζ −z γ1 γ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 1
(4.6)
f (ζ ) − f (z) f (ζ ) dζ . dζ = ∫ dζ − f (z) ∫ ζ −z ζ −z ζ −z γ2 γ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 2
(4.7)
2πi
and ∫g = ∫
γ2
get
γ2
0
Computing the value of f (z) from (4.6), and using the relations (4.5) and (4.7), we
f (z) =
f (ζ ) f (ζ ) f (ζ ) 1 1 1 1 dζ − dζ − dζ . ∫ ∫g = ∫ ∫ 2πi ζ − z 2πi 2πi ζ − z 2πi ζ − z γ1
γ1
γ1
(4.8)
γ2
(Note that the formula (4.8) is called the Cauchy formula for circular rings.) In the first integral of the righthand side of (4.8), we will use the same computation as to the proof of the Taylor series expansion theorem: 1 1 1 , = ζ − z ζ − z0 1 − z−z0 ζ −z 0
z − z 0 where ζ ∈ γ1 , and thus < 1, ζ − z0
hence n
f (ζ ) ∞ z − z0 f (ζ ) = ) . ∑( ζ − z ζ − z0 n=0 ζ − z0
126  4 Sequences and series of holomorphic functions The series of functions from the righthand side is uniformly convergent on the compact {γ1 }, consequently ∞ (z − z0 )n f (ζ ) 1 1 ∞ dζ = an (z − z0 )n , dζ = ∑ ∑ ∫ f (ζ ) ∫ 2πi ζ − z 2πi n=0 (ζ − z0 )n+1 n=0 γ1
γ1
where an =
f (ζ ) 1 dζ . ∫ 2πi (ζ − z0 )n+1 γ1
Now we will calculate the last integral that appeared in the formula (4.8). Since ζ −z ζ ∈ {γ2 }, it follows that  z−z0  < 1, and then we deduce that 0
n
−
∞ ζ − z0 1 1 1 1 1 = ( ) , = = ∑ ζ −z z−ζ z − z0 1 − ζ −z0 z − z0 n=0 z − z0 z−z 0
where the power series from the last term is uniformly convergent (with the variable ζ ) on the compact {γ2 }. Hence, −
∞ f (ζ ) 1 1 1 = ∑ ∫ ∫ f (ζ )(ζ − z0 )n dζ n+1 2πi ζ − z n=0 (z − z0 ) 2πi γ2
γ2
∞
= ∑ a−m (z − z0 )−m , m=1
where a−m =
f (ζ ) 1 dζ . ∫ 2πi (ζ − z0 )−m+1 γ2
Since the integrals that appear in the definitions of an and a−m are independent on the values of r1 and r2 , we have that an =
f (ζ ) 1 dζ , ∫ 2πi (ζ − z0 )n+1 γ
∀n ∈ ℤ,
where γ = 𝜕U(z0 ; ρ), with r < ρ < R, and ρ is the above choose arbitrary number. Finally, from the relation (4.8) we conclude that ∞
f (z) = ∑ an (z − z0 )n , n=−∞
where the coefficients an are given by the formulas (4.9).
(4.9)
4.8 Isolated singular points  127
4.8 Isolated singular points Definition 4.8.1. 1. Let f ∈ H(G), where G ⊂ ℂ is an open set. The number z0 ∈ ℂ is said to be an ̇ 0 ; r) ⊂ isolated singular point for f , if z0 ∉ G, but there exists r > 0 such that U(z ̇ 0 ; r) = U(z0 ; r) \ {z0 }. G, where U(z 2. An isolated singular point z0 for f ∈ H(G) is said to be a removable isolated singũ where G ̃ = G ∪ {z } such that ̃f  = f , lar point, if there exists a function ̃f ∈ H(G), 0 G (it means that f has a holomorphic extension (continuation) in z0 ). 3. The points of G together with all the removable isolated singular points of f are called regular points for f . Theorem 4.8.1 (The first removability criterion). An isolated singular point z0 ∈ ℂ for the function f ∈ H(G), where G ⊂ ℂ is an open set, is removable if and only if there exists limz→z0 f (z) = l ∈ ℂ. ̃ where G ̃ = G ∪ {z } and Proof. If z0 is removable, then there exists a function ̃f ∈ H(G), 0 ̃ ̃ ̃f  = f . It follows that ̃f is continuous on z , hence lim G 0 z→z0 f (z) = f (z0 ) = l ∈ ℂ, i. e., limz→z0 f (z) = l ∈ ℂ. Suppose that there exists the limit from the assumption of the theorem. Then, if we define ̃f (z) = { f (z), l = limz→z0 f (z),
if z ∈ G, if z = z0 ,
̃ \ {z }) and ̃f ∈ C(G), ̃ where G ̃ = G ∪ {z }. According to a wellit follows that ̃f ∈ H(G 0 0 ̃ ̃ ̃ known result, we obtain that f ∈ H(G) and f G = f , hence z0 is removable. Examples 4.8.1. 1. Let f (z) = sinz z , z ∈ ℂ∗ = ℂ \ {0}. Since sin z = z −
sin z 1 1 1 3 1 5 z + z − ⋅⋅⋅ ⇒ = 1 − z2 + z4 − ⋅ ⋅ ⋅ 3! 5! z 3! 5!
and
sin z = 1, z→0 z lim
2. 3.
it follows that z0 = 0 is a removable isolated singular point. Let f (z) = z1 , z ∈ ℂ∗ . Then limz→0 f (z) = ∞ ∉ ℂ, hence z0 = 0 is not a removable isolated singular point. Let f (z) = sin z1 , z ∈ ℂ∗ . Like we know, ∄ limz→0 f (z), because ∄ lim x→0 sin x1 , hence z0 = 0 is not a removable isolated singular point.
x∈ℝ
Theorem 4.8.2 (The second removability criterion). Let f ∈ H(G), where G ⊂ ℂ is an open set. An isolated singular point z0 ∈ ℂ for the function f is removable, if and only if
128  4 Sequences and series of holomorphic functions the main part of the Laurentseries expansion ∞
f (z) = ∑ an (z − z0 )n ,
̇ 0 ; R), z ∈ U(z
n=−∞
(4.10)
vanishes, i. e., an = 0 for all n < 0. Proof. If z0 is removable and ̃f is the holomorphic extension (continuation) of f to the ̃ = G ∪ {z }, then ̃f can be expanded in Taylorseries in a U(z ; R) ⊂ G ̃ disc. This set G 0
0
Taylor expansion need to be identical to the Laurentseries expansion given by (4.10) (according to the uniqueness of Laurentseries coefficients), since ̃f = f in the circular ̇ 0 ; R). ring U(z Conversely, if the Laurent expansion (4.10) has only Taylor part, then its sum is an ̇ 0 ; R). holomorphic function in a disc U(z0 ; R), and it is equal to f on the whole set U(z So, the function f has an holomorphic extension on z0 , thus z0 is a removable isolated singular point. Theorem 4.8.3 (The third, or Cauchy–Riemann, removability criterion). Let f ∈ H(G), where G ⊂ ℂ is an open set. An isolated singular point z0 ∈ ℂ for the function f is removable, if and only if there exists M > 0 and R > 0 such that f (z) < M, ∀z ∈ ̇ 0 ; R) ⊂ G. U(z ̇ 0 ; R) Proof. Let 0 < r < R. Then the function f may be expanded in the circular ring U(z in a Laurent expansion of the form (4.10), and according to a wellknown theorem, these coefficients are given by an =
f (ζ ) 1 dζ , ∫ 2πi (ζ − z0 )n+1 γ
∀n ∈ ℤ.
Using the same proof as for Cauchy inequalities, for all integers n ∈ ℤ we have an  ≤
M . rn
If n < 0, then r −n → 0, if r → 0, hence an  ≤
M → 0, rn
if r → 0.
Since we can choose the arbitrary number r as small is possible, it follows that an  = 0,
∀n < 0,
hence the main part of (4.10) is identically zero. ̃ Conversely, if z0 is a removable isolated singular point for f , there exists ̃f ∈ H(G) ̃ ̃ ̃ such that f G = f and G = G ∪ {z0 }. Since ∃R > 0 such that U(z0 ; R) ⊂ G, the function ̃f ̇ 0 ; R). is bounded on the compact set U(z0 ; R), hence f is bounded on U(z
4.8 Isolated singular points  129
Corollary 4.8.1. If z0 ∈ ℂ is not a removable isolated singular point for the function f ∈ H(G), where G ⊂ ℂ is an open set, then there exists a sequence (zn )n∈ℕ∗ ⊂ G, zn → z0 , such that limn→+∞ f (zn ) = ∞. Definition 4.8.2. Let z0 ∈ ℂ be a nonremovable isolated singular point for f ∈ H(G), where G ⊂ ℂ is an open set. 1. If limz→z0 f (z) = ∞, then z0 is said to be a pole for f ; 2. if ∄ limz→z0 f (z), then z0 is said to be an essential isolated singular point for f . Theorem 4.8.4 (Characterization theorem for the poles). Let f ∈ H(G) where G ⊂ ℂ is an open set, and suppose that z0 ∈ ℂ is an isolated singular point for f . The next statements are equivalent: (i) z0 is a pole for f ; 1 (ii) z0 is a zero for f1 , in the sense that limz→z0 f (z) = 0; ̇ 0 ; r) ⊂ G and (iii) there exists a unique n ∈ ℕ∗ , and a number r > 0, such that U(z f (z) =
a−n a + ⋅ ⋅ ⋅ + −1 + a0 + a1 (z − z0 ) + ⋅ ⋅ ⋅ n (z − z0 ) z − z0 ̇ 0 ; r), + an (z − z0 )n + ⋅ ⋅ ⋅ , ∀z ∈ U(z
where a−n ≠ 0; ̃ where G ̃ = G ∪ {z }, such (iv) there exists a unique n ∈ ℕ∗ , and a function g ∈ H(G), 0 that g(z0 ) ≠ 0 and f (z) =
g(z) , (z − z0 )n
∀z ∈ G.
Proof. (i) ⇒ (ii) Since z0 is a pole, limz→z0 f (z) = ∞, then there exists a number R > 0 ̇ 0 ; R)) and ̇ 0 ; R) ⊂ G, and f (z) > 1, if z ∈ U(z ̇ 0 ; R). Hence, h = 1 ∈ H(U(z such that U(z f ̇ h(z) < 1, if z ∈ U(z0 ; R). Using Theorem 4.8.3, the number z0 is a removable isolated
singular point for the function h = f1 , and limz→z0 h(z) = 0 = h(z0 ). (ii) ⇒ (iii) From the fact that limz→z0 f (z) = ∞, there exists a number R > 0 such ̇ 0 ; R). Hence, h = 1 ∈ H(U(z ̇ 0 ; R)). Since limz→z 1 = 0, the that f (z) > 1, ∀z ∈ U(z f 0 f (z) number z0 is a removable isolated singular point for the function f1 . Also, for the holo
morphic extension h = f1 the number z0 is a zero. From the theorem of characterization of holomorphic function zeros, there exists n ∈ ℕ∗ and h1 ∈ H(U(z0 ; R)), such that h(z) =
1 = h1 (z)(z − z0 )n , f (z)
∀z ∈ U(z0 ; R), where h1 (z0 ) ≠ 0.
Since h1 (z0 ) ≠ 0, the function h1 does not vanish in a small enough neighborhood of z0 . In this neighborhood, the function h1 is holomorphic, hence it can be expanded in 1 a Taylor series: 1 = a−n + a−n+1 (z − z0 ) + ⋅ ⋅ ⋅ + a0 (z − z0 )n + ⋅ ⋅ ⋅ , h1 (z)
and
0 ≠
1 = a−n . h1 (z0 )
130  4 Sequences and series of holomorphic functions Hence, we have f (z) =
a−n a−n+1 1 + ⋅⋅⋅ = + h1 (z)(z − z0 )n (z − z0 )n (z − z0 )n−1 + a0 + a1 (z − z0 ) + a2 (z − z0 )2 + ⋅ ⋅ ⋅ .
From the theorem of characterization of the zeros of holomorphic functions, the number n ∈ ℕ∗ is unique. (iii) ⇒ (iv) Lets define g(z) = {
(z − z0 )n f (z), a−n ,
if z ∈ G, if z = z0 .
From the expansion of f about the point z0 , given in the previous point of the theorem, ̃ where G ̃ = G ∪ {z }. Hence, from a wellknown result, it follows that g ∈ H(G) ∩ C(G), 0 ̃ we obtain that g ∈ H(G) and g(z0 ) = a−n ≠ 0. (iv) ⇒ (i) From limz→z0 g(z) = g(z0 ) ≠ 0, we have that limz→z0 f (z) = ∞, hence z0 is a pole for f . Definition 4.8.3. The natural number n ∈ ℕ∗ that appears in the previous theorem is called the order of pole z0 . Corollary 4.8.2. Let f ∈ H(G), where G ⊂ ℂ is an open set. The number z0 ∈ ℂ is an nth order pole for the function f , if and only if z0 is an nth order zero for the holomorphic extension (continuation) at z0 of the function f1 . Theorem 4.8.5. Let z0 ∈ ℂ be an essential isolated singular point for the function f ∈ H(G), where G ⊂ ℂ is an open set. Then z0 is an essential isolated singular point for the function f1 , or it is an accumulation point for the poles of f1 . Proof. 1. Suppose that z0 is not an accumulation point for the zeros of f . Then ∃r > 0 ̇ 0 ; r) ⊂ G and f (z) ≠ 0, ∀z ∈ U(z ̇ 0 ; r). Hence, g = 1 is holomorphic in such that U(z f ̇ 0 ; r). U(z
Now we will prove that the number z0 is an essential isolated singular point for g = f1 . Supposing that there exists limz→z0 g(z) = l ∈ ℂ∞ , then we have the next two cases: (i) If l ∈ ℂ∗∞ , then limz→z0 f (z) = 1l ∈ ℂ, hence z0 is a removable isolated singular point for f , which contradicts the assumption; (ii) If l = 0, then limz→z0 f (z) = ∞, hence z0 is a pole for f , which also contradicts the assumption. In conclusion, it follows that in such a case the number z0 is an essential isolated singular point for g = f1 . 2. Suppose that z0 is an accumulation point for the zeros of f . These zeros of f are poles for g = f1 , hence z0 is an accumulation point for the poles of f1 .
4.9 Meromorphic functions  131
Example 4.8.1. If f (z) = 1 { kπ
1 sin
1 z
, then z0 = 0 is an accumulation point for the poles of f ,
1 because f ∈ H(ℂ \ : k ∈ ℤ∗ }), where zk = kπ , k ∈ ℤ∗ , are poles for f and z0 = 0 is ∗ an accumulation point for these zk , k ∈ ℤ points. ∗
Theorem 4.8.6 (Casorati–Weierstrass theorem). If z0 ∈ ℂ is an essential isolated singular point for the function f ∈ H(G), where G ⊂ ℂ is an open set, then ∀w0 ∈ ℂ∞ , ∃(zn )n∈ℕ∗ ⊂ G sequence, such that limn→+∞ zn = z0 and limn→+∞ f (zn ) = w0 . Proof. 1. If w0 = ∞, the conclusion follows from Corollary 4.8.1. 2. If w0 ∈ ℂ, according to the Theorem 4.8.5, the number z0 is an essential isolated singular point for the function g, or it is an accumulation point for the poles of g, where g(z) =
1 f (z) − w0
(because z0 is an essential isolated singular point also for the function f − w0 ). (i) If z0 is an accumulation point for the poles of g, then z0 is an accumulation point for the zeros of f − w0 . Hence ∃(zn )n∈ℕ∗ ⊂ G, zn → z0 , such that f (zn ) − w0 = 0, ∀n ∈ ℕ∗ , i. e., limn→+∞ f (zn ) = w0 . (ii) If z0 is an essential isolated singular point for the function g, from Corollary 4.8.1, it follows that ∃(zn )n∈ℕ∗ ⊂ G, zn → z0 , such that limn→+∞ g(zn ) = ∞. Hence, lim (f (zn ) − w0 ) = 0 ⇒ lim f (zn ) = w0 .
n→+∞
n→+∞
Definition 4.8.4. 1. Let z0 = ∞ and f ∈ H(G), where G ⊂ ℂ is an open set. We say that ∞ is an isolated singular point for f , if ∃R > 0 such that {z ∈ ℂ : z > R} ⊂ G. 2. Making the substitution ζ = z1 , we see that for z0 = ∞ corresponds the point ζ0 = 0. Define the function 1 φ(ζ ) = f ( ). ζ We say that the point z0 = ∞ is a removable isolated singular point, a pole or an essential isolated singular point for f , if ζ0 = 0 is a removable isolated singular point, a pole or an essential isolated singular point for φ, respectively.
4.9 Meromorphic functions ̃ ⊂ ℂ be an open set. We say that the function f ∈ H(G ̃ \ E) is Definition 4.9.1. Let G ∞ ̃ ̃ meromorphic on G, if the set E ⊂ G consists only in the removable isolated singular points and the poles of the function f .
132  4 Sequences and series of holomorphic functions Remarks 4.9.1. 1. If we denote by G the set of the points where f is regular, and by B the set of the f ̃ = G ∪ B. function poles, then G ̃ is open, and G is connected if G ̃ is connected. 2. Evidently, the set G is open if G Remarks 4.9.2. 1. The set E is at most countable. ̃ 2. The set B is at most countable, and cannot have any accumulation points in G. ̃ 3. The function f : G → ℂ∞ is continuous as a function with the images in the whole set ℂ∞ , where f (b) = ∞, if b ∈ B. ̃ → ℂ is meromorphic, if and only if about ∀z ∈ G ̃ the function 4. The function f : G ∞ 0 f can be expanded in such a Laurentseries, whose main part contains a finite number of terms. Every point z0 ∈ G is characterized by the fact that the Laurent expansion of f about z0 has a vanishing main part. ̃ the set of all meromorphic functions on G. ̃ Notation. We denote by M(G) ̃ then this function has a continuous extension ̃f : G ̃→ℂ . Remark 4.9.3. If f ∈ M(G), ∞ ̃ ̃ ̃ It is usual to write f instead of f , and we will use this notation. Evidently, H(G) ⊂ M(G). Examples 4.9.1. 1. Every rational function (the quotient of two polynomials) is meromorphic on ℂ∞ . 2. The function cot z is meromorphic on ℂ and the points zk = kπ, k ∈ ℤ, are its poles. The ∞ point is an accumulation point for the poles zk , hence cot z cannot be a meromorphic function on ℂ∞ . 3. The function tan z1 is meromorphic on ℂ∞ \ {0}, because ∞ is a regular point, and 2 , k ∈ ℤ, are its poles, and 0 is the accumulation point for the poles zk . zk = (2k+1)π ̃ is a commutative field with respect to the addition and Theorem 4.9.1. The set M(G) multiplication. ̃ then their product fg has only regular points or poles Proof. Evidently, if f , g ∈ M(G), ̃ in G. ̃ hence 1 ∈ ̃ then the function 1 has only regular points or poles in G, If f ∈ M(G), f f ̃ M(G). Theorem 4.9.2. A function is meromorphic on ℂ∞ , if and only if it is a rational function (the quotient of two polynomials). Proof. 1. Evidently, any rational function f is meromorphic on ℂ∞ . 2. First, we will see that any meromorphic function on ℂ∞ could have only a finite number of poles. Contrary, if there exists a meromorphic function with an infinite
4.9 Meromorphic functions  133
number of poles on ℂ∞ , then those poles will have an accumulation point in ℂ∞ , and this accumulation point cannot be regular point or pole. Let B = {b1 , . . . , bn } denote the set of all poles of the function f . (i) Suppose that ∞ ∉ B. The Laurent expansion of f about the point bk ∈ B has the form f (z) = Pk (z)+Tk (z), where Pk is the main part. For the function g = f − ∑nk=1 Pk , the points bk are removable isolated singular points, because about every bk point the function g can be expanded in a Taylor series (its Laurent expansion about bk consists only from the Taylor part). Since all these bk are regular points for g, if g̃ is the holomorphic extension of g, then g̃ is holomorphic on ℂ∞ . But ∞ is a regular point for g̃ , and using Definition 4.8.4 and Theorem 4.8.3 it follows that g̃ is a bounded function in a neighborhood of ∞. From here, we deduce that g̃ is a bounded function on ℂ, and according to the Liouville theorem the function g̃ will be constant. Hence, f − ∑nk=1 Pk is a constant function, and if we denote this constant by c, then n
f = c + ∑ Pk . k=1
Since Pk are rational fractions ∀k = 1, n, it follows that f is also a rational function. (ii) Suppose now, that ∞ ∈ B. That means ∞ is a pole for f , and if a ∈ ℂ is a regular point for f , then for the function 1 h(z) = f ( + a) z the point z = ∞ is a regular point, and h is meromorphic on ℂ∞ . From the previous part of the proof, the function h will be a rational function, and thus we proved that f (z) = h(
1 ) z−a
is also a rational function. Remarks 4.9.4. P(z) , where P and Q are such polynomials which 1. Let R : ℂ∞ → ℂ∞ , R(z) = Q(z) don’t have any common zeros. Hence, R is a rational function and R ∈ H(ℂ \ {b1 , b2 , . . . , bm }), where {b1 , b2 , . . . , bm } is the set of all zeros of the Q polynomial. If R(z) =
α0 + α1 z + ⋅ ⋅ ⋅ + αn z n , β0 + β1 z + ⋅ ⋅ ⋅ + βm z m
then the function 1 S(z) = R( ) z
α0 ≠ 0, β0 ≠ 0,
(4.11)
134  4 Sequences and series of holomorphic functions will have the form S(z) = z m−n 2.
α0 z n + α1 z n−1 + ⋅ ⋅ ⋅ + αn , β0 z m + β1 z m−1 + ⋅ ⋅ ⋅ + βm
Hence, from Definition 4.8.4 we have R(∞) = S(0). Every rational function has the property that the number of its zeros and the number of its poles on ℂ∞ are the same, every zero and every pole will be counted together with its order. This number is called the order of the rational function. The next table shows the discussion about the number of the zeros and of the poles on ℂ∞ , of a rational function of the form (4.11):
Degrees comparison
The number of zeros in ℂ
The order of ∞ zero
The number of zeros in ℂ∞
The number of poles in ℂ
The order The number of of ∞ pole poles in ℂ∞
m>n m=n m − 2 .
= ∑∞ n=1
1+z =
Exercise 4.10.4. Prove the formulas: cos nφ φ 1. ∑∞ = − ln(2 sin 2 ), 0 < φ < 2π; n=1 n
2. 3.
4.
∑∞ n=1 ∑∞ n=0 ∑∞ n=0
π−φ sin nφ = 2 , 0 < φ < 2π; n cos(2n+1)φ φ = 21 ln(cot 2 ), 0 2n+1 sin(2n+1)φ = π4 , 0 < φ < π. 2n+1
< φ < π;
4.10.2 Taylor and Laurent series Exercise 4.10.5. Expand in a Taylor series, respectively, in the Laurent series, about the point z0 , the following functions: z , z0 = 1; 1. f (z) = z+1 2.
f (z) =
z2 , z0 (z+1)2
= 1;
136  4 Sequences and series of holomorphic functions 3.
f (z) =
4. f (z) = 5.
f (z) =
6.
f (z) =
7.
f (z) =
8. f (z) = 9.
f (z) =
10. f (z) =
11. f (z) =
z−1 , z = 0 and z0 = i; z−2 0 1 , z = 1 and z0 = ∞; z 2 +z+1 0 z+3 , z = 4 and z0 = ∞; z 2 −8z+15 0 z sin 1−z , z0 = 0; cosh2 z, z0 = 0; cos3 z, z0 = 0; sin3 z, z0 = 0; ez sin z, z0 = 0; √ 3 √z, where f (1) = −1+i2 3 , z0 =
12. f (z) = √z − 2i, where f (0) = √2e 5
5
13. f (z) = log z, where f (1 + i) = 14. f (z) =
log(1+z) , 1+z
1 2
i 7π 10
1;
, z0 = 0 and z0 = 2;
ln 2 − i 7π , z0 = −i; 4
where f (0) = −4πi, z0 = 0.
Exercise 4.10.6. Expand in the Laurent series about the point z0 = 0, in the corresponding domains, for the functions: 1.
f (z) =
2.
f (z) =
3.
f (z) =
4. f (z) =
1 , (a) z < 2, (b) 2 < z < 3, (c) z z 2 −5z+6 1 , (a) 0 < z < 1, (b) 1 < z < 2, z(z+1)(z+2)
> 3; (c) z > 2;
2
2z −3z+3 , (a) z < 1, (b) 1 < z < 2, (c) z > 2; z 3 −2z 2 +z−2 1−z log 1+z , where f (0) = 0, (a) z < 1, (b) z > 1.
Exercise 4.10.7. In the corresponding domains, expand in the Laurent series about z0 = ∞ the following functions:
1.
2.
f (z) =
f (z) =
1 , z 5 −z 2
z > 1;
15
z , (z 2 +1)5 (z 4 −16)
z > 2;
3.
f (z) = a0 + a1 z + ⋅ ⋅ ⋅ + an z n , z ∈ ℂ;
5.
z−a , where log 1 = 2kπi, z > max{a, b}; f (z) = log z−b
6.
f (z) = e z+1 , z > 1.
4. f (z) = z 3 sin z1 , z > 0; 2z
Exercise 4.10.8. Expand the following functions in the Laurent series about the point z0 , and then determine the type of the singularity z0 : 1.
f (z) =
2.
f (z) =
3.
f (z) =
1 , z0 = 1 and z0 = ∞; (z 2 −1)2 1 sin 1−z , z0 = 1 and z0 = ∞; 1 , z0 = 0 and z0 = π; z sin z
4.10 Exercises  137
2z
4. f (z) = e z+1 , z0 = ∞; 5.
z+i
f (z) = eiπ z−i , z0 = i;
6.
f (z) = cot z, z0 = 0 and z0 = kπ, k ∈ ℤ;
7.
f (z) =
sin2 z , z0 z5
= 0.
Exercise 4.10.9. Expand in the Laurent series about the point z0 = 0, in the domain 1 < z < 2, the next functions: 1. f (z) = (z−1)21(z+2) ; 2.
f (z) =
z . (z 2 +1)2 (z+2)
Exercise 4.10.10. In the given circular ring D, expand in the Laurent series about the corresponding point z0 in the following functions: 1 1. f (z) = z(z−3) 2 , z0 = 1, D = U(1; 1, 2);
2.
f (z) =
1 , z0 (z 2 −1)(z 2 +4)2 3
= 0, D = U(0; 2, ∞);
1 z
3.
f (z) = z e , z0 = 0, D = U(0; 0, ∞);
5.
f (z) =
6.
f (z) =
1 4. f (z) = z 3 cos z−2 , z0 = 2, D = U(2; 0, ∞); 1 , z0 = 0, D = U(0; 0, 1); z 2 (1−z) sin z 2 1 cos z , z0 = 0, D = U(0; 0, ∞).
Exercise 4.10.11. Write the first three terms of the Laurent series expansion about the point z0 = 0 for the next functions: 1. f (z) = sin1 z ;
2.
f (z) =
1 . ez −1
Exercise 4.10.12. Define on the set ℂ \ [−i, i] the holomorphic branch of the multivalued function f (z) = √1 + z 2 , for which f ( 43 ) = 45 . Expand this branch of the function f in a Laurent series around the point z0 = 0, on the circular ring z > 1. Exercise 4.10.13. Prove that 1
√z 2 − 1
=
−1 1 + ∑ z n=−∞
1 2
⋅
3 2
⋅ . . . ⋅ ( 21 − n − 1) (−n)!
z 2n−1 ,
z > 1,
where we considered the branch of the multivalued function f (z) = √z 2 − 1 that satisfies f (√5) = 2. Exercise 4.10.14. Prove that the expansion in the Laurent series of the function f (z) = 1 n about the point z0 = 0 has the form f (z) = ∑∞ n=0 an z , where 1−z−z 2 an =
n+1
1 1 + √5 [( ) √5 2
n+1
−(
1 − √5 ) 2
].
138  4 Sequences and series of holomorphic functions Exercise 4.10.15. Determine the main part of the Laurent series expansion about the point z0 for the functions: z 1. f (z) = (z+2) 2 , z0 = −2; 2. 3.
f (z) =
f (z) =
ez +1 , z = 0 and z0 ez −1 0 z−1 , z0 = 0. sin2 z
= 2πi;
4.10.3 Isolated singular points Exercise 4.10.16. Prove that z0 is a removable isolated singular point for the corresponding functions: 2 −1 1. f : ℂ \ {1} → ℂ, f (z) = zz−1 , z0 = 1; 2.
3.
sin z , z0 = z π { 2 + kπ : k
f : ℂ \ {0} → ℂ, f (z) =
f : ℂ \ ({kπ : k ∈ ℤ} ∪
4. f : ℂ \
{ π2
0;
+ kπ : k ∈ ℤ} → ℂ, f (z) = 1−cos z , z0 z2
1 cos2 z
−
= 0;
1 , (z− π2 )2
5.
f : ℂ \ {0} → ℂ, f (z) =
6.
f : ℂ \ ({kπ : k ∈ ℤ} ∪ {2kπi : k ∈ ℤ}) → ℂ, f (z) =
7.
z , z0 tan z π z0 = 2 ;
∈ ℤ}) → ℂ, f (z) =
3
f : ℂ \ {z ∈ ℂ : z + 1 = 0} → ℂ, f (z) =
z 2 −1 ,z z 3 +1 0
1 ez −1
−
= 0;
1 , z0 sin z
= 0;
= −1.
Exercise 4.10.17. Prove that the point z0 is a pole for the corresponding functions: 1. f : ℂ \ {0} → ℂ, f (z) = z1 , z0 = 0;
2. 3.
f : ℂ \ {−i, i} → ℂ, f (z) = f : ℂ \ {−1} → ℂ, f (z) =
1 , z0 (z 2 +1)2
2
z +1 , z0 z+1
= −1;
4. f : ℂ \ {2kπ : k ∈ ℤ} → ℂ, f (z) =
5.
6. 7.
f : ℂ \ {−i, i} → ℂ, f (z) =
1 , z0 1−cos z
1 , z0 (z 2 +1)2
= 0;
= −i;
z , z0 = 2π; 1−cos z f (z) = ezz+1 , z0 = πi.
f : ℂ \ {2kπ : k ∈ ℤ} → ℂ, f (z) =
f : ℂ \ {z ∈ ℂ : ez + 1 = 0} → ℂ,
= i;
Exercise 4.10.18. Prove that the point z0 is an essential isolated singular point for the corresponding functions: 1. f : ℂ \ {0} → ℂ, f (z) = sin zπ2 , z0 = 0;
2.
π f : ℂ \ {1} → ℂ, f (z) = (z − 1)2 cos z−1 , z0 = 1;
3.
f : ℂ \ {−1} → ℂ, f (z) = e z+1 , z0 = −1;
5.
f : ℂ \ {−i, i} → ℂ, f (z) = sin z 2π+1 , z0 = i;
1
z , z0 = −1; 4. f : ℂ \ {−1} → ℂ, f (z) = cos z+1
6.
1
f : ℂ \ {0} → ℂ, f (z) = sin e z , z0 = 0.
4.10 Exercises  139
Exercise 4.10.19. Let D ⊂ ℂ be a domain, and let f , g ∈ H(D) be such a function that satisfies f (a) = g(a) = 0, and g (a) ≠ 0. Prove that the point a is a removable isolated singular point of the function h : D \ {z ∈ D : g(z) = 0} → ℂ,
h(z) =
f (z) . g(z)
Exercise 4.10.20. Let f : U(a; r) \ {a} → ℂ∗ a holomorphic function, such that the point a is a pole for f . Prove that the function h : U(a; r) → ℂ,
h(z) = {
1 , f (z)
0,
if z ∈ U(a; r) \ {a}, if z = a
is holomorphic in the disc U(a; r).
4.10.4 The module maximum of the holomorphic functions Exercise 4.10.21. Let f ∈ H(D) be a nonconstant function, where D ⊂ ℂ is a domain. Prove that the functions g = Re f and h = Im f have no locally maximum or minimum points in the domain D. Exercise 4.10.22 (Carathéodory inequality). Let f ∈ H(U(0; r)), such that f (0) = 0 and Re f (z) ≤ A, ∀z ∈ U(0; r). Prove that 2Az , f (z) ≤ r − z
∀z ∈ U(0; r).
Exercise 4.10.23. Let f ∈ H(U(0; 1)), such that f (z) < 1, ∀z ∈ U(0; 1). If there exist the numbers a, b ∈ U(0; 1), a ≠ b, such that f (a) = a and f (b) = b, then prove that f (z) = z, ∀z ∈ U(0; 1). Exercise 4.10.24. Suppose that f ∈ H(U(0; 1)) and there exists α ∈ U(0; 1), such that f (α) = 0 and f (z) ≤ 1, ∀z ∈ U(0; 1). Prove that z − α , f (z) ≤ 1 − αz
∀z ∈ U(0; 1).
Exercise 4.10.25. Did there exist functions f ∈ H(ℂ \ {−1}), such that 1 1 , f( ) = n n+1
∀n ∈ ℕ∗ ?
If there exist, how many functions with this property did there exist? Exercise 4.10.26. Did there exist functions f ∈ H(U( 21 ; 65 )), such that its values on the points zn = n1 , n ∈ ℕ∗ , are given by the next sequences:
140  4 Sequences and series of holomorphic functions 1.
2.
0, 1, 0, 21 , 0, 31 , 0, 41 , . . . , 0, k1 , . . . 1 1 1 1 1 1 , , , , , , . . . k1 , k1 , . . . ? 2 2 3 3 4 4
Exercise 4.10.27. Did there exist functions f ∈ H(U(0; 1)), such that 1. f ( n1 ) = f (− n1 ) = n12 , ∀n ∈ ℕ∗ , 2.
f ( n1 ) = f (− n1 ) =
1 , n3
∀n ∈ ℕ∗ ?
5 Residue theory 5.1 Residue theorem One of the most important consequences of the Laurent series expansion theorem is the method of the calculation of the integrals on a closed rectifiable curve that is given by the residue theorem. Using this result, we can elegantly compute some famous integrals that appeared in physics and engineering. On the other side, the residue theorem has some profound theoretical consequences. In this part of our book, we will present both important implications. Let G ⊂ ℂ be an open set, let f ∈ H(G) and let S the set of all nonremovable isolated ̃ = G ∪ S is also an open set. The function f singular points of the function f . Then G ̃ i. e., ∀z ∈ G, ̃ ∃r > 0 may be expanded in Laurent series about every point z0 of G, 0 ̇ 0 ; r) ⊂ G and such that U(z ∞
f (z) = ∑ an (z − z0 )n , n=−∞
̇ 0 ; r). ∀z ∈ U(z
(5.1)
Definition 5.1.1. The coefficient a−1 that appears in the Laurent series expansion (5.1) is called the residue of f at the point z0 , and it is denoted by Res(f ; z0 ) = a−1 . ̃ is a regular point for f , then Res(f ; z ) = 0. Remark 5.1.1. If z0 ∈ G 0 Theorem 5.1.1 (The residue theorem). Let f ∈ H(G), where G ⊂ ℂ is an open set, and ̃ = G ∪ B and let γ be a let B the set of all isolated singular points of the function f . Let G ̃ that is γ ∼ 0. closed rectifiable curve (path) in G, i. e., homotopic with zero in G, Then the sum ∑z∈G̃ n(γ, z) Res(f ; z) is finite, and
̃ G
∫ f = 2πi ∑ n(γ, z) Res(f ; z). γ
̃ z∈G
̃ be the continuous deformation of γ to Proof. Let S = T = [0, 1] and let φ : S × T → G the constant path γ0 . The set K = φ(S × T) is compact, and thus ε = 21 d(K, ℂ \ G) > 0. It follows that the set D = ⋃{U(z; ε) : z ∈ K} is bounded, hence D− is compact, and from ̃ (Figure 5.1). its definition we have K ⊂ D ⊂ D− ⊂ G We see that the set D− ∩B is finite. Contrary, if D− ∩B is not finite, then D− ∩B needs ̃ and this point cannot be an isolated singular point to have an accumulation point in G, for f . Let D∩B = {b1 , . . . , bn }, and denote by Pk (z) the main part of the Laurent expansion of f about bk . (Pk exists and it is holomorphic on ℂ \ {bk }, because it exists and it is holomorphic in a small enough neighborhood of bk , or for enough small values of z −bk , such that for greater values of z −bk  it also exists and is holomorphic, because these terms appear to be the fractions’ denominators.) https://doi.org/10.1515/9783110657869005
142  5 Residue theory
Figure 5.1: Proof of Theorem 5.1.1.
The function g = f − ∑nk=1 Pk has a holomorphic extension to bk point, hence g can be considered as a holomorphic function on D, and γ ∼ 0. According to the Cauchy D
integral theorem, we have ∫γ g = 0, hence n
∫ f = ∑ ∫ Pk . γ
k=1 γ
Let a(k) −m . m m=1 (z − bk ) ∞
Pk (z) = ∑
Since Pk converges uniformly on each anact of ℂ \ {bk } (because Pk is holomorphic on the set ℂ \ {bk }), it can be integrated term by term on the curve γ. But ∫ γ
dz = 0, (z − bk )m
if m > 1
(since in this case the function has a primitive) and ∫ γ
dz = 2πin(γ, bk ). z − bk
Hence, ∫ Pk = 2πin(γ, bk ) Res(f ; bk ). γ
To complete the proof of the theorem, we need to show that n(γ, z0 ) Res(f ; z0 ) = 0, ̃ \ (D ∩ B). For the case z ∉ B, we have that z is a regular point and then ∀z0 ∈ G 0 0 Res(f ; z0 ) = 0. For the case z0 ∉ D, then ℂ \ {γ} has a such a connected component that contains the ∞ point, hence n(γ, z0 ) = 0. It follows that n
n
∫ f = ∑ ∫ Pk = 2πi ∑ n(γ, bk ) Res(f ; bk ) = 2πi ∑ n(γ, z) Res(f ; z). γ
k=1 γ
k=1
̃ z∈G
5.1 Residue theorem
 143
̃ is a pole of order k for the function f , Theorem 5.1.2 (Residue value in a pole). If z0 ∈ G then Res(f ; z0 ) =
1 (k−1) lim [(z − z0 )k f (z)] . (k − 1)! z→z0
Proof. It is well known that, in a neighborhood of z0 we have f (z) =
a−k a + ⋅ ⋅ ⋅ + −1 + a0 + a1 (z − z0 ) + ⋅ ⋅ ⋅ , k z − z0 (z − z0 )
̇ 0 ; r), ∀z ∈ U(z
hence (z − z0 )k f (z) = a−k + a−k+1 (z − z0 ) + ⋅ ⋅ ⋅ + a−1 (z − z0 )k−1 + ⋅ ⋅ ⋅ ,
̇ 0 ; r), ∀z ∈ U(z
and [(z − z0 )k f (z)]
(k−1)
= (k − 1)!a−1 + (z − z0 )(k!a0 + ⋅ ⋅ ⋅),
̇ 0 ; r). ∀z ∈ U(z
From here, it follows that a−1 =
1 (k−1) lim [(z − z0 )k f (z)] . (k − 1)! z→z0
Corollary 5.1.1. If z0 ∈ G is a simple pole for f , and f has the form f (z) =
g(z) , h(z)
where g
and h are holomorphic functions on G, with g(z0 ) ≠ 0, h(z0 ) = 0, and h (z0 ) ≠ 0, then Res(f ; z0 ) =
g(z0 ) = a−1 . h (z0 )
Proof. In this special case, a g(z) = −1 + a0 + a1 (z − z0 ) + ⋅ ⋅ ⋅ , h(z) z − z0
̇ 0 ; r), ∀z ∈ U(z
hence a−1 = lim (z − z0 ) z→z0
g(z) = lim h(z) z→z0
g(z)
h(z)−h(z0 ) z−z0
=
g(z0 ) . h (z0 )
Definition 5.1.2 (The residue in the z0 = ∞). Let z0 = ∞ be an isolated singular point for the holomorphic function f . Then there exists r > 0 such that f is holomorphic in the circular ring U(0; r, +∞), so f may be expanded in Laurent series on this circular ring, i. e., f (z) = ⋅ ⋅ ⋅ +
a−n a + ⋅ ⋅ ⋅ + −1 + a0 + a1 z + ⋅ ⋅ ⋅ + an z n + ⋅ ⋅ ⋅ , zn z
∀z ∈ U(0; r, +∞).
144  5 Residue theory Let γ − = 𝜕U(0; ρ), where r < ρ < +∞. Then γ runs with direct orientation about the ∞ point (because it runs with inverse orientation around the point 0), hence ∞ 1 1 ∞ 1 ∑ an ∫ ζ n dζ = −a−1 , ∫f = ∫( ∑ an ζ n )dζ = 2πi 2πi 2πi n=−∞ n=−∞ γ
γ
γ
because of the fact that the Laurent series converges uniformly on the compact {γ}, and since the orientation of γ is inverse. In order that the Theorem 5.1.1 will be true also in this case, we need to denote Res(f ; ∞) = −a−1 . Theorem 5.1.3. Let f be a holomorphic function on the domain D = ℂ \ {z1 , . . . , zn }, and let zn+1 = ∞. Then n+1
∑ Res(f ; zk ) = 0.
k=1
(5.2)
Proof. Let γ = 𝜕U(0; r), where r > max{z1 , . . . , zn }. Since the γ closed curve has the property n(γ, zj ) = 1, j = 1, n, and thus, according to Theorem 5.1.1 we have n 1 ∫ f = ∑ Res(f ; zk ). 2πi k=1 γ
From the definition of the residue in the ∞ point, we have 1 ∫ f = − Res(f ; ∞), 2πi γ
and using the above relations we obtain the conclusion (5.2).
5.2 Applications of the residue theorem to the calculation of the integrals Exercise 5.2.1. Let R(u, v) a rational function, where the numerator and the denominator are two real variables polynomials with real coefficients, such that the denominator does not vanish on the unit circle u2 + v2 = 1. Then the integral 2π
I = ∫ R(sin x, cos x)dx 0
can be calculated by using the next formula: I = 2π ∑ Res(g; z), z 0.
γr
Proof. The proof is similar to the proof of point 3 of Lemma 5.2.1. iπt π , where n ≥ 2, then γr (t) = re np , t ∈ [0, 1], hence If θ1 = 0 and θ2 = np 1 π π t imr p (cos πn t+i sin πn t) π i np t i np imz p e ri )e dt = dz f (z)e f (re ∫ ∫ np γ 0
r
1
π π p π i π t iπt ≤ ∫f (re np )emr (− sin n t+i cos n t) ri e np dt np 0
π np
= ∫ f (reiu )e−mr 0
π n
= M(r)r ∫ e−mr
p
0
Since
2 π
≤
sin φ φ π n
p
sin pu
sin φ
π np
rdu ≤ M(r)r ∫ e−mr
p
sin pu
du
0
π n
p M(r)r 1 dφ = ∫ e−mr sin φ dφ. p p
0
≤ 1, φ ∈ [0, πn ] ⊆ [0, π2 ] and m > 0, we get
∫ e−mr
p
sin φ
π n
dφ ≤ ∫ e−mr
0
p 2φ π
0 π − 2mrπ p φ π e , ≤ p π 2mr n 2mr p
dφ =
0
hence πM(r) imz p dz ≤ . ∫ f (z)e 2mpr p−1 γ r
Using the assumption lim z→∞ z 1−p f (z) = limz→∞ z 1−p f (z) = 0, we deduce that limr→+∞
M(r) r p−1
z∈{γr }
= 0, and from the above inequality we conclude that p
lim ∫ f (z)eimz dz = 0.
r→+∞
γr
5.2 Applications of the residue theorem to the calculation of the integrals  149
Figure 5.3: Proof of Problem 5.2.2. +∞
P(x) Exercise 5.2.2. Compute the integrals of the type ∫−∞ Q(x) dx, where P and Q are two real variables polynomials with real coefficients, such that the denominator Q has no real zeros (roots), and deg Q ≥ deg P + 2. Then the above integral may be computed using the formula +∞
∫ −∞
P P(x) dx = 2πi ∑ Res( ; zk ). Q(x) Q Im z >0 k
Solution. Let z1 , . . . , zm be the roots of the Q polynomial (Figure 5.3). We will choose the number r > 0 such that it satisfies r > max{zk  : k = 1, m}. Let γr (t) = reiπt , t ∈ [0, 1], and let us denote by γ(t) = −r(1 − t) + tr, t ∈ [0, 1], the linear path that connects the point −r with r. Thus, r
∫
γ1 =γ∪γr
f (z)dz = ∫ −r
P P(x) dx + ∫ f = 2πi ∑ Res( ; zk ), Q(x) Q Im z >0 γr
k
since the index the γ1 = γ ∪ γr curve with respect to every zk point is equal to 1 (this curve is a boundary curve), where the paths are presented in Figure 5.3. Hence, lim zf (z) = lim
z→∞
z→∞ +∞
⇒ ∫ −∞
Lemma zP(z) = 0 ⇒ lim ∫ f = 0 r→∞ Q(z) γr
P P(x) dx = 2πi ∑ Res( ; zk ). Q(x) Q Im z >0 k
Remark 5.2.4. If P and Q are two real variables polynomials with real coefficients, such that the denominator Q has no real zeros (roots), and deg Q ≥ deg P + 2, then by using a similar proof we obtain the formula: +∞
∫ −∞
P(x) P dx = −2πi ∑ Res( ; zk ), Q(x) Q Im z 0 k
Solution. There exist the numbers η > ξ > 0 such that the function R = QP is monotone on the [ξ , η] interval. Using now the second meanvalue theorem for the integral calculation, there exist the points ξ1 , ξ2 ∈ [ξ , η] such that η
ix
η
η
∫ R(x)e dx = ∫ R(x) cos xdx + i ∫ R(x) sin xdx ξ
ξ
ξ1
ξ
ξ2
η
η
= R(ξ ) ∫ cos xdx + R(η) ∫ cos xdx + i[R(ξ ) ∫ sin xdx + R(η) ∫ sin xdx]. ξ
ξ1
ξ
ξ2
Since the modules of all the integrals are bounded, and since limξ →+∞ R(ξ ) = +∞
limη→+∞ R(η) = 0, it follows that the integral ∫0 0 ∫−∞ R(x)eix dx
R(x)eix dx is convergent. Similarly, +∞
also converges, hence ∫−∞ R(x)eix dx is we may prove that the integral convergent. Let z1 , . . . , zm the zeros (roots) of the Q polynomial. Let us choose a number r > 0, such that r > max{zk  : k = 1, m}. Let γr (t) = reiπt , t ∈ [0, 1], and denote by γ(t) = −r(1 − t) + tr, t ∈ [0, 1] the linear path that connects −r with r (Figure 5.3). Thus, r
∫
γ1 =γ∪γr
R(z)eiz dz = ∫ −r
P(x) ix e dx + ∫ R(z)eiz dz Q(x) γr
= 2πi ∑ Res( Im zk >0
P(z) iz e ; zk ), Q(z)
since the index the γ1 = γ ∪ γr curve with respect to every zk point is equal to 1 (this curve is a boundary curve).
5.2 Applications of the residue theorem to the calculation of the integrals 
151
Figure 5.5: Proof of Problem 5.2.3.
Using point 3 of the lemma that extends the Jordan lemma, for the special case p = 1 and m = 1, we deduce that limz→∞ R(z) = 0, so it follows that limr→+∞ ∫γ R(z) × r
eiz dz = 0, hence
+∞
∫ −∞
P(x) ix P(z) iz e dx = 2πi ∑ Res( e ; zk ). Q(x) Q(z) Im z >0 k
Remark 5.2.5. If P and Q are two real variables polynomials with real coefficients, such that the denominator Q has no real zeros (roots), and deg Q ≥ deg P + 1, then +∞
∫ −∞
P(x) ix P(z) iz e dx = −2πi ∑ Res( e ; zk ), Q(x) Q(z) Im z 0 zQ(z) k
Proof. Let z1 , . . . , zm be the zeros (roots) of the Q polynomial. Let r > 0 such a number, that r > max{zk  : k = 1, m}, and let ε > 0 a number with ε < min{zk  : k = 1, m}. Define the curve γ3 = γ[−r,−ε] ∪ γε− ∪ γ[ε,r] ∪ γr , where γ[−r,−ε] (t) = −(1 − t)r − tε, γε (t) = εetπi , γ[ε,r] (t) = (1 − t)ε + tr and γr (t) = retπi , t ∈ [0, 1] (Figure 5.5). P(z) iz e , we have If we use the notation g(z) = zQ(z) r
−ε
P(x) ix P(z) iz P(z) iz P(x) ix e dx − ∫ e dz + ∫ e dx + ∫ e dz ∫g = ∫ xQ(x) zQ(z) xQ(x) zQ(z)
γ3
γε
−r
= 2πi ∑ Res(g; zk ). Im zk >0
ε
γr
152  5 Residue theory
Figure 5.6: Proof of Remark 5.2.6 and Problem 5.2.4.
Using point 3 of the lemma that extends the Jordan lemma, we get lim ∫
r→+∞
γr
P(z) iz e dz = 0, zQ(z)
and from point 4 of the same lemma we have lim ∫
ε→0
γε
P(z) iz P(z) iz πiP(0) e dz = πi Res( e ; 0) = . zQ(z) zQ(z) Q(0)
If we make in the first integral the variable changing y = −x, we conclude that +∞
∫[ 0
P(z) iz 1 P(0) P(x) ix P(−x) −ix dx e − e ] = 2πi[ + ∑ Res( e ; zk )]. Q(x) Q(−x) x 2 Q(0) Im z >0 zQ(z) k
Remark 5.2.7. For the special case P(z) = Q(z) = 1, the result of Remark 5.2.6 becomes +∞
∫ 0
sin x π dx = . x 2
Exercise 5.2.4. Prove the following identity: +∞
+∞
1 π ∫ cos x dx = ∫ sin x2 dx = √ . 2 2 2
0
0
(The above integrals are called Fresneltype integrals.) π
Solution. Let us define the curve γ4 = γ[0,r] ∪ γr ∪ λr , where γ[0,r] (t) = rt, γr (t) = reti 4 π
and λr (t) = (1 − t)rei 4 , t ∈ [0, 1] (Figure 5.6). 2 Denoting g(z) = eiz ∈ H(ℂ), then r
2
2
2
∫ eix dx + ∫ eiz dz + ∫ eiz dz = 0. 0
γr
λr
5.2 Applications of the residue theorem to the calculation of the integrals 
153
Now we will use the generalization of the Jordan lemma for the special case n = 2, p = 2, m = 1 and f (z) = 1, and we get 2
lim ∫ eiz dz = 0.
r→+∞
γr
Since 1
2
2
π
2 2
π
r
2
∫ eiz dz = ∫ ei (1−t) r (−r)ei 4 dt = −ei 4 ∫ e−u du, 0
λr
0
it follows that +∞
ix 2
∫ e dx = e
i π4
0
+∞
π
2
∫ e−u du = ei 4 0
√π . 2
Using the fact that the above inequality holds if and only if both real, and imaginary parts are respectively identical, we obtain the required Fresneltype formulas. +∞
P(x) dx, where P and Q are two Exercise 5.2.5. Compute the integrals of the type ∫0 Q(x) real variables polynomials with real coefficients, such that the denominator Q has no nonnegative zeros (roots), and deg Q ≥ deg P + 2. Then the above integral may be computed using the formula +∞
∫ 0
P(x) P(z) dx = − ∑ Res( log z; zk ), Q(x) Q(z) z ∈ℂ∗
where log 1 = 0.
k
P(z) Solution. Define the function f (z) = Q(z) log z, where log z = ln z + i arg z, 0 < arg z < 2π. Let z1 , . . . , zm be the zeros (roots) of the Q polynomial. Then there exist the numbers ε > 0 and r > 0, such that
0 < ε < min{zk  : k = 1, m} ≤ max{zk  : k = 1, m} < r. Let γ5 = λ∪γr ∪λ− ∪γε− , where this curve is presented in Figure 5.7, i. e., λ(t) = (1−t)ε+tr, γr (t) = re2πit , γε (t) = εe2πit , t ∈ [0, 1]. Thus, r
ε
P(x) P(x) ln xdx + ∫ f + ∫ [ln x + 2πi]dx − ∫ f ∫f = ∫ Q(x) Q(x)
γ5
γr
ε
γε
r
m
r
k=1
ε
= 2πi ∑ Res(f ; zk ) = −2πi ∫
P(x) dx + ∫ f − ∫ f . Q(x) γr
γε
154  5 Residue theory
Figure 5.7: Proof of Problem 5.2.5 and Problem 5.2.7.
But z 2 P(z) log z zP(z) log z = → 0, zf (z) = Q(z) z Q(z)
if z → ∞,
and P(z) zf (z) = z log z → 0, Q(z)
if z → 0,
because the ratio QP is continuous at 0 and limz→0 z log z = 0. Taking ε → 0 and r → ∞, from points 1 and 2 of the lemma that extends the Jordan lemma, we get +∞
∫ 0
m P(x) dx = − ∑ Res(f ; zk ). Q(x) k=1 +∞
P(x) Exercise 5.2.6. Compute the integrals of the type ∫0 Q(x) ln xdx, where P and Q are two real variables polynomials with real coefficients, such that the denominator Q has +∞ P(x) no nonnegative zeros (roots), and deg Q ≥ deg P + 2. Then the integral ∫0 Q(x) ln xdx is convergent, and +∞
∫ 0
P(x) 1 P(z) ln xdx = − Re[ ∑ Res( (log z)2 ; zk )], Q(x) 2 Q(z) ∗ z ∈ℂ
where log 1 = 0.
k
Solution. Since deg Q ≥ deg P + 2, it follows that the given integral is convergent. Let us define the function g(z) = R(z)(log z)2 , where log z = ln z + i arg z, 0 < arg z < 2π P(z) and R(z) = Q(z) . Let z1 , . . . , zm be the zeros (roots) of the Q polynomial. Then there exist the numbers ε > 0 and r > 0, such that 0 < ε < min{zk  : k = 1, m} ≤ max{zk  : k = 1, m} < r.
5.2 Applications of the residue theorem to the calculation of the integrals 
155
Let γ5 = λ∪γr ∪λ− ∪γε− , where this curve is presented in Figure 5.7, i. e., λ(t) = (1−t)ε+tr, γr (t) = re2πit , γε (t) = εe2πit , t ∈ [0, 1]. Thus, r
ε
∫ R(x) ln2 xdx + ∫ g + ∫ R(x)[ln x + 2πi]2 dx − ∫ g = 2πi ∑ Res(g; zk ). γr
ε
zk ∈ℂ∗
γε
r
But z 2 P(z) (log z)2 zP(z) (log z)2 = zg(z) = → 0, Q(z) Q(z) z
if z → ∞,
and P(z) 2 z(log z) → 0, zg(z) = Q(z)
if z → 0,
because the ratio QP is continuous at 0 and limz→0 z(log z)2 = 0. Letting now ε → 0 and r → ∞, from points 1 and 2 of the lemma that extends the Jordan lemma, we get +∞
2
+∞
∫ R(x) ln xdx − ∫ R(x)[ln x + 2πi]2 dx = 2πi ∑ Res(g; zk ). 0
zk ∈ℂ∗
0
Identifying in this last relation the real and the imaginary parts, we finally deduce that +∞
∫ 0
1 P(z) P(x) ln xdx = − Re[ ∑ Res( (log z)2 ; zk )] Q(x) 2 Q(z) z ∈ℂ∗ k
and +∞
∫ 0
P(z) 1 P(x) dx = − Im[ ∑ Res( (log z)2 ; zk )]. Q(x) 2π Q(z) z ∈ℂ∗ k
Remark 5.2.8. The last relation obtained in the above proof shows that if P and Q are two real variables polynomials with real coefficients, such that the denominator Q has no nonnegative zeros (roots), and deg Q ≥ deg P + 2, then +∞
∫ 0
P(x) P(z) 1 dx = − Im[ ∑ Res( (log z)2 ; zk )], Q(x) 2π Q(z) z ∈ℂ∗
where log 1 = 0.
k
+∞
P(x) 1 Exercise 5.2.7. Compute the integrals of the type ∫0 Q(x) dx, where P and Q are two xα real variables polynomials with real coefficients, such that the denominator Q has no
156  5 Residue theory nonnegative zeros (roots), and deg Q ≥ deg P +1 and α ∈ (0, 1). Then the above integral may be computed using the formula +∞
∫ 0
P(x) 1 πeαπi P(z) 1 dx = ; z ), ∑ Res( α Q(x) x sin(απ) z ∈ℂ∗ Q(z) z α k
where log 1 = 0.
k
P(z) Solution. Let g(z) = R(z) , where log z = ln z + i arg z, 0 < arg z < 2π and R(z) = Q(z) . zα Let z1 , . . . , zm be the zeros (roots) of the Q polynomial. Then there exist the numbers ε > 0 and r > 0, such that
0 < ε < min{zk  : k = 1, m} ≤ max{zk  : k = 1, m} < r. Let γ5 = λ∪γr ∪λ− ∪γε− , where this curve is presented in Figure 5.7, i. e., λ(t) = (1−t)ε+tr, γr (t) = re2πit , γε (t) = εe2πit , t ∈ [0, 1]. Thus, r
∫ ε
ε
R(x) R(x) dx + ∫ g + ∫ α e−2απi dx − ∫ g = 2πi ∑ Res(g; zk ), xα x z ∈ℂ∗ γr
γε
r
k
hence ∫ g − ∫ g + (1 − e
γr
−2απi
γε
r
)∫ ε
R(x) dx = 2πi ∑ Res(g; zk ). xα z ∈ℂ∗ k
But zP(z) 1 zg(z) = → 0, Q(z) z α
if z → ∞,
P(z) 1−α z → 0, zg(z) = Q(z)
if z → 0,
and
because the ratio QP is continuous at 0 and limz→0 z 1−α  = 0. Letting now ε → 0 and r → ∞, from the points 1 and 2 of the lemma that extends the Jordan lemma, we get +∞
(1 − e
−2απi
) ∫ 0
R(x) dx = 2πi ∑ Res(g; zk ), xα z ∈ℂ∗ k
i. e., +∞
∫ 0
πeαπi P(z) 1 P(x) 1 dx = ; z ). ∑ Res( α Q(x) x sin(απ) z ∈ℂ∗ Q(z) z α k k
5.2 Applications of the residue theorem to the calculation of the integrals 
157
Figure 5.8: Proof of Problem 5.2.8.
b
P(x) Exercise 5.2.8. Compute the integrals of the type ∫a Q(x)(x−a) α (b−x)β dx, where P and Q are two real variables polynomials with real coefficients, such that the denominator Q has no zeros (roots) on the interval [a, b], deg Q ≥ deg P + 2, and α, β > 0 satisfy the condition α + β = 1. Then the above integral may be computed using the formula b
∫ a
P(z) πe2απi P(x) dx = ; zk ), ∑ Res( α β sin(2απ) Q(z)(x − a) (b − x) Q(z)(z − a)α (b − z)β z ∈ℂ\[a,b] k
where we consider the main branch for all the powers, i. e., log 1 = 0. Solution. Let g(z) = P(z) . Q(z)
R(z) , (z−a)α (b−z)β
where log z = ln z + i arg z, 0 < arg z < 2π and
R(z) = Let z1 , . . . , zm be the zeros (roots) of the Q polynomial. Then there exist the numbers ε > 0 and r > 0, such that all the roots of Q lies in those bounded domain, with the boundary given by the γ6 curve, where this curve is presented in Figure 5.8, i. e., λ(t) = (1−t)(a+ε)+t(b−ε), γr (t) = re2πit , γa,ε (t) = a + εe2πit , γb,ε (t) = b+εe2πit , t ∈ [0, 1]. Let μ be a such a rectifiable curve that connects two arbitrary points from {λ} and {γr }, and don’t contain any root of Q. Thus, b−ε
∫g − ∫ g − ∫ g + ∫
γr
γa,ε
b−ε
− ∫
a+ε
γb,ε
a+ε
P(x) dx Q(x)(x − a)α (b − x)β
P(x) e2βπi e−2απi dx = 2πi ∑ Res(g; zk ). Q(x)(x − a)α (b − x)β z∈ℂ\[a,b]
158  5 Residue theory Letting ε → 0 and r → ∞, from points 1 and 2 of the lemma that extends the Jordan lemma, we get lim ∫ g = lim ∫ g = 0
ε→0
γa,ε
ε→0
and
γb,ε
lim ∫ g = 0.
r→+∞
γr
Hence, it follows that b
∫ a
P(x) πe2απi P(z) dx = ; zk ). ∑ Res( α β sin(2απ) z ∈ℂ\[a,b] Q(z)(x − a) (b − x) Q(z)(z − a)α (b − z)β k
5.3 The study of meromorphic functions with the residue theorem ̃ ⊂ ℂ an open set, and let f ∈ M(G). ̃ The order of the function f Definition 5.3.1. Let G ̃ point is the biggest integer number n ∈ ℤ that satisfies the next property: in the z0 ∈ G ∃r > 0, ∃g ∈ H(U(z0 ; r)) such that g(z0 ) ≠ 0 and ̇ 0 ; r), f (z) = (z − z0 )n g(z). ∀z ∈ U(z This integer number is denoted by the symbol Θ(f , z0 ). Remarks 5.3.1. From the theorem of characterization of holomorphic function zeros and the theorem of characterization of the poles, it follows that: ̃ is an nth order zero for the function f ⇔ Θ(f , z ) = n; 1. The point z0 ∈ G 0 ̃ is a pth order pole for the function f ⇔ Θ(f , z ) = −p; 2. The point z0 ∈ G 0 ̃ with f (z ) ≠ 0 is a regular point for the function f ⇔ 3. The point z0 ∈ G 0 Θ(f , z0 ) = 0. ̃ is an nth order zero for the function f , then Proof. If z0 ∈ G f (z) = (z − z0 )n (an + an+1 (z − z0 ) + ⋅ ⋅ ⋅),
∀z ∈ U(z0 ; r) and an ≠ 0,
hence f (z0 ) = f (z0 ) = ⋅ ⋅ ⋅ = f (n−1) (z0 ) = 0, f (n) (z0 ) ≠ 0. Then it follows that Θ(f , z0 ) = n. ̃ is a pth order pole for the function f , then If z0 ∈ G f (z) =
a−p
(z − z0 )p
+ ⋅⋅⋅ +
a−1 + a0 + ⋅ ⋅ ⋅ , z − z0
̇ 0 ; r) and a−p ≠ 0, ∀z ∈ U(z
hence Θ(f , z0 ) = −p. ̃ is a regular point for the function f such that f (z ) ≠ 0, then If z0 ∈ G 0 f (z) = a0 + an (z − z0 ) + ⋅ ⋅ ⋅ , so we get Θ(f , z0 ) = 0.
and a0 = f (z0 ) ≠ 0,
5.3 The study of meromorphic functions with the residue theorem
 159
Evidently, the next results hold. ̃ where G ̃ ⊂ ℂ is an open set and for all z ∈ G, ̃ the Theorem 5.3.1. For all f , g ∈ M(G), 0 following relations are true: Θ(f ⋅ g, z0 ) = Θ(f , z0 ) + Θ(g, z0 ),
f Θ( , z0 ) = Θ(f , z0 ) − Θ(g, z0 ). g
̃ where G ̃ ⊂ ℂ is an open set, and let D ⊂ G. ̃ If the sum Definition 5.3.2. Let f ∈ M(G), ∑z∈D Θ(f , z) is finite, then this sum represents the order of the function f in the set D and it is denoted by the symbol Θ(f , D). Remarks 5.3.2. 1. We have Θ(f , D) = N −P, where N represents the number of all zeros of the function f that lie in D, where every zero will be counted together with its order, and P represents the number of all poles of the function f that lie in D, where every pole will be counted together with its order. 2. If pn is a nth order polynomial, then pn ∈ H(ℂ) ⊂ M(ℂ) and Θ(pn , ℂ) = n. ̃ where G ̃ ⊂ ℂ is an open set, let g ∈ H(G), ̃ a ∈ G, ̃ and denote Lemma 5.3.1. Let f ∈ M(G), f h = f g. Then Res(h; a) = Θ(f , a)g(a). ̃ is an mth order pole of the function f , then there exists a function f Proof. 1. If a ∈ G 1 f1 (z) holomorphic in a neighborhood of a, such that f1 (a) ≠ 0 and f (z) = (z−a) m . Hence, m m−1 1 f (z) f1 (z)(z − a) − m(z − a) f1 (z) (z − a)m f1 (z) ⋅ = = −m , f (z) f1 (z) f1 (z) z−a (z − a)2m
where
f1 (z) f1 (z)
Res( ff ; a)
is holomorphic in the point a, then the point a is a firstorder pole of
= −m. Hence, it follows that Res(
f f
and
f g; a) = −mg(a). f
̃ is an nth order zero of the function f , then there exists a function f 2. If a ∈ G 1 holomorphic in a neighborhood of a, such that f1 (a) ≠ 0 and f (z) = (z − a)n f1 (z). Hence, n−1 n f1 (z) n f (z) n(z − a) f1 (z) + (z − a) f1 (z) = = + , f (z) (z − a)n f1 (z) z − a f1 (z)
160  5 Residue theory where
f1 f1
is holomorphic in the point a. then the point a is a firstorder pole of
Res( ff ; a)
= n. Thus, we get Res(
f f
and
f g; a) = ng(a). f
We obtained that in the both cases Res(h, a) = Θ(f , a)g(a). 3. If a is a regular point, then we may suppose f (a) ≠ 0, and taking in the second part of our proof n = 0 we deduce the required result. ̃ ⊂ ℂ an open Theorem 5.3.2 (Cauchy theorem related to the zeros and poles). Let G ̃ ̃ set, f ∈ M(G), g ∈ H(G), where f ≢ 0 and let γ a such a rectifiable closed curve (path) ̃ which does not contain nor any zero nor any pole of the function f , and γ ∼ 0. Then in G, ̃ G
the sum ∑ g(z)Θ(f , z)n(γ, z) is finite, and
̃ z∈G
f 1 ∫ g = ∑ g(z)Θ(f , z)n(γ, z). 2πi f ̃ z∈G
γ
̃ and the set of all singular Proof. Let us denote h = g ff . Then, h is meromorphic in G, points of h, denoted by S = S(h). It coincides with A ∪ B, where A is the set of all the zeros of the function f , and B is the set of all the poles of the function f . The above remarks follow from the proof of the Lemma 5.3.1. ̃ \ S. Then G is an open set and h ∈ H(G). Since {γ} lies in the G set, Let G = G applying the residue theorem for the function h, we have
∫g γ
f = ∫ h = 2πi ∑ Res(h; z)n(γ, z). f ̃
(5.5)
z∈G
γ
The index n(γ, z) does not vanish only on the bounded domain with the boundary given by γ, and in rest we have Res(h; z) ≠ 0 ⇒ z ∈ A∪B. Since in the above mentioned bounded domain with the boundary given by γ, the set A ∪ B has only a finite number of points; the sum (5.5) is finite. From Lemma 5.3.1, it follows that if z0 ∈ A ∪ B, then Θ(f , z0 ) ≠ 0, and Res(h; z0 ) = Θ(f , z0 )g(z0 ). Hence, 1 f ∫ g = ∑ g(z)Θ(f , z)n(γ, z). 2πi f ̃ γ
z∈G
̃ where G ̃ ⊂ ℂ is an open set and (γ)− ⊂ G ̃ Theorem 5.3.3. Let γ be a boundary curve in G, (where (γ) denotes that bounded domain, which is bounded by the curve {γ}). Let f a such ̃ which have nor any zero nor any pole in the {γ} set. Let a meromorphic function on G, a1 , . . . , aN denote the all zeros of the function f that lie in (γ), and let b1 , . . . , bP denote
5.3 The study of meromorphic functions with the residue theorem
 161
the all poles of the function f that lie in (γ) (where every zero and every pole is counted ̃ then together with its order). If g ∈ H(G), P N 1 f ∫ g = ∑ g(aj ) − ∑ g(bk ). 2πi f j=1 k=1 γ
Proof. Since γ is a boundary curve, then n(γ, aj ) = n(γ, bk ) = 1, ∀j = 1, N, ∀k = 1, P, and the proof follows from Theorem 5.3.2 and from Lemma 5.3.1. Corollary 5.3.1. From Theorem 5.3.3, we have N −P =
f 1 ∫ = Θ(f , D), 2πi f γ
where D = (γ)
and N represents the number of all the zeros of the function f that lie in D, where every zero is counted together with its order, and P represents the number of all the poles of the function f that lie in D, where every pole is counted together with its order. ̃ be a non constant function, where Theorem 5.3.4 (Argument principle). Let f ∈ M(G) ̃ ̃ G ⊂ ℂ is an open set, let γ a curve (path) in G, such that there exists γ which is continuous, and {γ} does not contain any pole of the function f . Let Γ = f ∘ γ, and let define the function I : ℂ \ {Γ} → ℂ,
I(w) =
f 1 . ∫ 2πi f − w γ
Thus, 1. I(w) = n(Γ, w); 2. If Γ is a closed curve (path), then I is locally constant, and I(w) ∈ ℤ, ∀w ∈ ℂ \ {Γ}; ̃ then I(w) = n(Γ, w) = Θ(f − w, D). 3. If D = U(z0 ; r) and γ = 𝜕D, D− ⊂ G, Proof. 1. From the definition, we have 1
dζ 1 Γ (t) 1 n(Γ, w) = = dt ∫ ∫ 2πi ζ − w 2πi Γ(t) − w 0
Γ
1
=
f (γ(t))γ (t) 1 f 1 dt = = I(w). ∫ ∫ 2πi f (γ(t)) − w 2πi f − w γ
0
2. Since I(w) = n(Γ, w), from the Index theorem it follows that I is locally constant, and I(w) ∈ ℤ, ∀w ∈ ℂ \ {Γ}. 3. Since I(w) =
1 f , ∫ 2πi f − w γ
162  5 Residue theory from Corollary 5.3.1 we get 1 f = Θ(f − w, D), ∫ 2πi f − w γ
hence I(w) = n(Γ, w) = Θ(f − w, D). ̃ Corollary 5.3.2 (Special case of the principle of argument variation). Let f ∈ M(G), ̃ ⊂ ℂ is an open set, and let γ be a boundary curve in G ̃ such that there exists γ where G which is continuous, and {γ} does not contain any zero or any pole of the function f . Let N be the number of all the zeros of the function f that lie in (γ) (every zero is counted together with its order), and let P be the number of all the poles of the function f that lie in (γ) (every pole is counted together with its order), and let Γ = f ∘ γ. Then the next relation holds Θ(f , D) = N − P = n(Γ, 0),
where D = (γ),
which we usually write as 1 f 1 {Arg f (z)}γ , ∫ = n(Γ, 0) = 2πi f 2π γ
where the righthand side symbol represents “the variation” of the argument of the function f when z runs over the curve (path) γ. Remark 5.3.3. The name of the previous corollary follows from the fact that log f (z) = ln f (z) + i arg f (z), where log f is the primitive of the function ff . Using the fact that γ(0) = γ(1) together with ln f (γ(0)) = ln f (γ(1)), it follows that
1 1 1 f 1 γ(1) log f (z)γ(0) = i[arg f (γ(1)) − arg f (γ(0))] = {Arg f (z)}γ . ∫ = 2πi f 2πi 2πi 2π γ
Theorem 5.3.5 (Rouché theorem). Let f , g ∈ H(G), where G ⊂ ℂ is an open set, and let γ be a boundary curve in G such that there exists γ which is continuous. Let D = (γ) (where (γ) denotes that bounded domain, which is bounded by the curve {γ}), and suppose that D− ⊂ G. If g(ζ ) < f (ζ ),
∀ζ ∈ {γ},
then the next two equations have the same numbers of zeros in D: 1. f (z) = 0; 2. f (z) + g(z) = 0,
5.3 The study of meromorphic functions with the residue theorem
 163
i. e., Θ(f , D) = Θ(f + g, D). Proof. We need to prove that Θ(f , D) = Θ(f + g, D). Since for all the points ζ ∈ {γ} = 𝜕D, we have g(ζ ) < f (ζ ). It follows that f (ζ ) ≠ 0, ∀ζ ∈ {γ}. On the other hand, h = gf ∈ M(D1 ), where D1 is a connected component of G that contains the set D. Hence, Θ(f + g, D) − Θ(f , D) = Θ(
f +g , D) = Θ(h + 1, D). f
If Γ = h ∘ γ, from Theorem 5.3.4 we have Θ(h + 1, D) = n(Γ, −1). If t ∈ [0, 1], then Γ(t) = h(γ(t)) =  g(γ(t))  < 1, hence {Γ} ⊂ U(0; 1), and from here f (γ(t)) we deduce F = ℂ \ U(0; 1) ⊂ ℂ \ {Γ}. Since F is not bounded connected set, then F it is a subset of the ℂ \ {Γ} unbounded component, and from the index theorem we get n(Γ, w) = 0, for all w ∈ F. Thus, we have n(Γ, −1) = 0, because −1 ∈ F, and hence Θ(h + 1, D) = 0. Corollary 5.3.3 (Fundamental theorem of the algebra). Let p(z) = a0 + a1 z + ⋅ ⋅ ⋅ + an z n ,
n ∈ ℕ∗ , an ≠ 0.
Then the polynomial p has n roots in ℂ. Proof. Let us consider the next two equations: an z n = 0;
(5.6)
p(z) = 0.
Let us denote f (z) = an z n , and g(z) = a0 +a1 z +⋅ ⋅ ⋅+an−1 z n−1 . Since limz→∞ it follows that ∃r0 > 0
(5.7) g(z) f (z)
= 0,
g(ζ ) such that ∀r > r0 , and ∀ζ ∈ {γ}, γ = 𝜕U(0; r) we have < 1, f (ζ )
i. e., g(ζ ) < f (ζ ),
∀ζ ∈ {γ}.
According to the Rouché theorem, equations (5.6) and (5.7) have the same number of zeros in U(0; r). But the equation (5.6) has exactly n zeros in 0.
164  5 Residue theory Theorem 5.3.6 (The principle of the domain conservation). Let f ∈ H(D), where D ⊂ ℂ is a domain (i. e., an open connected set). If the function f is nonconstant, then Δ = f (D) is also a domain. Proof. Since the function f is continuous and the set D is connected, it follows easily that Δ is also connected. We need to prove that Δ = f (D) is an open set. We will use two different methods to prove this. 1. Let w0 ∈ f (D). Then ∃z0 ∈ D such that f (z0 ) = w0 , hence z0 is a zero of the function f − w0 . But f − w0 is a nonconstant function, because f is nonconstant, hence the point z0 is an isolated zero. Thus, ∃r > 0 such that U − (z0 ; r) ⊂ D and ∀ζ ∈ {γ}, where γ = 𝜕U(z0 ; r), we have f (ζ ) − w0 ≠ 0. Since the set {γ} is compact and the function f is continuous, we deduce that ρ = min f (ζ ) − w0 > 0. ζ ∈{γ}
Let w1 ∈ U(w0 ; ρ) be an arbitrary number, and consider the two equations: f (z) − w0 = 0;
(5.8)
f (z) − w1 = 0.
(5.9)
The second one may be written in the equivalent form: (f − w0 ) + (w0 − w1 ) = 0.
(5.10)
From the definition of the number ρ and from the way we have choose w1 , we get w0 − w1 < ρ ≤ f (ζ ) − w0 ,
∀ζ ∈ {γ}.
Now, according to Rouché theorem, equations (5.8) and (5.10) have the same number of zeros in the set (γ) = U(z0 ; r). Since z0 ∈ U(z0 ; r) is the zero of the equation (5.8), it follows that the equation (5.9) has at least one zero in U(z0 ; r), hence ∃z1 ∈ U(z0 ; r) such that f (z1 ) = w1 . Since w1 is an arbitrary element of the open disc U(w0 ; ρ), then U(w0 ; ρ) ⊂ f (U(z0 ; r)) ⊂ f (D). Hence, the point w0 is an interior point of f (D), thus Δ = f (D) is an open set. 2. Let w0 ∈ f (D). Then ∃z0 ∈ D such that f (z0 ) = w0 and ∃r1 > 0 such that U(z0 ; r1 ) ⊂ D. Hence, the number z0 is a zero of the function f − w0 , and the function f − w0 is a nonconstant function, because f is nonconstant. Thus, the point z0 is an isolated zero, hence ∃r > 0 : U − (z0 ; r) ⊂ U(z0 ; r1 ) and f (ζ ) − w0 ≠ 0, ∀ζ ∈ {γ}, where γ = 𝜕U(z0 ; r). It follows that w0 ∈ ℂ \ {Γ}, where Γ = f ∘ γ.
5.4 Exercises  165
According to the principle of the argument variation (Theorem 5.3.6), we have that the function I(w) = Θ(f − w, U(z0 ; r)) is locally constant, an since I(w0 ) > 0, because f (z0 ) = w0 , it follows that ∃r0 > 0 : Θ(f − w, U(z0 ; r)) = Θ(f − w0 , U(z0 ; r)) = I(w0 ) > 0,
∀w ∈ U(w0 ; r0 ).
Thus, w ∈ f (U(z0 ; r)) ⊂ f (D),
∀w ∈ U(w0 ; r0 ) ⇔ U(w0 ; r0 ) ⊂ f (U(z0 ; r)) ⊂ f (D).
Hence, the point w0 is an interior point of f (D), so Δ = f (D) is an open set. Similarly, we can prove the following generalization of the above theorem. Theorem 5.3.7. Let f ∈ H(G), where G ⊂ ℂ is an open set. Suppose that the function f is not locally constant. Then, for all open subset G1 ⊂ G, its image f (G1 ) is also an open set.
5.4 Exercises 5.4.1 Residue theorem Exercise 5.4.1. Calculate the residue of the following functions in the corresponding points: z 1. f : ℂ \ {2} → ℂ, f (z) = sin , z0 = 2; z−2 2.
3.
z , z0 = 1 and z1 = 2; (z−1)(z−2)2 z {z1 , z2 } → ℂ, f (z) = (z−z )p (z−z ) , where p ∈ ℕ∗ , z1 and z2 ; 1 2 1 {−1, 0, 1} → ℂ, f (z) = z 3 −z 5 , z0 = −1, z1 = 0 and z2 = 1; 2 {−i, i} → ℂ, f (z) = z 2z+1 , z0 = −i and z1 = i;
f : ℂ \ {1, 2} → ℂ, f (z) =
f :ℂ\
4. f : ℂ \ 5.
f :ℂ\
6.
f : ℂ \ {−1} → ℂ, f (z) =
7.
f : ℂ \ {2kπi : k ∈ ℤ} → ℂ, f (z) =
z 2n , z0 (z+1)n
= −1 and z1 = ∞; 1 , zk 1−ez
= 2kπi, k ∈ ℤ.
Exercise 5.4.2. Calculate the residue of the following functions in their isolated singular points: 1. 3.
f (z) =
z 2n , 1 + zn
f (z) = e
z− z1
;
n ∈ ℕ∗ ;
2. f (z) =
eaz
z
(1 + e 2 )2
,
a ∈ ℂ;
1
zne z 4. f (z) = , 1+z
n ∈ ℕ∗ .
166  5 Residue theory Exercise 5.4.3. Calculate the residue of the following functions in their isolated singular points: 1. f (z) =
1
eiaz , sinh z
4. f (z) =
1
z 2 sin z
a ∈ ℂ;
2. f (z) = 5.
;
ez ; (z − 1)2 1
f (z) = z 3 e 1−z ;
3. f (z) = 6. f (z) =
1 , (z 2 + 1)n eiaz
cosh2 z
,
n ∈ ℕ∗ ; a ∈ ℂ.
Exercise 5.4.4. Using the residue theorem, calculate the next integrals on the corresponding paths: 1. ∫γ z(zz+2 2 +4)2 dz, where
2.
(a) {γ} = 𝜕U(0; 1), (b) {γ} = 𝜕U(0; 5) are directly oriented; ∫γ (z−1)21(z 2 +1) dz, where {γ} = 𝜕U(1 + i; 2) is directly oriented;
3.
∫γ
1 dz, z n +2
n ∈ ℕ∗ , where γ(t) = (3 + ε)e2πit + 1, t ∈ [0, 1], and ε > 0;
6.
1 dz, n ∈ ℕ∗ , where {γ} = 𝜕U(0; r), r < 1, is directly (z n −1)(z 3 −1) 1 √2e2πit + (1 + i), t ∈ [0, 1]; ∫γ (z−1)(z 2 +1) dz, where γ(t) = 1 dz, where {γ} = 𝜕U(0; 2) is directly oriented; ∫γ z 3 (z 10 −1)
7.
∫γ e 1−z dz, where {γ} = 𝜕U(0; 2) is directly oriented;
4. ∫γ 5.
oriented;
1
8. ∫γ tan πzdz, where {γ} = 𝜕U(0; n), n ∈ ℕ∗ , is directly oriented; 9. 10. 11.
tan z dz, where {γ} = 𝜕U(0; 2) is directly oriented; z2 ∫γ z 41+1 dz, where {γ} = 𝜕U(1; 1) is directly oriented; 1 ∫γ z1 sin (z−1) 2 dz, where {γ} = 𝜕U(0; 2) is directly oriented;
∫γ
12. ∫γ 13.
where {γ} = 𝜕U(0; 2) is directly oriented;
∫γ z 111 +z111 +z+1 dz,
where {γ} = 𝜕U(0; r) is directly oriented, and r > 0 is enough big,
such that the disc U(0; r) contains all the zeros of the function z 111 + z 11 + z + 1;
14. ∫γ 15. ∫γ 16.
1 (z−1)3 z−1 e dz, z
1 1 e z−1 dz, z−1
1 zn e z
where {γ} = 𝜕U(0; r), r ≠ 1, is directly oriented;
dz, n ∈ ℕ∗ , where {γ} = 𝜕U(0; 3) is directly oriented;
z 2 −1 ∫γ sinn z1 dz,
n ∈ ℕ∗ , where {γ} = 𝜕U(0; r), r > 0, is directly oriented; 1
1
1
17. ∫γ (1 + z + z 2 )(e z + e z−1 + e z−2 )dz, where {γ} = 𝜕U(0; 3) is directly oriented;
18. ∫γ (1+e1 z )2 dz, where {γ} = 𝜕U(0; r), (2n − 1)π < r < (2n + 1)π, n ∈ ℕ∗ , is directly oriented. Exercise 5.4.5. Calculate the following integrals, where the corresponding paths are directly oriented:
5.4 Exercises  167
1.
∫γ
2.
∫γ
3.
∫γ
cosh πz 2 (z+i)4 z
100 iπz
e z 2 +1
5 z+3i √ 3−z
dz, where {γ} = {z ∈ ℂ : z + 2i = 2}; dz, where {γ} = {z = x + iy ∈ ℂ : 4x2 + y2 − 4 = 0}; 19π
10  = √ 2ei 20 ; dz, where {γ} = 𝜕U(0; r), r < 3, where √5 z+3i 3−z z=3i
4.
z3 1 dz, ∫γ z cos z2
5.
∫γ z 2 e z+1 dz, where {γ} = {z = x + iy ∈ ℂ : x2 + y2 + 2x = 0};
6.
∫γ
7.
∫γ
8.
where {γ} = {z = x + iy ∈ ℂ : x2 + y2 − 2y − 3 = 0};
2z
z 13 dz, (z−2)4 (z 5 +3)2
where {γ} = {z = x + iy ∈ ℂ : 4x2 + 9y2 − 36 = 0};
1 dz, where {γ} = 𝜕U(0; r), 0 < r z √4z 2 +12z+13 dz, where {γ} = {z ∈ ℂ : z = Reiθ , 0 ∫γ log(z−a) z2
a > 0, and log 1 = 0.
Exercise 5.4.6. Using the residue theorem calculate the following integrals on the corresponding directly oriented paths: 1 1. ∫γ z sin dz, where {γ} = {z ∈ ℂ : z = r}, r ≠ kπ, k ∈ ℤ; z
2. 3.
∫γ
1 dz, 3 sin z−sin 3z
∫γ
(z−1)2
1 ez
where {γ} = {z ∈ ℂ : z = 4};
dz, where
(a) {γ} = {z ∈ ℂ : z = 22 }, (b) {γ} = {z ∈ ℂ : z = 2}; √
4. ∫γ
5. 6. 7.
z−1 log z+1 dz, (z 2 +1)(z 2 −4)
where
(a) {γ} = {z ∈ ℂ : z = 21 }, (b) {γ} = {z ∈ ℂ : z = √2}, (c) {γ} = {z = x + iy ∈ ℂ : 3x2 + y2 − 2 = 0},  = πi; (d) {γ} = {z ∈ ℂ : z = 3}, where log z−1 z+1 z=0 z dz, where {γ} = {z ∈ ℂ : z = 2}; ∫γ z 2sin (z 4 +1) 1 dz, where {γ} = {z ∈ ℂ : z = r}, nπ z 2 (z−1) sin z z ∫γ sin z(1−cos z) dz, where {γ} = {z ∈ ℂ : z = 4};
∫γ
8. ∫γ
1
z3 e z z+1
dz, where
(a) {γ} = {z ∈ ℂ : z = 22 }, (b) {γ} = {z ∈ ℂ : z = √2}; √
1
∫γ
zn e z z 2 −1
10. ∫γ
π e z−i z 2 +1
9.
dz, n ∈ ℕ, where
(a) {γ} = {z ∈ ℂ : z = 21 }, (b) {γ} = {z ∈ ℂ : z = 2}, (c) {γ} = {z = x + iy ∈ ℂ : x2 + y2 − 2x − dz, where {γ} = {z ∈ ℂ : z = 2};
5 4
= 0};
< r < (n + 1)π, n ∈ ℕ;
168  5 Residue theory 1 dz, where {γ} = {z z(z 2 +a2 )2 1 ∫γ z n +1 dz, n ∈ ℕ∗ , where {γ} 1 dz, where {γ} = {z ∈ ∫γ z 2 sin z
11. ∫γ
∈ ℂ : z = r}, r > a > 0;
13.
ℂ : z = r}, nπ < r < (n + 1)π, n ∈ ℕ∗ ;
12.
14. ∫γ
√1+z 2 dz, (1−z 2 )(z 2 −4)
where
(a) {γ} = {z ∈ ℂ : z =
√2 }, 2
(b) {γ} = {z = x + iy ∈ ℂ : 1 = 0, x < 0}; 4
15. ∫γ
√z 2 −1 log
(a) (b) (c) (d)
z−1 z+1
(z 2 +1)(z 2 +4)
= {z ∈ ℂ : z = r}, r ≠ 1;
x2 2
+ 4y2 − 1 = 0, x ≥ 0} ∪ {z = x + iy ∈ ℂ : x2 + y2 −
dz, where
{γ} = {z ∈ ℂ : z = 22 }, {γ} = {z = x + iy ∈ ℂ : 36x 2 + 4y2 − 9 = 0}, {γ} = {z = x + iy ∈ ℂ : 10x 2 + y2 − 5 = 0}, {γ} = {γ} = {z = x + iy ∈ ℂ : x2 + y2 − 5y = 0}, where √z 2 − 1z=0 = i and  = πi. log z−1 z+1 z=0 √
5.4.2 Applications of the residue theorem to the calculation of the trigonometric integrals Exercise 5.4.7. Calculate the following real integrals by using the residue theorem: 2π 1 1. ∫0 2 dθ; 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
2−sin θ 2π 1+sin θ dθ; 2+cos θ 2π 2+sin θ ∫0 2+cos θ dθ; 2π 1 dθ, a > 1; ∫0 a+cos θ 2π 1 ∫0 1+a sin θ dθ, −1 < a < 1; 2π 1 dθ, a > b > 0; ∫0 (a+b cos θ)2 2π 1+cos θ dθ; ∫0 (13−5 cos θ)2 2π 1 ∫0 (17+8 cos θ)2 dθ; 2π 1 dθ, 0 < p < 1; ∫0 1−2p cos θ+p2 2 2π 2θ dθ, 0 < p < 1; ∫0 1−2pcos cos θ+p2 2 2π 3θ dθ, 0 < p < 1. ∫0 1−2pcos cos 2θ+p2
∫0
Exercise 5.4.8. Using the residue theorem, calculate the following real integrals: 2π 2π nθ nθ 1. ∫0 1−2acos dθ, ∫0 1−2acos dθ, a > 1, n ∈ ℕ∗ ; cos θ+a2 sin θ+a2 2.
2π 2π sin θ sin nθ θ cos nθ dθ, ∫0 sin dθ, 5−4 cos θ 5−4 cos θ
∫0
n ∈ ℕ∗ .
5.4 Exercises  169
Exercise 5.4.9. Prove the formulas: n n π cos nθ 1. ∫0 b−ia dθ = √ πi2 2 √ 2 a 2 2 n , a > 0, b > 0, n ∈ ℕ; cos θ 2.
2π
∫0
1 dθ a+b cos2 θ
a +b ( a +b +b )
=
2π , √a(a+b)
a > 0, b > 0.
5.4.3 Applications of the residue theorem to the calculation of the improper integrals Exercise 5.4.10. Calculate the following real improper integrals: +∞ 1. ∫0 sinx x dx; 2.
3. 4. 5. 6. 7. 8.
+∞ 1 dx; x4 +1 +∞ 1 ∫0 (a+bx2 )n dx, a, b > 0, n ∈ ℕ∗ ; +∞ ∫−∞ 1+x1 2n dx, n ∈ ℕ∗ ; +∞ ln x ∫0 (x+a) 2 +b2 dx, a, b > 0; +∞ cos ax ∫0 (x2 +b2 )2 dx, a, b ∈ ℝ, b > 0; +∞ bx dx, a, b ∈ ℝ; ∫0 cos ax−cos x2 2 +∞ −ax cos bxdx, a > 0 (Poissontype ∫−∞ e
∫−∞
integral).
5.4.4 The study of meromorphic functions using the residue theorem Exercise 5.4.11. Determine the number of the roots of the following equations that lies in the disc U(0; 1): 1. z 8 − 4z 5 + z 2 − 1 = 0; 2. z 8 − 11z + 1 = 0; 3. 23z 5 − 10z 3 + 6z − 5 = 0; 4. ez − 3iz = 0; 5. z n + 8z 2 + 1 = 0, n ∈ ℕ, n ≥ 3; 6. z n + 3z 2 + 1 = 0, n ∈ ℕ, n ≥ 3; 7. ez − 4z n + 1 = 0, n ∈ ℕ∗ ; 8. z n + pz 2 + qz + r = 0, n ∈ ℕ, n ≥ 3, where p > q + r + 1; 9. z 10 − 9z 6 + 3z 3 + z 2 − 2 = 0. Exercise 5.4.12. Determine the number of the roots of the equation az n = ez , that belong to the disc U(0; r), if a >
er rn
n ∈ ℕ∗ ,
and r > 0 are given numbers.
170  5 Residue theory Exercise 5.4.13. Determine the number of the roots of the following equations that lies in the corresponding circular ring: 1. z 4 − 9z + 1 = 0, U(0; 1, 3); 2. 3z 5 − 7z 3 + z 2 − 2 = 0, U(0; 1, 2); 3. z 4 + z 3 − 4z + 1 = 0, U(0; 1, 3). Exercise 5.4.14. If 0 < a < e1 , prove that: 1. the equation z = aez has only one root in the disc U(0; 1); 2. the equation z 2 = aez has two roots in the disc U(0; 1).
6 Conformal representations 6.1 Special classes of holomorphic functions Definition 6.1.1. 1. Denote by C(G) the class of all continuous functions in an open set G ⊂ ℂ. We say that the subclass ℱ ⊂ C(G) is uniformly continuous at the point z0 ∈ G, if ∀ε > 0, ∃δ = δ(ε) > 0,
such that ∀z ∈ G, z − z0  < δ ⇒ f (z) − f (z0 ) < ε, ∀f ∈ ℱ .
2.
(Here, the number δ > 0 is independent of the function f .) We say that the subset ℱ ⊂ C(G) is uniformly continuous (on the set G), if ℱ is uniformly continuous in all the points z ∈ G.
Theorem 6.1.1. Let ℱ ⊂ C(G) a set of uniformly continuous functions on the open subset E
G ⊂ ℂ. Suppose that (fn )n∈ℕ ⊂ ℱ and fn → f , where E ⊂ G ⊂ E (E is the set of all accumulation points of E). G
Then the function f has an extension to G, such that fn → f , f ∈ C(G) and fn K f , ∀K ⊂ G, where K is compact (i. e., the sequence (fn )n∈ℕ uniformly converges on the compacts to f ). Proof. 1. First, we will prove that (fn )n∈ℕ converges on G. Let z0 ∈ G and ε > 0 be arbitrary. Since ℱ is uniformly continuous, we get ∃r > 0,
ε such that U(z0 ; r) ⊂ G and fn (z) − fn (z0 ) < , ∀n ∈ ℕ, ∀z ∈ U(z0 ; r). 3
The set E is dense in G, then there exists z1 ∈ E ∩ U(z0 ; r). Since (fn (z1 ))n∈ℕ converges, then ε ∃n0 ∈ ℕ : ∀n, m > n0 ⇒ fn (z1 ) − fm (z1 ) < 3 ⇒ fn (z0 ) − fm (z0 ) = fn (z0 ) − fn (z1 ) + fn (z1 ) − fm (z1 ) + fm (z1 ) − fm (z0 ) ≤ fn (z0 ) − fn (z1 ) + fn (z1 ) − fm (z1 ) + fm (z1 ) − fm (z0 ) ε ε ε < + + = ε, ∀n, m > n0 , 3 3 3 hence the sequence (fn (z0 ))n∈ℕ converges. Let f (z0 ) = limn→∞ fn (z0 ). 2. Second, we will prove that the function f is continuous on G. Let ε > 0 and z0 ∈ G be arbitraries. From the fact that ℱ is uniformly continuous, it follows that ∃r > 0 : U(z0 ; r) ⊂ G
ε and fn (z) − fn (z0 ) < , 3
https://doi.org/10.1515/9783110657869006
∀n ∈ ℕ, ∀z ∈ U(z0 ; r).
172  6 Conformal representations Letting n → +∞, then ε f (z) − f (z0 ) ≤ , 3
∀z ∈ U(z0 ; r),
hence f ∈ C(G). 3. Finally, we will prove that fn K f , ∀K ⊂ G, where K is compact. Let K ⊂ G be a compact set and z0 ∈ K. For a given number ε > 0, as in the previous G
steps, it corresponds the number r > 0 such that U(z0 ; r) ⊂ G. Since fn → f , we have ε ∃n0 ∈ ℕ : ∀n > n0 , fn (z0 ) − f (z0 ) < 3 ⇒ fn (z) − f (z) = fn (z) − fn (z0 ) + fn (z0 ) − f (z0 ) + f (z0 ) − f (z) ε ε ε ≤ fn (z) − fn (z0 ) + fn (z0 ) − f (z0 ) + f (z0 ) − f (z) < + + = ε, 3 3 3 ∀z ∈ U(z0 ; r), ∀n > n0 .
For all the points zj ∈ K, it corresponds to a number rj > 0 with the above property, and an index nj ∈ ℕ. All the discs of the form U(zj ; rj ) represent a covering of the set K, and from the fact that K is compact it follows that it contains a finite covering U(z1 ; r1 ), . . . , U(zm ; rm ). If n > max{n1 , . . . , nm }, then fn (z) − f (z) < ε, ∀z ∈ K, hence fn K f . Definition 6.1.2. We say that the function set ℱ ⊂ C(G) is bounded (i. e., all the functions of ℱ are uniformly bounded on the compacts of G), where G ⊂ ℂ is an open set, if such that ∀z ∈ K, ∀f ∈ ℱ ⇒ f (z) ≤ M.
∀K ⊂ G, K compact, ∃M = M(K) > 0,
(The number M is the same for all the functions of ℱ .) Theorem 6.1.2. If ℱ ⊂ H(G) is bounded, where G ⊂ ℂ is an open set, then the set ℱ = {f : f ∈ ℱ } is also bounded. Proof. Let K ⊂ G be an arbitrary compact, and let 0 < R < d(K, 𝜕G), 0 < r < R, where R and r are fixed numbers. Thus, U(z0 ; R) ⊂ G, ∀z0 ∈ K, hence γ = 𝜕U(z0 ; R) ⊂ G. Letting an arbitrary z ∈ U(z0 ; r), then ∀ζ ∈ {γ} we have ζ − z ≥ R − r > 0. From the Cauchy formula, we get f (z) =
f (ζ ) 1 dζ . ∫ 2πi (ζ − z)2 γ
Since the set ℱ is bounded, then f (ζ ) ≤ M, ∀ζ ∈ {γ}, ∀f ∈ ℱ , and from here it follows that M 1 2πR = M , f (z) ≤ 2π (R − r)2
∀z ∈ U(z0 ; r).
6.1 Special classes of holomorphic functions  173
Since K is compact, it may be covered by a finite number of the discs of the form U(zj ; r), with f (z) ≤ Mj , ∀z ∈ U(zj ; r). Hence, the set {f : f ∈ ℱ } = ℱ is uniformly bounded on the compacts of G, and thus the function set ℱ is bounded. Theorem 6.1.3. If ℱ ⊂ H(G) is a bounded function set, where G ⊂ ℂ is an open set, then ℱ is uniformly continuous in all the points of G. Proof. Let z0 ∈ G and ε > 0 be arbitraries, and let r > 0 such that U(z0 ; r) ⊂ G. Since ℱ is bounded, from the above theorem the function set ℱ is also bounded. Hence, ∃M > 0 such that f (z) ≤ M, ∀z ∈ U(z0 ; r), ∀f ∈ ℱ . Let us consider the linear path λ(t) = (1 − t)z0 + tz, t ∈ [0, 1], connecting the points z ∈ U(z0 ; r) and z0 . Then z
f (z) − f (z0 ) = ∫ f (ζ )dζ z0
and f (z) − f (z0 ) ≤ Mz − z0 ,
hence, for all points z ∈ U(z0 ; r), with z − z0  < δ = ∀f ∈ ℱ .
ε , M
we have f (z) − f (z0 ) < ε,
Corollary 6.1.1. Let ℱ ⊂ H(G), where ℱ is bounded and G ⊂ ℂ is an open set. If (fn )n∈ℕ ⊂ ℱ converges to f on the set E, where E ⊂ G ⊂ E , then the function f has a such an holomorphic extension to G that satisfies the condition fn K f ,
∀K ⊂ G, K compact.
Proof. Since the function set ℱ ⊂ H(G) is bounded, from the above theorem, we have that ℱ is uniformly continuous on G. Now, according to Theorem 6.1.1, there exists the continuous function f : G → ℂ that satisfies the above uniformly convergence on the compacts sets. But according to the Weierstrass theorem, the function f will be holomorphic on G. Definition 6.1.3. Let ℱ ⊂ C(G), where G ⊂ ℂ is an open set. We say that the function set ℱ is relatively compact (in C(G)), if ∀(fn )n∈ℕ ⊂ ℱ sequence ∃(fnk )k∈ℕ ⊂ (fn )n∈ℕ , such that (fnk )k∈ℕ uniformly converges on compact sets to f ∈ C(G). Remark 6.1.1. From the Bolzano–Weierstrass theorem, we get: in the Euclidian spaces the relatively compact sets coincide to the bounded sets. This property also holds for the set H(G) of holomorphic functions on the open set G. Theorem 6.1.4 (Montel theorem). Let ℱ ⊂ H(G), where G ⊂ ℂ is an open set. Then ℱ is bounded ⇔ ℱ is relatively compact in H(G).
Proof. “⇒” Let E be the subsets of all the points of G with rational coordinates. Then E ⊂ G ⊂ E and the set E is countable, so it may be written as E = {zj : j ∈ ℕ∗ }. Let (fn )n∈ℕ∗ ⊂ ℱ be an arbitrary function sequence. Since ℱ is bounded, then (fn (z1 ))n∈ℕ∗
174  6 Conformal representations is a bounded sequence in ℂ, hence, according to the Bolzano–Weierstrass theorem, it contains a convergent subsequence f11 (z1 ), f21 (z1 ), . . . , fn1 (z1 ), . . . . The sequence (fn1 (z2 ))n∈ℕ∗ is a bounded sequence, hence it contains a convergent subsequence f11 (z2 ), f22 (z2 ), f32 (z2 ), . . . , fn2 (z2 ), . . . . The sequence (fn2 (z3 ))n∈ℕ∗ is a bounded sequence, hence it contains a convergent subsequence, and we continue this algorithm. After n steps, we may choose the subsequence f11 , f22 , f33 , . . . , fnn , fn+1n , fn+2n , . . . that converges on the {z1 , z2 , . . . , zn } set. Continuing in the same way, we obtain the subsequence (fnn )n∈ℕ∗ that converges in all the points of the E set. Hence, (fnn )n∈ℕ∗ ⊂ E
H(G), fnn → f , where f (zk ) = limn→∞ fnn (zk ), ∀k ∈ ℕ∗ . From Corollary 6.1.1 of Theorem 6.1.3, the function f has such a holomorphic extension on G, which satisfies that (fnn )n∈ℕ∗ uniformly converges on compacts to f . “⇐” Suppose that ℱ is not bounded. Then there exist K ⊂ G, K compact, and there exists fn ∈ ℱ , such that ∃zn ∈ K : fn (zn ) > n. From the sequence (fn )n∈ℕ∗ , we may choose such a subsequence (fnk )k∈ℕ∗ that uniformly converges on compacts to f ∈ H(G). Let z0 ∈ K be an accumulation point for the sequence (zn )n∈ℕ∗ ⊂ K. Thus, ∃(znk )k∈ℕ∗ ⊂ (zn )n∈ℕ∗ is a subsequence, such that limk→∞ znk = z0 and limk→∞ f (znk ) = f (z0 ). Now, since (fnk )k∈ℕ∗ uniformly converges on compacts to f , we get limk→∞ fnk (znk ) − f (znk ) = 0, hence fnk (znk ) ≤ fnk (znk ) − f (znk ) + f (znk ) → f (z0 ),
k → ∞.
But this last inequality cannot be true, because fnk (znk ) > nk → ∞, k → ∞. Examples 6.1.1. 1. Let G and G1 be two open sets in ℂ, such that G1 is bounded. Thus, ℱ = {f ∈ H(G) : f (G) ⊂ G1 } is bounded, hence it is relatively compact in H(G). 2. Let us consider the set U = U(0; 1), and let
ℱ = {f ∈ H(U) : f (z) ≤
Thus ℱ ≠ 0, because g(z) = we have
1 1+z
1 , z ∈ U}. 1 − z
∈ ℱ . The set ℱ is bounded, because for all r ∈ [0, 1)
∀z ∈ U(0; r),
1 ∀f ∈ ℱ ⇒ f (z) ≤ , 1−r
and according to the Montel theorem the set ℱ is relatively compact in H(U).
6.2 Univalent functions  175
3.
If f ∈ H(G), where G ⊂ ℂ is an open set, and ℱ = {f + c : c ∈ ℂ}, then ℱ is not bounded, hence it is not relatively compact in H(G).
Theorem 6.1.5 (Vitali theorem). Let D ⊂ ℂ be a domain, and let (fn )n∈ℕ ⊂ H(D) be a E
bounded sequence such that fn → f , where E ⊂ D and E ∩ D ≠ 0. Then fn K f ,
∀K ⊂ D, K compact
(i. e., the sequence (fn )n∈ℕ uniformly converges to f on compact sets in D). D
Proof. According to the Corollary 6.1.1, it is sufficient to prove that fn → f . If this is not true, then there exists the point z0 ∈ D such that the sequence (fn (z0 ))n∈ℕ diverges. Since (fn )n∈ℕ is a bounded sequence, the set (fn (z0 ))n∈ℕ has two different limit points w1 and w2 . It follows that the sequence (fn (z0 ))n∈ℕ contains two subsequences (fn1 (z0 ))n1 ∈ℕ and (fn2 (z0 ))n2 ∈ℕ , such that lim f (z ) nj →∞ nj 0
= wj ,
j = 1, 2.
From the Montel theorem (Theorem 6.1.4), the sequences (fn1 )n1 ∈ℕ and (fn2 )n2 ∈ℕ have such subsequences that uniformly converges on compacts in D to g, respectively to h, where g and h are holomorphic functions. Since (fn1 )n1 ∈ℕ , (fn2 )n2 ∈ℕ ⊂ (fn )n∈ℕ and (fn )n∈ℕ converges on E, it follows that gE = hE . But E ∩ D ≠ 0, and according to the theorem of holomorphic functions identity we get g = h, that contradicts the fact that (from the definitions of g and h) g(z0 ) = w1 ≠ w2 = h(w0 ).
6.2 Univalent functions Definition 6.2.1. An holomorphic and injective function on the domain D ⊂ ℂ is said to be an univalent function. The set of all univalent functions on the domain D is denoted by Hu (D). Examples 6.2.1. 1. Any circular transform is univalent on the set ℂ \ {z0 }, where z0 is the pole of the function. z 2. The function f (z) = (1+z) 2 is univalent in the unit disc U = U(0; 1), and 1 f (U) = ℂ \ [ , +∞). 4 Since f (eiθ ) =
1 , (2 cos θ2 )2
θ ∈ ℝ, it follows that f (𝜕U) = [ 41 , +∞) ∪ {∞}.
3. If f1 , f2 ∈ Hu (D), then f1 ∘ f2 ∈ Hu (D). 1−z 4. If f (z) = log 1+z , where log 1 = 0, then f ∈ Hu (U) and D = f (U) = {w ∈ ℂ : −
π π < Im w < }. 2 2
176  6 Conformal representations Letting g(z) = 1−z ∈ Hu (U), then g(U) = Δ = {ζ ∈ ℂ : Re ζ > 0} and h(ζ ) = log ζ ∈ 1+z Hu (Δ), h(Δ) = D, thus f = h ∘ g ∈ Hu (U) and f (U) = D. Theorem 6.2.1. If f ∈ Hu (D), where D ⊂ ℂ is a domain, then f (z) ≠ 0, ∀z ∈ D. Proof. Suppose that there exists a point z0 ∈ D, such that f (z0 ) = 0. From the univalence of f it follows that f and f are not constant functions, and from the theorem of holomorphic functions zeros, there exists a number r > 0, such that U(z0 ; r) ⊂ D, ̇ 0 ; r) and f (z) ≠ f (z0 ), ∀z ∈ 𝒞 = 𝜕U(z0 ; r). f (z) ≠ 0, ∀z ∈ U(z Since f − f (z0 ) is a continuous function, and since f (z) − f (z0 ) ≠ 0, ∀z ∈ 𝒞 , where 𝒞 is a compact set, it follows that min{f (z) − f (z0 ) : z ∈ 𝒞 } = m > 0. ̇ 0 ; r) such that From the continuity of the function f in the point z0 , there exists z1 ∈ U(z f (z1 ) − f (z0 ) < m, hence f (z1 ) − f (z0 ) < f (z) − f (z0 ), ∀z ∈ 𝒞 . From here, according to the Rouché theorem, we obtain that the functions f − f (z0 ) and f − f (z1 ) = f − f (z0 ) + [f (z0 ) − f (z1 )] have the same number of zeros in the U(z0 ; r) disc. Since f (z0 ) = 0, the function f − f (z0 ) has at least two zeros in U(z0 ; r), hence the function f − f (z1 ) also has at least two zeros in U(z0 ; r). From the fact that f (z1 ) ≠ 0, the number z1 is a simple zero for f − f (z1 ). Hence, there exists a point z2 ∈ U(z0 ; r) \ {z1 } which is a zero for f − f (z1 ), and thus f (z2 ) = f (z1 ), with z2 ≠ z1 . These last relations contradict the injectivity of f . Remarks 6.2.1. 1. If f ∈ Hu (D), where D ⊂ ℂ is a domain, then the function f is a conformal mapping of D. 2. Let D ⊂ ℂ be a domain. Contrary as in the case of real functions, the condition f (z) ≠ 0, ∀z ∈ D is a necessary, but not a sufficient condition for the injectivity of the function f . For example, the function f (z) = ez , z ∈ ℂ, is not injective in ℂ, but f (z) ≠ 0, ∀z ∈ ℂ. Theorem 6.2.2 (Pompeiu theorem). Let f ∈ H(D), where D ⊂ ℂ is a domain, and let z0 ∈ D. Then, in any disc U(z0 ; r) ⊂ D there exist two distinct points z1 , z2 ∈ U(z0 ; r), such that f (z1 ) − f (z2 ) = f (z0 )(z1 − z2 ). Proof. 1. Suppose that f (z0 ) = 0. If f ≡ 0, then the conclusion is evident. If f ≢ 0, then f and f are not constant functions, and similarly like in the proof of Theorem 6.2.1 there exist two distinct points z1 , z2 ∈ U(z0 ; r), such that f (z1 ) − f (z2 ) = 0.
6.2 Univalent functions  177
2. Suppose that f (z0 ) ≠ 0. Let us define the function g(z) = f (z) − f (z0 )z, z ∈ D. Thus, g (z0 ) = 0. According to the previous step, there exist two points z1 , z2 ∈ U(z0 ; r), z1 ≠ z2 , such that g(z1 ) = g(z2 ), so it follows that f (z1 ) − f (z2 ) = f (z0 )(z1 − z2 ). Theorem 6.2.3. Let f ∈ H(U(z0 ; r)), where z0 ∈ ℂ and r > 0. If f (z) − f (z0 ) < f (z0 ),
∀z ∈ U(z0 ; r) \ {z0 },
(6.1)
then the function f is univalent in U(z0 ; r). Proof. Let z1 , z2 ∈ U(z0 ; r) such that z1 ≠ z2 . Let us denote by γ the linear path that connects the points z1 and z2 , i. e., γ(t) = (1 − t)z1 + tz2 , t ∈ [0, 1]. Hence, {γ} = [z1 , z2 ] ⊂ U(z0 ; r), and f (z1 ) − f (z2 ) = f (γ(0)) − f (γ(1)) = ∫ f (ζ )dζ γ
≥ ∫ f (z0 )dζ − ∫(f (ζ ) − f (z0 ))dζ . γ γ
Since
and
(6.2)
∫ f (z0 )dζ = f (z0 )z1 − z2  γ ∫(f (ζ ) − f (z0 ))dζ ≤ z1 − z2  sup{f (ζ ) − f (z0 ) : ζ ∈ [z1 , z2 ]}, γ
from the inequality (6.2) it follows that f (z1 ) − f (z2 ) ≥ z1 − z2 (f (z0 ) − sup{f (ζ ) − f (z0 ) : ζ ∈ [z1 , z2 ]}).
(6.3)
1. If there exists a ∈ [z1 , z2 ] \ {z0 } such that f (a) − f (z0 ) = sup{f (ζ ) − f (z0 ) : ζ ∈ [z1 , z2 ]}, from the relations (6.3) and (6.1) we get f (z1 ) − f (z2 ) ≥ z1 − z2 (f (z0 ) − f (a) − f (z0 )) > 0, hence f is univalent in U(z0 ; r). 2. If z0 ∈ [z1 , z2 ] and 0 = f (z0 ) − f (z0 ) = sup{f (ζ ) − f (z0 ) : ζ ∈ [z1 , z2 ]}, then f (ζ ) = f (z0 ), ∀ζ ∈ [z1 , z2 ]. From here, according to a previous theorem, it follows that f (ζ ) = f (z0 ), ∀ζ ∈ U(z0 ; r), thus f (ζ ) = f (z0 ) + f (z0 )(ζ − z0 ),
∀ζ ∈ U(z0 ; r).
Since from (6.1) we have f (z0 ) ≠ 0, using the above relation we deduce that f is univalent in U(z0 ; r).
178  6 Conformal representations Theorem 6.2.4. Let f ∈ H(D), where D ⊂ ℂ is a simply connected domain. Suppose that there exists a function g ∈ Hu (D), such that (i)
g(D)
and (ii)
Re
is a convex domain
f (z) > 0, g (z)
∀z ∈ D.
Then the function f is univalent in D, i. e., f ∈ Hu (D). Proof. Let us denote Δ = g(D). Since g ∈ Hu (D), there exists g −1 : Δ → D and g −1 ∈ Hu (Δ). Thus, h = f ∘ g −1 ∈ H(Δ), and Re h (w) = Re
f (z) > 0, g (z)
∀w = g(z) ∈ Δ.
Let us consider w1 , w2 ∈ Δ, w1 ≠ w2 . The set Δ = g(D) is convex, hence [w1 , w2 ] ⊂ Δ. Computing the next integral on this segment, we have w2
1
h(w2 ) − h(w1 ) = ∫ h (w)dw = (w2 − w1 ) ∫ h [w1 + t(w2 − w1 )]dt, w1
0
which implies 1
Re
h(w2 ) − h(w1 ) = ∫ Re h [w1 + t(w2 − w1 )]dt > 0. w2 − w1 0
Hence, the function h is injective in Δ, so it follows that f = h ∘ g is also injective in D. Corollary 6.2.1. If D ⊂ ℂ is a convex domain and the function f ∈ H(D) satisfies the condition Re f (z) > 0, ∀z ∈ D, then f is injective in D, i. e., f ∈ Hu (D). Corollary 6.2.2. Let U = U(0; 1) and let f ∈ H(U), such that Re[(1 − z 2 )f (z)] > 0,
∀z ∈ U.
Then the function f is univalent in U, i. e., f ∈ Hu (U). 1+z , z ∈ U, where we choose the principal Proof. Let us define the function g(z) = log 1−z branch of the logarithmic function, i. e., log 1 = 0. According to the fourth point of Example 6.2.1, we have g ∈ Hu (U) and the image
g(U) = {w ∈ ℂ : −
π π < Im w < } 2 2
6.3 The problem of conformal representation 
179
is a convex domain. Since Re
f (z) 1 = Re[(1 − z 2 )f (z)] > 0, g (z) 2
∀z ∈ U,
from the above corollary we deduce f ∈ Hu (U). Theorem 6.2.5 (Hurwitz theorem). Let D ⊂ ℂ be a domain, and let (fn )n∈ℕ ⊂ Hu (D) be a sequence of univalent functions in D, that uniformly converges on compacts to f in D, and the function f is not constant. Then f ∈ Hu (D), i. e., the function f is univalent in D. Proof. From the Weierstrass theorem, we have f ∈ H(D). Suppose that the function f is not injective, then there exist the points z1 , z2 ∈ D, z1 ≠ z2 , such that f (z1 ) = f (z2 ) = w0 . Since f is not constant, according to the theorem of the zeros of holomorphic functions, there exists a number r > 0 such that U(z1 ; r) ∪ U(z2 ; r) ⊂ D,
U(z1 ; r) ∩ U(z2 ; r) = 0,
and ∀z ∈ 𝒞1 ∪ 𝒞2 ,
where 𝒞1 = 𝜕U(z1 ; r), 𝒞2 = 𝜕U(z2 ; r) ⇒ f (z) − w0 ≠ 0.
Let m = min{f (z)−w0  : z ∈ 𝒞1 ∪ 𝒞2 } > 0. Since the function sequence (fn )n∈ℕ uniformly converges on compacts to f in D, there exists n0 ∈ ℕ such that ∀n > n0 ,
∀z ∈ 𝒞1 ∪ 𝒞2 ⇒ fn (z) − f (z) < m.
Hence, ∀n > n0 and ∀z ∈ 𝒞1 ∪ 𝒞2 we have fn (z) − f (z) < f (z) − w0 , and according to the Rouché theorem, we get that ∀n > n0 , the functions f − w0 and fn − w0 = f − w0 + (fn − f ) have the same number of zeros in U(z1 ; r) and U(z2 ; r), respectively. It follows that ∀n > n0 the function fn − w0 has zeros in both of the above discs. From here, if ∀n > n0 then the functions fn cannot be injective, which contradicts the assumption.
6.3 The problem of conformal representation According to the principle of the domain conservation, we have: if D is a domain and f ∈ H(D), then Δ = f (D) is also a domain. The function f is conformal in every points z ∈ D with f (z) ≠ 0. From the geometric interpretation of the derivative, It is well known that if f belongs to the class C 1 (i. e., the real and the imaginary parts of f have continuous firstorder partial derivatives), and it is first and second type conformally in a point z, then we have f (z) ≠ 0. If the function f is univalent in the domain D, then f is a homeomorphism between D and Δ = f (D), and f −1 is univalent in the domain Δ. From a previous result, for all f ∈ Hu (D) the relation f (z) ≠ 0, ∀z ∈ D holds, hence (f −1 ) (w) ≠ 0, ∀w ∈ Δ. So, we deduce that f is a conformal mapping in all the points of D, while f −1 is a conformal mapping in all the point of Δ.
180  6 Conformal representations Definition 6.3.1. 1. The univalent function f in the domain D ⊂ ℂ is said to be a conformal isomorphism between the domains D and Δ = f (D). 2. The domains D and Δ are called to be conformally equivalent, if there exists a conformal isomorphism between D and Δ. 3. The conformal isomorphisms of the domain D to itself are called the conformal automorphisms of D. 4. The set of all the conformal automorphisms of the domain D together with the function composition represents a group, is called the conformal group of D, and it is denoted by A(D). Remarks 6.3.1. 1. The direct problem of the conformal representation is: for a given function f ∈ Hu (D), determine the image f (D) (i. e., the image of the domain D by the function f ). 2. The inverse problem of the conformal representation, that in fact is the problem of the conformal representation: Let D and Δ two given domains. Determine, if there exists, a conformal representation between these domains, i. e., determine, if there exists, a function f ∈ Hu (D) such that f (D) = Δ. A such kind of conformal mapping does not exist in every case. For example, if D is a simply connected domain, but the domain Δ is not simply connected, the above problem has no solution, because for all D simply connected domain the image f (D) is also simply connected, whenever f ∈ Hu (D). Really, if γ is an arbitrary closed path in f (D), then f −1 ∘ γ is a closed path in D, because f −1 is a homeomorphism. If φ is the continuous deformation in D of the closed path f −1 ∘ γ to the constant path ez0 , then f ∘ φ is the continuous deformation in f (D) of the path γ to the constant path ef (z0 ) . Hence, the domain f (D) is a simply connected domain. Theorem 6.3.1. If the function f0 is a conformal representation of the domain D onto the domain Δ, then the set of all the conformal representations of D onto Δ is given by {f : f = φ ∘ f0 , φ ∈ A(Δ)}, where A(Δ) represents the set of the conformal automorphisms of Δ. Proof. Let the function f be such a representation. Then φ = f ∘ f0−1 ∈ A(Δ), hence f = φ ∘ f0 . Conversely, if φ ∈ A(Δ), then the function f = φ ∘ f0 is a conformal representation of D onto Δ. Theorem 6.3.2. Every conformal representation f : D → Δ induces an isomorphism function f ∗ between the groups A(D) and A(Δ), where f ∗ : A(D) → A(Δ),
f ∗ (φ) = f ∘ φ ∘ f −1 ,
φ ∈ A(D).
6.4 The Riemann mapping theorem
 181
Proof. If φ ∈ A(D), then f ∗ (φ) ∈ A(Δ). Conversely, if ψ ∈ A(Δ), then φ = f −1 ∘ψ∘f ∈ A(D) and f ∗ (φ) = ψ. If f ∗ (φ1 ) = f ∗ (φ2 ), then φ1 = φ2 , hence the function f ∗ is a bijection. If φ1 , φ2 ∈ A(D), then f ∗ (φ1 ) ∘ f ∗ (φ2 ) = (f ∘ φ1 ∘ f −1 ) ∘ (f ∘ φ2 ∘ f −1 ) = f ∘ (φ1 ∘ φ2 ) ∘ f −1 = f ∗ (φ1 ∘ φ2 ), hence the function f ∗ is a homomorphism. Theorem 6.3.3. Denote by U = U(0; 1) the unit open disc. Then the functions of the conformal group A(U) (i. e., all the conformal isomorphisms of the unit disc U onto itself) are of the form: φ(z) = eiθ
z−a , 1 − az
θ ∈ ℝ, a ∈ U.
(6.4)
Proof. All the functions φ of the form (6.4) are conformal automorphisms of U like we know from the theory of the circular transforms. Let f ∈ A(U), and denote f (0) = w0 . Let us define the function g = φ ∘ f , where φ ∈ A(U) is given by φ(w) =
w − w0 , 1 − w0 w
w ∈ U.
Since g(0) = 0 and g(z) < 1, ∀z ∈ U, from the Schwarz lemma we get g(z) ≤ z, for all z ∈ U. The function g −1 = f −1 ∘ φ−1 also satisfies the conditions of the assumption of the Schwarz lemma, hence g −1 (w) ≤ w, ∀w ∈ U. Letting w = g(z), then z ≤ g(z), ∀z ∈ U, and hence g(z) = z, ∀z ∈ U. According to the Schwarz lemma, the last equality holds if and only if g(z) = eiθ z, θ ∈ ℝ. Since g(z) = φ(f (z)), then f (z) = φ−1 (eiθ z). Letting a = −e−iθ w0 , from the relation z+w φ−1 (z) = 1+w 0z we have that 0
φ−1 (eiθ z) =
−iθ eiθ z + w0 z−a iθ z + w0 e = e = eiθ , 1 − az −iθ 1 + w0 eiθ z 1 + w0 e z
and it follows f (z) = φ−1 (eiθ z) = eiθ
z−a . 1 − az
6.4 The Riemann mapping theorem Let us denote by U the open unit disc, i. e., U = U(0; 1), and let D ⊂ ℂ be a simply connected domain. Let us put the next natural question: did there exist a conformal representation of D onto U?
182  6 Conformal representations Theorem 6.4.1. The complex plane ℂ and the unit disc U = U(0; 1) are not conformally equivalent. Proof. If f ∈ H(ℂ) and f (ℂ) = U, then f (z) < 1, ∀z ∈ ℂ, and from the Liouville theorem we have f (z) = c, ∀z ∈ ℂ, hence the function f is not univalent. Theorem 6.4.2. If D is a simply connected domain of ℂ, and if the function f0 : D → U is a conformal representation of D onto U, then all the conformal representations of the domain D onto the unit disc U have the form: f (z) = eiθ
f0 (z) − a , 1 − af0 (z)
θ ∈ ℝ, a ∈ U.
Proof. This result is an immediate consequence of Theorem 6.3.1 and Theorem 6.3.3. Theorem 6.4.3 (Riemann theorem). Every simply connected domain D ⊂ ℂ, where D ≠ ℂ, is conformally equivalent with the unit disc U. Proof. 1. Let z0 ∈ D be a fixed point, and let us define the ℱ set by ℱ = {f ∈ Hu (D) : f (z0 ) = 0, f (D) ⊂ U}.
We will prove that ℱ is a nonempty set. Since D ≠ ℂ, let a ∈ ℂ \ D. From the multivalued functions analytical branches existence theorem, there exists a function g ∈ H(D) such that (g(z))2 = z − a, ∀z ∈ D 1 (i. e., the function g is a branch of the multivalued function (z −a) 2 , where the function z−a does not vanish in D). If z1 , z2 ∈ D and g(z1 ) = ±g(z2 ), then (g(z1 ))2 = (g(z2 ))2 , hence z1 = z2 . From here, it follows that g ∈ Hu (D). Since g(D) is a domain, there exists a number r > 0 such that U(g(z0 ); r) ⊂ g(D). We will prove that U(−g(z0 ); r) ∩ g(D) = 0. Contrary, if it is not true, i. e., U(−g(z0 ); r) ∩ g(D) ≠ 0, then ∃z1 ∈ D : g(z1 ) ∈ U(−g(z0 ); r) ⇒ g(z1 ) + g(z0 ) < r ⇔ − g(z1 ) − g(z0 ) < r ⇔ −g(z1 ) ∈ U(g(z0 ); r) ⊂ g(D). Then ∃z2 ∈ D such that g(z2 ) = −g(z1 ), and from the above proved property of g we deduce that z1 = z2 . Hence, g(z1 ) = −g(z1 ), i. e., g(z1 ) = 0, and from here we get 0 = (g(z1 ))2 = z1 − a, so z1 = a. This contradicts the fact that a ∈ ℂ \ D. r is welldefined on the domain D (since It follows that the function g1 = g+g(z 0) g(z) ≠ −g(z0 ), ∀z ∈ D, because g(D) ∩ U(−g(z0 ); r) = 0). Moreover, the function g1 is univalent on D, because g is univalent. Further, g1 (D) ⊂ U, because g(z) + g(z0 ) ≥ r, ∀z ∈ D, thus g1 (z) ≤ 1, ∀z ∈ D. Since g1 (D) is an open set, it follows that g1 (D) ⊂ U.
6.4 The Riemann mapping theorem
 183
Let us choose the function φ ∈ A(U) such that φ(w) =
w − g1 (z0 )
1 − g1 (z0 )w
.
Then φ(g1 (z0 )) = 0, and the function g2 = φ ∘ g1 is univalent on the domain D. Also, g2 (z0 ) = 0, g2 (D) ⊂ U, i. e., g2 ∈ ℱ , and thus ℱ ≠ 0. 2. To all functions f ∈ ℱ , it corresponds the strictly positive number f (z0 ). Since this number represents the coefficient of the linear deformation of f in z0 , we expect that the domain f (D) will be maximal if f (z0 ) will be maximal. This reason will be used for the following steps of the proof. Let M = sup{f (z0 ) : f ∈ ℱ }. From the definition, the set ℱ is a bounded function set, and by a known theorem the set ℱ = {f : f ∈ ℱ } will be also bounded. Hence, M < +∞. From the definition of M, we have ∀n ∈ ℕ∗ , ∃fn ∈ ℱ : M −
1 < f (z ) ≤ M, n n 0
hence limn→+∞ fn (z0 ) = M. Since ℱ is bounded, from the Montel theorem, we have that ℱ is relatively compact in H(D). Thus, the function sequence (fn )n∈ℕ∗ contains the subsequence (fnk )k∈ℕ∗ that uniformly converges on compacts to a function f0 ∈ H(D). Thus, let f0 (z0 ) = limk→+∞ fnk (z0 ) = 0. According to the Weierstrass theorem, the function sequence (fnk )k∈ℕ∗ uniformly converges on compacts to the function f0 , hence M = limk→+∞ fnk (z0 ) = f0 (z0 ). The function f0 cannot be constant on D, since contrary, from f0 (z0 ) = 0 it follows that f0 ≡ 0. Thus, the relations f0 ≡ 0 and f0 (z0 ) = M > 0 are in contradiction. From the Hurwitz theorem, we get that f0 ∈ Hu (D). On the other hand, fnk (z) < 1, ∀z ∈ D, ∀k ∈ ℕ∗ , and by letting k → +∞ we deduce that f0 (z) ≤ 1, ∀z ∈ D. Since the function f0 is not constant, from the theorem of the module maximum of the holomorphic functions, we obtain that the function f0  cannot take its maximum value on the domain D, thus it follows that f0 (z) < 1, ∀z ∈ D, hence f0 (D) ⊂ U. So, we obtained that f0 ∈ ℱ and f0 (z0 ) = M = sup{f (z0 ) : f ∈ ℱ }. 3. In the last step, we will prove f0 (D) = U. Let us suppose that this is not true. Then ∃α ∈ U \ f0 (D), and let us define the function ψ ∈ A(U) as follows: ψ(w) =
w−α , 1 − αw
w ∈ U.
The function ψ ∘ f0 does not vanish in D, because f0 (z) ≠ α, ∀z ∈ D. So, we may use the multivalued functions analytical branches existence theorem, which implies
184  6 Conformal representations that: there exists a function h holomorphic in the domain D, such that h2 = ψ ∘ f0 . Since (ψ ∘ f0 )(D) ⊂ U, it follows that h(D) ⊂ U. Moreover, h ∈ Hu (D), because if this is not true, then the function ψ ∘ f0 will not be univalent, which is a contradiction with the facts that ψ ∈ A(U) and f0 ∈ ℱ . Further, we have h(z0 )2 =  − α = α and 2h(z0 )h (z0 ) = ψ (f0 (z0 ))f0 (z0 ) = ψ (0)f0 (z0 ) w − α =( ) f0 (z0 ) = (1 − α2 )f0 (z0 ), 1 − αw w=0 i. e., 2 1 − α2 1 − α 1 ψ (0)f0 (z0 ) = M. f0 (z0 ) = h (z0 ) = 2h(z0 ) 2√α 2√α
(6.5)
We will define the function χ ∈ A(U) as follows: χ(w) =
w − h(z0 )
1 − h(z0 )w
,
w ∈ U.
The function f = χ ∘ h will be univalent in the domain D, and f (D) ⊂ U, f (z0 ) = 0. Hence, f ∈ ℱ . On the other hand, f (z0 ) = χ (h(z0 ))h (z0 ) =
1 − h(z0 )w + h(z0 )(w − h(z0 )) 1 h (z0 ) = h (z0 ), 2 1 − h(z0 )2 w=h(z0 ) (1 − h(z0 )w)
and thus, using the inequality (6.5), we have f (z0 ) =
1 1 − α2 1 + α M= M. 1 − α 2√α 2√α
But 1 + α > 2√α, and we deduce that f (z0 ) > M, which contradicts the choice of M. This contradiction proves that f0 (D) = U. Corollary 6.4.1. If D is a simply connected domain in ℂ with D ≠ ℂ, and if z0 ∈ D, then there exists a unique function f ∈ Hu (D) such that f (z0 ) = 0 and f (z0 ) = 1, which conformally maps the domain D onto an open disc with the center at the origin 0. The radius of this open disc is called the conformal radius of the domain D at the z0 point. Proof. Let the function f0 be a conformal representation of the domain D onto the unit disc U that satisfies the conditions f0 (z0 ) = 0 and f0 (z0 ) > 0. We will show that there exists a such a function, and this is the unique function with these properties. Let g ∈ Hu (D) a function with g(z0 ) = 0 and g(D) = U. Then
6.4 The Riemann mapping theorem
 185
all the other functions that satisfy the above properties are of the form f = eiθ g, θ ∈ ℝ g−a (because from Theorem 6.4.2, we have f = eiθ 1−ag , and from f (z0 ) = 0 we get a = 0). On
the other hand, f (z0 ) = eiθ g (z0 ) and g (z0 ) ≠ 0, because the function g is univalent. Supposing that f (z0 ) > 0, then arg(eiθ g (z0 )) = 0, hence θ + arg g (z0 ) = 0, or θ = − arg g (z0 ). From here, it follows immediately that there exists a such a function f0 , and this is the unique function with these properties. Let R = f (z1 ) . Then the function f = Rf0 conformally maps the domain D onto 0 0 the disc U(0; R). The uniqueness of the function f follows from the uniqueness of f0 , because if f satisfies the assumptions of the Corollary 6.4.1, then the function R1 f will satisfy the previous conditions on f0 , hence R1 f = f0 . Let D be a simply connected domain, such that 0 ∈ D and D ≠ ℂ. Let us define the function set ℱ = {f ∈ H(D) : f (0) = 0, f (0) = 1},
which is nonempty, because in contains the function f (z) = z. Let us introduce the functional M : ℱ → [0, +∞], by M(f ) = sup{f (z) : z ∈ D},
f ∈ ℱ.
Corollary 6.4.2. The infimum value of the functional M on the domain D is the conformal radius of the domain D in the origin 0, denoted by R. The infimum of the functional M is attained for that function f1 ∈ Hu (D) which conformally transforms the domain D onto the disc U(0; R), i. e., M(f1 ) = R = min{M(f ) : f ∈ ℱ }. Proof. Let f0 ∈ Hu (D) that unique conformal mapping of the domain D onto the unit disc U, which satisfies f0 (0) = 0 and f0 (0) > 0 (see the proof of Corollary 6.4.1). Letting g = f0−1 , we see that it is sufficient to search the infimum of the functional M on the subset of ℱ for which M(f ) < +∞, i. e., ℱ0 = {f ∈ ℱ : M(f ) < +∞}, because ∃f0 ∈ ℱ such that f0 (D) = U(0; R). 1 Let f ∈ ℱ0 be arbitrary, and let us define the function h = M(f f ∘ g. The function h ) is holomorphic in the unit disc U, and satisfies the conditions h(0) = 0 and 1 1 f (g(w)) = f (z) ≤ 1, h(w) = M(f ) M(f )
∀w ∈ U,
since g(w) = z ∈ D. From the fact that the set h(D) is open, we deduce that h(w) < 1, ∀w ∈ U. Thus, h(0) = 0 and h(w) < 1, ∀w ∈ U, and using the Schwarz lemma we get h(w) ≤ w,
∀w ∈ U and h (0) ≤ 1.
186  6 Conformal representations From here, using the relation h (0) = it follows that
1 1 1 f (g(0))g (0) = f (0) , M(f ) M(f ) f0 (0)
1 1 1 ≤ 1 ⇒ M(f ) ≥ = R, h (0) = f (0) M(f ) f0 (0) f0 (0)
where R is the conformal radius of the domain D in the origin 0. Hence, we obtained that ∀f ∈ ℱ0 we have M(f ) ≥ R, and this inequality holds also for all f ∈ ℱ , since ℱ0 ⊂ ℱ . But, the function f determined in the Corollary 6.4.1 belongs to ℱ , and for this function the relation M(f ) = R holds, because sup{f (z) : z ∈ D} = R. Remarks 6.4.1. 1. From the Riemann theorem, it follows that any two simply connected domains of ℂ, different to ℂ, is conformally equivalent (conformally isomorph). The simply connected domains of ℂ can be divided into two classes: the first one is the entire set ℂ, while the second ones consist of the all simply connected domains of ℂ, different to ℂ. 2. From the Riemann theorem and Theorem 6.3.2, we deduce the next: if D ≠ ℂ is a simply connected domain, then the group of all the conformal automorphisms of D, denoted by A(D), is isomorphic with A(U). Hence, according to the point 1 of these remarks, we deduce that the groups of conformal automorphisms of any two simply connected domains of ℂ, not equal to ℂ, are isomorph groups. Theorem 6.4.4. The functions of the group A(ℂ) of all the conformal automorphisms of ℂ, have the form f : ℂ → ℂ, where f (z) = az + b, a, b ∈ ℂ, a ≠ 0. Proof. 1. All the functions of f of the above form is a conformal automorphism of ℂ. 2. Conversely, let f ∈ A(ℂ). Then the function f is an entire function, so it may be expanded in Taylor series around the origin 0, i. e., f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ ,
∀z ∈ ℂ.
Let us consider the neighborhood V = ℂ∞ \ U(0; r) of the ∞ point, where r > 0 is an arbitrary number. Since f −1 (U(0; r)) is compact, then it is closed and bounded, hence we get that W = ℂ∞ \ f −1 (U(0; r)) is a neighborhood of ∞. But f (W ∩ ℂ) = f (ℂ \ f −1 (U(0; r))) = f (ℂ) \ U(0; r) ⊂ V ⇒ lim f (z) = ∞, z→∞
hence z = ∞ is a pole for f , and hence we deduce that 0 is a pole for the function φ(z) = f ( z1 ). But 1 1 1 φ(z) = f ( ) = a0 + a1 + a2 2 + ⋅ ⋅ ⋅ , z z z
∀z ∈ ℂ∗ ,
6.5 Exercises  187
and since z = 0 is its pole, the sum needs to have a finite number of nonvanishing terms, which implies that f (z) = a0 + a1 z + ⋅ ⋅ ⋅ + an z n ,
∀z ∈ ℂ,
hence f is a polynomial. If the degree of f is greater than 1, i. e., deg f ≥ 2, then deg f ≥ 1, hence the equation f (z) = 0 has at least a zero in ℂ. Then, according to a wellknown theorem, we have f ∉ A(ℂ). It follows that f is a first degree polynomial, i. e., f (z) = az + b, a ≠ 0.
6.5 Exercises Exercise 6.5.1. Consider the halfplanes D = {z ∈ ℂ : Re z > 0} and Ω = {z ∈ ℂ : Im z > 0}. 1. Prove that the function f : D → Ω, f (z) = iz maps conformally the halfplane D onto the halfplane Ω; 2. Prove that the function g : Ω → D, g(z) = −iz maps conformally the halfplane Ω onto the halfplane D. Exercise 6.5.2. Let D = {z ∈ ℂ : z < 1, Re z > 0} and
Ω = U(0; 1) \ [−1, 0].
Prove that the function f : D → Ω, f (z) = z 2 maps conformally the domain D onto the domain Ω. z−i Exercise 6.5.3. Let D = {z ∈ ℂ : Im z > 0}, Ω = U(0; 1) and g : ℂ \ {−i} → ℂ, g(z) = z+i . Prove that the function f = gD conformally maps the halfplane D onto the disc Ω.
Exercise 6.5.4. Prove that the function f (z) = i 1+z conformally maps the disc D = 1−z U(0; 1) onto the halfplane Ω = {z ∈ ℂ : Im z > 0}. z−i Exercise 6.5.5. Prove that the function f (z) = z+i conformally maps the domain D = {z ∈ ℂ : Re z < 0, Im z > 0} onto the domain Ω = {z ∈ ℂ : z < 1, Im z > 0}.
Exercise 6.5.6. Prove that the Joukowski function f (z) = 21 (z + z1 ) conformally maps the domain D = {z ∈ ℂ : z < 1, Im z > 0} onto the halfplane Ω = {z ∈ ℂ : Im z < 0}. Exercise 6.5.7. Let g : ℂ∗ → ℂ, g(z) = z + z1 , and let us define the domain D = {z ∈ ℂ : z > 1, Im z > 0}. Prove that the function f = gD conformally maps the domain D onto the halfplane Ω = {z ∈ ℂ : Im z > 0}.
188  6 Conformal representations Exercise 6.5.8. Determine a function that conformally maps the domain D = {z ∈ ℂ∞ : z > 1, Im z > 0} onto the disc Ω = U(0; 1). Exercise 6.5.9. Let us define the domains D = {z ∈ ℂ : z > 1} and Ω = ℂ \ [−1, 1]. Prove that the Joukowski function f : D → Ω, f (z) = 21 (z + z1 ), conformally maps the domain D onto the domain Ω. Exercise 6.5.10. Let D = {z ∈ ℂ : Re z > 0} and
Ω = ℂ \ {z ∈ ℂ : Im z = 0, Re z ≤ 0}.
Prove that the function f : D → Ω, f (z) = z 2 , conformally maps the halfplane D onto the domain Ω. Exercise 6.5.11. Determine a function that conformally maps the domain D = {z ∈ ℂ : z < 1, Im z > 0} onto the unit disc Ω = U(0; 1). Exercise 6.5.12. Determine one function that conformally maps the domain 3
D = {z ∈ ℂ : z < √2, 0 < arg z
0}. Exercise 6.5.13. Let a, b ∈ ℝ such that a < b and D = {z ∈ ℂ : a < Re z < b}. Determine a function that conformally maps the domain D onto the halfplane Ω = {z ∈ ℂ : Im z > 0}. Exercise 6.5.14. Let r > 0, a ∈ ℂ, with a = r, and let us define the domains D = {z ∈ ℂ : z < 2r, z − a > r} and Prove that the function f (z) =
z z−2a
1 Ω = {z ∈ ℂ : 0 < Re z < }. 2
conformally maps the domain D onto the domain Ω.
Exercise 6.5.15. Determine a function that conformally maps the domain D = {z ∈ ℂ : z < 2, z − i > 1} onto the halfplane Ω = {z ∈ ℂ : Im z > 0}. Exercise 6.5.16. Determine a function that conformally maps the domain D = {z ∈ ℂ : z < r, z + onto the halfplane Ω = {z ∈ ℂ : Im z > 0}.
r r > , Im z > 0} 2 2
6.5 Exercises  189
Exercise 6.5.17. Determine a function that conformally maps the domain D = {z ∈ ℂ : z < r, z −
r r > ,} 2 2
onto the halfplane Ω = {z ∈ ℂ : Im z > 0}. Exercise 6.5.18. Determine a function that conformally maps the domain D = {z ∈ ℂ : z < 1, 0 < arg z
0}. Exercise 6.5.20. Determine a function that conformally maps the domain D = ℂ \ {z ∈ ℂ : Im z = 0, Re z ≥ 1} onto the disc Ω = U(0; 1). Exercise 6.5.21. 1. Determine a function that conformally maps the set D = ℂ\T onto the set Ω = {z ∈ ℂ : Im z > 0} \ {i}, where T = [a, b] is a closed interval, that connects the points a and b, with a, b ∈ ℂ, a ≠ b; 2. Determine a function that conformally maps the set D = ℂ \ T onto the set Ω = {z ∈ ℂ : Im z > 0} \ {i}, where T = [1 + i, −3 − i] is a closed interval, that connects the points z0 = 1 + i and z1 = −3 − i; 3. Determine a function that conformally maps the set D = ℂ\T onto the set Ω = {z ∈ ℂ : Im z > 0} \ {i}, where T = [−i, i] is a closed interval, that connects the points z0 = −i and z1 = i; 4. Determine a function that conformally maps the set D = ℂ \ [−1, 1] onto the set Ω = {z ∈ ℂ : Im z > 0} \ {i}, where T = [−1, 1] is a closed interval, that connects the points z0 = −1 and z1 = 1. Exercise 6.5.22. Determine a function that conformally maps the domain D = {z ∈ ℂ : z − i > 1, Im z > 0} onto the halfplane Ω = {z ∈ ℂ : Im z > 0}.
190  6 Conformal representations Exercise 6.5.23. Determine a function that conformally maps the domain D = {z ∈ ℂ : 0 < arg z < π3 } onto the domain Ω = {z ∈ ℂ : 0 < Im z < π2 }. Exercise 6.5.24. Let a ∈ ℂ and α ∈ U(a; r). Let us denote by α∗ the inverse of the point α with respect to the circle 𝜕U(a; r). z−α 1. Prove that the function f (z) = z−α ∗ conformally maps the disc D = U(a; r) onto the ); disc Ω = U(0; α−a r z−α ∗ 2. Prove that the function f (z) = z−α ∗ conformally maps the set D = ℂ\(U(a; r)∪{α }) ) ∪ {1}). onto the set Ω = ℂ \ (U(0; α−a r Exercise 6.5.25. Prove that the Joukowski function f (z) = 21 (z + z1 ) conformally maps the domain D = U(0; 1) \ {0} onto the domain Ω = ℂ \ [−1, 1]. Exercise 6.5.26. Determine a function that conformally maps the domain D = U(0; 1)\ ({0} ∪ [ 21 , 1)) onto the domain Ω = ℂ \ [−1, 45 ]. Exercise 6.5.27. For a given a ∈ (0, 1), determine a function that conformally maps the domain D = U(0; 1) \ ((−1, 0] ∪ [a, 1)) onto the halfplane Ω = {z ∈ ℂ : Im z > 0}. Exercise 6.5.28. Determine a function that conformally maps the set D = U(0; 1) \ ({0} ∪ [ 21 , 1)) onto the set Ω = {z ∈ ℂ : Im z > 0} \ {i}. Exercise 6.5.29. Determine a function that conformally maps the set D = {z ∈ ℂ : z − 3 > 9, z − 8 < 16} onto the circular ring Ω = U(0; 32 , 1). Exercise 6.5.30. Determine a function that conformally maps the set D = {z ∈ ℂ : z > 1, z +
1 5 0} \ (0, i] onto the halfplane Ω = {z ∈ ℂ : Im z > 0}. Exercise 6.5.32. Let α ∈ ℝ, with 0 < α < 1, and let D = {z ∈ ℂ : z < 1, Im z > 0} \ (0, αi]. Determine a function that conformally maps the domain D onto the halfplane Ω = {z ∈ ℂ : Im z > 0}.
7 Solutions to the chapterwise exercises 7.1 Solutions to the exercises of Chapter 1 Solution of Exercise 1.6.1 1. We will use the algebraic form of the complex numbers, i. e., z = x+iy, where x, y ∈ ℝ. Since z + i = x + (y + 1)i, it follows that Im(z + i) = y + 1, and thus {z ∈ ℂ : Im(z + i) = 0} = {z = x + iy ∈ ℂ : y = −1}. Hence {z ∈ ℂ : Im(z + i) = 0} is the line parallel with Ox that contains the point M(−i). 2. Since z − 2i = x + (y − 2)i, it follows that arg(z − 2i) =
π y−2 ⇔ = 1, 4 x
x > 0 ⇔ x − y + 2 = 0,
x > 0.
Hence {z ∈ ℂ : arg(z − 2i) =
π } = {z = x + iy ∈ ℂ : x − y + 2 = 0, x > 0}, 4
that represents an open halfline. 3. If {z ∈ ℂ : z − α = z − β}, where α = −i and β = 4 + i, then z − α = z − β ⇔ x + (y − 1)i = x + 4 + (y + 1)i ⇔ y + 2x + 4 = 0. Hence {z ∈ ℂ : z − α = z − β}, where α = −i and β = 4 + i, is the orthogonal line on the segment AB, where A(α), B(β), and contains the middle point of the segment [AB]. 4. Since z + i = 2 ⇔ x + (y + 1)i = 2 ⇔ x2 + (y + 1)2 = 4, it follows that {z ∈ ℂ : z + i = 2} = 𝜕U(−i; 2), i. e., x2 + (y + 1)2 = 4. 5. The set {z ∈ ℂ : z − 2i + z + 4i = 10} is an ellipse, because https://doi.org/10.1515/9783110657869007
192  7 Solutions to the chapterwise exercises x2 (y + 1)2 z − 2i + z + 4i = 10 ⇔ x + (y − 2)i + x + (y + 4)i = 10 ⇔ 2 + = 1. 4 52 From here, {z ∈ ℂ : z − 2i + z + 4i = 10} = {z = x + iy ∈ ℂ :
x2 (y + 1)2 + = 1}. 42 52
6. If z = x + iy, x, y ∈ ℝ, then z − 1 x − 1 + yi (x2 + y2 − 1) + 2yi = = , z + 1 x + 1 + yi (x + 1)2 + y2 and thus arg Since we have
2
1 z−1 π 2y 4 = √3 ⇒ x2 + (y − = ⇒ 2 ) = . 2 √3 z+1 3 3 x +y −1
2y (x+1)2 +y2
{z ∈ ℂ : arg
> 0, we deduce that y > 0. Consequently, we obtain 2
z−1 π 1 4 ) = , y > 0} = } = {z = x + iy ∈ ℂ : x2 + (y − √3 z+1 3 3
that is the arc of the circle x2 + (y − 7. Clearly, {z ∈ ℂ : −
1 2 ) √3
=
4 3
contained in the upper halfplane.
π π π π < arg(z − i) < } = Si (− , ). 4 6 4 6
Since arg(z − i) =
π y − 1 √3 ⇔ = , 6 x 3
x > 0 ⇔ 3y − √3x − 3 = 0,
x > 0,
and arg(z − i) = −
π y−1 ⇔ = −1 4 x
x > 0 ⇔ y + x − 1 = 0,
x > 0,
it follows that Si (− π4 , π6 ) is the angular sector with the apex in i, bounded by the halflines 3y − √3x − 3 = 0, x > 0 and y + x − 1 = 0, x > 0. 8. Since 1 − 2i1 (z − z) = 1 − Im z, it follows that 1 z < 1 − Im z ⇔ x2 + 2(y − ) < 0, 2
y ≤ 1,
and thus {z ∈ ℂ : z < 1 −
1 (z − z)} = {z = x + iy ∈ ℂ : x2 + 2y − 1 < 0}. 2i
7.1 Solutions to the exercises of Chapter 1  193
This is the open set of the complex plane ℂ bounded by the parabola x2 + 2y − 1 = 0, which contains the origin O. 9. We deduce immediately that z 3 z < 1 ⇔ z < z + 3i ⇔ y > − , − }. z + 3i 2 2 1 − z 9 5 2 > 3 ⇔ 1 − z > 31 + z ⇔ (x + ) + y < , 1 + z 4 16
2 1 − z 9 5 2 {z ∈ ℂ : > 3} = {z = x + iy ∈ ℂ : (x + ) + y < }. 1 + z 4 16
11. If z = x + iy, where x, y ∈ ℝ, then
(x + 1)2
z − 4 − z + 6 − 9 > 0 ⇔
92 2
−
y2 19 4
> 1,
x < −1.
From here, {z ∈ ℂ : z − 4 − z + 6 − 9 > 0} = {z = x + iy ∈ ℂ : that is the exterior of the hyperbola
(x+1)2 81 4
−
y2 19 4
(x + 1)2 81 4
−
y2 19 4
> 1, x < −1},
= 1 that lies on the halfplane x < −1.
Solution of Exercise 1.6.2 Using the fact that Im z =
1 (z 2i
− z), we obtain
z − a 2 z−az−a z−a z−a ( )−1= −1 −1= z − a z−a z−a z−az−a (a − a)(z − z) 2i Im a ⋅ 2i Im z −4 Im a Im z = = = . z − a2 z − a2 z − a2 Since z − a2 > 0 for all z ∈ ℂ \ {a}, we get z − a z − a 2 Im a Im z > 0 ⇔ − 1 < 0 ⇔ < 1. z − a z − a
We will use a similar method to prove the points 2 and 3.
194  7 Solutions to the chapterwise exercises Solution of Exercise 1.6.3 Like in the previous proof, we have z − a 2 z−a z−a z−az−a ( )−1= −1= −1= z + a z+a z+a z+az+a −(a + a)(z + z) −4 Re a Re z = = , z + a2 z + a2 where we used the relation Re z = 21 (z + z). Since z + a2 > 0 for all z ∈ ℂ \ {−a}, we get z − a 2 z − a < 1. Re a Re z > 0 ⇔ − 1 < 0 ⇔ z + a z + a The proofs of the next two points are similar. Solution of Exercise 1.6.4 1. Let D = {z ∈ ℂ : Im z > 0}. Then ∀z ∈ D ⇒ z ≠ a, because Im a > 0 ⇒ a ∈ D ⇒ a ∉ D. Hence from Exercise 1.6.2, z − a ∀z ∈ D ⇔ Im z Im a > 0⇔f (z) = < 1 ⇔ f (z) ∈ U, z − a so the function f is welldefined. (i) For the proof of the injectivity, we have f (z1 ) = f (z2 ),
∀z1 , z2 ∈ D ⇔
z1 − a z2 − a = z1 − a z2 − a
⇔ z1 (a − a) − z2 (a − a) = 0 ⇔ (a − a)(z1 − z2 ) = 0 ⇔ z1 = z2 , because a − a = 0 ⇔ a = a ⇔ Im a = 0, but from the assumption we have Im a > 0. (ii) For the proof of the surjectivity, we will show that for all w ∈ U ≡ U(0; 1) there exists z ∈ D, such that f (z) = w, i. e., f (z) = w ⇔
aw − a z−a =w⇔z= . z−a w−1
We need to prove that the values of z belong to D. Since Im z = =
1 aw − a aw − a z−z = ( − ) 2i 2i w − 1 w−1
(a − a)(w2 − 1) − Im a(w2 − 1) = > 0, 2iw − 12 w − 12
7.1 Solutions to the exercises of Chapter 1  195
because we have from the assumption that Im a > 0, w − 12 > 0, and w ∈ U ⇔ w < 1 ⇔ w2 − 1 < 0, ∀w ∈ U. Hence z ∈ D, and consequently the function f is bijective. 2. We will use the same method like in the previous point. Using the notation D = {z ∈ ℂ : Re z > 0} and U = U(0, 1), then ∀z ∈ D ⇒ z ≠ −a, because Re a > 0 ⇒ a ∈ D ⇒ a ∈ D ⇒ −a ∉ D. Hence ∀z ∈ D ⇔ Re z Re a > 0
Exercise
⇔
1.6.3
z − a f (z) = < 1 ⇔ f (z) ∈ U, z + a
so the function f is welldefined. For the proof of the injectivity, we have f (z1 ) = f (z2 ),
∀z1 , z2 ∈ D ⇔
z1 − a z2 − a = z1 + a z2 + a
⇔ az1 − az2 − az2 + az1 = 0 ⇔ (a + a)(z1 − z2 ) = 0 ⇔ z1 = z2 , because a + a = 0 ⇔ Re a = 0, but from the assumption Re a > 0. For the proof of the surjectivity, we will show that for all w ∈ U ≡ U(0; 1) there exists z ∈ D, such that f (z) = w, i. e., f (z) = w ⇔ z =
aw + a . w−1
Now we will prove that z ∈ D. Since Re z = =
z + z 1 aw + a aw + a = ( + ) 2 2 1−w 1−w
(a + a)(1 − w2 ) Re a(1 − w2 ) = > 0, 21 − w2 1 − w2
because from the assumption Re a > 0, w ∈ U ⇔ w < 1 ⇔ 1−w2 > 0 and 1−w2 > 0, ∀w ∈ U. Hence z ∈ D, and consequently the function f is bijective. Solution of Exercise 1.6.5 Let zk = xk + yk , k ∈ {1, 2} and z = x + iy. Then A ∈ A1 A2 ⇔
x − x1 y − y1 = ⇔ x2 − x1 y2 − y1
z+z 1 − z1 +z 2 2 z2 +z2 1 − z1 +z 2 2
=
z−z − z12i−z1 2i z2 −z2 − z12i−z1 2i
z − z1 + z − z1 z − z1 − z − z1 (7.1) z − z1 = = z2 − z1 + z2 − z1 z2 − z1 − z2 − z1 z2 − z1 z − z1 + z − z1 z − z1 z − z1 (7.2) z − z1 z − z1 ⇔ = = ⇔ = , z2 − z1 + z2 − z1 z2 − z1 z2 − z1 z2 − z1 z2 − z1 ⇔
196  7 Solutions to the chapterwise exercises where we used that z+z , 2
x = Re z =
y = Im z =
z−z 2i
and a c a+c = = b d b+d a c a−c = = . b d b−d
(7.1) (7.2)
Solution of Exercise 1.6.6 Using the fact z = z ⇔ z ∈ ℝ, it follows from Exercise 1.6.5: z − z1 z − z1 z3 − z1 z3 − z1 = =( 3 )⇔ 3 ∈ ℝ. z2 − z1 z2 − z1 z2 − z1 z2 − z1
A3 ∈ A1 A2 ⇔
Solution of Exercise 1.6.7 It is well known that A3 ∈ [A1 A2 ] ⇔ A3 ∈ A1 A2 Exercise
⇔
1.6.6
z3 − z1 = t, z2 − z1
and ‖A1 A3 ‖ + ‖A3 A2 ‖ = ‖A1 A2 ‖
t∈ℝ
⇔ z3 = (1 − t)z1 + tz2 ,
and z3 − z1  + z2 − z3  = z2 − z1 
t ∈ ℝ and t(z2 − z1 ) + (1 − t)(z2 − z1 ) = z2 − z1 ,
→ where we used the relation Ai Aj = zj − zi . If z2 − z1  = 0 ⇔ z1 = z2 , which is a contradiction, so it follows that z2 − z1  ≠ 0. From the above result, dividing the second equality by z2 − z1  ≠ 0, we obtain that A3 ∈ [A1 A2 ] ⇔ z3 = (1 − t)z1 + tz2
and t + 1 − t = 1, t ∈ ℝ,
and using the last relation, we conclude that A3 ∈ [A1 A2 ] ⇔ z3 = (1 − t)z1 + tz2 , Solution of Exercise 1.6.8 Since A1 A2 :
y − y1 x − x1 = x2 − x1 y2 − y1
t ∈ [0, 1].
7.1 Solutions to the exercises of Chapter 1  197
and x − x3 y − y3 = , x4 − x3 y4 − y3
A3 A4 : using the identities x = Re z = A1 A2 :
z+z 2
and y = Im z =
x − x1
z2 − z1 + (z2 − z1 )
z−z 2i
we get that y − y1
=
−i(z2 − z1 − (z2 − z1 ))
=
−i(z4 − z3 − (z4 − z3 ))
and A3 A4 :
x − x3
z4 − z3 + (z4 − z3 )
y − y3
.
1. From the above relations, A1 A2 ⊥A3 A4 ⇔ (z2 − z1 + (z2 − z1 ))(z4 − z3 + (z4 − z3 )) + (−i)(z2 − z1 − (z2 − z1 ))(−i)(z4 − z3 − (z4 − z3 )) = 0 ⇔ (z3 − z4 )(z1 − z2 ) = −(z3 − z4 )(z1 − z2 ) ⇔
z1 − z2 z − z2 z − z2 z − z2 =− 1 = −( 1 )⇔ 1 ∈ iℝ, z3 − z4 z3 − z4 z3 − z4 z3 − z4
because z = −z if and only if z ∈ iℝ. 2. Similarly, A1 A2 ‖A3 A4 ⇔
z2 − z1 + (z2 − z1 )
z4 − z3 + (z4 − z3 )
=
−i(z2 − z1 − (z2 − z1 ))
−i(z4 − z3 − (z4 − z3 ))
⇔ (z3 − z4 )(z1 − z2 ) = (z3 − z4 )(z1 − z2 ) ⇔ ⇔
z1 − z2 ∈ ℝ, z3 − z4
z − z2 z1 − z2 z − z2 = 1 =( 1 ) z3 − z4 z3 − z4 z3 − z4
because z = z if and only if z ∈ ℝ. Solution of Exercise 1.6.9 Denoting by Ω = Ω(a) the center of the circle, we have A(z) ∈ C(a; r) ⇔ ‖AΩ‖ = r ⇔ z − a = r ⇔ z − a2 = r 2
⇔ (z − a)(z − a) = r 2 ⇔ (z − a)(z − a) = r 2 ⇔ zz − az − az + aa = r 2 .
198  7 Solutions to the chapterwise exercises Solution of Exercise 1.6.10 From Exercise 1.6.9, the circle C(z1, z2 , z3 ) has the equation of the form zz −az −az +aa = r 2 , or equivalently zz − az − az + a2 − r 2 = 0. Denote u = −a, v = a and w = a2 − r 2 , then A(z) ∈ C(z1 , z2 , z3 ) ⇔ A1 (z1 ), A2 (z2 ), A3 (z3 ) and A(z) are the solutions of the following system: ℛ:
{ { { { { { { { {
tzz + uz − vz + w = 0 tz1 z1 + uz1 − vz1 + w = 0 tz2 z2 + uz2 − vz2 + w = 0 tz3 z3 + uz3 − vz3 + w = 0.
This system has the solution (1, −a, a, a2 − r 2 ), and all the other points that satisfy the equation of the circle will satisfy the condition det ℛ = 0, i. e., det ℛ = 0 ⇔
zz z1 z1 z2 z2 z3 z3
z z1 z2 z3
z z1 z2 z3
1 1 1 1
= 0.
Solution of Exercise 1.6.11 An arbitrary complex number will be real if and only if it coincides with its conjugate, and thus z2 − z1 z2 − z4 z − z1 z3 − z4 z2 − z1 z3 − z4 : ∈ℝ⇔ 2 ⋅ = ⋅ . z3 − z1 z3 − z4 z3 − z1 z2 − z4 z3 − z1 z2 − z4 A simple computation shows that z2 − z1 z3 − z4 z2 − z1 z3 − z4 ⋅ = ⋅ ⇔ z3 − z1 z2 − z4 z3 − z1 z2 − z4 Exercise
⇔
1.6.10
z1 z1
z2 z2
z3 z3
z4 z4
z1
z2
z3
z4
z1
z2
z3
z4
1 1 = 0 1 1
there exists a circle 𝒞 that contains the given points.
7.2 Solutions to the exercises of Chapter 2  199
Solution of Exercise 1.6.12 From (1), it follows that f (0) = 0. Since the function f is continuous, from (1) and (2), we deduce that f (tz) = tf (z), ∀z ∈ ℂ.
∀t ∈ ℝ,
Let ∀z = a + ib, where a, b ∈ ℝ, and let ∀z ∈ ℂ. Then f (az) = af (z),
f (ibz) = bf (iz) = bf (i)f (z),
and thus f (z )f (z) = f (z z) = f (az + ibz) = [a + bf (i)]f (z),
∀z ∈ ℂ, ∀z = a + ib ∈ ℂ,
or equivalently f (z)[f (z ) − (a + bf (i))] = 0,
∀z ∈ ℂ, ∀z = a + ib ∈ ℂ.
(7.3)
From the above relation, it follows that (i) or
f (z) = 0,
∀z ∈ ℂ,
(ii) ∃z0 ∈ ℂ : f (z0 ) ≠ 0. If ∃z0 ∈ ℂ such that f (z0 ) ≠ 0, then the equality (7.3) holds if and only if f (z ) = a+bf (i), ∀z = a + ib ∈ ℂ. Since −1 = f (−1) = f (i2 ) = f (i)2 , we obtain that f (i) = i or f (i) = −i, and thus: (iii) if f (i) = i,
or
(iv) if f (i) = −i,
then f (z ) = z , ∀z ∈ ℂ, then f (z ) = z , ∀z ∈ ℂ.
7.2 Solutions to the exercises of Chapter 2 Solution of Exercise 2.9.1 1. If f (t) = −3 + i + (1 + i)t, t ∈ ℝ, and f (t) = α(t) + iβ(t), then {
x = α(t) y = β(t),
t∈ℝ
⇔{
x = −3 + t ⇔ y = x + 4, y = 1 + t, t ∈ ℝ
x ∈ ℝ.
200  7 Solutions to the chapterwise exercises 2. Since f (t) = 2 − 3i + (1 − 2i)t, t ∈ [1, 3], we obtain similarly that x =2+t y = −3 − 2t,
{ 3. If f (t) =
16 , 3−i+(1+2i)t
f (t) =
t ∈ [1, 3]
⇔ y = 1 − 2x,
x ∈ [3, 5].
t ∈ ℝ, then
16(t + 3) 16(1 − 2t) +i , (t + 3)2 + (2t − 1)2 (t + 3)2 + (2t − 1)2
t ∈ ℝ,
and { x= { y= {
16(t+3) (t+3)2 +(2t−1)2 16(1−2t) , (t+3)2 +(2t−1)2
t∈ℝ
⇒
t+3 x − 3y x = ⇔t= . y 1 − 2t 2x + y
From here, using the first (or the second) equality we obtain that x2 + y2 − + i 87 ; 8 7 5 ). i. e., 𝜕U( 16 7
32 16 x − y = 0, 7 7
√
4. Since f (t) =
−12+2i−(1−3i)t , 1+3i−(2−i)t
{ x= { { y= ⇔ ⇒
t ∈ ℝ, then 5t 2 +34t−6 5t 2 +2t+10
−5t 2 +14t+38 , t∈ℝ 5t 2 +2t+10 x − 1 = 5t16(2t−1) 2 +2t+10 { 16(t+3) y + 1 = 5t 2 +2t+10 ,
t∈ℝ
x − 1 2t − 1 3x + y − 2 = ⇔t= . y+1 t+3 −x + 2y + 3
If we replace this value into the first or into the second relation, then the required curve has the equation 2 18 34 x2 + y2 + x − y − = 0, 7 7 7 i. e., 𝜕U(− 71 + i 97 ; 8 7 5 ). √
Solution of Exercise 2.9.2 We will use the Cauchy–Riemann theorem. Since f (z) = z, it follows that f (x+iy) = x−iy, x, y ∈ ℝ. If we will write the function f in the algebraic form f (x + iy) = u(x, y) + iv(x, y),
7.2 Solutions to the exercises of Chapter 2  201
then we get u(x, y) = x, v(x, y) = −y, hence 𝜕u(z) 𝜕v(z) = 1 ≠ = −1, ∀z ∈ ℂ, 𝜕x 𝜕y 𝜕u(z) 𝜕v(z) =0=− , ∀z ∈ ℂ 𝜕y 𝜕x so there does not exist any point in ℂ in which the function f is differentiable. Solution of Exercise 2.9.3 If f (z) = f (x + iy) = u(x, y) + iv(x, y), x, y ∈ ℝ, then u(x, y) = x2 + y2 + x,
v(x, y) = 4xy + 3y.
Both of the u and v functions have continuously partial derivatives in ℂ, i. e., f ∈ C 1 (ℂ), hence f is ℝdifferentiable on ℂ. Using the Cauchy–Riemann relations, we will determine those points of ℂ, such that { { {
𝜕u(z) 𝜕x 𝜕u(z) 𝜕y
= =
𝜕v(z) 𝜕y − 𝜕v(z) 𝜕x
⇔{
2x + 1 = 4x + 3 ⇔ z = −1. 2y = −4y
So, we get that at the only point z0 = −1 the function f is differentiable. Thus, f (−1) =
𝜕v(−1) 𝜕u(−1) +i = −1. 𝜕x 𝜕x
We can obtain the same result if we will use the equivalent form of the Cauchy– Riemann system, i. e., 𝜕f𝜕z(z) = 0, hence 𝜕f (z) = 0 ⇔ z − 2z − 1 = 0, 𝜕z and then we will solve this last equation. Solution of Exercise 2.9.4 We will use a similar method within the previous problem. If f (z) = u(x, y) + iv(x, y), x, y ∈ ℝ, then u(x, y) = x2 ,
v(x, y) = xy
and the functions u and v belong to the class C 1 (ℂ), so the function f is ℝdifferentiable in ℂ.
202  7 Solutions to the chapterwise exercises Since
𝜕u(z) 𝜕x
= 2x,
𝜕u(z) 𝜕y
= 0,
𝜕v(z) 𝜕x
= y and
{ { {
𝜕u(z) 𝜕x 𝜕u(z) 𝜕y
= =
𝜕v(z) 𝜕y
= x, the Cauchy–Riemann system
𝜕v(z) 𝜕y − 𝜕v(z) 𝜕x
has the solution z0 = 0. The function is differentiable only at this point, and f (0) =
𝜕u(0) 𝜕v(0) +i = 0. 𝜕x 𝜕x
Solution of Exercise 2.9.5 If f (z) = u(x, y) + iv(x, y), x, y ∈ ℝ, then u(x, y) = x2 − y2 + 2x,
v(x, y) = 2xy − 2y.
From here, we have that u, v ∈ C 1 (ℂ), hence the function f is ℝdifferentiable in ℂ. = 2x + 2, 𝜕u(z) = −2y, 𝜕v(z) = 2y and 𝜕v(z) = 2x − 2, the Cauchy–Riemann Since 𝜕u(z) 𝜕x 𝜕y 𝜕x 𝜕y system { { {
𝜕u(z) 𝜕x 𝜕u(z) 𝜕y
= =
𝜕v(z) 𝜕y − 𝜕v(z) 𝜕x
has no solutions. It follows that there does not exist any point in ℂ in which the function f is differentiable. Solution of the Exercise 2.9.6 We write immediately the real and the imaginary part of f , i. e., u(x, y) = ln √x2 + y2 ,
y v(x, y) = arctan , x
z = x + iy ∈ ℂ∗ .
Now we will check the conditions of the Cauchy–Riemann theorem: 1. The functions u and v are continuously differentiable in ℂ∗ , because 𝜕u(z) x , = 2 𝜕x x + y2
𝜕u(z) y , = 2 𝜕y x + y2
𝜕v(z) y , =− 2 𝜕x x + y2
𝜕v(z) x . = 2 𝜕y x + y2
2. Since 𝜕u(z) 𝜕v(z) = , 𝜕x 𝜕y
𝜕u(z) 𝜕v(z) =− , 𝜕y 𝜕x
∀z ∈ ℂ,
it follows that f is differentiable in ℂ∗ , i. e., f ∈ H(ℂ∗ ), and the derivative will be f (z) =
𝜕u(z) 𝜕v(z) 1 +i = , 𝜕x 𝜕x z
∀z ∈ ℂ∗ .
7.2 Solutions to the exercises of Chapter 2  203
Solution of Exercise 2.9.7 If f (z) = u(x, y) + iv(x, y), x, y ∈ ℝ, then u(x, y) = (ex − e−x ) cos y,
v(x, y) = (ex + e−x ) sin y.
Since u, v ∈ C 1 (ℂ), and 𝜕u(z) 𝜕v(z) = = (ex + e−x ) cos y, 𝜕x 𝜕y 𝜕u(z) 𝜕v(z) =− = −(ex − e−x ) sin y, 𝜕y 𝜕x
∀z ∈ ℂ,
it follows that the function satisfies the conditions from the Cauchy–Riemann theorem in the whole complex plane ℂ, and thus f ∈ H(ℂ). Further, f (z) =
𝜕v(z) 𝜕u(z) +i = (ex + e−x ) cos y + i(ex − e−x ) sin y = ez + e−z , 𝜕x 𝜕x
∀z ∈ ℂ.
Solution of Exercise 2.9.8 1. Since f = u + iv is an entire function, the function will satisfy the conditions from the Cauchy–Riemann theorem. We have u(x, y) = x + ay ∈ C 1 (ℂ) and v(x, y) = bx + cy ∈ C 1 (ℂ). Using the Cauchy–Riemann system, we will obtain the required values, i. e., { { {
𝜕u(x,y) 𝜕x 𝜕u(x,y) 𝜕y
= =
𝜕v(x,y) 𝜕y − 𝜕v(x,y) , 𝜕x
∀(x, y) ∈ ℝ2
⇔{
c=1 a = −b,
hence f (z) = x + ay + i(−ax + y), a ∈ ℝ. The rest of the points will be treated similarly. 2. Since f = u + iv, it follows that u(x, y) = x2 + axy + by2 ,
v(x, y) = cx 2 + dxy + y2 ,
hence u, v ∈ C 1 (ℂ). Further, { { {
𝜕u(x,y) 𝜕x 𝜕u(x,y) 𝜕y
⇔{
= =
𝜕v(x,y) 𝜕y 𝜕v(x,y) − 𝜕x ,
∀(x, y) ∈ ℝ2
x(2 − d) + y(a − 2) = 0 x(2c + a) + y(d + 2b) = 0,
hence f (z) = x2 + 2xy − y2 + i(−x2 + 2xy + y2 ).
∀(x, y) ∈ ℝ2
{ { { { ⇔{ { { { {
a=2 b = −1 c = −1 d = 2,
204  7 Solutions to the chapterwise exercises 3. Since f = u + iv, we have that u(x, y) = cos x(cosh y + a sinh y),
v(x, y) = sin x(cosh y + b sinh y),
hence u, v ∈ C 1 (ℂ). From here, { { {
𝜕u(x,y) 𝜕x 𝜕u(x,y) 𝜕y
= =
𝜕v(x,y) 𝜕y 𝜕v(x,y) − 𝜕x ,
∀(x, y) ∈ ℝ
2
⇔{
a = −1 b = −1,
and thus f (z) = cos x(cosh y − sinh y) + +i sin x(cosh y − sinh y). Solution of Exercise 2.9.9 1.
Method 1. If z = x + iy = reiθ , x, y ∈ ℝ, then x = x(r, θ) = r cos θ,
y = y(r, θ) = r sin θ
and f (z) = u(x(r, θ), y(r, θ)) + iv(x(r, θ), y(r, θ)). From here, { { {
𝜕u 𝜕r
=
𝜕u 𝜕x 𝜕x 𝜕r
+
𝜕u 𝜕θ
=
𝜕u 𝜕x 𝜕x 𝜕θ
+
⇔
𝜕u 𝜕y 𝜕y 𝜕r 𝜕u 𝜕y 𝜕y 𝜕θ
=
𝜕u 𝜕x
cos θ +
𝜕u 𝜕y
= − 𝜕u r sin θ + 𝜕x
𝜕u sin θ 𝜕u 𝜕u = cos θ − 𝜕x 𝜕r 𝜕θ r
sin θ 𝜕u r cos θ 𝜕y
and
𝜕u 𝜕u 𝜕u cos θ = sin θ + . 𝜕y 𝜕r 𝜕θ r
(7.4)
Identically, we obtain similar results for the imaginary part v, i. e., { { {
𝜕v 𝜕r 𝜕v 𝜕θ
⇔
=
𝜕v 𝜕x 𝜕x 𝜕r
=
𝜕v 𝜕x 𝜕x 𝜕θ
+ +
𝜕v 𝜕y 𝜕y 𝜕r 𝜕v 𝜕y 𝜕y 𝜕θ
cos θ +
=
𝜕v 𝜕x
=
𝜕v − 𝜕x r sin θ
𝜕v 𝜕y
sin θ
+
𝜕v r cos θ 𝜕y
𝜕v 𝜕v 𝜕v sin θ = cos θ − 𝜕x 𝜕r 𝜕θ r
and
𝜕u 1 𝜕v − , 𝜕r r 𝜕θ
B=
𝜕v 𝜕v 𝜕v cos θ = sin θ + . 𝜕y 𝜕r 𝜕θ r
Denoting A=
𝜕v 1 𝜕u + , 𝜕r r 𝜕θ
(7.5)
7.2 Solutions to the exercises of Chapter 2  205
from (7.4), (7.5) according to the Cauchy–Riemann system it follows that {
𝜕u 𝜕x 𝜕u 𝜕y
= =
𝜕v 𝜕y 𝜕v − 𝜕x
⇔{
( 𝜕u − 𝜕r − ( 𝜕u 𝜕r
1 r 1 r
𝜕v ) cos θ = ( 𝜕v + 1r 𝜕u ) sin θ 𝜕θ 𝜕r 𝜕θ 𝜕v 𝜕v 1 𝜕u ) sin θ = −( 𝜕r + r 𝜕θ ) cos θ. 𝜕θ
A cos θ − B sin θ = 0 A=0 ⇔{ A sin θ + B cos θ = 0 B=0 𝜕u 1 𝜕v 𝜕v 1 𝜕u ⇔ = , =− . 𝜕r r 𝜕θ 𝜕r r 𝜕θ
⇔{
Using the relations (7.4) and (7.5), we conclude that f (z) =
𝜕v(z) (7.6) −iθ 𝜕u 𝜕v (7.6) e−iθ 𝜕v 𝜕u 𝜕u(z) = e ( +i +i ) = ( − i ), 𝜕x 𝜕x 𝜕r 𝜕r r 𝜕θ 𝜕θ
hence f (z) = e−iθ (
𝜕u(z) 𝜕v(z) e−iθ 𝜕v(z) 𝜕u(z) +i )= ( −i ), 𝜕r 𝜕r r 𝜕θ 𝜕θ
z = reiθ .
Method 2. If z = reiθ and z0 = r0 eiθ0 , then f (z) − f (z0 ) { ={ z − z0 { { { { { ={ { { { {
f (reiθ0 )−f (r0 eiθ0 ) , eiθ0 (r−r0 ) f (r0 eiθ )−f (r0 eiθ0 ) , r0 (eiθ −eiθ0 )
if θ = θ0 if r = r0
f (reiθ0 )−f (r0 eiθ0 ) , eiθ0 (r−r0 )
if θ = θ0
1 r0
if r = r0 .
f (r0 eiθ )−f (r0 eiθ0 ) θ−θ0 eiθ −eiθ0 θ−θ0
,
Since f is differentiable at z0 , we get
f (z0 ) = e−iθ0 (
𝜕u(z0 ) 𝜕v(z0 ) +i ) 𝜕r 𝜕r
and f (z0 ) =
𝜕u(z0 ) 1 −iθ0 𝜕v(z0 ) e ( −i ), r0 𝜕θ 𝜕θ
thus 𝜕u(z0 ) 1 𝜕v(z0 ) = , 𝜕r r0 𝜕θ
𝜕v(z0 ) 1 𝜕u(z0 ) =− . 𝜕r r0 𝜕θ
2. If f (z) = u + iv = ReiΦ , z = x + iy, where x, y ∈ ℝ, then u = u(R(x, y), Φ(x, y)) = R cos Φ,
v = v(R(x, y), Φ(x, y)) = R sin Φ
and f (z) = u(R(x, y), Φ(x, y)) + iv(R(x, y), Φ(x, y)).
(7.6)
206  7 Solutions to the chapterwise exercises Similarly, as in the above, the Cauchy–Riemann system becomes: { { {
𝜕u 𝜕x 𝜕u 𝜕y
𝜕u 𝜕R 𝜕u 𝜕Φ 𝜕v 𝜕R 𝜕v 𝜕Φ { 𝜕R 𝜕x + 𝜕Φ 𝜕x = 𝜕R 𝜕y + 𝜕Φ 𝜕y ⇔ { 𝜕u 𝜕R 𝜕u 𝜕Φ 𝜕v 𝜕R 𝜕v 𝜕Φ = + = − 𝜕R − 𝜕Φ 𝜕x 𝜕x { 𝜕R 𝜕y 𝜕Φ 𝜕y 𝜕R 𝜕Φ 𝜕R 𝜕Φ cos Φ − R sin Φ = sin Φ + R cos Φ { 𝜕x 𝜕x 𝜕y 𝜕y ⇔{ 𝜕R 𝜕Φ 𝜕R cos Φ 𝜕y − R sin Φ 𝜕y = − sin Φ 𝜕x − R cos Φ 𝜕Φ 𝜕x { 𝜕R 𝜕Φ 𝜕R 𝜕Φ { cos Φ( 𝜕x − R 𝜕y ) = sin Φ( 𝜕y + R 𝜕x ) ⇔{ sin Φ( 𝜕R − R 𝜕Φ ) = − cos Φ( 𝜕R + R 𝜕Φ ). 𝜕x 𝜕y 𝜕y 𝜕x {
=
𝜕v 𝜕y 𝜕v − 𝜕x
Denoting A=
𝜕Φ 𝜕R −R , 𝜕x 𝜕y
B=
𝜕R 𝜕Φ +R , 𝜕y 𝜕x
then the last equation system is equivalent to the next one: {
A cos Φ − B sin Φ = 0 A=0 ⇔{ A sin Φ + B cos Φ = 0 B=0 ⇔
𝜕R 𝜕Φ =R , 𝜕x 𝜕y
𝜕R 𝜕Φ = −R . 𝜕y 𝜕x
The derivative f will be f (z) =
𝜕v(z) (7.7) iΦ 𝜕R 𝜕Φ (7.7) iΦ 𝜕Φ 𝜕R 𝜕u(z) = e ( +i + iR ) = e (R − i ), 𝜕x 𝜕x 𝜕x 𝜕x 𝜕y 𝜕y
i. e., f (z) = eiΦ(z) (
𝜕Φ(z) 𝜕Φ(z) 𝜕R(z) 𝜕R(z) + iR ) = eiΦ(z) (R(z) −i ). 𝜕x 𝜕x 𝜕y 𝜕y
3. If f (z) = u + iv = ReiΦ , then u = u(R(r, θ), Φ(r, θ)) = R cos Φ,
v = v(R(r, θ), Φ(r, θ)) = R sin Φ
and f (z) = u(R(r, θ), Φ(r, θ)) + iv(R(r, θ), Φ(r, θ)). Using the results obtained in point 1 of this problem, we get {
𝜕u 𝜕r 𝜕v 𝜕r
= =
⇔{
1 𝜕v r 𝜕θ − 1r 𝜕u 𝜕θ
⇔{
− R sin Φ 𝜕Φ = 1r (sin Φ 𝜕R + R cos Φ 𝜕Φ ) cos Φ 𝜕R 𝜕r 𝜕r 𝜕θ 𝜕θ
sin Φ 𝜕R + R cos Φ 𝜕Φ = − 1r (cos Φ 𝜕R − R sin Φ 𝜕Φ ) 𝜕r 𝜕r 𝜕θ 𝜕θ
R 𝜕Φ ) = sin Φ( 1r 𝜕R + R 𝜕Φ ) r 𝜕θ 𝜕θ 𝜕r R 𝜕Φ 1 𝜕R 𝜕Φ 𝜕R − sin Φ( 𝜕r − r 𝜕θ ) = cos Φ( r 𝜕θ + R 𝜕r ).
cos Φ( 𝜕R − 𝜕r
(7.7)
7.2 Solutions to the exercises of Chapter 2  207
Denoting 𝜕R R 𝜕ϕ − , 𝜕r r 𝜕θ
A=
B=
𝜕ϕ 1 𝜕R +R , r 𝜕θ 𝜕r
then the last system is equivalent to the next one: {
A cos ϕ − B sin ϕ = 0 A=0 ⇔{ A sin ϕ + B cos ϕ = 0 B=0 ⇔
𝜕R R 𝜕Φ = , 𝜕r r 𝜕θ
𝜕R 𝜕Φ = −rR . 𝜕θ 𝜕r
(7.8)
Using the result obtained in point 1 of the problem, we deduce that f (z) = e−iθ ( (7.8)
=
𝜕u 𝜕v 𝜕R 𝜕Φ + i ) = e−iθ eiΦ ( + iR ) 𝜕r 𝜕r 𝜕r 𝜕r
e−iθ eiΦ 𝜕Φ 𝜕R (R − i ), r 𝜕θ 𝜕θ
thus f (z) = e−iθ eiΦ(z) ( =
𝜕R(z) 𝜕Φ(z) + iR(z) ) 𝜕r 𝜕r
e−iθ eiΦ(z) 𝜕Φ(z) 𝜕R(z) (R(z) −i ), r 𝜕θ 𝜕θ
z = reiθ .
Solution of Exercise 2.9.10 We will use the result obtained in point 1 of Exercise 2.9.9 for the Cauchy–Riemann system. For the function f given in the algebraic form, we will write the exponential form f (z) = r n cos nθ + ir n sin nθ = u(r, θ) + iv(r, θ). We get that u = u(r, θ) = r n cos nθ, v = v(r, θ) = r n sin nθ and u, v ∈ C 1 (ℂ). The relations (7.6) of the Cauchy–Riemann system are fulfilled in ℂ, because 𝜕u 1 𝜕v = = nr n−1 cos nθ, 𝜕r r 𝜕θ
𝜕v 1 𝜕u =− = nr n−1 sin nθ, 𝜕r r 𝜕θ
hence f ∈ H(ℂ). The derivative of the function will be f (z) = e−iθ (
𝜕u(z) 𝜕v(z) +i ) = nz n−1 , 𝜕r 𝜕r
∀z ∈ ℂ.
208  7 Solutions to the chapterwise exercises Solution of Exercise 2.9.11 Suppose that ∃a, b ∈ ℂ, such that f (D) = [a, b]. Let us define the function f1 (z) = e−iθ f (z), where θ is the angle between A(a) and B(b) line and the real axis, i. e., θ = arg(b − a). Since f is differentiable on D, it follows that f1 is also differentiable on D. The function f1 satisfies the fact that its image f1 (D) = [a, c], where c = e−iθ b. It means that the imaginary part v1 of the function f1 is constant on the domain D, because Im f1 (z) = Im a, ∀z ∈ D. Since f1 ∈ H(D), and D ⊂ ℂ is a domain, we deduce that f1 is a constant function on D that contradicts the fact that f is a nonconstant function on D. Solution of Exercise 2.9.12 Define the function φ : G \ {z} → ℂ,
as φ(ζ ) =
f (ζ ) − f (z) , ζ −z
and let use the notation ζ = ζ1 + iζ2 , z = x + iy. 1. We know that u(ζ ) − u(z) + i[v(ζ ) − v(z)] ∃ lim Re φ(ζ ) = lim Re ζ −z ζ →z ζ →z [u(ζ ) − u(z)](ζ1 − x) + [v(ζ ) − v(z)](ζ2 − y) = lim = l1 . ζ →z (ζ1 − x)2 + (ζ2 − y)2
that
u(ζ )−u(z) ζ1 −x v(ζ )−v(z) limζ2 →y ζ −y 2
If ζ2 = y ⇒ [ζ → z ⇔ ζ1 → x], thus ∃l1 = limζ1 →x
=
If ζ1 = x ⇒ [ζ → z ⇔ ζ2 → y], thus ∃l1 =
=
𝜕u(z) 𝜕x
=
𝜕v(z) . 𝜕y
𝜕u(z) . 𝜕x 𝜕v(z) , hence 𝜕y
2. Similarly,
∃ lim Im φ(ζ ) = lim ζ →z
ζ →z
−[u(ζ ) − u(z)](ζ2 − y) + [v(ζ ) − v(z)](ζ1 − x) = l2 . (ζ1 − x)2 + (ζ2 − y)2 v(ζ )−v(z) = 𝜕v(z) . ζ1 −x 𝜕x u(ζ )−u(z) limζ2 →y ζ −y = − 𝜕u(z) , 𝜕y 2
If ζ2 = y ⇒ [ζ → z ⇔ ζ1 → x], thus ∃l2 = limζ1 →x 𝜕u(z) 𝜕x
we deduce
If ζ1 = x ⇒ [ζ → z ⇔ ζ2 → y], thus ∃l2 = = − 𝜕v(z) . 𝜕y
Solution of Exercise 2.9.13 Let ζn = (1 + nz )n = rn (cos θn + i sin θn ), where rn = ζn , θn = arg ζn . Then n
n
n 2 2 2 2x x2 + y2 2 x + iy x y rn = 1 + ) + = [(1 + ) + ( ) ] = (1 + n n n n n2
and we get
7.2 Solutions to the exercises of Chapter 2  209
and θn = n arctan
y n
1+
x n
.
Taking the limits for n → ∞, we have n
2x x2 + y2 2 lim r = lim (1 + + ) = ex n→∞ n n→∞ n n2 and lim θn = lim n arctan
n→∞
n→∞
y n
1+
x n
= y.
From the above relations, we conclude that ez = lim ζn = lim (1 + n→∞
n→∞
n
z ) = ex (cos y + i sin y). n
Solution of Exercise 2.9.14 1. Using the wellknown definitions, sin z =
eiz − e−iz , 2i
cos z =
eiz + e−iz , 2
we obtain that eiz − e−iz eiz + e−iz − =i 2i 2 ⇔ (1 − i)e2iz + 2eiz − 1 − i = 0
sin z − cos z = i ⇔
⇔ (1 − i)t 2 + 2t − 1 − i = 0,
not.
t = eiz ⇔ t1,2 =
−1 ± √3 . 1−i
Thus, −1 ± √3 −1 ± √3 ⇔ z ∈ −iLog 1−i 1−i √3 − 1 √3 + 1 π 3π = { + 2kπ − i ln + 2kπ − i ln : k ∈ ℤ} ∪ {− : k ∈ ℤ}. √2 √2 4 4
eiz =
2. From the next two definitions, cosh z =
ez + e−z , 2
sinh z =
ez − e−z , 2
210  7 Solutions to the chapterwise exercises we have that cosh z − i sinh z = 1 ⇔
ez + e−z ez − e−z −i =1 2 2
⇔ (1 − i)e2z − 2ez + 1 + i = 0 ⇔ (1 − i)t 2 − 2t + 1 + i = 0,
not.
t = ez ⇔ t1 = 1,
t2 = i.
Thus, we get z ∈ Log 1 ∪ Log i = {2kπi : k ∈ ℤ} ∪ {(
π + 2kπ)i : k ∈ ℤ}. 2
3. Since sin z =
eiz − e−iz , 2i
it follows that sin z =
eiz − e−iz 4i 4i ⇔ = ⇔ 3t 2 + 8t − 3 = 0, 3 2i 3 1 ⇔ z ∈ (−iLog ) ∪ (−iLog (−3)) 3
not.
t = eiz ⇔ t1 =
1 , 3
t2 = −3
= {2kπ + i ln 3 : k ∈ ℤ} ∪ {(2k + 1)π − i ln 3 : k ∈ ℤ}.
4. Since cos z =
eiz + e−iz , 2
we have cos z =
eiz + e−iz 3 + i 3+i not. ⇔ = ⇔ 2t 2 − (3 + i)t + 2 = 0, t = eiz 4 2 4 1−i −i + 1 ⇔ z ∈ (−iLog (1 + i)) ∪ (−iLog ) ⇔ t1 = i + 1, t2 = 2 2 π π = { + 2kπ − i ln √2 : k ∈ ℤ} ∪ {− + 2kπ + i ln √2 : k ∈ ℤ}. 4 4
5. We easily deduce that sinh z =
i ez − e−z i i ± √3 not. ⇔ = ⇔ t 2 − it − 1 = 0, t = ez ⇔ t1,2 = 2 2 2 2 π 5π i ± √3 = {( + 2kπ)i : k ∈ ℤ} ∪ {( + 2kπ)i : k ∈ ℤ}. ⇔ z ∈ Log 2 6 6
6. Since tan z =
sin z eiz − e−iz = , cos z i(eiz + e−iz )
7.2 Solutions to the exercises of Chapter 2  211
we deduce that 5i i not. ⇔ 4t 2 + 1 = 0, t = eiz ⇔ t1,2 = ± 3 2 i π ⇔ z ∈ −iLog (± ) = {± + 2kπ + i ln 2 : k ∈ ℤ}. 2 2
tan z =
7. We may easily obtain that i π ei3z = −1 ⇔ 3iz ∈ Log (−1) ⇔ z ∈ − Log (−1) = {(2k + 1) : k ∈ ℤ}. 3 3 8. A simple computation shows that 1
1 1 ∈ Log 1 = {2kπi : k ∈ ℤ} ⇔ 2 = 2kπi, z2 z 1 ⇔ zk = , k ∈ ℤ∗ √2kπi
e z2 = 1 ⇔
π
k∈ℤ
π
ei( 4 +lπ) e−i( 4 +lπ) : k ∈ ℤ, k < 0, l ∈ {0, 1}}. : k ∈ ℤ, k > 0, l ∈ {0, 1}} ∪ { ⇔z∈{ √2kπ √−2kπ 9. We easily deduce the following equivalence: ez − e−z = 2 ⇔ e2z = −3 ez + e−z 1 ln 3 π ⇔ z ∈ Log (−3) ⇔ z ∈ { + i(2k + 1) : k ∈ ℤ}. 2 2 2
tanh z = 2 ⇔
Solution of Exercise 2.9.15 1. ei = cos 1 + i sin 1. 2i −2i 2. sinh 2i = e −e = i sin 2. 2
2 −2 2 −2 e2+3i +e−2−3i = e +e cos 3 + i e −e sin 3 = cosh 2 cos 3 + i sinh 2 sin 3. 2 2 2 e+e−1 e−e−1 e1+i +e−1−i = 2 cos 1 + i 2 sin 1 = cosh 1 cos 1 + i sinh 1 sin 1. 4. cos(1 − i) = 2 sin(1−2i) sin 1 cosh 2−i cos 1 sinh 2 5. tan(1 − 2i) = cos(1−2i) = cos = sin 1 cos 1−i sinh 2 cosh 2 . 1 cosh 2+i sin 1 sinh 2 cos2 1 cosh2 2+sin2 1 sinh2 2 6. log(−2i) = ln 2 − π2 i. 7. log(−3 + 4i) = ln 5 + i(π − arctan 43 ).
3. cosh(2 + 3i) =
8. Since
ζk =
π θ 1−i 4 2 = √ e−i( 4 + 2 +kπ) , √3 + i 5
1 k ∈ {0, 1}, where θ = arctan , 3
it follows that log ζk =
1 2 π θ ln − i( + + kπ), 4 5 4 2
k ∈ {0, 1}.
212  7 Solutions to the chapterwise exercises Solution of Exercise 2.9.16 π
π
π
1. i1−i = e(1−i)Log i = e(1−i){i( 2 +2kπ):k∈ℤ} = e{(1+i)( 2 +2kπ):k∈ℤ} = {e(1+i)( 2 +2kπ) : k ∈ ℤ} = π {ie 2 +2kπ : k ∈ ℤ}. π π π √ 2. (1+i√3)i = eiLog (1+i 3) = ei{ln 2+i( 3 +2kπ):k∈ℤ} = e{−( 3 +2kπ)+i ln 2:k∈ℤ} = {e−( 3 +2kπ)+i ln 2 : π k ∈ ℤ} = {e−( 3 +2kπ) (cos ln 2 + i sin ln 2) : k ∈ ℤ}. 3. 1−i = e−iLog 1 = e−i{2kπi:k∈ℤ} = e{2kπ:k∈ℤ} = {e2kπ : k ∈ ℤ}. π
π
π
π
π
4. e i = e{cos( 4 +kπ)+i sin( 4 +kπ):k∈ℤ} = {ecos( 4 +kπ)+i sin( 4 +kπ) : k ∈ ℤ} = {ecos( 4 +kπ) ⋅ (cos sin( π4 + kπ) + i sin sin( π4 + kπ)) : k ∈ ℤ}. √
Solution of Exercise 2.9.17 iz −iz −e−iz = e −e = sin z and sin z = −i sinh(iz) ⇔ sinh(iz) = i sin z. 2 2i e−z +ez 2. cos(iz) = 2 = cosh z and cos(i(iz)) = cosh(iz) ⇔ cos z = cosh(iz). −z1 z2 z1 −z1 z2 −(z1 +z2 ) z1 z1 +z2 e +e−z2 e −e−z2 3. sinh z1 cosh z2 + cosh z1 sinh z2 = e −e + e +e = e −e2 2 2 2 2 −z1 z2 z1 −z1 z2 z1 e +e−z2 e −e−z2 sinh(z1 + z2 ) and sinh z1 cosh z2 − cosh z1 sinh z2 = e −e − e +e 2 2 2 2 ez1 −z2 −e−(z1 −z2 ) = sinh(z − z ). 1 2 2
1. −i sinh(iz) = −i e
iz
4. cos z1 cos z2 − sin z1 sin z2 =
eiz1 +e−iz1 eiz2 +e−iz2 2 2
cos(z1 + z2 ) and cos z1 cos z2 + sin z1 sin z2 = ei(z1 −z2 ) +e−i(z1 −z2 ) 2
= cos(z1 − z2 ).
5. 2 sin z cos z = 2 e
6.
2 tan z 1−tan2 z
=
=
e2z +e−2z 2
2
8. Since
−e−iz eiz +e−iz 2i 2
iz −iz 2 e iz−e −iz i(e +e ) iz −iz 1−( e iz−e −iz )2 i(e −e )
7. sinh2 z = ( e e2z +e−2z +2 4
iz
z
+1
−e−z 2 ) 2
=
=
=
=
ei2z −e−i2z 2i
ei2z −e−i2z i(ei2z +e−i2z )
e2z +e−2z −2 4
i(z1 +z2 ) eiz1 −e−iz1 eiz2 −e−iz2 +e−i(z1 +z2 ) = e 2i 2i 2 −iz1 iz2 iz1 −iz2 −iz1 iz2 e −e e +e e −e−iz2 e +e + 2 2 2i 2i
−
iz1
= =
= tan 2z.
=
e2z +e−2z 2
2
−1
=
cosh 2z−1 2
and cosh2 z = ( e
z
+e−z 2 ) 2
cosh 2z+1 . 2
z = x + iy,
it follows that  sin z2 = sin2 x cosh2 y + cos2 x sinh2 y = (1 − cos2 x) cosh2 y + cos2 x sinh2 y = cosh2 y − cos2 x(− sinh2 y + cosh2 y) = cosh2 y − cos2 x 7.
=
= sin 2z.
sin z = sin x cosh y + i cos x sinh y,
=
=
cosh 2y + 1 cos 2x + 1 1 − = (cosh 2y − cos 2x), 2 2 2
and thus  sin z = √ 21 (cosh 2y − cos 2x).
=
7.2 Solutions to the exercises of Chapter 2  213
Similarly,  sin z2 = sin2 x cosh2 y + cos2 x sinh2 y
= sin2 x(1 + sinh2 y) + (1 − sin2 x) sinh2 y = sin2 x + sinh2 y,
and thus  sin z = √sin2 x + sinh2 y. 9. Since cos z = cos x cosh y + i sin x sinh y,
z = x + iy,
it follows that  cos z2 = cos2 x cosh2 y + sin2 x sinh2 y = cos2 x(1 + sinh2 y) + sin2 x sinh2 y 7.
= cos2 x + sinh2 y =
cos 2x + 1 cosh 2y − 1 1 + = (cosh 2y + cos 2x), 2 2 2
hence  cos z = √cos2 x + sinh2 y and  cos z = √ 21 (cosh 2y − cos 2x). z
z
sin(iz) = i tanh z and tan z = − tan(i(iz)) = = i(ee −z−e = i eez −e 10. tan(iz) = cos(iz) +ez ) +e−z −i tanh(iz). 11. Using the result of the point 7, we obtain that cosh2 z + sinh2 z = cosh22z+1 + −z
cosh 2z−1 2
2
2
= cosh 2z and cosh z + sinh z =
−z
2
2
cosh2 z(1+ sinh 2 z ) cosh z 2
cosh2 z−sinh z
=
cosh2 z(1+ sinh 2 z ) cosh z 2 cosh2 z(1− sinh 2 z ) cosh z
=
1+tanh2 z . 1−tanh2 z
Solution of Exercise 2.9.18 Since we know that (z1 , z2 , z3 , z4 ) = (T(z1 ), T(z2 ), T(z3 ), T(z4 )), denoting by z1 = z, T(z) = w and T(zj ) = wj , j = 2, 3, 4, we obtain z − 1 − i ∞ w − i −1 − i : = : z−1 ∞ ∞ ∞ w−i z−i z−1−i = ⇔w= . ⇔ z−1 −1 − i −z + 1
(z, 1 + i, ∞, 1) = (w, i, −1, ∞) ⇔
1−i From w (z) = (−z+1) 2 ≠ 0, ∀z ∈ ℂ \ {1}, we obtain that the function w is a conformal mapping. Further, we have
w(1) = ∞,
w(i) = 0,
1 1 w(1 + 2i) = − + i, 2 2
214  7 Solutions to the chapterwise exercises where {1, i, 1 + 2i} ⊂ {z = x + iy ∈ ℂ : (x − 1)2 + (y − 1)2 = 1}, hence w({z = x + iy ∈ ℂ : (x − 1)2 + (y − 1)2 = 1}) = {w = u + iv ∈ ℂ : u + v = 0}. Since 1 + i ∈ {z = x + iy ∈ ℂ : (x − 1)2 + (y − 1)2 < 1} and w(1 + i) = i ∈ {w = u + iv ∈ ℂ : u + v > 0}, it follows that w({z = x + iy ∈ ℂ : (x − 1)2 + (y − 1)2 < 1}) = {w = u + iv ∈ ℂ : u + v > 0}.
Solution of Exercise 2.9.19 We will determine the inverse of w, and then we will prove that it maps the domain {w ∈ ℂ : w < 1} onto the domain {z ∈ ℂ : Re z > 0}, which will prove the required result. A simple computation shows that g(z) = w−1 (z) = − z+1 , hence z−1 g(eiθ ) = i From v (θ) =
u = u(θ) = 0 sin θ = u(θ) + iv(θ) ⇔ { sin θ 1 − cos θ v = v(θ) = 1−cos , θ
1 cos θ−1
θ ∈ (0, 2π).
< 0, we have that v is a continuous decreasing function, and since lim v(θ) = +∞, θ↓0
lim v(θ) = −∞,
θ↑2π
we deduce g({w ∈ ℂ : w = 1}) = {z ∈ ℂ : Re z = 0}. Since g(0) = 1 ∈ {z ∈ ℂ : Re z > 0}, we conclude that g({w ∈ ℂ : w < 1}) = {z ∈ ℂ : Re z > 0}, i. e., w({z ∈ ℂ : Re z > 0}) = {w ∈ ℂ : w < 1}. From w (z) =
2 (z+1)2
≠ 0, ∀z ∈ ℂ \ {−1}, the function w is a conformal mapping.
7.2 Solutions to the exercises of Chapter 2  215
Solution of Exercise 2.9.20 Let us denote by D = {z ∈ ℂ : Re z < 1} and by Δ = {w ∈ ℂ : w < R}. It is well known ̂ into that a bilinear transform maps two inverse points with respect of a circle from ℂ, two inverse points with respect to the image of the given circle. From here, since z1 = −1 and z2 = 3 are inverse points with respect to the circle 𝜕D = {z ∈ ℂ : Re z = 1}, it follows that w1 = f (z1 ) = 0 and w2 = f (z2 ) are inverse points with respect to the circle 𝜕Δ = {w ∈ ℂ : w = R}, hence we deduce that w2 = f (z2 ) = ∞. In order to determine the required function, we need to know the image of three points. From the assumptions f (−1) = 0 and f (3) = ∞, it follows that f (z) = k
z+1 , z−3
k ∈ ℂ,
and by using the assumption f (−1) = 1 we deduce that f (−1) = Hence, our function is w(z) = −4
−k 4
= 1 ⇔ k = −4.
z+1 . z−3
To determine the value of the radius R, it is sufficient to find the module of the image of a chosen point. Since 1 ∈ 𝜕D, it follows that f (1) ∈ 𝜕Δ, hence f (1) = R ⇔ R = 4. 16 Since f (z) = (z−3) 2 ≠ 0, ∀z ∈ ℂ \ {3}, the function f is a conformal mapping. Solution of Exercise 2.9.21 Since f is a bilinear transform, the problem will be solved by using the wellknown previous technique: we will choose three distinct points on the boundary of the given domain (which is a circle), and the circle determined by its images will be the boundary of the image. Thus, f (1) = β, f (i) = β + i(β − 1), f (−i) = β + i(1 − β) and f (0) = 1. From here, since 2(β−1) f (z) = (z+1)2 ≠ 0, ∀z ∈ ℂ \ {−1}, it follows that the function f conformally maps the disc {z ∈ ℂ : z < 1} onto the halfplane {w ∈ ℂ : Re w > β}.
Solution of Exercise 2.9.22 Suppose that ∃z1 , z2 ∈ {z ∈ ℂ : z
1 −
1 1 − = 0, 2 2
∀z1 , z2 ∈ U(0;
1 ), 2a
1 thus 1 + az1 + az2 ≠ 0, ∀z1 , z2 ∈ U(0; 2a ), hence the function f is injective on the disc
{z ∈ ℂ : z
0}, then ∀z1 , z2 ∈ D ⇒ Re z1 > 0, Re z2 > 0 ⇒ Re(5 + z1 + z2 ) > 5, hence 5 + z1 + z2 ≠ 0, ∀z1 , z2 ∈ D. If D = {z ∈ ℂ : Re z > − 52 }, then ∀z1 , z2 ∈ D ⇔ Re z1 > − 52 , Re z2 > − 52 ⇒ Re(5 + z1 + z2 ) > 0, hence 5 + z1 + z2 ≠ 0, ∀z1 , z2 ∈ D. If D = {z ∈ ℂ : Re z < − 52 }, then ∀z1 , z2 ∈ D ⇔ Re z1 < − 52 , Re z2 < − 52 , ⇒ Re(5 + z1 + z2 ) < 0, hence 5 + z1 + z2 ≠ 0, ∀z1 , z2 ∈ D. It follows that the function w is injective in all the three cases. 1. Let z = x + iy ∈ 𝜕D = d1 ∪ d2 ∪ d3 , where d1 : {
x=0 y = t,
t ∈ [0, +∞)
d2 : {
x=1 y = t,
t ∈ [0, +∞)
d3 : {
x=t y = 0,
t ∈ [0, 1].
Then ∀z ∈ d1 ⇔ z = it,
t ∈ [0, +∞) ⇒ w(z) = 6 − t 2 + 5it,
t ∈ [0, +∞).
Hence, if d1 = w(d1 ), then d1 : {
u = 6 − t2 v2 ⇔u=6− , v = 5t, t ∈ [0, +∞) 25
v ∈ [0, +∞).
Now, ∀z ∈ d2 ⇔ z = 1 + it,
t ∈ [0, +∞) ⇒ w(z) = 12 − t 2 + 7it,
t ∈ [0, +∞).
7.2 Solutions to the exercises of Chapter 2  217
Hence, if d2 = w(d2 ), then d2 : {
u = 12 − t 2 v2 ⇔ u = 12 − , v = 7t, t ∈ [0, +∞) 49
v ∈ [0, +∞).
Similarly, we will determine the image of d3 as follows: ∀z ∈ d3 ⇔ z = t,
t ∈ [0, 1] ⇒ w(z) = t 2 + 5t + 6,
t ∈ [0, 1].
Hence, if d3 = w(d3 ), then d3 : {
u = t 2 + 5t + 6 ⇔ u = 6t + 6, v = 0, t ∈ [0, 1]
t ∈ [0, 1],
v = 0.
Since z0 = 21 + 2i ∈ D ⇒ w(z0 ) = 172 + 3i ∈ w(D), we deduce that w(D) is the subset of ℂ that contains the point M( 172 + 3i) and has the boundary 𝜕w(D) = d1 ∪ d2 ∪ d3 . 2. Let z = x + iy ∈ 𝜕D = d, where d:{
x = − 52 y = t, t ∈ ℝ.
We have 5 ∀z ∈ d ⇔ z = − + it, 2
t ∈ ℝ ⇒ w(z) = −t 2 −
1 , 4
t ∈ ℝ.
Hence, if d = w(d), then d : {
u = −t 2 − 41 v = 0, t ∈ ℝ,
thus w(D) = ℂ \ {w ∈ ℂ : Im w = 0, Re w ≤ − 41 } = E. It follows that the function w : D → E, w(z) = z 2 + 5z + 6 is bijective, and then ∀w ∈ E,
w(z) = w ⇔ z = √w +
1 5 − . 4 2
From here, since w(−2) = 0, we deduce that w−1 : E → D,
w−1 (w) = √w +
1 5 − , 4 2
where w−1 (0) = −2.
3. Similarly, since the function w can be written as w(z) = (z − 52 )2 − 41 , all the computations are the same like in the previous point. Hence w(D) = ℂ \ {w ∈ ℂ : Im w = 0, Re w ≤ − 41 } = E and w−1 : E → D,
w−1 (w) = √w +
1 5 − , 4 2
where w−1 (0) = −3.
218  7 Solutions to the chapterwise exercises Solution of Exercise 2.9.24 Let D = {z ∈ ℂ : 0 < Re z < a}, a > 0, and Δ = {w ∈ ℂ : Im w > 0}. It is evident that: (i) If g(z) = iz, then g(D) = {z ∈ ℂ : 0 < Im z < a} = Δ1 ; (ii) If h(z) = πa z, then h(Δ1 ) = {z ∈ ℂ : 0 < Im z < π} = Δ2 ; (iii) If k(z) = ez , then k(Δ2 ) = Δ. From here, it follows that the function will be πi
f (z) = (k ∘ h ∘ g)(z) = e a z , because f (D) = Δ.
7.3 Solutions to the exercises of Chapter 3 Solution of Exercise 3.10.1 In each of the above cases, first we will determine the parametric equations of the path, then by applying the definition of the complex integral we will compute the given integral. 1. From the equation of the circle, C(O; 1) : z = 1 ⇔ z = eiθ ,
θ ∈ [0, 2π],
it follows that π π θ ∈ [− , ]. 2 2
⌢
AB : z = eiθ , Letting θ(t) = a + bt, t ∈ [0, 1], then {
θ(0) = − π2 θ(1) = π2
⇔{
a = − π2 b=π
⇔ θ(t) =
π (2t − 1), 2
and the parametric equation of the path is ⌢
π
AB : γ(t) = ei 2 (2t−1) ,
t ∈ [0, 1].
Then the value of the integral is 1
1 1 γ (t)dt = πi. I = ∫ dz = ∫ z γ(t) γ
0
t ∈ [0, 1],
7.3 Solutions to the exercises of Chapter 3  219
2. The equation of the arc is π 3π θ ∈ [− , − ]. 2 2
⌢
AB : z = eiθ , Letting θ(t) = a + bt, t ∈ [0, 1], then {
θ(0) = − π2
θ(1) =
⇔{
− 3π 2
a = − π2 b = −π
π ⇔ θ(t) = − (2t + 1), 2
t ∈ [0, 1].
Hence the parametric equation of the path is ⌢
π
AB : γ(t) = e−i 2 (2t+1) ,
t ∈ [0, 1],
and the integral is given by 1
1 1 I = ∫ dz = ∫ γ (t)dt = −πi. z γ(t) γ
0
3. Since C(O; √2) : z = √2 ⇔ z = √2eiθ ,
θ ∈ [0, 2π],
we deduce that π π θ ∈ [− , ]. 4 4
⌢
CD : z = √2eiθ , Letting θ(t) = a + bt, t ∈ [0, 1], then {
θ(0) = − π4 θ(1) = π4
⇔{
a = − π4 b = π2
⇔ θ(t) =
π (2t − 1), 4
The parametric equation of the path is ⌢
π
CD : γ(t) = √2ei 4 (2t−1) ,
t ∈ [0, 1],
hence 1
1 π 1 γ (t)dt = i. I = ∫ dz = ∫ z γ(t) 2 γ
0
4. Since the equation of the path is CD : γ(t) = 1 − i + 2it,
t ∈ [0, 1],
it follows that 1
1
1 1 1 I = ∫ dz = ∫ dt γ (t)dt = ∫ z γ(t) t + 1−i γ
0
1
=∫ 0
t2
t−
1 2
−t+
1
1 2
dt +
2i
0
i π 1 dt = i. ∫ 2 t2 − t + 1 2 0
2
t ∈ [0, 1].
220  7 Solutions to the chapterwise exercises Solution of Exercise 3.10.2 Since γ = γ1 ∪ γ2 ∪ γ3 , it follows that I=∫ γ
z z z z dz = ∫ dz + ∫ dz + ∫ dz. z z z z γ1
γ2
γ3
Now we will compute the value of each of these integrals. Since γ1 (t) = t − 3,
t ∈ [0, 1],
then 1
I1 = ∫ γ1
γ (t) z dz = ∫ 1 γ1 (t)dt = −1. z γ1 (t) 0
For the path γ2 , we have C(O; 2) : z = 2 ⇔ z = 2eiθ ,
θ ∈ [0, 2π],
then γ2 : z = 2eiθ ,
θ ∈ [π, 0].
Letting θ(t) = a + bt, t ∈ [0, 1], we get {
θ(0) = π a=π ⇔{ ⇔ θ(t) = −π(t − 1), θ(1) = 0 b = −π
We conclude that the path γ2 has the parametric equation γ2 (t) = 2e−iπ(t−1) ,
t ∈ [0, 1],
and the second integral is 1
γ (t) z I2 = ∫ dz = ∫ 2 γ2 (t)dt = 0. z γ2 (t) γ2
0
For the path γ3 , we have γ3 (t) = t + 2,
t ∈ [0, 1],
and the value of the third integral is 1
γ (t) z I3 = ∫ dz = ∫ 3 γ3 (t)dt = 1. z γ3 (t) γ3
0
Now, we conclude that I = I1 + I2 + I3 = 0.
t ∈ [0, 1],
7.3 Solutions to the exercises of Chapter 3  221
Solution of Exercise 3.10.3 From the definition of the multivalued function Log , i. e., Log z = {ln z + i(arg z + 2kπ) : k ∈ ℤ}, it is easy to determine the main branch log, since log 1 = 0 ⇒ log z = ln z + i arg z. The integral will be 1
1
π π ∫ log zdz = ∫ log γ(t)γ (t)dt = ∫ i(2t − 1) iπei(2t−1) 2 dt = −2i, 2
γ
0
0
where we used that π log γ(t) = i(2t − 1) , 2
t ∈ [0, 1].
Solution of Exercise 3.10.4 Using the definition of the multivalued function Log , i. e., Log z = {ln z + i(arg z + 2kπ) : k ∈ ℤ}, 1
since √z = e 2 Log z , from the condition √1 = 1 we get k = 0, and thus √z = √zei
arg z 2
,
arg z ∈ (−π, π).
The value of the integral is 1
1
πt
πi i 2 e 1 1 dz = ∫ γ (t)dt = ∫ 2 πt dt = √2 − 2 + i√2, ∫ √z √γ(t) ei 4 γ 0 0
where we used the fact that πt
√γ(t) = ei 4 ,
t ∈ [0, 1].
Solution of Exercise 3.10.5 1. From the definition, Log z = {ln z + i(arg z + 2kπ) : k ∈ ℤ},
222  7 Solutions to the chapterwise exercises 1
and √z = e 2 Log z , since √1 = 1 it follows that k = 0. Then the main branch of the root function is given by √z = √zei
arg z 2
z ∈ ℂ∗ .
,
The parametric equation of the path is γ(t) = r 2 e2πit ,
t ∈ [0, 1],
and the integral will be computed as follows:
∫ γ
1
1
0
0
πi 4 1 1 2 dz = ∫ γ (t)dt = ∫ πit dt = , r e r z√z γ(t)√γ(t)
where we used the fact that √γ(t) = reiπt ,
t ∈ [0, 1].
2. Since √1 = −1, it follows that the corresponding branch of the root function has the form √z = √zei
arg z+2π 2
,
z ∈ ℂ∗ .
The integral is given by
∫ γ
1
1
0
0
1 4 πi 1 γ (t)dt = ∫ πi(2t+1) dt = 0, dz = ∫ r e z√z γ(t)√γ(t)
where we used the relation √γ(t) = reiπ(2t+1) ,
t ∈ [0, 1].
Solution of Exercise 3.10.6 Like in the previous problem, since the parametric equations of the path are given, we will use the definition of the complex integral to compute the given integral. 1. Since γ(t) = 2t − 1, t ∈ [0, 1], it follows that 1
∫ zdz = ∫ 22t − 1dt = 1. γ
0
7.3 Solutions to the exercises of Chapter 3  223
2. If γ(t) = eπit , t ∈ [0, 1], then 1
1
0
0
∫ zdz = ∫ γ(t)γ (t)dt = ∫ πieiπt dt = −2. γ
3. If γ(t) = e
πi(t−1)
, t ∈ [0, 1], then 1
1
0
0
∫ zdz = ∫ γ(t)γ (t)dt = ∫ πieiπ(t−1) dt = 2. γ
Solution of Exercise 3.10.7 We will use the definition of the complex integral to compute the below integrals. 1. 1
1
∫ zdz = ∫ γ(t)γ (t)dt = ∫ 2(2t − 1)dt = 0. γ
0
0
2. 1
1
∫ zdz = ∫ γ(t)γ (t)dt = ∫ πie−i γ
0
(2t−1)π 2
ei
(2t−1)π 2
dt = iπ.
0
3. The parametric equation of the path is γ(t) = e2πit ,
t ∈ [0, 1],
and thus 1
1
∫ zdz = ∫ γ(t)γ (t)dt = ∫ 2πie−2πit e2πit dt = 2πi. γ
0
0
Solution of Exercise 3.10.8 In all of these three cases, we will use the definition of the complex integral. 1. 1
2 2 γ (t) + γ k (t) z+z z+z dz = ∑ ∫ dz = ∑ ∫ k γk (t)dt ∫ Re zdz = ∫ 2 2 2 k=1 k=1 γ
γ1 ∪γ2 1
=∫ 0
γk
0
1
2t −2(1 − t) (1 + i)dt + ∫ (1 − i)dt = −i. 2 2 0
224  7 Solutions to the chapterwise exercises 2. 1
∫ Re zdz = ∫
γ3
γ3
1
=∫ 0
z+z πi e−i dz = ∫ 2 2
π(t−1) 2
0
+ ei 2
π(t−1) 2
ei
π(t−1) 2
dt
πi iπ(t−1) 2 + πi (e + 1)dt = . 4 4
3. Using results obtained to points 1 and 2, we deduce that ∫ Re zdz =
Re zdz = ∫ Re zdz +
∫
γ
γ3
γ3 ∪(γ1 ∪γ2 )−
= ∫ Re zdz − ∫ Re zdz = γ3
∫
Re zdz
(γ1 ∪γ2 )−
γ1 ∪γ2
2 + (π + 4)i . 4
Solution of Exercise 3.10.9 1. The parametric equation of the path is C(α; r) : z − α = r ⇔ z = α + reiθ ,
θ ∈ [0, 2π],
hence γ(θ) = α + reiθ ,
θ ∈ [0, 2π],
that is γ(t) = α + re2πit ,
t ∈ [0, 1].
We will compute the integral by using the wellknown definition of the complex integral, i. e., 1
1 1 I=∫ dz = ∫ γ (t)dt = 2πi. z−α γ(t) − α γ
0
1 2. Since f (z) = z−α ∈ H(D), where D = ℂ \ {α}, we will discuss two cases. The first one is thus when α ∈ ℂ belongs to the interior of the triangle T, and the second one is thus when it belongs to the exterior of the triangle. Case 1. If α ∈ Ext(T), then γ ∼ 0, and thus D
∫ γ
1 dz = 0. z−α
Case 2. If α ∈ Int(T), then we will replace the integration path with the homotopic circle 𝜕U(α; r). Since ∃r > 0 : 𝜕U(α; r) ∈ Int(T), let γ∗ (t) = α + re2πit ,
t ∈ [0, 1].
7.3 Solutions to the exercises of Chapter 3  225
Then γ∗ ∼ γ, and thus, according to the Cauchy theorem we get D
1
∫ γ
1 1 1 dz = ∫ dz = ∫ γ (t)dt = 2πi. z−α z−α γ∗ (t) − α ∗ γ∗
0
Solution of Exercise 3.10.10 1. Using the definition of the complex integral, we deduce that 1
n
n
In = ∫(z − a) dz = ∫(γ(t) − a) γ (t)dt = r γ
1
n+1
∫ πieπi(n+1)t dt.
0
(7.9)
0
Now we will discuss the next two cases according to the values of the parameter n ∈ ℤ: Case 1. If n ∈ ℤ \ {−1}, from the relation (7.9) we have that In =
r n+1 πi(n+1) (e − 1). n+1
Case 2. If n = −1, from the relation (7.9) it follows that 1
I−1 = ∫ πidt = πi. 0
2. The parametric equation of the path γ will be γ(t) = a + re2πit ,
t ∈ [0, 1],
hence the integral is given by 1
n
1
In = ∫(z − a)n dz = ∫(γ(t) − a) γ (t)dt = r n+1 ∫ 2πie2πi(n+1)t dt. γ
0
0
Similarly, we will discuss the next two cases according to the values of the parameter n ∈ ℤ: Case 1. If n ∈ ℤ \ {−1}, then In =
r n+1 2πi(n+1) (e − 1). n+1
Case 2. If n = −1, then 1
I−1 = ∫ 2πidt = 2πi. 0
226  7 Solutions to the chapterwise exercises 3. Since n ∈ ℤ \ {−1}, let g(z) =
(z − a)n+1 . n+1
This new function satisfies the relation g (z) = f (z). Case 1. If n ∈ ℕ, then g ∈ H(ℂ) and g (z) = f (z), ∀z ∈ ℂ, and thus the function f has a primitive in ℂ. Then ∫(z − a)n dz = g(z1 ) − g(z0 ) = γ
(z1 − a)n+1 − (z0 − a)n+1 , n+1
∀γ ∈ 𝒟(z0 , z1 ).
Case 2. If n ∈ ℤ− \{−1}, and a ∉ {γ}, then g ∈ H(ℂ\{a}) and g (z) = f (z), ∀z ∈ ℂ\{a}, and thus the function f has a primitive in ℂ \ {a}. It follows that in this case we also have ∫(z − a)n dz = g(z1 ) − g(z0 ) = γ
4. Since n = which
1 , 2
(z1 − a)n+1 − (z0 − a)n+1 , n+1
∀γ ∈ 𝒟(z0 , z1 ) : a ∉ {γ}.
then the square root function has two branches: the branch for and thus √z = √zei
√1 = 1,
arg z 2
z ∈ ℂ∗ ,
,
and the branch for which and thus √z = √zei
√1 = −1,
arg z+2π 2
,
z ∈ ℂ∗ .
Case 1. If √1 = 1, then the value of the integral is 1
1
0
0
∫ √z − adz = ∫ √γ(t) − aγ (t)dt = 2πir √r ∫ e3πit dt = − γ
4r √r , 3
because √re2πit = √reπit . Case 2. If √1 = −1, then the value of the integral is 1
1
0
0
∫ √z − adz = ∫ √γ(t) − aγ (t)dt = 2πir √reπi ∫ e3πit dt = γ
because √re2πit = √reπi(t+1) .
4r √r , 3
7.3 Solutions to the exercises of Chapter 3  227
Solution of Exercise 3.10.11 First, we will determine the poles of the function, i. e., sin z + cos z = 0 ⇔
1−i 1−i eiz − e−iz eiz + e−iz + = 0 ⇔ e2iz = ⇔ 2iz ∈ Log , 2i 2 1+i 1+i
and thus the poles of the function are zk = −
π + kπ, 4
k ∈ ℤ.
It follows that g(z) =
ez ∈ H(ℂ \ {zk : k ∈ ℤ}). sin z + cos z
Since we are interested only about the poles that lie in the disc U(0, 41 ), we have that 1 1 zk ∈ U(0, ) ⇔ zk  < ⇔ k ∈ 0, 4 4 hence 1 g ∈ H(U(0, )), 4 and thus, according to the Cauchy theorem we have that ∫ γ
ez dz = 0. sin z + cos z
Solution of Exercise 3.10.12 We have ez sin z ∈ H(ℂ), since this function is obtained from the product of two elementary functions, and thus from the Cauchy theorem we get ∫ ez sin zdz = 0. γ
The equation of the integrating path γ is γ(t) = 2e2πit , t ∈ [0, 1]. From the above result, using the definition of the complex integral we deduce that z
1
z
∫(z + e sin z)dz = ∫ zdz + ∫ e sin zdz = ∫ zdz = ∫ 8πie−2πit e2πit dt = 8πi. γ
γ
γ
γ
0
228  7 Solutions to the chapterwise exercises Solution of Exercise 3.10.13 The function g(z) =
ez cos z cos iz
is holomorphic in the whole complex plane ℂ, except the zeros of the cos z and cos iz functions. Since π cos z = 0 ⇔ eiz + e−iz = 0 ⇔ e2iz = −1 ⇔ zk = + kπ, k ∈ ℤ, 2
and
cos iz = 0 ⇔ e−z + ez = 0 ⇔ e2z = −1 ⇔ zk = i
π + kπi, 2
k ∈ ℤ.
it follows that zk  ≥ π2 > 1, and zk  ≥ π2 > 1, ∀k ∈ ℤ, and then g ∈ H(U(0, 1)). From the Cauchy theorem, we deduce that the value of the integral is zero. Solution of Exercise 3.10.14 We will determine the zeros of the denominator, i. e., z 2 + 4 = 0 ⇔ z = ±2i and thus g(z) =
and z 2 + 16 = 0 ⇔ z = ±4i,
1 ∈ H(D), (z 2 + 4)(z 2 + 16)
D = ℂ \ {−4i, −2i, 2i, 4i}.
Since any of these zeros does not belong to the interior of the triangle T, then γ ∼ 0. D
Hence, from the Cauchy theorem we conclude that the value of the integral is zero. Solution of Exercise 3.10.15 Since the elementary functions are holomorphic, we have 1
e z−3 ∈ H(ℂ \ {3})
and
cos z, (z − 2)n ∈ H(ℂ).
It follows that 1
g(z) =
e z−3 cos z ∈ H(D), (z − 2)n
D = ℂ \ {2, 3},
has the nth order pole z0 = 2, and has the essential isolated singular point z1 = 3. Both of these singularities does not belong to the disc U(0; 1), and thus γ ∼ 0. Hence, from the Cauchy theorem we obtain that the value of the integral is zero.
D
7.3 Solutions to the exercises of Chapter 3  229
Solution of Exercise 3.10.16 Since the function is rational, it will be holomorphic on the whole complex plane except the zeros of the denominator, i. e., g(z) =
z ∈ H(D), z2 − 1
D = ℂ \ {−1, 1}.
We have to discuss the two cases according to the values of the parameter a > 0, with a ≠ 1. Case 1. If 0 < a < 1, then γ ∼ 0. Using the Cauchy theorem, it follows that the D
integral is zero. Case 2. If a > 1, then both of the above zeros belong to the disc U(0; a). Method 1. We will decompose the given integral into a sum of integrals, where the integration path is disjoint and where the integrated function has only one pole. These kinds of decompositions are useful also for some applications of the Cauchy formula for the discs. Thus, ∃r1 , r2 > 0 : U(−1; r1 ) ⊂ Int U(0; a),
U(1; r2 ) ⊂ Int U(0; a),
U(−1; r1 ) ∩ U(1; r2 ) = 0.
If γ1 (t) = −1 + r1 e2πit ,
t ∈ [0, 1],
γ2 (t) = 1 + r2 e2πit ,
t ∈ [0, 1],
then {γ1 } = 𝜕U(−1; r1 ), {γ2 } = 𝜕U(1; r2 ) and ∫ γ
z2
z z z dz = ∫ 2 dz + ∫ 2 dz. −1 z −1 z −1 γ1
γ2
To compute the integrals from the righthand side, we see that g(z) =
1 z 1 1 = ( + ). z2 − 1 2 z − 1 z + 1
Thus, according to Exercise 3.10.9 we get 1 1 1 1 z 1 dz + ∫ dz) = ∫ dz = πi, dz = (∫ 2 z−1 z+1 2 z+1 −1
∫
z2
∫
1 1 1 1 1 z dz = (∫ dz + ∫ dz) = ∫ dz = πi. 2 z−1 z+1 2 z−1 z2 − 1
γ1
γ2
γ1
γ1
γ2
γ2
From here, using the relation (7.10) we conclude that ∫ γ
z dz = 2πi. z2 − 1
γ1
γ2
(7.10)
230  7 Solutions to the chapterwise exercises Method 2. Similarly, ∃r1 , r2 > 0 : U(−1; r1 ) ⊂ Int U(0; a),
U(1; r2 ) ⊂ Int U(0; a),
U(−1; r1 ) ∩ U(1; r2 ) = 0.
If we introduce the notation γ1 (t) = −1 + r1 e2πit ,
γ2 (t) = 1 + r2 e2πit ,
t ∈ [0, 1],
t ∈ [0, 1],
then {γ1 } = 𝜕U(−1; r1 ), {γ2 } = 𝜕U(1; r2 ), and ∫ γ
z z z dz = ∫ 2 dz + ∫ 2 dz. z2 − 1 z −1 z −1 γ1
γ2
Now, by using the Cauchy formula for the discs we get z
∫
z 1 dz = ∫ z−1 dz = 2πi = πi, z+1 2 z2 − 1
∫
z 1 dz = ∫ z+1 dz = 2πi = πi. z−1 2 z2 − 1
γ1
γ1
and z
γ2
γ2
Replacing these two relations in (7.11), we conclude that ∫ γ
z dz = 2πi. z2 − 1
Solution of Exercise 3.10.17 The function under the integral is obtained from the elementary functions, i. e., sin z, z −
π 2 , z + 5 ∈ H(ℂ), 2
hence g(z) = Since only z0 =
π 2
sin z ∈ H(D), (z − π2 )(z 2 + 5)
π where D = ℂ \ { , −i√5, i√5}. 2
∈ U(0; 2), it follows that ∃r > 0 : U( π2 , r) ⊂ U(0, 2), and γr (t) =
π + re2πit , 2
t ∈ [0, 1] ⇒ γ ∼ γr . D
(7.11)
7.3 Solutions to the exercises of Chapter 3  231
From the Cauchy integral theorem, it follows that sin z z 2 +5 z − π2 γr
∫ g(z)dz = ∫ g(z)dz = ∫ γ
γr
and using the Cauchy formula for the disc, where f (z) = sin z z 2 +5 z − π2 γr
∫ g(z)dz = ∫ γ
dz =
dz, sin z ,z z 2 +5 0
=
π 2
and k = 0, we get
2πi π 8πi . f( ) = 0! 2 20 + π 2
Solution of Exercise 3.10.18 Since the function under the integral is obtained from the elementary functions, we have g(z) =
z5 ∈ H(ℂ \ {−1}). + 1)6
ez (z
Because z0 = −1 ∈ U(0; 2), from the Cauchy formula for closed paths we deduce that ∫ g(z)dz = ∫ γ
γ
z5 ez
(z + 1)6
dz =
2πi (k) f (z0 ), k!
5
where f (z) = ez z , z0 = −1 and k = 5. Computing the fifthorder derivative of the function f , we have f (5) (z) = 120e−z − 600ze−z + 600z 2 e−z − 200z 3 e−z + 25z 4 e−z − z 5 e−z , then f (5) (−1) = 1546e, and thus ∫ g(z)dz = γ
773 2πi (5) f (−1) = πie. 5! 30
Solution of Exercise 3.10.19 Consider the function g(z) =
zez (z − 2)n
defined on the maximal domain, where g is holomorphic.
232  7 Solutions to the chapterwise exercises If n = 0, then g ∈ H(ℂ). Since γ ∼ 0, from the Cauchy integral theorem it follows ℂ
that the integral is zero. If n ∈ ℕ∗ , then g ∈ H(D), where D = ℂ \ {2}, thus the point z0 = 2 is a nth order pole for g and γ ≁ 0. Since
D
∫ g(z)dz = ∫ γ
γ
zez dz, (z − 2)n
from the Cauchy formula for closed paths it follows that ∫ γ
2πi (k) zez f (z0 ), dz = (z − 2)n k!
where f (z) = zez , z0 = 2 and k = n − 1. If we compute the n − 1th order derivative of the function f , then f (n−1) (z) = (n − 1)ez + zez , and using this result we get ∫ γ
2πi 2(n + 1)e2 πi zez dz = ((n − 1)e2 + 2e2 ) = . n (z − 2) (n − 1)! (n − 1)!
Solution of the Exercise 3.10.20 Case 1. If a = 0, in this case the zero of the denominator is a zero for the function sin z. Since ez sin z = 1 ∈ ℂ, z→0 z lim
from the first removability criterion the point z0 = 0 will be removable. From the Cauchy integral theorem, it follows that the required integral is zero, for all r > 0. Case 2. If a ≠ 0, then g(z) =
ez sin z ∈ H(D), z−a
D = ℂ \ {a},
where z0 = a is a firstorder pole for the function g. Case 2.1. If r < a, then γ ∼ 0, and from the Cauchy integral theorem it follows
that
D
∫ 𝜕U(0;r)
ez sin z dz = 0. z−a
7.3 Solutions to the exercises of Chapter 3  233
Case 2.2. If r > a, then γ ≁ 0, and from the Cauchy formula for closed paths it D
follows that
∫ g(z)dz = ∫ γ
γ
ez sin z 2πi (k) dz = f (z0 ), z−a k!
where f (z) = ez sin z, z0 = a and k = 0. Thus ∫ γ
ez sin z dz = 2πif (a) = 2πiea sin a. z−a
Solution of the Exercise 3.10.21 The function under the integral may be written into the form g(z) =
1 1 1 1 − ), = ( z 2 + 1 2i z − i z + i
(7.12)
that is holomorphic on D = ℂ \ {−i, i}. 1. Method 1. If {γ} = 𝜕U(i; 1) is a directly oriented path, then the equation of the path is γ(t) = i + e2πit ,
t ∈ [0, 1].
Using the above decomposition, since γ does not turn around the point −i, the function 1 is holomorphic on the domain bounded by {γ}, hence z+i ∫ γ
1 dz = 0. z+i
(7.13)
From the relations (7.12) and (7.13), according to Exercise 3.10.9 we conclude that ∫ γ
1 1 1 1 1 1 dz − ∫ dz) = ∫ dz = π. dz = (∫ 2i z−i z+i 2i z − i z2 + 1 γ
γ
γ
Method 2. If the Cauchy formula for the disc using then 1
∫ γ
1 1 dz = ∫ z+i dz = 2πi = π, z−i 2i z2 + 1 γ
the other points of the problem will be solved similarly.
234  7 Solutions to the chapterwise exercises 2. Method 1. If {γ} = 𝜕U(−i; 1) is a directly oriented path, then the equation of the path is γ(t) = −i + e2πit ,
t ∈ [0, 1].
The value of the integral is ∫ γ
z2
1 1 1 1 1 1 dz − ∫ dz) = − ∫ dz = −π, dz = (∫ 2i z−i z+i 2i z + i +1 γ
γ
γ
where we used the result of Exercise 3.10.9. Method 2. If we use the Cauchy formula for the disc, then 1
∫ γ
1 1 dz = ∫ z−i dz = 2πi(− ) = −π. z+i 2i z2 + 1 γ
3. Method 1. We decompose our integral into the sum of two integrals, such that the new integration paths are disjoints, and each of them turns around only one pole. Then we will use directly the Cauchy formula for the disc. Thus, we have ∃r1 , r2 > 0 : U(−i; r1 ) ⊂ Int U(0; 2),
U(i; r2 ) ⊂ Int U(0; 2),
U(−i; r1 ) ∩ U(i; r2 ) = 0.
If γ1 (t) = −i + r1 e2πit ,
γ2 (t) = i + r2 e2πit ,
t ∈ [0, 1],
t ∈ [0, 1],
then {γ1 } = 𝜕U(−i; r1 ), {γ2 } = 𝜕U(i; r2 ). Since γ1 ∼ γ1 ,
where γ1 (t) = −i + e2πit ,
t ∈ [0, 1],
γ2 ∼ γ2 ,
where γ2 (t) = i + e2πit ,
t ∈ [0, 1],
D
and D
we will use the results from the first two points of the problem to calculate our integral: ∫ γ
z2
1 1 1 dz = ∫ 2 dz + ∫ 2 dz +1 z +1 z +1 γ1
γ2
γ1
γ2
1 1 dz + ∫ 2 dz = −π + π = 0. =∫ 2 z +1 z +1
7.3 Solutions to the exercises of Chapter 3  235
To obtain the above results, we used the fact that ∫ γ
1 dz = 2πi, z−a
where {γ} = 𝜕U(a; r) is a directly oriented path (Exercise 3.10.9). Method 2. Similarly, ∃r1 , r2 > 0 : U(−i; r1 ) ⊂ Int U(0; 2),
U(i; r2 ) ⊂ Int U(0; 2),
U(−i; r1 ) ∩ U(i; r2 ) = 0.
If we use the notation γ1 (t) = −i + r1 e2πit ,
t ∈ [0, 1],
γ2 (t) = i + r2 e2πit ,
t ∈ [0, 1],
then {γ1 } = 𝜕U(−i; r1 ), {γ2 } = 𝜕U(i; r2 ). From the Cauchy formula for the disc, we get ∫ γ
1 1 1 dz = ∫ 2 dz + ∫ 2 dz z2 + 1 z +1 z +1 γ1
=∫ γ1
1 z−i
z+i
γ2
dz + ∫ γ2
1 z+i
z−i
1 1 dz = 2πi(− ) + 2πi = −π + π = 0. 2i 2i
Solution of Exercise 3.10.22 The elementary functions ez , cos z and z 3 + 8 are holomorphic on the whole complex plane, hence f (z) =
ez cos z ∈ H(D), z3 + 8
D = ℂ \ A,
where A = {z ∈ ℂ : z 3 + 8 = 0} = {2ei
(2k+1)π 3
: k ∈ {0, 1, 2}}.
Since ∀z ∈ A, z = 2 > 1, the bounded domain with the boundary {γ} does not contain any pole of the function f . Also, the path γ(t) = e2πit , t ∈ [0, 1], has the image {γ} = 𝜕U(0; 1), and thus γ ∼ 0. From the Cauchy integral theorem, it follows that the integral
will be zero, i. e.,
D
∫ γ
ez cos z dz = 0. z3 + 8
236  7 Solutions to the chapterwise exercises Solution of Exercise 3.10.23 According to the values of the parameter k ∈ ℤ, we will discuss the next cases: Case 1. If k ∈ ℤ ∩ (−∞, 0], then f (z) =
sin z ∈ H(ℂ), zk
and since a γ(t) = e2πit , t ∈ [0, 1], the integration path is homotopic with 0 in ℂ. It follows that ∫ γ
sin z dz = 0. zk
Case 2. If k = 1, then f ∈ H(ℂ∗ ). Since lim f (z) = 1 ∈ ℂ,
z→0
form the first removability criterion the point z0 = 0 is removable. Using the Cauchy integral theorem, it follows that the integral is equal to zero. Case 3. If k ∈ ℕ ∩ [2, +∞), then f ∈ H(ℂ∗ ). From the Cauchy formula for the disc for f (z) = sin z, z0 = 0 and n = k − 1, we obtain ∫ γ
sin z 2πi 2πi sin z (k−1) (0) = u(k), dz = k (k − 1)! (k − 1)! z
where −1, 0, 1, 0,
if k if k if k if k
g(z) =
ez z−a
{ { { { (k−1) u(k) = sin (0) = { { { { {
= 4p, p ∈ ℕ∗ , = 4p + 1, p ∈ ℕ∗ , = 4p + 2, p ∈ ℕ, = 4p + 3, p ∈ ℕ.
Solution of Exercise 3.10.24 The function
has the unique pole z0 = a ∈ ℂ∗ , that is a simple pole. 1. If we use the Cauchy formula for the disc, we obtain ∫ γ
2πi (k) ez dz = f (z0 ), z−a k!
7.3 Solutions to the exercises of Chapter 3  237
where f (z) = ez , z0 = a and k = 0, and thus ez dz = 2πiea . z−a
∫ γ
2. If {γ} = 𝜕U(−a; a), a ∈ ℂ∗ is a directly oriented path, then γ(t) = −a + ae2πit ,
t ∈ [0, 1].
Since g ∈ H(D), D = ℂ \ {a}, it follows that γ ∼ 0, and from the Cauchy integral theorem D
we get
∫ γ
ez dz = 0. z−a
Solution of Exercise 3.10.25 If we use the Cauchy formula for the disc, using then ∫ γ
where f (z) = sinh z, z0 =
πi 2
sinh z z−
πi 2
dz =
2πi (k) f (z0 ), k!
and k = 0, and thus ∫ γ
sinh z z−
πi 2
dz = 2πi sinh
πi = −2π. 2
Solution of Exercise 3.10.26 According to the values of the parameter k ∈ ℤ, we need to discuss a few cases. First, we will denote In = ∫(z − α)n dz. γ
Case 1. If n ∈ ℕ, then the function f (z) = (z − α)n is holomorphic on ℂ, and γ ∼ 0.
From the Cauchy integral theorem, the integral is equal to zero, i. e., In = 0,
ℂ
∀n ∈ ℕ.
Case 2. If n ∈ ℤ ∩ (−∞, 0), then the function f is holomorphic on ℂ except the point z0 = α, i. e., f ∈ H(D), where D = ℂ \ {α}.
238  7 Solutions to the chapterwise exercises Let us see if the integration path turns around this pole. Case 2.1. If α > r, the integration path does not turn around the pole z0 = α, and thus γ ∼ 0. Now, according to the Cauchy integral theorem it follows that the integral D
is zero. Case 2.2. If α < r, the integration path turns around the pole z0 = α. From the Cauchy formula for the disc, we obtain that In = ∫ γ
1 2πi (k) dz = f (z0 ), −n (z − α) k!
where f (z) = 1, z0 = α, and k = −n − 1. The function f is constant, thus all its derivatives are zero. The integral will be not zero only in the case n = −1, i. e., I−1 = 2πi. From the above results, we conclude that I−1 = 2πi, and In = 0 if n ∈ ℤ ∩ (−∞, −2]. Solution of Exercise 3.10.27 The function g(z) =
1 z(z 2 + z − 2)
has the simple poles z1 = 0, z2 = 1 and z3 = −2, thus g ∈ H(D), where D = ℂ \ {−2, 0, 1}. 1. The path {γ} turns around the points z1 = 0 and z2 = 1. Let γ1 (t) = r1 e2πit ,
t ∈ [0, 1]
and γ2 (t) = 1 + r2 e2πit ,
t ∈ [0, 1],
where 0 < r1 < 21 , 0 < r2 < 21 . Then ∫ g(z)dz = ∫ g(z)dz + ∫ g(z)dz. γ
γ1
γ2
Using the Cauchy formula for the disc, we have 1 dz = ∫ ∫ 2 z(z + z − 2)
1 (z 2 +z−2)
γ1
γ1
z
dz = 2πi
1 = −πi −2
and 1
2 1 1 2πi dz = ∫ z +2z dz = 2πi = . ∫ z−1 3 3 z(z 2 + z − 2)
γ2
γ2
(7.14)
7.3 Solutions to the exercises of Chapter 3  239
From the above results, according to (7.14) we obtain that ∫ γ
2πi 1 πi dz = −πi + =− . 3 3 z(z 2 + z − 2)
(7.15)
2. In this case, the path {γ} turns around all the three points z1 = 0, z2 = 1 and z3 = −2. In order to use the relation (7.15), we will decompose our integral in the next sum: ∫ γ
1 dz = z(z 2 + z − 2)
∫ 𝜕U(0; 32 )
1 1 dz + ∫ dz, z(z 2 + z − 2) z(z 2 + z − 2)
(7.16)
γ3
where γ3 (t) = −2 + r3 e2πit , t ∈ [0, 1] and 0 < r3 < 21 . To calculate the second integral, we will use the Cauchy formula for the disc, i. e., 1
∫
γ3
2 1 1 πi dz = ∫ z −z dz = 2πi = . z+2 6 3 z(z 2 + z − 2)
γ3
Replacing this in (7.16), and using (7.15), we conclude that ∫
z(z 2
γ
πi πi 1 dz = − + = 0. 3 3 + z − 2)
Solution of Exercise 3.10.28 The function has the simple pole z1 = 0, while z2 = i and z3 = −i are nth order poles. Hence, the function is holomorphic on ℂ except at these points, i. e., g(z) =
f (z) ∈ H(D), z(z 2 + 1)n
D = ℂ \ {0, −i, i}.
1. Method 1. The function may be written like 1 1 1 1 − , = − z(z 2 + 1) z 2(z − i) 2(z + i) thus ∫ γ
1 1 1 1 1 1 dz = ∫ dz − ∫ dz − ∫ dz. z 2 z−i 2 z+i z(z 2 + 1) γ
γ
γ
The integration path is an ellipse that turns around only the point z1 = 0. The fractions 1 1 and 2(z+i) are holomorphic functions on the bounded domain, which has the 2(z−i)
240  7 Solutions to the chapterwise exercises boundary this ellipse, and from the Cauchy formula for closed curves it follows that these last two integrals are equal to zero. Using the Cauchy formula for the disc for the third integral, we deduce that ∫ γ
1 1 dz = ∫ dz = 2πi. z z(z 2 + 1) γ
Method 2. From the Cauchy formula for closed curves, it follows that 1
∫ γ
2 1 dz = ∫ z +1 dz = 2πi. 2 z z(z + 1)
γ
2. Method 1. Similar to the point 1, the function may be written like 1 1 1 1 = − − . z(z 2 + 1) z 2(z − i) 2(z + i) The integration path is a circle, and the bounded domain bounded by this circle contains all the poles of the function. We will decompose the integral into the sum of three integrals, i. e., ∫ γ
1 1 1 1 1 1 dz = ∫ dz − ∫ dz − ∫ dz. z 2 z−i 2 z+i z(z 2 + 1) γ
γ
γ
We will use the Cauchy formula for the disc for each of the above integrals (or the result of Exercise 3.10.9), and we conclude that ∫ γ
1 dz = 2πi − πi − πi = 0. z(z 2 + 1)
Method 2. Let γ1 (t) = r1 e2πit ,
γ2 (t) = i + r2 e2πit ,
γ3 (t) = −i + r3 e2πit ,
t ∈ [0, 1],
where 0 < rk < 21 , k ∈ {1, 2, 3}. Now, using the Cauchy formula for the disc for each of the integrals we get ∫ γ
3 1 1 dz = dz ∑ ∫ 2 + 1) z(z 2 + 1) z(z k=1 γk
=∫ γ1
1 z 2 +1
z
dz + ∫ γ2
1 z(z+i)
z−i
dz + ∫ γ3
1 z(z−i)
z+i
1 1 dz = 2πi + 2πi(− ) + 2πi(− ) = 0. 2 2
7.3 Solutions to the exercises of Chapter 3  241
3. Method 1. We will use the same method like to the point 1 of this problem. The function may be decomposed as follows: eiz eiz eiz eiz − − , = z 2(z − i) 2(z + i) z(z 2 + 1) and the integration path is an ellipse that turns around only the pole z1 = 0. If we use the Cauchy formula for the disc for f (z) = ez , z1 = 0, and k = 0, then ∫ γ
eiz eiz dz = dz = 2πi. ∫ z z(z 2 + 1)
(7.17)
γ
Method 2. Using the Cauchy formula for closed curves, we have eiz
2 eiz dz = ∫ z +1 dz = 2πi. ∫ 2 z z(z + 1)
γ
γ
4. Method 1. This point is similar with the point 2 of this problem. If we write eiz eiz eiz eiz = − − , z 2(z − i) 2(z + i) z(z 2 + 1) we obtain that ∫ γ
eiz eiz eiz eiz 1 1 dz = ∫ dz − ∫ dz − ∫ dz. 2 z 2 z−i 2 z+i z(z + 1) γ
γ
γ
Using the Cauchy formula for the disc for each of these integrals, where f (z) = eiz , k = 0, z1 = 0, z2 = i, and z3 = −i, respectively, we conclude that ∫ γ
1 πi − πie = πi(2 − e−1 − e). dz = 2πi − e z(z 2 + 1)
Method 2. Let γ1 (t) = r1 e2πit ,
γ2 (t) = i + r2 e2πit ,
γ3 (t) = −i + r3 e2πit ,
t ∈ [0, 1],
242  7 Solutions to the chapterwise exercises where 0 < rk < 21 , k ∈ {1, 2, 3}. Now, using the Cauchy formula for the disc for each of the integrals we get ∫ γ
3 eiz eiz dz = dz ∑ ∫ z(z 2 + 1) z(z 2 + 1) k=1 γk
=∫ γ1
eiz z 2 +1
z
dz + ∫ γ2
= 2πi + 2πi(−
eiz z(z+i)
z−i
dz + ∫ γ3
eiz z(z−i)
z+i
dz
1 e ) + 2πi(− ) = πi(2 − e−1 − e). 2e 2
5. Method 1. We see that zeiz eiz zeiz eiz eiz eiz (z 3 + 2z) − 2 − − 2 = = . 2 2 z z + 1) z + 1 (z + 1) (z 2 + 1)2
z(z 2
Since the integration path is an ellipse that turns around only the pole z1 = 0, we have ∫ γ
eiz (z 3 + 2z) eiz eiz eiz dz − ∫ dz, dz = ∫ dz = ∫ 2 2 2 z z + 1) (z + 1)
z(z 2
γ
γ
γ
because the function under the second integral is holomorphic on the bounded domain which is bounded by the ellipse, and from the Cauchy integral theorem this integral will be zero. If we use the formula (7.17) to calculate the last integral, we get ∫ γ
eiz dz = 2πi. + 1)2
z(z 2
Method 2. From the Cauchy formula for closed curves, we deduce that eiz dz = ∫ ∫ 2 z(z + 1)2 γ
eiz (z 2 +1)2
γ
z
dz = 2πi.
6. Method 1. If we decompose the function as follows, eiz iz eiz i i iz = − ), − eiz (z 3 + 2z)( − + 2 z 2(z + i) 2(z − i) 4(z − i)2 4(z + i)2 + 1)
z(z 2
the integral can be calculated as the sum of the following three integrals.
7.3 Solutions to the exercises of Chapter 3  243
The value of the integral I1 = ∫ γ
eiz dz = 2πi, z
is the same with those given by (7.17). For the integral I2 = ∫ γ
−eiz i(z 3 + 2z) dz, 2(z + i)
we use the Cauchy formula for the disc, where f (z) = Thus, we have
−eiz i(z 3 +2z) , z0 2
= −i and k = 0.
e I2 = 2πif (−i) = 2πi(− ) = −πie. 2 For the integral, I3 = ∫ γ
eiz i(z 3 + 2z) dz, 2(z − i)
using the Cauchy formula for the disc where f (z) = obtain I3 = 2πif (i) = 2πi(−
γ
z0 = i and k = 0, we
πi 1 )=− . 2e e
Using the Cauchy formula for the disc for f (z) = have I4 = ∫
eiz i(z 3 +2z) , 2
−eiz iz(z 3 +2z) , z0 4
= i and k = 1, we
−eiz iz(z 3 + 2z) 1 πi dz = 2πif (i) = 2πi(− ) = − . 2 4e 2e 4(z − i)
If we use the Cauchy formula for the disc for f (z) = obtain that I5 = ∫ γ
eiz iz(z 3 +2z) , z0 4
= −i and k = 1, we
eiz iz(z 3 + 2z) e πie . dz = 2πi = 4 2 4(z + i)2
Combining the above results, we conclude that ∫ γ
eiz e 3 dz = I1 + I2 + I3 + I4 + I5 = πi(2 − − ). 2 2e z(z 2 + 1)2
244  7 Solutions to the chapterwise exercises Method 2. Using the notation, γ1 (t) = r1 e2πit ,
γ2 (t) = i + r2 e2πit ,
γ3 (t) = −i + r3 e2πit ,
t ∈ [0, 1],
where 0 < rk < 21 , k ∈ {1, 2, 3}. Now, using the Cauchy formula for the disc for each of the integrals, we have ∫ γ
3 eiz eiz dz = dz ∑ ∫ z(z 2 + 1)2 z(z 2 + 1)2 k=1 γk
=∫
eiz (z 2 +1)2
z
γ1
dz + ∫ γ2
eiz z(z+i)2
(z − i)2
dz + ∫ γ3
eiz z(z−i)2
(z + i)2
dz
= 2πif1 (0) + 2πif2 (i) + 2πif3 (−i), where f1 (z) =
eiz , f (z) (z 2 +1)2 2
=
eiz z(z+i)2
∫ γ
and f3 (z) =
eiz , z(z−i)2
thus
e eiz 3 dz = πi(2 − − ). 2 2e + 1)2
z(z 2
Solution of Exercise 3.10.29 Since the function is obtained from elementary functions, we have that it is holomorphic on the set D = ℂ \ {−1, 1}. The integration path is a circle with the radius a > 0, and the center in a ∈ ∗ ℝ \ {− 21 , 21 }, i. e., {γ} = 𝜕U(a; a) with direct orientation. We need to make the following discussion according to the values of the parameter a. Case 1. If a < 21 , then the integration path does not turn around any of the poles of the function, hence γ ∼ 0. From here, according to the Cauchy integral theorem, the D
value of the integral is zero, i. e.,
∫ γ
cos πz dz = 0. z2 − 1
Case 2. If a > 21 , then the integration path turns around only one of the poles of the function, as follows. Case 2.1. If a < − 21 , then −1 ∈ U(a; a) and 1 ∉ U(a; a). Using the Cauchy formula πz for closed curves, where f (z) = cos , z0 = −1, and k = 0, we have z−1 cos πz
∫ γ
cos πz dz = ∫ z−1 dz = 2πif (−1) = πi. z+1 z2 − 1 γ
7.3 Solutions to the exercises of Chapter 3  245
Case 2.2. If a > 21 , then 1 ∈ U(a; a) and −1 ∉ U(a; a). Similarly, using the Cauchy πz , z0 = 1 and k = 0, we get formula for closed curves, where f (z) = cos z+1 cos πz
∫ γ
cos πz dz = ∫ z+1 dz = 2πif (1) = −πi. z−1 z2 − 1 γ
Solution of Exercise 3.10.30 If n ∈ ℤ ∩ (−∞, 0], then f (z) =
1 ∈ H(ℂ), (z 2 + 1)n
and since γ ∼ 0, from the Cauchy integral theorem it follows that the value of the ℂ
integral is zero. If n ∈ ℕ∗ , similarly as in the previous problem we need to determine only the zeros of the denominator. Hence, the function f is holomorphic on the set D = ℂ \ {−i, i}. The integration path is a circle with the radius  a2  and the center i a2 , i. e., {γ} = 𝜕U(i a2 ;  a2 ) with direct orientation. We need to make the following discussion according to the values of the parameter a. Case 1. If a < 1, the function is holomorphic in the disc bounded by the integration circle, hence γ ∼ 0. From the Cauchy integral theorem, we have that the value of D
the integral is zero, i. e.,
∫ γ
(z 2
1 dz = 0. + 1)n
Case 2. If a > 1, then the integration path turns around only one of the poles of the function, as follows. Case 2.1. If a < −1, then −i ∈ U(i a2 ;  a2 ) and i ∉ U(i a2 ;  a2 ). From the Cauchy formula 1 for closed curves, where f (z) = (z−i) n , z0 = −i, and k = n − 1, we get 1
∫ γ
2πi (n−1) 1 (z−i)n f (−i) dz = dz = ∫ 2 n (z + i)n (n − 1)! (z + 1) γ
=−
π
22(n−1) (n
− 1)!
n(n + 1) ⋅ . . . ⋅ (2n − 1),
because f (n−1) (z) = (−1)n−1 n(n + 1)(n + 2) ⋅ . . . ⋅ (2n − 1)(z − i)−2n+1 .
246  7 Solutions to the chapterwise exercises Case 2.2. If a > 1, then i ∈ U(i a2 ;  a2 ) and −i ∉ U(i a2 ;  a2 ). Similarly, from the Cauchy 1 formula for closed curves, where f (z) = (z+i) n , z0 = i, and k = n − 1, we have 1
∫ γ
2πi (n−1) 1 (z+i)n dz = f (i) dz = ∫ 2 n (z − i)n (n − 1)! (z + 1) γ
=
π
22(n−1) (n
− 1)!
n(n + 1) ⋅ . . . ⋅ (2n − 1),
because f (n−1) (z) = (−1)n−1 n(n + 1)(n + 2) ⋅ . . . ⋅ (2n − 1)(z + i)−2n+1 . Solution of Exercise 3.10.31 The function is holomorphic on the set D = ℂ \ {1}. We need to make the following discussion according to the values of the parameter a. Case 1. If 0 < a < 1, then the function is holomorphic in the disc with the center in O and radius a. According to the Cauchy integral theorem, the value of the integral is zero, i. e., iπ
ze 2 z dz = 0. ∫ z−1 γ
Case 2. If a > 1, then the disc with the center in O and radius a contains the simple iπ pole z0 = 1. If we use the Cauchy formula for closed curves, where f (z) = ze 2 z , z0 = 1 and k = 0, then iπ
∫ γ
iπ ze 2 z dz = 2πie 2 = −2π. z−1
Solution of Exercise 3.10.32 The function is holomorphic on the set D = ℂ \ {a}. We need to make the following discussion according to the values of the parameter a. Case 1. If a < 1, then point a belongs to the unit disc. Using the Cauchy formula for closed curves, where f (z) = z n , z0 = a and k = 0, we obtain that ∫ γ
zn dz = 2an πi. z−a
7.3 Solutions to the exercises of Chapter 3  247
Case 2. If a > 1, then the function is holomorphic in the unit disc, and from the Cauchy integral theorem the value of the integral will be zero, i. e., ∫ γ
zn dz = 0. z−a
Solution of Exercise 3.10.33 Since the function is obtained from elementary functions, we need to determine only the zeros of the denominator. Thus, the function f is holomorphic on the set D = ℂ \ {0, i}. We need to make the following discussion according to the values of the parameter a. Case 1. If 0 < a < 1, then the disc with the center in O and radius a > 0 contains eπz , z0 = 0 only the pole z1 = 0. Using the Cauchy formula for the disc, where f (z) = z−i and k = 0, we get eπz
eπz 1 dz = ∫ z−i dz = 2πi(− ) = −2π. ∫ z(z − i) z i γ
(7.18)
γ
Case 2. If a > 1, then both of the poles of the function belong to the disc with the center in O and radius a > 0. We decompose the integral into the sum of two integrals, where the integration paths are disjoints and each of them turns around only one of the poles. Then, from a corollary of the Cauchy integral theorem we have eπz
eπz
eπz dz = ∫ z−i dz + ∫ z dz, ∫ z(z − i) z z−i γ
γ1
γ2
where γ1 (t) = r1 e2πit ,
γ2 (t) = i + r2 e2πit ,
and 0 < rk < 21 , k ∈ {1, 2, }.
t ∈ [0, 1],
Using now the Cauchy formula for the disc, where f (z) = together with (7.18), we conclude that ∫ γ
eπz , z0 z
eπi eπz dz = −2π + 2πi = −2π − 2π = −4π. z(z − i) i
Solution of Exercise 3.10.34 The function is holomorphic on the set D = ℂ \ {−π, π}.
= i and k = 0,
248  7 Solutions to the chapterwise exercises The discussion will be similar to that of Exercise 3.10.33. Case 1. If 0 < a < π, then the function is holomorphic in the disc with the center in O and radius a, and from the Cauchy integral theorem the value of the integral is zero, i. e., ∫ γ
z2
eiz dz = 0. − π2
Case 2. If a > π, then both of the poles belong to the disc with the center in O and radius a. We decompose the integral into the sum of two integrals, where the integration paths are disjoint and each of them turns around only one of the poles. Then, from a corollary of the Cauchy integral theorem we have eiz
eiz
eiz dz = ∫ z−π dz + ∫ z+π dz, ∫ 2 z+π z−π z − π2 γ
γ1
γ2
where γ1 (t) = −π + r1 e2πit ,
γ2 (t) = π + r2 e2πit ,
t ∈ [0, 1],
and 0 < rk < a − π, k ∈ {1, 2}.
Using the Cauchy formula for the disc for the functions f (z) =
k = 0 and respectively f (z) = ∫ γ
eiz , z+π
where z0 = π, k = 0, we have
eiz , where z0 z−π
= −π,
e−πi eiz eπi dz = 2πi(− ) + 2πi = i − i = 0. 2π 2π z2 − π2
Solution of Exercise 3.10.35 The function is obtained from elementary functions, hence we need to determine only the zeros of the denominator. Thus, the function f is holomorphic on the set D = ℂ \ {−2, 2}. The integration path is a circle with radius then  a2  and the center a2 , i. e., {γ} = 𝜕U( a2 ;  a2 ) with direct orientation. We need to make the following discussion according to the values of the parameter a. Case 1. If a < 2, then the function is holomorphic in the disc with radius the  a2  and the center a2 , and by the Cauchy integral theorem the value of the integral is zero, i. e., ∫ γ
cos πz dz = 0. (z 2 − 4)2
7.3 Solutions to the exercises of Chapter 3  249
Case 2. If a > 2, then the integration path turns around only one of the poles of the function, as follows: Case 2.1. If a < −2, then we will use the Cauchy formula for closed curves, where cos πz f (z) = (z−2) 2 , z0 = −2 and k = 1. Since f (z) = −
πz sin πz − 2π sin πz + 2 cos πz , (z − 2)3
it follows that cos πz
πi 1 cos πz (z−2)2 dz = ∫ dz = 2πif (z0 ) = 2πi = . ∫ 2 32 16 (z − 4)2 (z + 2)2 γ
γ
Case 2.2. If a > 2, then we will use the Cauchy formula for closed curves, where cos πz f (z) = (z+2) 2 , z0 = 2 and k = 1. Since f (z) = −
πz sin πz + 2π sin πz + 2 cos πz , (z + 2)3
it follows that cos πz
1 cos πz πi (z+2)2 dz = ∫ dz = 2πif (z0 ) = 2πi(− ) = − . ∫ 2 2 2 32 16 (z − 4) (z − 2) γ
γ
Solution of Exercise 3.10.36 The function is holomorphic on the set D = ℂ \ {−i, i}, while the integration part is an ellipse that turns around both of the poles z1 = i, z2 = −i. We decompose the integral into the sum of two integrals, where the integration paths are disjoints and each of them turns around only one of the poles, i. e., z 100 eiπz
z 100 eiπz
z 100 eiπz dz = ∫ z+i dz + ∫ z−i dz, ∫ 2 z−i z+i z +1 γ1
γ
γ2
where γ1 (t) = i + r1 e2πit ,
γ2 (t) = −i + r1 e2πit ,
t ∈ [0, 1]
and 0 < rk < 1, k ∈ {1, 2}. Using the Cauchy formula for the disc for f (z) =
z0 = i, k = 0, and for f (z) = ∫ γ
z 100 eiπz , z−i
where z0 = −i, k = 0, we deduce that
z 100 eiπz , where z+i
z 100 eiπz i100 e−π (−i)100 eπ dz = 2πi + 2πi = πe−π − πeπ = −2π sinh π. 2i −2i z2 + 1
250  7 Solutions to the chapterwise exercises Solution of Exercise 3.10.37 The function is holomorphic on D = ℂ\{−i}, and the integration path turns around the pole z = i. Using the Cauchy formula for closed curves for the function f (z) = cosh πz , 2 where z0 = −i and k = 3, since f (3) (z) =
π3 πz sinh , 8 2
it follows that cosh πz 2
∫
(z +
γ
i)4
dz =
2πi (3) π4 f (−i) = . 3! 24
Solution of Exercise 3.10.38 If z = rei
π(2t−1) 2
, with t ∈ [0, 1], since limz→∞ zf (z) = 0 it follows that lim rf (γr (t)) = 0,
∀t ∈ [0, 1].
r→∞
(7.19)
Since f is continuous, there exists M(r) = max{f (z) : z ∈ {γr }} = max{f (γr (t)) : t ∈ [0, 1]} < +∞. From (7.19), we obtain that lim rM(r) = 0.
(7.20)
r→∞
From the definition of the integral, we have 1 1 aγr (t) az f (γr (t))γr (t)dt ≤ ∫eaγr (t) f (γr (t))γr (t)dt ∫ e f (z)dz = ∫ e γ 0
r
0
1
1
π(2t−1) ≤ M(r) ∫eaγr (t) γr (t)dt = rM(r)π ∫ ear cos 2 dt,
0
0
i. e., 1 az ar cos π(2t−1) 2 dt. ∫ e f (z)dz ≤ rM(r)π ∫ e γ 0
r
But 1
∫ ear cos 0
π(2t−1) 2
π 2
2 dt = ∫ ear cos u du, π 0
(7.21)
7.3 Solutions to the exercises of Chapter 3  251
and from a ≤ 0 we have that the function φ(u) = ear cos u is decreasing and convex in [0, π2 ], and thus π 2
∫ ear cos u du ≤ 0
(1 + ear )π , 4
hence 1
∫ ear cos
π(2t−1) 2
dt ≤
0
1 + ear . 2
From here, using the inequality (7.21) it follows that (1 + ear )π az . ∫ e f (z)dz ≤ rM(r) 2 γ r
Since a ≤ 0, from the above inequality and from (7.20) we deduce that lim ∫ eaz f (z)dz = 0. r→∞ γ r
Solution of Exercise 3.10.39 The proof is similar with the proof of the previous problem. If z = rei [0, 1], since limz→∞ zf (z) = 0 it follows that lim rf (γr (t)) = 0,
r→∞
π(2t+1) 2
∀t ∈ [0, 1].
, with t ∈ (7.22)
Since f is continuous, there exists M(r) = max{f (z) : z ∈ {γr }} = max{f (γr (t)) : t ∈ [0, 1]} < +∞. From (7.22), we obtain that lim rM(r) = 0.
(7.23)
r→∞
From the definition of the integral, we have 1 1 aγr (t) az f (γr (t))γr (t)dt ≤ ∫eaγr (t) f (γr (t))γr (t)dt ∫ e f (z)dz = ∫ e γ r
0
1
0
1
π(2t+1) ≤ M(r) ∫eaγr (t) γr (t)dt = rM(r)π ∫ ear cos 2 dt,
0
0
252  7 Solutions to the chapterwise exercises i. e., 1 az ar cos π(2t+1) 2 dt. ∫ e f (z)dz ≤ rM(r)π ∫ e γ
(7.24)
0
r
But 1
∫e
ar cos
π(2t+1) 2
0
π 2
2 dt = ∫ e−ar cos u du, π 0
and from a ≥ 0 we have that the function φ(u) = e−ar cos u is decreasing and convex on [0, π2 ], and thus π 2
∫ e−ar cos u du ≤ 0
(1 + e−ar )π , 4
hence 1
∫ ear cos
π(2t+1) 2
0
dt ≤
1 + e−ar . 2
From here, using the inequality (7.24) it follows that (1 + e−ar )π az . ∫ e f (z)dz ≤ rM(r) 2 γ r
Since a ≥ 0, from the above inequality and from (7.23) we deduce that lim ∫ eaz f (z)dz = 0. r→∞ γ r
Solution of Exercise 3.10.40 If f = u + iv ∈ H(G), then 𝜕2 u 𝜕2 u + = 0, 𝜕x 2 𝜕y2 i. e., u and v are harmonic functions.
𝜕2 v 𝜕2 v + = 0, 𝜕x 2 𝜕y2
7.3 Solutions to the exercises of Chapter 3  253
If z = x + iy and z0 = x0 + iy0 , it is well known that z
z
𝜕u(z) 𝜕u(z) dx + dy 𝜕x 𝜕y
u(z) − u(z0 ) = ∫ du(z) = ∫ z0
z0
z
x
y
x0
y0
𝜕v(x, y0 ) 𝜕v(x, y) 𝜕v(z) 𝜕v(z) dx − dy = ∫ dx − ∫ dy =∫ 𝜕y 𝜕x 𝜕y 𝜕x z0
and z
z
v(z) − v(z0 ) = ∫ dv(z) = ∫ z0
z0
z
𝜕v(z) 𝜕v(z) dx + dy 𝜕x 𝜕y x
y
x0
y0
𝜕u(x, y0 ) 𝜕u(z) 𝜕u(x, y) 𝜕u(z) = ∫− dx + dy = − ∫ dx + ∫ dy. 𝜕y 𝜕x 𝜕y 𝜕x z0
From here, we deduce that the corresponding real and imaginary parts of the function f are given by y
x
𝜕v(x, y0 ) 𝜕v(x, y) dx − ∫ dy, u(z) − u(z0 ) = ∫ 𝜕y 𝜕x x0
y0
x
y
𝜕u(x, y0 ) 𝜕u(x, y) dx + ∫ dy. v(z) − v(z0 ) = − ∫ 𝜕y 𝜕x x0
y0
In the next special cases, we will use the above methods to determine the real or the imaginary part of the functions. 1. Since v(x, y) = ex sin y, it follows that 𝜕v = ex sin y, 𝜕x 2
𝜕v 𝜕2 v = ex cos y ⇒ 2 = ex sin y, 𝜕y 𝜕x
𝜕2 v = −ex sin y, 𝜕y2
2
𝜕v 𝜕v thus 𝜕x 2 + 𝜕y 2 = 0, hence v is a harmonic function. Let z0 = x0 + iy0 = 0 and z = x + iy. We will determine the real part u using the above method: x
u(z) − u(0, 0) = ∫ 0
y
𝜕v(x, 0) 𝜕v(x, y) dx − ∫ dy = ex cos y − 1, 𝜕y 𝜕x 0
thus u(x, y) = ex cos y − 1 + u(0, 0) = ex cos y + k, k ∈ ℝ, hence f (z) = ex cos y + k + iex sin y,
k ∈ ℝ.
254  7 Solutions to the chapterwise exercises From the assumption f (0) = 1, we will determine the constant k ∈ ℝ as follows: f (0) = 1 ⇔ k = 0. Hence, the function will be f (z) = ex cos y + iex sin y = ez . 2. Since v(x, y) = x2 − y2 + xy, it follows that 𝜕v 𝜕2 v = −2y + x ⇒ 2 = 2, 𝜕y 𝜕x
𝜕v = 2x + y, 𝜕x 2
𝜕2 v = −2, 𝜕y2
2
𝜕v 𝜕v thus 𝜕x 2 + 𝜕y 2 = 0, hence v is a harmonic function. Let z0 = x0 + iy0 = 0 and z = x + iy. Then x
u(z) − u(0, 0) = ∫ 0
thus u(x, y) =
2
x 2
− 2xy − f (z) =
2
y 2
y
𝜕v(x, 0) 𝜕v(x, y) x2 y2 dx − ∫ dy = − 2xy − , 𝜕y 𝜕x 2 2 0
+ u(0, 0) =
2
x 2
y2 2
− 2xy −
+ k, k ∈ ℝ, hence
y2 x2 − 2xy − + k + i(x 2 − y2 + xy), 2 2
k ∈ ℝ.
From the assumption f (0) = 0, we will determine the constant k ∈ ℝ as follows: f (0) = 0 ⇔ k = 0. Hence, the function will be f (z) = (1 + 2i)
z2 . 2
3. Since v(x, y) = 2xy, it follows that 𝜕v 𝜕2 v 𝜕2 v = 2x ⇒ 2 = 2 = 0, 𝜕y 𝜕x 𝜕y
𝜕v = 2y, 𝜕x 2
2
𝜕v 𝜕v thus 𝜕x 2 + 𝜕y 2 = 0, hence v is a harmonic function. Let z0 = x0 + iy0 = 0 and z = x + iy. Then x
u(z) − u(0, 0) = ∫ 0
y
𝜕v(x, 0) 𝜕v(x, y) dx − ∫ dy = x2 − y2 , 𝜕y 𝜕x 0
thus u(x, y) = x2 − y2 + u(0, 0) = x2 − y2 + k, k ∈ ℝ, hence f (z) = x2 − y2 + k + i2xy,
k ∈ ℝ.
From the assumption f (0) = 0, we will determine the constant k ∈ ℝ as follows: f (0) = 0 ⇔ k = 0. Hence, the function will be f (z) = x2 − y2 + i2xy = z 2 .
7.3 Solutions to the exercises of Chapter 3  255
4. If u(x, y) = ex (x cos y − y sin y), we have that 𝜕u = ex (x cos y + cos y − y sin y), 𝜕x ⇒ 2
𝜕u = −ex (x sin y + sin y + y cos y) 𝜕y
𝜕2 u = −ex (x cos y + 2 cos y − y sin y), 𝜕y2
𝜕2 u = ex (x cos y + 2 cos y − y sin y), 𝜕x 2 2
hence 𝜕𝜕xu2 + 𝜕𝜕yu2 = 0, and thus u is a harmonic function. Let z0 = x0 + iy0 = 0 and z = x + iy. We will determine the imaginary part v using the above method: x
v(z) − v(0, 0) = − ∫ 0
y
𝜕u(x, 0) 𝜕u(x, y) dx + ∫ dy = ex (x sin y + y cos y), 𝜕y 𝜕x 0
x
hence v(x, y) = e (x sin y + y cos y) + v(0, 0) = ex (x sin y + y cos y) + k, k ∈ ℝ, thus f (z) = ex (x cos y − y sin y) + i[ex (x sin y + y cos y) + k],
k ∈ ℝ.
From the assumption f (0) = 0, we will determine the constant k ∈ ℝ as follows: f (0) = 0 ⇔ k = 0. Now, the function will be f (z) = ex (x cos y − y sin y) + iex (x sin y + y cos y) = zez . Solution of Exercise 3.10.41 The solution is similar to the above problem, with the difference that the starting point z0 ∈ ℂ should not be chosen z0 = 0 (because 0 ∉ D), but (for example) it could be z0 = 1 + 0i = 1 ∈ D. Since −2xy 𝜕v , = 𝜕x (x2 + y2 )2 ⇒
𝜕v x2 − y2 = 2 𝜕y (x + y2 )2
𝜕2 v 2y(3x 2 − y2 ) = , 𝜕x 2 (x2 + y2 )3
𝜕2 v 2y(3x2 − y2 ) = − , 𝜕y2 (x2 + y2 )3
the function v is harmonic. Now, the real part u may be found as follows: x
y
1
0
𝜕v(x, y) 𝜕v(x, 0) x , dx − ∫ dy = 1 − 2 u(z) − u(1, 0) = ∫ 𝜕y 𝜕x x + y2 hence u(x, y) = 1 −
x x + u(1, 0) = − 2 + k, x2 + y2 x + y2
k ∈ ℝ.
The constant k ∈ ℝ can be found by using the assumption f (1) = 0, i. e., f (1) = 0 ⇔ k = 1. Finally, we get f (z) = 1 − z1 .
256  7 Solutions to the chapterwise exercises Solution of Exercise 3.10.42 1. The function u is welldefined if and only if (0, 0) ∉ D. Since 𝜕u −2x(x2 − 3y2 ) = , 𝜕x (y2 + x2 )3
𝜕u −2y(3x2 − y2 ) = , 𝜕y (y2 + x2 )3
it follows that 𝜕2 u 𝜕2 u −6(x4 + y4 − 6y2 x2 ) = − = , 𝜕y2 𝜕x 2 (x2 + y2 )4 thus u is a harmonic function. If we choose the starting point z0 = 1 = 1 + 0i ∈ D, then x
v(x, y) − v(1, 0) = − ∫ y
1
=∫ 0
y
𝜕u(x, y) 𝜕u(x, 0) dx + ∫ dy 𝜕y 𝜕x 0
2
2
−2x(x − 3y ) 2xy dy = − 2 , 2 2 3 (y + x ) (x + y2 )2
hence f (z) =
1 + ki, z2
k ∈ ℝ,
and f ∈ H(ℂ∗ ). 2. From the definition of the function v, we have that (0, 0) ∉ D, and 2x 𝜕v = − 2x, 𝜕x y2 + x2
2y 𝜕v = + 2y, 𝜕y y2 + x2
then 𝜕2 v 2(y2 − x2 ) 𝜕2 v =− 2 = 2 − 2, 2 𝜕x 𝜕y (x + y2 )2 and thus v is a harmonic function. If we choose the starting point z0 = 1 = 1 + 0i ∈ D, then y
x
𝜕v(x, y) 𝜕v(x, 0) dx − ∫ dy u(x, y) − u(1, 0) = ∫ 𝜕y 𝜕x 1
y
= − ∫( 0
0
2x y − 2x)dy = −2 arctan + 2xy, x y2 + x2
7.3 Solutions to the exercises of Chapter 3  257
and thus y + 2xy + k + i(ln(x 2 + y2 ) − x2 + y2 ) x = i(2 log z − z 2 ) + k, k ∈ ℝ,
f (z) = −2 arctan
with D = ℂ \ {z ∈ ℂ : Re z ≤ 0, Im z = 0}. 3. Ha z = x + iy ∈ Δ = ℂ \ {z ∈ ℂ : Re z = 0}, then y y 𝜕u = − 2 ex , 𝜕x x
y 1 𝜕u = ex , 𝜕y x
hence y y(2x + y) 𝜕2 u = ex , x4 𝜕x 2
y 1 𝜕2 u = ex 2 . 𝜕y2 x
From here, we have that u is harmonic only if y y(2x + y) y 1 𝜕2 u 𝜕2 v + 2 = 0 ⇔ ex + e x 2 = 0 ⇔ (x + y)2 = 0 4 2 x 𝜕x 𝜕y x ̃ = {z = x + iy ∈ ℂ : y = −x, x ≠ 0}. ⇔z∈D
But ̃ D ⊂ ℂ is not a domain, thus there does not exist any holomorphic function in a y domain D ⊂ ℂ, of the form f = u + iv, such that u(x, y) = e x . Solution of Exercise 3.10.43 y 1. If v(x, y) = x2 +y 2 , first we will check that it is a harmonic function. A simple calculation shows that
𝜕v 2xy , =− 2 𝜕x (x + y2 )2
𝜕v 𝜕2 v 𝜕2 v 2y(3x2 − y2 ) x2 − y2 ⇒ = − = , = 2 𝜕y (x + y2 )2 𝜕x 2 𝜕y2 (x2 + y2 )3
hence the function v is harmonic. Since z0 = 0 ∉ D, let choose z0 = −1 = −1 + 0i ∈ D to be the starting point. We have y
x
𝜕v(x, 0) 𝜕v(x, y) u(x, y) − u(−1, 0) = ∫ dx − ∫ dy 𝜕y 𝜕x −1 x
=∫ −1
thus u(x, y) = 1 −
x x2 +y2
y
0
1 2xy x dx − ∫ 2 dy = −1 − 2 , x2 (x + y2 )2 x + y2 0
x + u(−1, 0) = − x2 +y 2 + k, k ∈ ℝ. From here, we deduce that
f (z) = −
x y +k+i 2 , x2 + y2 x + y2
k ∈ ℝ.
258  7 Solutions to the chapterwise exercises The constant k ∈ ℝ will be determined by using the assumption f (−2) = 0, i. e., k = − 21 . Finally, we get f (z) = − 2. If v(x, y) = 3 + x2 − y2 −
1 1 1 y x − +i 2 =− − . z 2 x2 + y2 2 x + y2
y , 2(x2 +y2 )
similarly 𝜕v x2 − y2 = −2y − 𝜕y 2(x2 + y2 )2
𝜕v xy , = 2x + 2 𝜕x (x + y2 )2
𝜕2 v 𝜕2 v y(y2 − 3x 2 ) = − = 2 + , 𝜕x 2 𝜕y2 (x2 + y2 )3
⇒
thus v is a harmonic function. For the starting point z0 = 2 = 2 + 0i ∈ D, we obtain x
u(x, y) − u(2, 0) = ∫ 2 x
y
𝜕v(x, y) 𝜕v(x, 0) dx − ∫ dy 𝜕y 𝜕x 0 y
= ∫(− 2
thus u(x, y) =
x 2(x2 +y2 )
1 xy 1 x , )dx − ∫(2x + 2 )dy = − − 2xy + 2 2 2 2 4 2x (x + y ) 2(x + y2 ) 0
− 2xy + k, k ∈ ℝ. The function f will be of the form
f (z) = −2xy +
x y + k + i(3 + x2 − y2 − ), 2(x2 + y2 ) 2(x2 + y2 )
k ∈ ℝ,
and since f (2) = 41 + k + 7i ≠ 0, ∀k ∈ ℝ, we conclude that there not exists any holomorphic function which satisfies the assumptions. 3. Since u(x, y) = 21 ln(x 2 + y2 ), we have x 𝜕u = 2 , 𝜕x x + y2
y 𝜕u 𝜕2 u 𝜕2 u y2 − x2 = 2 ⇒ = − = , 𝜕y x + y2 𝜕x 2 𝜕y2 (x2 + y2 )2
thus the function u is harmonic. Let choose z0 = 1 = 1 + 0i ∈ D to be the starting point. Then x
v(x, y) − v(1, 0) = − ∫ 1
y
y
0
0
𝜕u(x, y) x y 𝜕u(x, 0) dy = arctan , dx + ∫ dy = ∫ 2 𝜕y 𝜕x x x + y2
hence v(x, y) = arctan xy + k, k ∈ ℝ. From here, we get f (z) = ln √x2 + y2 + i(arctan
y + k) = log z + ik, x
k ∈ ℝ.
7.3 Solutions to the exercises of Chapter 3  259
4. If v(x, y) = ln(x2 + y2 ) + x − 2y, then 𝜕v 𝜕2 v 𝜕2 v 2(y2 − x2 ) 2y ⇒ 2 =− 2 = 2 , = −2 + 2 2 𝜕y x +y 𝜕x 𝜕y (x + y2 )2
𝜕v 2x , =1+ 2 𝜕x x + y2
hence the function v is harmonic. Since z0 = 0 ∉ D, choose the starting point to be z0 = 1 = 1 + 0i ∈ D. Then y
x
𝜕v(x, 0) 𝜕v(x, y) u(x, y) − u(1, 0) = ∫ dx − ∫ dy 𝜕y 𝜕x 1 x
0
y
= ∫(−2)dx − ∫( 1
0
hence u(x, y) = −2x − y − 2 arctan
y x
y 2x + 1)dy = −2x + 2 − y − 2 arctan , x x2 + y2
+ k, k ∈ ℝ. The required function will be
f (z) = 2i log z − (2 − i)z + k,
k ∈ ℝ.
5. If v(x, y) = 1 + xy, then 𝜕v = y, 𝜕x
𝜕v 𝜕2 v 𝜕2 v = x ⇒ 2 = 2 = 0, 𝜕y 𝜕x 𝜕y
thus v is a harmonic function. Let z0 = 0 ∈ D be the starting point. Then x
y
x
y
0
0
0
0
𝜕v(x, y) 1 𝜕v(x, 0) dx − ∫ dy = ∫ xdx − ∫ ydy = (x 2 − y2 ), u(x, y) − u(0, 0) = ∫ 𝜕y 𝜕x 2 hence f (z) = y , x2 +y2
6. Since v(x, y) = y −
z2 + k + i, 2
k ∈ ℝ.
it follows that
𝜕v 𝜕2 v 𝜕2 v 2y(3x2 − y) x2 − y2 ⇒ = − = , =1− 2 𝜕y (x + y2 )2 𝜕x 2 𝜕y2 (x2 + y2 )3
𝜕v 2xy , = 𝜕x (x2 + y2 )2
thus v is a harmonic function. If we choose the starting point z0 = i = 0 + i ∈ D, we get x
y
0 x
1
𝜕v(x, y) 𝜕v(x, 1) dx − ∫ dy u(x, y) − u(0, 1) = ∫ 𝜕y 𝜕x = ∫(1 − 0
y
x2 − 1 2xy x )dx − ∫ 2 dy = x + 2 . (x2 + 1)2 (x + y2 )2 x + y2 1
260  7 Solutions to the chapterwise exercises Hence, the function will be of the form f (z) = z + 7. If u(x, y) =
x x2 +(y+1)2
+
x , x2 +(y−1)2
1 + k, z
k ∈ ℝ.
then
𝜕u (y − 1)2 − x2 (y + 1)2 − x2 + , = 2 𝜕x (x + (y + 1)2 )2 (x2 + (y − 1)2 )2 𝜕u −2(y + 1)x −2(y − 1)x = + , 𝜕y (x2 + (y + 1)2 )2 (x2 + (y − 1)2 )2 hence 𝜕2 u 𝜕2 u 2x(3(y + 1)2 − x2 ) 2x(3(y − 1)2 − x2 ) = − = + , 𝜕y2 𝜕x2 (x2 + (y + 1)2 )3 (x2 + (y − 1)2 )3 and the function u is harmonic. If we let z0 = 0 ∈ D be the starting point, then y
x
v(x, y) − v(0, 0) = − ∫ y
0
=−
0
(y − 1)2 − x2 (y + 1) − x + )dy (x2 + (y + 1)2 )2 (x2 + (y − 1)2 )2 2
= ∫( 0
𝜕u(x, y) 𝜕u(x, 0) dx + ∫ dy 𝜕y 𝜕x 2
y+1 y−1 − , (y + 1)2 + x2 (y − 1)2 + x2
thus f (z) =
x x + x2 + (y + 1)2 x2 + (y − 1)2 + i(−
y+1 y−1 − + k), (y + 1)2 + x2 (y − 1)2 + x2
From the assumption f (1) = 1, it follows that k = 0, and then f (z) = 8. From v(x, y) = − (x+1)y2 +y2 −
y , (x−1)2 +y2
2z . z2 + 1
we have that
2(x − 1)y 2(x + 1)y 𝜕v + 2 , = 2 2 2 𝜕x (y + (x + 1) ) (y + (x − 1)2 )2
𝜕v (x − 1)2 − y2 (x + 1)2 − y2 − , =− 2 𝜕y (y + (x + 1)2 )2 (y2 + (x − 1)2 )2
k ∈ ℝ.
7.3 Solutions to the exercises of Chapter 3  261
hence 𝜕2 v 𝜕2 v −2y(3(x + 1)2 − y2 ) −2y(3(x − 1)2 − y2 ) =− 2 = + , 2 𝜕x 𝜕y (y2 + (x + 1)2 )3 (y2 + (x − 1)2 )3 thus, the function v is harmonic. Choosing z0 = 0 ∈ D, we obtain x
y
0 x
0
𝜕v(x, y) 𝜕v(x, 0) dx − ∫ dy u(x, y) − u(0, 0) = ∫ 𝜕y 𝜕x = ∫(− 0
1 1 − )dx (x + 1)2 (x − 1)2
y
− ∫( 0
=
2(x + 1)y 2(x − 1)y + )dy (y2 + (x + 1)2 )2 (y2 + (x − 1)2 )2
x−1 x+1 + , (x + 1)2 + y2 (x − 1)2 + y2
hence f (z) =
x+1 x−1 + +k 2 2 (x + 1) + y (x − 1)2 + y2 + i(−
y2
y y − 2 ), 2 + (x + 1) y + (x − 1)2
k ∈ ℝ.
The assumption f (0) = 0 yields that k = 0, and f (z) =
2z . −1
z2
2xy 9. If v(x, y) = − (x2 −y2 −1) 2 +4x 2 y 2 , then
2y((x2 − y2 − 1)2 − 4x4 + 4x 2 ) 𝜕v =− 𝜕x ((x2 − y2 − 1)2 + 4x2 y2 )2
𝜕v 2x((x2 − y2 − 1)2 − 4y4 + 4y2 ) . =− 𝜕y ((x2 − y2 − 1)2 + 4x2 y2 )2 The relation 𝜕2 v 𝜕2 v =− 2 2 𝜕x 𝜕y =
−8xy(3 − 14x2 y2 + 3y2 − 3x2 − 3x4 − 3y4 + 3x6 − 3y6 + 3x4 y2 − 3x2 y4 ) (x2 − 2x + y2 + 1)3 (x2 + 2x + y2 + 1)3
shows that v is a harmonic function.
262  7 Solutions to the chapterwise exercises Letting z0 = 0 ∈ D, we have x
u(x, y) − u(0, 0) = ∫ 0 x
y
𝜕v(x, 0) 𝜕v(x, y) dx − ∫ dy 𝜕y 𝜕x 0
y
2x 2y((x2 − y2 − 1)2 − 4x4 + 4x 2 ) = ∫(− 2 )dx + dy ∫ (x − 1)2 ((x2 − y2 − 1)2 + 4x2 y2 )2 0
=
2
2
0
x −y −1 + 1, (x2 − y2 − 1)2 + 4x2 y2
thus f (z) =
x2 − y2 − 1 2xy +k−i 2 , 2 2 2 2 2 2 (x − y − 1) + 4x y (x − y − 1)2 + 4x2 y2
k ∈ ℝ.
From the assumption f (i) = − 21 , we obtain k = 0, hence f (z) =
1 . z2 − 1
Solution of Exercise 3.10.44 1. We will write the function in the exponential form. If z = r1 eiθ1 and z − 2i = r2 eiθ2 , where r1 , r2 > 0, then 2i − z = r2 ei(θ2 +π) . We obtain that r θ1 −θ2 −π+2kπ 3 , wk (z) = √3 1 ei r2
k ∈ {0, 1, 2}.
If the function w is holomorphic in D ⊂ ℂ, i. e., w ∈ H(D), then for any arbitrary closed curve that lies in D, i. e., γ ∈ 𝒟D (z0 ), ∀z0 ∈ D, the branch of the multivalued function w remains the same whenever the point z ∈ {γ} runs over all the path γ. We will discuss the next three cases: Case 1. If the closed curve γ turns around only the point 0. If z ∈ {γ} runs once over the curve γ, then we obtain the value r θ1 −θ2 −π+2(k+1)π 3 wk+1 (z) = √3 1 ei , r2
k ∈ {0, 1, 2}.
Case 2. If the closed curve γ turns around, we have only the point 2i. If z ∈ {γ} runs once over the curve γ, then we obtain the value r θ1 −θ2 −π+2(k−1)π 3 wk−1 (z) = √3 1 ei , r2
k ∈ {0, 1, 2}.
7.3 Solutions to the exercises of Chapter 3  263
Case 3. If the closed curve γ turns around both of the points 0 and 2i. If z ∈ {γ} runs once over the curve γ, then we obtain the value r θ1 −θ2 −π−2kπ 3 wk (z) = √3 1 ei , r2
k ∈ {0, 1, 2}.
It follows that the domain D cannot contain only one of these points, but if D contains one of them, then it is necessary to contain both of these points. Hence D = ℂ \ {z ∈ ℂ : 0 ≤ Im z ≤ 2, Re z = 0}, or D = ℂ \ {z ∈ ℂ : Im z ≥ 0, Re z = 0},
D = ℂ \ {z ∈ ℂ : Im z ≤ 2, Re z = 0}.
The other points of this problem will be solved similarly. 2. The exponential form of the function is i 3 wk (z) = √r 1e
θ1 +2kπ 3
k ∈ {0, 1, 2},
,
if we denote z + 1 = z − (−1) = r1 eiθ1 ,
r1 > 0.
Here, we will discuss only one case. If the closed curve γ turns once around the point −1, then we obtain the value i 3 wk+1 (z) = √r 1e
θ1 +2(k+1)π 3
,
k ∈ {0, 1, 2}.
It follows that the domain D cannot contain the point −1, hence D = ℂ \ {z ∈ ℂ : Im z = 0, Re z ≤ −1}, or D = ℂ \ {z ∈ ℂ : Im z = 0, Re z ≥ −1}. 3. We will use the exponential form for the argument. If z+i = r1 eiθ1 and z−i = r2 eiθ2 , where r1 , r2 > 0, then z 2 + 1 = r1 r2 ei(θ1 +θ2 ) . Thus, we have wk (z) = ln(r1 r2 ) + i(θ1 + θ2 + 2kπ),
k ∈ ℤ.
Similar to the first point of the problem, we will discuss the next three cases: Case 1. If the closed curve γ turns around only the point i, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk+1 (z) = ln(r1 r2 ) + i(θ1 + θ2 + 2(k + 1)π),
k ∈ ℤ.
264  7 Solutions to the chapterwise exercises Case 2. If the closed curve γ turns around only the point −i, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk+1 (z) = ln(r1 r2 ) + i(θ1 + θ2 + 2(k + 1)π),
k ∈ ℤ.
Case 3. If the closed curve γ turns around both of the points i and −i, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk+2 (z) = ln(r1 r2 ) + i(θ1 + θ2 + 2(k + 2)π),
k ∈ ℤ.
It follows that the domain D cannot contain any of these points, hence D = ℂ \ {z ∈ ℂ :  Im z ≥ 1, Re z = 0}, or D = ℂ \ {z ∈ ℂ : Im z ≥ −1, Re z = 0},
D = ℂ \ {z ∈ ℂ : Im z ≤ 1, Re z = 0}.
4. We will use the exponential form for the argument, i. e., z − 1 = r1 eiθ1 and z + 1 = z−1 = rr1 ei(θ1 −θ2 ) , and we get r2 e , where r1 , r2 > 0. Thus z+1 iθ2
2
wk (z) = ln
r1 + i(θ1 − θ2 + 2kπ), r2
k ∈ ℤ.
Since two singular points are involved, we will discuss the next three cases: Case 1. If the closed curve γ turns around only the point −1, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk−1 (z) = ln
r1 + i(θ1 − θ2 + 2(k − 1)π), r2
k ∈ ℤ.
Case 2. If the closed curve γ turns around only the point 1, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk+1 (z) = ln
r1 + i(θ1 − θ2 + 2(k + 1)π), r2
k ∈ ℤ.
Case 3. If the closed curve γ turns around both of the points −1 and 1, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk (z) = ln
r1 + i(θ1 − θ2 + 2kπ), r2
k ∈ ℤ.
It follows that in the domain D we cannot turn around to only one of these points, but we can turn around both of them, hence D = ℂ \ [−1, 1] = ℂ \ {z ∈ ℂ :  Re z ≤ 1, Im z = 0},
7.3 Solutions to the exercises of Chapter 3  265
or D = ℂ \ [−1, +∞) = ℂ \ {z ∈ ℂ : Re z ≥ −1, Im z = 0}, D = ℂ \ (−∞, 1] = ℂ \ {z ∈ ℂ : Re z ≤ 1, Im z = 0}.
5. The exponential form of the function is ln r + i(θ + 2kπ) , reiθ
wk (z) =
k ∈ ℤ,
whenever z = reiθ , with r > 0. Here, we will discuss only one case. If the closed curve γ turns once around the point 0, then we obtain the value wk+1 (z) =
ln r + i(θ + 2(k + 1)π) , reiθ
k ∈ ℤ,
because rei(θ+2π) = reiθ . It follows that the domain D cannot contain the point 0, hence D = ℂ \ {z ∈ ℂ : Im z = 0, Re z ≤ 0}, or D = ℂ \ {z ∈ ℂ : Im z = 0, Re z ≥ 0}. 6. We will use the exponential form of the function. If z − i = r1 eiθ1 and z + i = r2 eiθ2 , where r1 , r2 > 0, we obtain wk,l (z) = √r1 ei
θ1 +2kπ 2
+ √r2 ei
θ2 +2lπ 2
k, l ∈ {0, 1}.
,
Since two singular points are involved, we will discuss the next three cases: Case 1. If the closed curve γ turns around only the point i, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk+1,l (z) = √r1 ei
θ1 +2(k+1)π 2
+ √r2 ei
θ2 +2lπ 2
k, l ∈ {0, 1}.
,
Case 2. If the closed curve γ turns around only the point −i, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk,l+1 (z) = √r1 ei
θ1 +2kπ 2
+ √r2 ei
θ2 +2(l+1)π 2
k, l ∈ {0, 1}.
,
Case 3. If the closed curve γ turns around both of the points i and −i, and z ∈ {γ} runs once over the curve γ, then we obtain the value wk+1,l+1 (z) = √r1 ei
θ1 +2(k+1)π 2
+ √r2 ei
θ2 +2(l+1)π 2
,
k, l ∈ {0, 1}.
266  7 Solutions to the chapterwise exercises It follows that the domain D cannot contain any of these points, hence D = ℂ \ {z ∈ ℂ :  Im z ≥ 1, Re z = 0}, or D = ℂ \ {z ∈ ℂ : Im z ≥ −1, Re z = 0}, D = ℂ \ {z ∈ ℂ : Im z ≤ 1, Re z = 0}.
Solution of Exercise 3.10.45 1. The exponential form of the function is r θ1 −θ2 +2kπ wk (z) = √3 1 ei 3 , r2
k ∈ {0, 1, 2},
where z + 1 = r1 eiθ1 , z − 1 = r2 eiθ2 , and r1 , r2 > 0. It follows that the domain ̃ D where the function w is holomorphic, cannot contain only one of the points −1 or 1, but necessary both of them. Hence, this domain does not contain any of these points, or both of them, for instance ̃ D = ℂ \ [−1, 1]. Since D ⊂ ̃ D, we deduce that there exists a branch of the function w, i. e., holomorphic on the domain D. 2. We will write the exponential form of the function wk (z) = 2 ln
r1 + 2i(θ1 − θ2 + 2kπ), r2
k ∈ ℤ,
where z + 1 = r1 eiθ1 , z − 1 = r2 eiθ2 and r1 , r2 > 0. It follows that the domain ̃ D where the function w is holomorphic, cannot contain only one of the points −1 or 1, but necessary both of them. Hence, this domain does not contain any of these points, or both of them, for instance ̃ D = ℂ \ [−1, 1]. D, we deduce that there exists a branch of the function w, i. e., holomorphic Since D ⊂ ̃ on the domain D. 3. The exponential form of the function is wk (z) = √r1 r2 r3 r4 ei
θ1 +θ2 +θ3 +θ4 +2kπ 2
,
k ∈ {0, 1},
7.3 Solutions to the exercises of Chapter 3  267
where z +1 = r1 eiθ1 ,
z −2 = r4 eiθ4
z +2 = r3 eiθ3 ,
z −1 = r2 eiθ2 ,
and rk > 0, k ∈ {1, 2, 3, 4}.
If z ∈ D, then Re z > 0, then the points −1 and −2 does not belong to D. If γ ∈ 𝒟D (z0 ), z0 ∈ D, is a closed curve, and if γ turns around the points 1 and 2, i. e., γ turns around both of these points, then we obtain wk (z) = wk+2 (z) = √r1 r2 r3 r4 ei
θ1 +θ2 +θ3 +θ4 +2(k+2)π 2
,
k ∈ {0, 1}.
It follows that there exists a branch of the function w that is holomorphic on the domain D.
Solution of Exercise 3.10.46 1. First, we will write the function in the exponential form fk (z) = √3 rei
θ+2kπ 3
k ∈ {0, 1, 2},
,
where z + 1 = reiθ , r > 0, θ ∈ [0, 2π].
According to the point 2 of Exercise 3.10.44, the function has an holomorphic branch on those domains that does not contain the point. The given domain satisfies this condition, because D = ℂ \ [−1, +∞). Since −2 ∈ D, then f0 (−2) = −1 ⇔ ei
π+2kπ 3
= −1 ⇔ k = 1,
thus f0 (z) = √3 rei
θ+2π 3
,
where z + 1 = reiθ , r > 0, θ ∈ (0, 2π).
2. We will determine the parameters that appeared in the previous exponential form, replacing the variable by i and −i, i. e., z = i ⇒ r = √2,
θ=
π , 4
z = −i ⇒ r = √2,
θ=
7π . 4
and
Then the required values will be 3π
f0 (i) = √2ei 4 = √2(− 6
6
√2 √2 +i ) 2 2
268  7 Solutions to the chapterwise exercises and 5π
f0 (−i) = √2ei 4 = √2(− 6
6
√2 √2 − i ). 2 2
The value of the derivative is f0 (z) =
1 , 3(f0 (z))2
z ∈ D,
then we get f0 (i) =
1
3π 3(√2ei 3 )2 6
=
i 3√3 2
and f0 (−i) =
1
5π 3(√2ei 4 )2 6
=−
i . 3√3 2
3. Since ∀zf ∈ Tf ⇔ zf + 1 = rei0 ⇔ r = zf + 1,
θ = 0,
the values of the function ̃f on the superior border Tf of T are 2π
i3 3 ̃f (z ) = √z f + 1e , f
∀zf ∈ Tf .
since ∀za ∈ Ta ⇔ za + 1 = rei2π ⇔ r = za + 1,
θ = 2π,
the values of the function ̃f on the border Ta of T are 4π
̃f (z ) = √3 z + 1ei 3 , a a
∀za ∈ Ta .
Solution of Exercise 3.10.47 1. To the point 6 of Exercise 3.10.44 we already obtained that the function has a holomorphic branch on the given domain, that has the form fk,l (z) = √r1 ei
θ1 +2kπ 2
+ √r2 ei
θ2 +2lπ 2
,
k, l ∈ {0, 1},
where z − i = r1 eiθ1 ,
r1 > 0, θ1 ∈ (−
3π π , ) and z + i = r2 eiθ2 , 2 2
π 3π r2 > 0, θ2 ∈ (− , ). 2 2
Using the given assumption, we will determine now the required branch. We have z = 0 ⇔ −i = r1 eiθ1 ⇒ r1 = 1,
π θ1 = − , 2
7.3 Solutions to the exercises of Chapter 3  269
and z = 0 ⇔ i = r2 eiθ2 ⇒ r2 = 1,
θ2 =
π , 2
and since f0 (0) = −i√2 ⇔ ei
− π2 +2kπ
+ ei
2
π +2lπ 2 2
= −i√2 ⇔ k = 0,
l = 1,
we obtain θ1
f0 (z) = f0,1 (z) = √r1 ei 2 + √r2 ei
θ2 +2π 2
,
where z − i = r1 eiθ1 , z + i = r2 eiθ2 .
2. We will determine the parameters that appeared in this last exponential form, replacing the variable by √13 and 2. Thus z1 =
2 1 ⇒ r1 = , √3 √3
π θ1 = − , 3
r2 =
2 , √3
θ2 =
π , 3
and z2 = 2 ⇒ r1 = √5,
1 θ1 = − arctan , 2
r2 = √5,
1 θ2 = arctan . 2
The required values are f0 (
7π π 1 2 i√6√4 3 , ) = √ (e−i 6 + ei 6 ) = − √3 √3 3
and i
1
f0 (2) = √5(e− 2 arctan 2 + ei 4
arctan 21 +2π 2
1 1 4 ) = −2i√5 sin( arctan ). 2 2
3. Since ∀z ∈ T1j ⇔ r1 = z − i,
π θ1 = − , 2
r2 = z + i,
π θ2 = − , 2
the values of the function ̃f on the righthand side border T1j of T1 are given by π
3π
̃f (z) = √z − ie−i 4 + √z + iei 4 ,
∀z ∈ T1j .
Since π θ1 = − , 2
∀z ∈ T1b ⇔ r1 = z − i,
r2 = z + i,
θ2 =
3π , 2
the values of the function ̃f on the lefthand side border T1b of T1 are given by π
7π
̃f (z) = √z − ie−i 4 + √z + iei 4 ,
∀z ∈ T1b .
270  7 Solutions to the chapterwise exercises Since ∀z ∈ T2j ⇔ r1 = z − i,
θ1 =
π , 2
r2 = z + i,
θ2 =
π , 2
the values of the function ̃f on the righthand side border T2j of T2 are given by π
5π
̃f (z) = √z − iei 4 + √z + iei 4 ,
∀z ∈ T2j .
Since ∀z ∈ T2b ⇔ r1 = z − i,
θ1 = −
3π , 2
r2 = z + i,
θ2 =
π , 2
the values of the function ̃f on the lefthand side border T2b of T2 are given by 3π
5π
̃f (z) = √z − ie−i 4 + √z + iei 4 ,
∀z ∈ T2b .
Solution of Exercise 3.10.48 1. To point 3 of Exercise 3.10.44, we have already obtained that the function has a holomorphic branch on the given domain that has the form fk (z) = ln(r1 r2 ) + i(θ1 + θ2 + 2kπ),
k ∈ ℤ,
where z − i = r1 eiθ1 ,
r1 > 0, θ1 ∈ (−
3π π , ) and z + i = r2 eiθ2 , 2 2
π 3π ). r2 > 0, θ2 ∈ (− , 2 2
Using the given assumption, we will determine now the required branch. We have z = 0 ⇔ −i = r1 eiθ1 ⇒ r1 = 1,
θ1 = −
π 2
and z = 0 ⇔ i = r2 eiθ2 ⇒ r2 = 1,
θ2 =
π . 2
Using the fact f0 (0) = 0 ⇔ i2kπ = 0 ⇔ k = 0, we deduce f0 (z) = ln(r1 r2 ) + i(θ1 + θ2 ),
where z − i = r1 eiθ1 , z + i = r2 eiθ2 .
7.3 Solutions to the exercises of Chapter 3  271
2. We will determine the parameters that appeared in this last exponential form, replacing the variable by 2 + i and −2 − i. Thus z1 = 2 + i ⇒ r1 = 2,
θ1 = 0,
z2 = −2 − i ⇒ r1 = 2√2,
r2 = 2√2,
θ1 = −
3π , 4
θ2 =
r2 = 2,
π , 4 θ2 = π,
and z3 =
1 i ⇒ r1 = , 2 2
π θ1 = − , 2
r2 =
3 , 2
θ2 =
π . 2
The required values are π f0 (2 + i) = ln 4√2 + i , 4
f0 (−2 − i) = ln 4√2 − i
π 4
i 3 and f0 ( ) = ln . 2 4
3. The values of the function ̃f on the righthand side border T1j of T1 are given by ̃f (z) = ln(z − iz + i) − iπ,
∀z ∈ T1j ,
because ∀z ∈ T1j ⇔ r1 = z − i,
π θ1 = − , 2
r2 = z + i,
π θ2 = − . 2
The values of the function ̃f on the lefthand side border T1b of T1 are given by ̃f (z) = ln(z − iz + i) + iπ,
∀z ∈ T1b ,
because ∀z ∈ T1b ⇔ r1 = z − i,
π θ1 = − , 2
r2 = z + i,
θ2 =
3π . 2
The values of the function ̃f on the righthand side border T2j of T2 are given by ̃f (z) = ln(z − iz + i) + iπ,
∀z ∈ T2j ,
because ∀z ∈ T2j ⇔ r1 = z − i,
θ1 =
π , 2
r2 = z + i,
θ2 =
π . 2
The values of the function ̃f on the lefthand side border T2b of T2 are given by ̃f (z) = ln(z − iz + i) − iπ,
∀z ∈ T2b ,
because ∀z ∈ T2b ⇔ r1 = z − i,
θ1 = −
3π , 2
r2 = z + i,
θ2 =
π . 2
272  7 Solutions to the chapterwise exercises Solution of Exercise 3.10.49 The exponential form of the function is fk (z) = √r1 r2 ei
θ1 +θ2 +2kπ 2
,
k ∈ {0, 1},
where z − i = r1 eiθ1 ,
z + i = r2 eiθ2 ,
r1 , r2 > 0.
In each of the three cases, the argument θ belongs in the different intervals, as follows: 1. θ1 ∈ (−
3π π , ), 2 2
π 3π θ2 ∈ (− , ). 2 2
θ1 ∈ (−
3π π , ), 2 2
θ2 ∈ (−
2. 3π π , ). 2 2
3. π 3π θ1 ∈ [− , ], 2 2
π 3π π π θ2 ∈ [− , ] : (θ1 , θ2 ) ≠ (− , ), 2 2 2 2
(θ1 , θ2 ) ≠ (
3π π , ). 2 2
Solution of Exercise 3.10.50 To point 5 of Exercise 3.10.44, we have already obtained that the function has a holomorphic branch on the given domain that has the form fk (z) =
ln r + i(θ + 2kπ) , reiθ
k ∈ ℤ,
where z = reiθ ,
r > 0,
θ ∈ (−π, π).
Using the given assumption, we will determine now the required branch. If z > 0, then r = z > 0 and θ = 0. Since 1 z ∈ (0, +∞) ⇒ fk (z) = (ln r + i2kπ) ∈ ℝ, r we get k = 0, hence 1 f0 (z) = (ln r + iθ)e−iθ , r
where z = reiθ , r > 0, θ ∈ (−π, π).
7.3 Solutions to the exercises of Chapter 3  273
2. We will determine the parameters that appeared in this last exponential form, replacing the variable by i and −i. Since z = i ⇔ r = 1,
θ=
π , 2
and z = −i ⇔ r = 1,
π θ=− , 2
the required values are f0 (i) =
π , 2
f0 (−i) =
π . 2
Solution of Exercise 3.10.51 Letting z ∈ {Im z < 0} ∪ {Im z ≥ 0, z > r} = Δ, then 𝜕Δ = {γ}, and Δ is the unbounded component of the set ℂ \ {γ}, and according to the index theorem we have n(γ, z) = 0. Letting z ∈ {Im z > 0, z < r} = D, then 𝜕D = {γ}. Then there exists ρ > 0 such that U(z; ρ) ⊂ D. Denoting by γ̃(t) = z + ρe2πit , t ∈ [0, 1], then {γ̃} = 𝜕U(z; ρ) ⊂ D and γ̃ ∼ γ. From here, we obtain that ℂ\{z}
n(γ, z) =
dζ dζ 1 1 1 = = ∫ ∫ 2πi ζ − z 2πi ζ − z 2πi γ
γ̃
∫ 𝜕U(z;ρ)
dζ = 1. ζ −z
Solution of Exercise 3.10.52 Consider z ∈ {z ∈ ℂ : r < z < ρ, Im z > 0} = D. Then 𝜕D = {γ}, and there exists r > 0 such that U(z; r) ⊂ D. If γ̃(t) = z + re2πit , t ∈ [0, 1], then {γ̃} = 𝜕U(z; r) ⊂ D and γ̃ ∼ γ. ℂ\{z}
It follows that n(γ, z) =
dζ dζ 1 1 1 = = ∫ ∫ 2πi ζ − z 2πi ζ − z 2πi γ
γ̃
∫ 𝜕U(z;r)
dζ = 1. ζ −z
Solution of Exercise 3.10.53 We will prove first that the function f has primitives on the set G ⊂ ℂ, and according to the Morera theorem it follows that f ∈ H(G). Consider the line a d = AB, where A(a) and B(b), i. e., d : d(t) = (1 − t)a + tb,
t ∈ ℝ.
274  7 Solutions to the chapterwise exercises Since G \ d ⊂ G \ [a, b], then f ∈ C(G) ∩ H(G \ (G ∩ d)). Let z0 ∈ G be a given arbitrary point, and let r > 0 such that U(z0 ; r) ⊂ G. The disc U(z0 ; r) will intersect at most the line d, and according to the theorem related to the holomorphic functions and the primitive existence it follows that f has primitives on the disc U(z0 ; r). Then, from the Morera theorem we get f ∈ H(U(z0 ; r)). Since the point z0 ∈ G was arbitrary, we deduce that f ∈ H(G). Solution of Exercise 3.10.54 According to Exercise 3.10.53, the function f is entire, i. e., f ∈ H(ℂ). But f is bounded, hence from the Liouville theorem it follows that f is constant on ℂ. Solution of Exercise 3.10.55 1−z It is well known that g({z ∈ ℂ : Re z > 0}) = U(0; 1), where g(z) = 1+z . The upper halfplane is mapped onto the right halfplane by the rotation with the center at the origin and the rotation angle equal to − π2 , i. e.,
l : {z ∈ ℂ : Im z > 0} → {w ∈ ℂ : Re w > 0},
l(z) = −iz.
Let us define the function ψ = g ∘ l ∘ f ∈ H(ℂ), which is bounded, because ψ(z) = g(l(f (z)) < 1,
∀z ∈ ℂ.
Hence ψ is a bounded entire function, and from the Liouville theorem it will be constant, i. e., ψ(z) = c,
∀z ∈ ℂ,
where c < 1 ⇔ f (z) =
1c−1 = k, i c+1
∀z ∈ ℂ.
It follows that there not exists any nonconstant functions that satisfy the required properties. Solution of Exercise 3.10.56 It is well known that g({z ∈ ℂ : Re z > 0}) = U(0; 1), where g(z) = function ψ = g ∘ f ∈ H(ℂ),
1−z . Let us define the 1+z
7.3 Solutions to the exercises of Chapter 3  275
which is bounded, because ψ(z) = g(f (z)) < 1,
∀z ∈ ℂ.
Hence ψ is a bounded entire function, and from the Liouville theorem it will be constant, i. e., ψ(z) = c,
∀z ∈ ℂ,
where c < 1 ⇔ f (z) =
1−c = k, 1+c
∀z ∈ ℂ.
It follows that there not exists any nonconstant functions that satisfy the required properties.
Solution of Exercise 3.10.57 For the solution, we will use again the Liouville theorem. It is well known that g({z ∈ ℂ : z > 1}) = U(0; 1), where g(z) = z1 . Let us define the function ψ = g ∘ f ∈ H(ℂ), which is bounded, because ψ(z) = g(f (z)) < 1,
∀z ∈ ℂ.
Hence ψ is a bounded entire function, and from the Liouville theorem it will be constant, i. e., ψ(z) = c,
∀z ∈ ℂ,
where c < 1 ⇔ f (z) =
1 = k, c
∀z ∈ ℂ.
It follows that there not exists any nonconstant functions that satisfy the required properties.
Solution of Exercise 3.10.58 We will use the assumption as follows: lim f (z) = a ∈ ℂ ⇔ [∀ε > 0, ∃σ > 0 : ∀z ∈ ℂ, z > σ ⇒ f (z) − a < ε].
z→∞
Let ε = 1; hence ∃σ > 0 such that ∀z ∈ ℂ : z > σ ⇒ f (z) − a < 1. From here, we get f (z) ∈ U(a; 1),
∀z ∈ ℂ : z > σ.
276  7 Solutions to the chapterwise exercises If ∀z ∈ U(0; σ), that is a compact set, since f ∈ C(U(0; σ)) it follows that ∃M > 0 : f (z) ≤ M,
∀z ∈ U(0; σ).
(7.25)
If ∀z ∈ ℂ \ U(0; σ), then f (z) ∈ U(a; 1) ⇒ f (z) − a ≤ f (z) − a < 1 ⇒ −1 < f (z) − a < 1, thus f (z) < 1 + a,
∀z ∈ ℂ : z > σ.
(7.26)
Letting N = max{M; a + 1}, from the relations (7.25) and (7.26) we deduce that f (z) ≤ N, ∀z ∈ ℂ, hence f is a bounded entire function. Now, according to the Liouville theorem we conclude that f is a constant function on ℂ.
Solution of Exercise 3.10.59 If g = Re f is a bounded function, then ∃M > 0 : g(z) < M, ∀z ∈ ℂ. Let us define the function h : ℂ → ℂ,
h(z) = ef (z) .
Since f is holomorphic on ℂ, the function h will be also holomorphic on ℂ. Now, it will be sufficient to prove that h is a bounded function. But Re f (z) = eg(z) < eM , h(z) = e
∀z ∈ ℂ,
and according to the Liouville theorem we obtain that h(z) = k,
∀z ∈ ℂ ⇔ ef (z) = k,
∀z ∈ ℂ.
It follows f (z) ∈ {ln k + i(arg k + 2lπ) : l ∈ ℤ} = {cl : l ∈ ℤ}, and since f is continuous, then ∃ ! l ∈ ℤ : f (z) = cl . Hence, the function f is constant on ℂ.
7.4 Solutions to the exercises of Chapter 4  277
Solution of Exercise 3.10.60 Let denote K = 𝜕U(0; r), G = ℂ \ K, and let g : G × K → ℂ,
g(z, ζ ) =
f (ζ ) . ζ −z
f (ζ )
Since the function gz (z, ζ ) = − (ζ −z)2 is continuous on G × K (because f ∈ C(ℂ)), then the function F : G → ℂ,
F(z) =
∫ 𝜕U(0;r)
f (ζ ) dζ ζ −z
will be holomorphic on G = ℂ \ K, i. e., F ∈ H(ℂ \ K). Let a ∈ ℂ∗ an arbitrary fixed point. Then there exists r > 0 such that a > r. Since F ∈ H(ℂ \ K) and ∀z ∈ ℂ \ U(0; r) we have f (z) = F(z), then it follows that the function f is differentiable at the point a ∈ ℂ∗ , and thus f ∈ H(ℂ∗ ). From here, since f ∈ C(ℂ), we deduce that f ∈ H(ℂ). If r > 0 is fixed, since ∀z ∈ ℂ \ U(0; r) we have f (z) = F(z), it follows that lim f (z) = lim F(z) = 0,
z→∞
z→∞
i. e., f is a bounded entire function. Now, from the Liouville theorem we obtain that f is a constant function on ℂ, i. e., f (z) = c, ∀z ∈ ℂ. If γ(t) = re2πit , t ∈ [0, 1], then c=
∫ 𝜕U(0;r)
1 c dζ = c ∫ dζ = cn(γ, z) = 0, ζ −z ζ −z γ
∀z ∈ ℂ : z > r,
hence f (z) = 0, ∀z ∈ ℂ.
7.4 Solutions to the exercises of Chapter 4 Solution of Exercise 4.10.1 We will use the formulas from the Cauchy–Hadamard theorem, i. e., R=
1 n lim supn→+∞ √a n
or R = lim sup n→+∞
an  . an+1 
1. an = n1 , hence R=
1 1 1 = lim √n n = 1. = = n n→+∞ n 1 n 1 limn→+∞ √a  n limn→+∞ √ n  limn→+∞ √ n
278  7 Solutions to the chapterwise exercises 2. an = R= 3. an =
1 , n2
hence
1 1 1 n = = = lim √n2 = 1. n n→+∞ n 1 n 1 limn→+∞ √a  n limn→+∞ √ n2  limn→+∞ √ n2 1 , n!
hence
an  = lim n→+∞ a n→+∞ n+1 
R = lim
1 n! 1 (n+1)!
= lim
n→+∞
(n + 1)! = lim (n + 1) = +∞. n→+∞ n!
4. an = n!, hence R = lim
n→+∞
5. an =
an  n! n! 1 = lim = lim = lim = 0. an+1  n→+∞ (n + 1)! n→+∞ (n + 1)! n→+∞ n + 1
1⋅2⋅...⋅n , 3⋅5⋅...⋅(2n+1)
hence
1⋅2⋅...⋅n   3⋅5⋅...⋅(2n+1) an  2n + 3 = lim = 2. = lim 1⋅2⋅...⋅n(n+1) n→+∞ n→+∞ a  3⋅5⋅...⋅(2n+1)(2n+3)  n→+∞ n + 1 n+1 
R = lim 6. an =
1 , n√n
hence 1  n√n  an  (n + 1)√n + 1 = lim = 1. = lim 1 n→+∞ a n→+∞  n→+∞ n√n  n+1  √ (n+1) n+1
R = lim
7. an = eαn , α = a + ib, hence an  eαn  ean = lim = lim = lim e−a = e−a . n→+∞ eα(n+1)  n→+∞ a n→+∞ ea(n+1) n→+∞  n+1
R = lim
8. an = cos nα, α = a + ib, b ≠ 0, hence inα
  e +e an   cos nα 2 = lim = lim R = lim i(n+1)α +e−i(n+1)α n→+∞  cos(n + 1)α n→+∞ e n→+∞ a  n+1   −inα
2
eiα (e2nαi + 1) e−inα (e2nαi + 1) = lim −i(n+1)α 2(n+1)αi = lim 2(n+1)αi n→+∞ e + 1) n→+∞ e (e + 1 2nαi + 1 e = lim eiα 2(n+1)αi n→+∞ e + 1
= lim e−b √ n→+∞
e−4nb + 2e−2nb cos 2na + 1 . + 2e−2(n+1)b cos 2(n + 1)a + 1
e−4(n+1)b
If b > 0, then R = lim e−b √ n→+∞
e−4nb + 2e−2nb cos 2na + 1 = e−b . + 2e−2(n+1)b cos 2(n + 1)a + 1
e−4(n+1)b
7.4 Solutions to the exercises of Chapter 4  279
If b < 0, then R = lim e−b √ n→+∞
e−4nb (e4nb + 2e2nb cos 2na + 1) e−4(n+1)b (e4(n+1)b + 2e2(n+1)b cos 2(n + 1)a + 1)
= lim e−b √e4b n→+∞
e4nb + 2e2nb cos 2na + 1 = eb . + 2e2(n+1)b cos 2(n + 1)a + 1
e4(n+1)b
Combining the above two results, we conclude that R = e−b . 9. an = [3 + (−1)n ]n , hence R=
1
lim supn→+∞ √an  n
=
1 1 1 = , = n] n n n lim sup [3 + (−1) 4 lim supn→+∞ √[3 + (−1) ]  n→+∞
since 2, 4,
lim [3 + (−1)n ] = {
n→+∞
10. an =
n 2
+
1+(−1)n+1 , 4
if n is odd if n is even.
hence n+1
 n2 + 1+(−1)  an  4 R = lim = lim n+2 n+1 1+(−1) n→+∞ a n→+∞ n+1   2 +  4 { { limn→+∞ ={ { { limn→+∞ 11. an =
1+(−1)n n 2
+
1−(−1)n 2
  n+1 2  n+1  2
if n is odd
,
 n2   n2 +1
,
if n is even
= 1.
ln n, hence
R=
1 1 = = 1, n 1 lim supn→+∞ √a  n
because lim √n an  = {
n→+∞
limn→+∞ √n  ln n,
if n is odd
limn→+∞ √n n,
if n is even
and thus lim sup √n an  = 1. n→+∞
12. If an = ∑nk=1 k1 , then R=
1
n limn→+∞ √a n
= 1.
={
1, 1,
if n is odd if n is even,
280  7 Solutions to the chapterwise exercises Since n n 1 ≤ an ≤ n ⇒ 1 ≤ √a n ≤ √n,
taking to the limits in the above double inequality, we get n n n 1 = lim 1 ≤ lim √a n ≤ lim √n = 1 ⇒ lim √an = 1.
n→+∞
n→+∞
n→+∞
n→+∞
13. an = n + αn , α = a + ib, hence an  n + αn  = lim . n→+∞ n + 1 + αn+1  n→+∞ a n+1 
R = lim If α > 1, then
n 1 1 1 + αn  = , n→+∞ α 1 + n+1  α n+1
R = lim
α
because lim
n→+∞
n = 0, αn
if α > 1.
If α ≤ 1, then n
α n 1 + n  = 1, R = lim n+1 n→+∞ n + 1 1 + α  n+1
because αn = 0, n→+∞ n lim
if α ≤ 1.
From the above two relations, we conclude that 1, n + αn  ={ 1 n→+∞ n + 1 + αn+1  , α
R = lim
if α ≤ 1 if α > 1.
14. an = P(n), where P is a polynomial with deg P = p, i. e., P(n) = b0 + b1 n + b2 n2 + ⋅ ⋅ ⋅ + bp np ,
bp ≠ 0.
Then R = lim
n→+∞
= lim
n→+∞
an  an+1 
b0 + b1 n + b2 n2 + ⋅ ⋅ ⋅ + bp np 
b0 + b1 (n + 1) + b2 (n + 1)2 + ⋅ ⋅ ⋅ + bp (n + 1)p 
= 1.
7.4 Solutions to the exercises of Chapter 4  281
15. an =
(n!)2 , (3n)!
hence 1
R= =
=
1
limn→+∞
an+1  an 
limn→+∞
(n+1)2 (3n+1)(3n+2)(3n+3)
1
limn→+∞
(3n)! [(n+1)!]2 (n!)2 [3(n+1)]!
= +∞.
)n , hence 16. an = ( n+1 n R= 17. an =
1 1 1 = = n n n+1 n limn→+∞ √a  limn→+∞ n limn→+∞ √( n )
3n +(−2)n , n
n+1 n
= 1.
hence
an  = lim R = lim n→+∞ n→+∞ a n+1 
3n +(−2)n n 3n+1 +(−2)n+1 n+1
2 n n + 1 1 + (− 3 ) 1 = . n→+∞ n 3 − 2(− 32 )n 3
= lim
18. an = 3 + 2(−1)n , hence R=
1
lim supn→+∞ √an  n
=
1 = 1, 1
because lim √n an  = {
n→+∞
limn→+∞ √n 1, limn→+∞ √n 5,
if n is odd = 1, if n is even
and thus lim sup √n an  = 1. n→+∞
19. an = {
an , bn ,
n = 2m, m ∈ ℕ, n = 2m + 1, m ∈ ℕ
hence R=
1
lim supn→+∞ √an  n
= min{
1 1 , }, a b
because lim √n an  = {
n→+∞
b, a,
if n is odd if n is even,
282  7 Solutions to the chapterwise exercises hence lim sup √n an  = max{a, b}. n→+∞
20. 1 , n = 3m, m ≠ 0, { { n 1 n an = { (1 − n ) , n = 3m + 1, m ∈ ℕ, { n n = 3m + 2, m ∈ ℕ { 2 ,
hence R=
1 1 = , n 2 lim supn→+∞ √a  n
because limn→+∞ √n n1 , { { { { { { { { 1 { n lim √an  = { limn→+∞ (1 − n ), n→+∞ { { { { { { limn→+∞ 2, { { {
1, n = 3m, { { { { { m ≠ 0, { { { n = 3m + 1, = { 1, { { m ∈ ℕ, { { { { 2, { n = 3m + 2, { { m ∈ ℕ. {
n = 3m, m ≠ 0, n = 3m + 1, m ∈ ℕ, n = 3m + 2, m ∈ ℕ,
and thus lim sup √n an  = max{1, 1, 2} = 2. n→+∞
Solution of Exercise 4.10.2 1. According to point 1 of Exercise 4.10.1, the radius of convergence is R = 1, hence the set of convergence will be U(0; 1). Using the wellknown relation, ∞
log(1 + z) = ∑ (−1)n+1 n=1
zn , n
z ∈ U(0; 1),
where log 1 = 0,
we get zn , n=1 n ∞
log(1 − z) = − ∑
z ∈ U(0; 1),
where log 1 = 0.
Finally, we conclude that zn = 1 − log(1 − z), n=1 n ∞
1+ ∑
z ∈ U(0; 1).
(7.27)
7.4 Solutions to the exercises of Chapter 4  283
2. The radius of convergence is R=
1
1 = 1, n limn→+∞ √n
=
n limn→+∞ √a n
hence the set of convergence will be U(0; 1). Since ∞ 1 = ∑ zn, 1 − z n=0
z ∈ U(0; 1),
differentiating the both sides of this relation we have ∞ 1 = ∑ nz n−1 , 2 (1 − z) n=1
z ∈ U(0; 1),
(7.28)
and multiplying by z this last identity it follows that ∞
∑ nz n =
n=1
z , (1 − z)2
z ∈ U(0; 1).
(7.29)
3. The radius of convergence is 1
R=
limn→+∞ √an  n
=
1
n
limn→+∞ √n2
= 1,
hence the set of convergence will be U(0; 1). Now we will differentiate the relation (7.29), i. e., ∞
∑ nz n =
n=1
z , (1 − z)2
z ∈ U(0; 1),
and we obtain ∞
∑ n2 z n−1 =
n=1
1+z , (1 − z)3
z ∈ U(0; 1).
Multiplying this last identity by z, we obtain the required result ∞
∑ n2 z n =
n=1
z(1 + z) , (1 − z)3
z ∈ U(0; 1).
4. Since ∞
∞
n=2
m=0
∑ n(n − 1)z n−2 = ∑ (m + 2)(m + 1)z m ,
284  7 Solutions to the chapterwise exercises the radius of convergence is given by R = lim
m→+∞
am  (m + 2)(m + 1) = lim = 1, am+1  m→+∞ (m + 3)(m + 2)
hence the set of convergence will be U(0; 1). Now we will differentiate the relation (7.28), i. e., ∞ 1 = nz n−1 , ∑ (1 − z)2 n=1
z ∈ U(0; 1),
and we conclude that ∞
∑ n(n − 1)z n−2 =
n=2
2 , (1 − z)3
z ∈ U(0; 1).
5. The radius of convergence is (−1)n n2n−1  an  = lim n→+∞ (−1)n+1 n+1  n→+∞ a n+1  (n+1)2 −1 (−1)n n((n + 1)2 − 1) = lim = 1, n→+∞ (−1)n+1 (n2 − 1)(n + 1)
R = lim
hence the set of convergence will be U(0; 1). The sum may be written as ∞
∑ (−1)n
n=2
n2
n 1 ∞ zn zn 1 ∞ z n = ∑ (−1)n + ∑ (−1)n , 2 n=2 n − 1 2 n=2 n+1 −1
(7.30)
so we need to determine the sums of both of the terms from the righthand side. It is well known that ∞
log(1 + z) = ∑ (−1)n+1 n=1
zn , n
z ∈ U(0; 1).
(7.31)
Multiplying this relation by z, we get z n+1 , n
z ∈ U(0; 1),
zn , n−1
z ∈ U(0; 1).
log(1 + z) ∞ z n−1 = ∑ (−1)n+1 , z n n=1
z ∈ U(0; 1),
∞
z log(1 + z) = ∑ (−1)n+1 n=1
and the above identity is equivalent to ∞
z log(1 + z) = ∑ (−1)n n=2
Dividing the relation (7.31) by z, we have
(7.32)
7.4 Solutions to the exercises of Chapter 4  285
which is equivalent to log(1 + z) zn z ∞ = 1 − + ∑ (−1)n , z 2 n=2 n+1
z ∈ U(0; 1).
(7.33)
Using the identities (7.30), (7.32) and (7.33), we conclude that ∞
∑ (−1)n
n=2
n2
z 1 log(1 + z) z n z n = log(1 + z) + ( − 1 + ), 2 2 z 2 −1
z ∈ U(0; 1).
6. Since ∞
∞
n=1
m=0
∑ nz n−1 = ∑ (m + 1)z m ,
the radius of convergence is given by R=
1
limm→+∞ √am  m
=
1
m limm→+∞ √ m+1
= 1,
hence the set of convergence will be U(0; 1). The relation (7.28) obtained to point 2 shows that ∞ 1 = nz n−1 , ∑ (1 − z)2 n=1
z ∈ U(0; 1).
7. Since an = {
0,
if n = 2m, m ∈ ℕ, if n = 2m + 1, m ∈ ℕ,
1 , 2m+1
the radius of convergence is R=
1 1 = = 1, n 1 lim supn→+∞ √a  n √ 2m+1 limm→+∞ 2m+1
hence the set of convergence will be U(0; 1). Now we will use the relation (7.27) obtained to point 1, i. e., ∞
log(1 + z) = ∑ (−1)n+1 n=1 ∞
zn , n=1 n
− log(1 − z) = ∑
zn , n
z ∈ U(0; 1)
z ∈ U(0; 1).
Adding these two identities, the even terms vanishes, hence we deduce that 1 z 2n+1 = [log(1 + z) − log(1 − z)], 2n + 1 2 n=0 ∞
∑
z ∈ U(0; 1),
286  7 Solutions to the chapterwise exercises and thus z 2n+1 1 = [log(1 + z) − log(1 − z)] − z, 2n + 1 2 n=1 ∞
∑
z ∈ U(0; 1).
8. The radius of convergence is R=
1
lim supn→+∞ √an  n
=
1 n
lim supn→+∞ √n  (−1)  n
= 1,
hence the set of convergence will be U(0; 1). The required sum is given by the wellknown formula ∞
log(1 + z) = ∑ (−1)n+1 n=1
9. Substituting
z z+1
zn , n
z ∈ U(0; 1).
= ζ , we get n
∞
∑(
n=0
∞ z ) = ∑ ζ n, z+1 n=0
and the power series of the righthand side converges if and only if z 1 ζ  < 1 ⇔ < 1 ⇔ Re z > − . z + 1 2
It follows that the convergence set of the given series of functions will be the halfplane 1 D = {z ∈ ℂ : Re z > − }. 2 Starting from the relation ∞ 1 = ∑ ζ n, 1 − ζ n=0
and using the substitution ζ = ∞
∑(
n=0
z , z+1
ζ ∈ U(0; 1),
we deduce that
n
1 z ) = z = z + 1, z+1 1 − z+1
1 Re z > − . 2
Solution of Exercise 4.10.3 1. A simple computation shows that 1 1 1 = z 2 + a2 a2 1 + ( az )2 =
2n
1 ∞ z ∑ (−1)n ( ) a a2 n=0
∞
= ∑ (−1)n a−2n−2 z 2n , n=0
if z < a, a ≠ 0,
7.4 Solutions to the exercises of Chapter 4  287
because if z z 2 a ≠ 0 ⇔ < 1 ⇔ ( ) < 1, a a
z < a,
we can use the wellknown relation ∞ 1 = ∑ zn, 1 − z n=0
z ∈ U(0; 1).
2. We will use the relation obtained to point 3 of Exercise 4.10.2, i. e., ∞
∑ n2 z n =
n=1
z(1 + z) , (1 − z)3
z ∈ U(0; 1).
Since z < a, using the substitution z :=
z a
z a ≠ 0 ⇔ < 1, a
we get
az(z + a) ∞ n2 z n =∑ n , (a − z)3 n=1 a
if z < a, a ≠ 0.
Now, by dividing with a the above identity we conclude that z(z + a) ∞ n2 z n , =∑ (a − z)3 n=1 an+1
if z < a, a ≠ 0.
3. First, we will prove that z 1 Re z > − ⇔ < 1. z + 1 2 This equivalence holds, since if we denote z = x + iy, x, y ∈ ℝ, then z 1 2 2 2 2 < 1 ⇔ z < z + 1 ⇔ x + y < (x + 1) + y ⇔ x = Re z > − . 2 z + 1 Using the result from point 9 of Exercise 4.10.2, we deduce that ∞
1 + z = ∑( n=0
n
z ) , z+1
1 if Re z > − . 2
288  7 Solutions to the chapterwise exercises Solution of Exercise 4.10.4 1. Let z = 1 ⇔ z = eiφ ,
φ ∈ [0, 2π].
We easily see that ∞ ∞ cos nφ sin nφ einφ =∑ +i∑ , n n n n=1 n=1 n=1 ∞
∑
hence it is sufficient to obtain the sum ∑∞ n=1
einφ . n
zn , n=1 n ∞
− log(1 − z) = ∑
For this, we will use the relation z ∈ U(0; 1).
and we get φ φ−π einφ = − log(1 − eiφ ) = − ln(2 sin ) − i , 2 2 n=1 n ∞
∑
0 < φ < 2π,
(7.34)
because 1 − eiφ = 2 sin
φ i φ−π e 2 . 2
Identifying the real parts of the both sides of (7.34), we conclude that φ cos nφ = − ln(2 sin ), n 2 n=1 ∞
∑
0 < φ < 2π.
2. Identifying the imaginary parts of the both sides of (7.34), we obtain the required sum sin nφ π − φ = , n 2 n=1 ∞
∑
0 < φ < 2π.
3. We will use a similar proof with in point 1 of this problem. If we denoted z = 1 ⇔ z = eiφ ,
φ ∈ [0, 2π],
it is easy to see that ∞ ∞ cos(2n + 1)φ sin(2n + 1)φ ei(2n+1)φ = ∑ +i ∑ . 2n + 1 2n + 1 2n + 1 n=0 n=0 n=0 ∞
∑
7.4 Solutions to the exercises of Chapter 4  289
Hence it is sufficient to compute the sum ∑∞ n=0
ei(2n+1)φ . For this, we will use the relation 2n+1
1 z 2n+1 = [log(1 + z) − log(1 − z)], 2n + 1 2 n=0 ∞
z ∈ U(0; 1).
∑
Thus ei(2n+1)φ 1 = [log(1 + eiφ ) − log(1 − eiφ )] 2n + 1 2 n=0 ∞
∑
φ φ φ φ−π 1 ] = [ln(2 cos ) + i − ln(2 sin ) − i 2 2 2 2 2 φ 1 π = [ln cot + i ], if 0 < φ < π, 2 2 2
because 1 − eiφ = 2 sin
φ i φ−π e 2 2
and
1 + eiφ = 2 cos
(7.35)
φ i φ2 e . 2
Identifying the real parts of the both sides of (7.35), we deduce that φ cos(2n + 1)φ 1 = ln(cot ), 2n + 1 2 2 n=0 ∞
∑
0 < φ < π.
4. If we identify the imaginary parts of the both sides of the relation (7.35), it follows that sin(2n + 1)φ π = , 2n + 1 4 n=0 ∞
∑
0 < φ < π.
Solution of Exercise 4.10.5 1. We will write the function in the form ∞
f (z) = ∑ an (z − z0 )n , n=0
where z0 = 1.
In order to determine the coefficients from the above formula, we will decompose the given function by using functions with wellknown Taylor series expansions. Thus, we have f (z) =
z+1−1 1 z = =1− . z+1 z+1 z+1
Since the Taylor series expansion is about the point z0 = 1, we need to emphasize the expression z − 1, i. e., f (z) = 1 −
1 1 1 1 1 =1− . =1− =1− z+1 z−1+2 2 1 + z−1 ) 2(1 + z−1 2 2
290  7 Solutions to the chapterwise exercises Using the wellknown formula, ∞ 1 = ∑ (−1)n ζ n , 1 + ζ n=0
ζ ∈ U(0; 1),
(7.36)
we get f (z) = 1 −
n
z−1 1 ∞ 1 1 ) , = 1 − ∑ (−1)n ( 2 1 + z−1 2 n=0 2
z − 1 < 2,
2
where z − 1 < 2, since the function is not defined in the point z1 = −1. Thus n+1
∞ 1 f (z) = 1 + ∑ (− ) 2 n=0
that
(z − 1)n ,
where z − 1 < 2.
2. We will use a similar method like to the point 1 of the problem. It is well known ∞ 1 = (−1)n−1 nζ n−1 , ∑ (1 + ζ )2 n=1
ζ ∈ U(0; 1).
(7.37)
Since the Taylor series expansion is about the point z0 = 1, we will emphasize the expression z − 1, i. e., f (z) =
z2 (z − 1 + 1)2 (z − 1)2 + 2(z − 1) + 1 = = (z + 1)2 (z − 1 + 2)2 4(1 + z−1 )2 2
2
=
1 (z − 1) + 2(z − 1) + 1 . 4 (1 + z−1 )2 2
Using the relation (7.37), we have f (z) =
n−1
(z − 1)2 + 2(z − 1) + 1 ∞ z−1 ) ∑ (−1)n−1 n( 4 2 n=1
,
z − 1 < 2,
and thus n+1
∞ 1 f (z) = ∑ (− ) 2 n=1
n
∞ 1 n(z − 1)n+1 − ∑ (− ) n(z − 1)n 2 n=1
n+1
∞ 1 + ∑ (− ) 2 n=1
n(z − 1)n−1 ,
z − 1 < 2.
A simple computation gives us the final form of the expansion f (z) =
n+1
∞ 1 1 1 + (z − 1) + ∑ (− ) 4 4 2 n=3
(9n − 12)(z − 1)n−1 ,
where z − 1 < 2.
7.4 Solutions to the exercises of Chapter 4  291
3. The function is not defined in the point z1 = 2. Thus, the disc U(0; 2) is the largest disc with the center at the origin where the given function is holomorphic, and f (z) =
z−1 z−2+1 1 1 = =1+ =1+ z−2 z−2 z−2 2
z 2
1 1 1 =1− . 2 1 − z2 −1
Using the formula, ∞ 1 = ∑ ζ n, 1 − ζ n=0
ζ ∈ U(0; 1),
(7.38)
we get f (z) = 1 −
1 1 21−
z 2
=1−
n
1 ∞ z ∑( ) , 2 n=0 2
z < 2,
and thus n+1
∞ 1 f (z) = 1 − ∑ ( ) 2 n=0
zn,
where z < 2.
The expansion about the point z0 = i will be found similarly, but in this case we need to emphasize the expression z − i. The function is not defined in the point z1 = 2, and thus U(i; √5) is the maximal disc with the center in z0 = i where the given function is holomorphic, and 1 1 1 1 z−1 . =1+ =1+ =1+ z−2 z−2 z−i+i−2 i − 2 1 + z−i
f (z) =
i−2
Using the relation (7.36), we get f (z) = 1 +
n
1 1 z−i 1 ∞ ) , = 1 + ∑ (−1)n ( i − 2 1 + z−i i − 2 n=0 i−2 i−2
z − i < √5,
and thus (−1)n n z , n+1 n=0 (i − 2) ∞
f (z) = 1 + ∑
where z − i < √5. 2π
4. The function is not defined in points ε and ε, where ε = ei 3 is the third root of the unity. The disc U(1; ρ) is the maximal disc with the center in z0 = 1 where the given function is holomorphic, z0 − ε = z0 − ε = ρ = √3, and 1 i √3 1 1 1 = = ( − ) 3 z−ε z−ε + z + 1 (z − ε)(z − ε) 1 1 i √3 ( − ) = 3 (z − 1) + 1 − ε (z − 1) + 1 − ε 1 i√3 1 1 1 − ). = ( 3 1 − ε 1 + z−1 1 − ε 1 + z−1
f (z) =
z2
1−ε
1−ε
292  7 Solutions to the chapterwise exercises Using the relation (7.36), we get f (z) =
n n i√3 ∞ z−1 i √3 ∞ z−1 ) − ) , ∑ (−1)n ( ∑ (−1)n ( 3(1 − ε) n=0 1−ε 3(1 − ε) n=0 1−ε
z − 1 < ρ,
if z − 1 < 1 ⇔ z − 1 < 1 − ε = ρ, 1 − ε
From the above relations, we conclude that n+1
∞
i√3 1 [(− ) 3 1 − ε n=0
f (z) = ∑
z − 1 < 1 ⇔ z − 1 < 1 − ε = ρ. 1 − ε n+1
− (−
1 ) 1−ε
](z − 1)n ,
where z − 1 < √3.
We will use a similar method for the expansion about the point z0 = ∞, and we need to emphasize the expression z1 . The circular ring U(0; 1, ∞) is the maximal ring with the center at the origin where the function is welldefined. Since f (z) =
z2
1 1 i √3 1 1 i √3 1 ( − )= ( = 3 z−ε z−ε 3 z 1− +z+1
ε z
+
1 1 ), z 1 − zε
using the relation (7.38) we get f (z) =
n n i√3 ∞ ε i √3 ∞ ε ∑( ) − ∑( ) , 3z n=0 z 3z n=0 z
z > 1,
where
Thus,
ε < 1 ⇔ z > ε = 1, z
ε < 1 ⇔ z > ε = 1. z
i(ε n − εn )√3 1 , 3 z n+1 n=0 ∞
f (z) = ∑
where z > 1.
5. The solution is similar to that of point 4 of the problem. The function is not defined in points z1 = 3 and z2 = 5. The disc U(4; 1) is the maximal disc with the center in z0 = 4 where the function is welldefined, and z+3 z+3 = z 2 − 8z + 15 (z − 3)(z − 5) 3 4 3 4 − = − . = z − 5 z − 3 (z − 4) − 1 (z − 4) + 1
f (z) =
Using the relations (7.36) and (7.38), we get ∞
∞
n=0
n=0
f (z) = −4 ∑ (z − 4)n − 3 ∑ (−1)n (z − 4)n ,
z − 4 < 1.
7.4 Solutions to the exercises of Chapter 4  293
Consequently, we have that ∞
f (z) = − ∑ [4 + 3(−1)n ](z − 4)n , n=0
where z − 4 < 1.
For the expansion about z0 = ∞, we will emphasize the expression z1 . The circular ring U(0; 5, ∞) is the maximal ring with the center at the origin where the function is welldefined, and f (z) =
z2
z+3 4 3 4 1 = − = − 8z + 15 z − 5 z − 3 z 1 −
5 z
−
3 1 . z 1− 3 z
Using the relation (7.38), we get f (z) =
n
n
3 ∞ 3 4 ∞ 5 ∑( ) − ∑( ) , z n=0 z z n=0 z
z > 5,
and thus ∞
f (z) = ∑ [4 ⋅ 5n − 3n+1 ] n=0
1 , z n+1
where z > 5.
z 6. The function f (z) = sin 1−z is not defined in the point z1 = 1, and U(0; 1) is the maximal disc with the center at the origin where the function is welldefined. The coefficients from the expansion will be determined by using the formula
f (z) = f (0) +
f (0) f (0) z+ z + ⋅⋅⋅, 1! 2!
z ∈ U(0; 1).
We easily obtain that f (0) = 0
1 z cos ⇒ f (0) = 1 2 1−z (1 − z) z z z − 2 cos 1−z + 2z cos 1−z sin 1−z ⇒ f (0) = 2 f (z) = − 1 − 4z + 6z 2 − 4z 3 + z 4 z z z z z 5 cos 1−z − 6 sin 1−z + 6z sin 1−z − 12z cos 1−z + 6z 2 cos 1−z f (z) = 1 − 6z + 15z 2 − 20z 3 + 15z 4 − 6z 5 + z 6 ⇒ f (0) = 5,
f (z) =
hence
las,
5 f (z) = z + z 2 + z 3 + ⋅ ⋅ ⋅ , 6
where z < 1.
7. The function is defined in the whole complex plane. Using the following formucosh 2z = 2 cosh2 z − 1
and
cosh z =
ez + e−z , 2
294  7 Solutions to the chapterwise exercises we get cosh 2z + 1 1 e2z + e−2z = + . 2 2 4
f (z) = cosh2 z = Since
zn , n! n=0 ∞
ez = ∑
z ∈ ℂ,
the function may be written in the following form: f (z) = =
1 1 ∞ 2n z n 1 ∞ (−2)n z n 1 e2z + e−2z + = + ∑ + ∑ 2 4 2 4 n=0 n! 4 n=0 n! 1 1 ∞ n 1 + (−1)n n + ∑2 z , 2 4 n=0 n!
z ∈ ℂ.
Thus, f (z) =
1 ∞ 22n−1 2n +∑ z , 2 n=0 (2n)!
where z ∈ ℂ.
8. Using the following wellknown formula, cos 3z = 4 cos3 z − 3 cos z
and
∞
cos z = ∑ (−1)n n=0
z 2n , (2n)!
z ∈ ℂ,
we get cos 3z + 3 cos z 4 2n 1 ∞ z 2n 3 ∞ n (3z) = ∑ (−1) + ∑ (−1)n , 4 n=0 (2n)! 4 n=0 (2n)!
f (z) = cos3 z =
z ∈ ℂ.
Thus, ∞
f (z) = ∑ (−1)n n=0
las,
32n + 3 2n z , 4 ⋅ (2n)!
where z ∈ ℂ.
9. The function is defined in the whole complex plane. Using the following formusin 3z = 3 sin z − 4 sin3 z
and
∞
sin z = ∑ (−1)n n=0
z 2n+1 , (2n + 1)!
z ∈ ℂ,
we get 3 sin z − sin 3z 4 2n+1 (3z)2n+1 1 ∞ 3 ∞ n z − ∑ (−1)n , = ∑ (−1) 4 n=0 (2n + 1)! 4 n=0 (2n + 1)!
f (z) = sin3 z =
z ∈ ℂ,
7.4 Solutions to the exercises of Chapter 4  295
and thus ∞
f (z) = ∑ (−1)n n=0
3 − 32n+1 2n+1 z , 4 ⋅ (2n + 1)!
where z ∈ ℂ.
10. Using the following formulas, sin z =
eiz − e−iz 2i
zn , n! n=0 ∞
and ez = ∑
z ∈ ℂ,
we get eiz − e−iz e(1+i)z − e(1−i)z = 2i 2i 1 ∞ (1 + i)n z n 1 ∞ (1 − i)n z n + ∑ , z ∈ ℂ. = ∑ 2i n=0 n! 2i n=0 n!
f (z) = ez sin z = ez
Thus, (1 + i)n + (1 − i)n n z , 2i ⋅ n! n=0 ∞
f (z) = ∑
where z ∈ ℂ. −1+i√3 , z0 2
3 where f (1) = 11. The given function is f (z) = √z, function in the exponential form, i. e.,
∀z = reiθ ,
r > 0,
θ ∈ [0, 2π] ⇒ fk (z) = √3 rei
θ+2kπ 3
= 1. We will write the k ∈ {0, 1, 2},
,
and thus the function f has three branches. Using the condition f (1) = determine the required branch: z = 1 ⇔ z = 1ei⋅0 ⇒ fk (1) = ei
2kπ 3
−1+i√3 , 2
we
.
Since f (1) =
2π −1 + i√3 ⇔ f (1) = ei 3 ⇒ k = 1, 2
and thus f (z) = f1 (z) = √3 rei
θ+2π 3
.
The function is not defined at the point z1 = 0, and thus U(1; 1) is the maximal disc with the center in z0 = 1 where the function is welldefined, and 1
f (z) = √3 z = √3 1 + (z − 1) = [1 + (z − 1)] 3 .
296  7 Solutions to the chapterwise exercises From the relation, ∞ α(α − 1) ⋅ . . . ⋅ (α − n + 1) n (1 + ζ )α = (1 + ζ )α ζ =0 ⋅ [1 + ∑ ζ ], n! n=1
ζ ∈ U(0; 1),
(7.39)
we get ∞ 1 ( 1 − 1) ⋅ . . . ⋅ ( 1 − n + 1) −1 + i√3 3 [1 + ∑ 3 3 (z − 1)n ], 2 n! n=1
1
f (z) = [1 + (z − 1)] 3 =
for z − 1 < 1. 7π 5 12. The given function is f (z) = √z − 2i, where f (0) = √5 2ei 10 , z0 = 0 and z0 = 2. Letting z − 2i = reiθ ,
θ ∈ [0, 2π] ⇒ fk (z) = √5 rei
r > 0,
θ+2kπ 5
k ∈ {0, 1, 2, 3, 4},
,
and thus the function f has five branches. Using the given condition, we determine the required branch, i. e., z = 0 ⇔ −2i = 2ei
3π 2
⇒ fk (0) = √2ei 5
3π +2kπ 2 5
.
Since 7π
f (1) = √2ei 10 ⇒ k = 1, 5
we get f (z) = f1 (z) = √5 rei
θ+2π 5
.
The function is not defined at the point z1 = 2i, then U(0; 2) is the maximal disc with the center in z0 = 0 where the function is welldefined, and thus 5
f (z) = √z − 2i = √−2i(1 − 5
7π z z z 5 5 ) = √−2i ⋅ √5 1 − = √2ei 10 √5 1 − , 2i 2i 2i
where √5 1 −
z = 1. 2i z=0
From the relation (7.39) used to point 11, we get 5
f (z) = √2e
i 7π 10 7π
1
z 5 (1 − ) 2i ∞
5 = √2ei 10 (1 + ∑ (−1)n
n=1
1 1 ( 5 5
− 1) ⋅ . . . ⋅ ( 51 − n + 1) (2i)n n!
z n ),
z < 2.
7.4 Solutions to the exercises of Chapter 4  297
For the expansion about point z0 = 2, we need to write the given function into the form ∞
f (z) = ∑ an (z − 2)n , n=0
z ∈ U(2; ρ).
The disc U(2; 2√2) is the maximal disc with the center in z0 = 2 where the function is welldefined, i. e., ρ = 2√2, and 5
f (z) = √z − 2i = √5 (z − 2) + 2 − 2i 5
= √2 − 2i ⋅ √1 + 5
where √5 1 +
z − 2 √5 √ i 3π4 5 z−2 = 2 2e √1 + , 2 − 2i 2 − 2i z − 2 = 1. 2 − 2i z=2
From the relation (7.39) used to point 11, we get 1
3π 5 z−2 5 f (z) = √2√2ei 4 (1 + ) 2 − 2i 3
3π
∞ 1(1 5 5
= 2 10 ei 4 (1 + ∑
− 1) ⋅ . . . ⋅ ( 51 − n + 1)
n=1
(2 − 2i)n n!
(z − 2)n ),
13. The given function is f (z) = log z, where f (1 + i) = write the function in the exponential form, i. e., ∀z = reiθ ,
r > 0,
1 2
z − 2 < 2√2.
ln 2 − i 7π , z0 = −i. We will 4
θ ∈ (−π, π) ⇒ fk (z) = ln r + i(θ + 2kπ),
k ∈ ℤ.
Using the given condition we determine the required branch, i. e., π
z = 1 + i ⇔ z = √2ei 4 ⇒ fk (1 + i) = Hence f (1 + i) =
π 1 ln 2 + i( + 2kπ), 2 4
k ∈ ℤ.
7π 1 ln 2 − i ⇒ k = −1, 2 4
and thus f (z) = f−1 (z) = ln r + i(θ − 2π). The function has a holomorphic branch on the domain ℂ \ {z ∈ ℂ : Re z ≤ 0, z = 0}, and thus the disc U(−i; 1) is the maximal disc with the center in z0 = −i where the function is welldefined. A simple computation shows that z+i )] i z+i 5πi z+i = log−1 (−i) + log(1 − )=− + log(1 − ), i 2 i
f (z) = log z = log[(z + i) − i] = log[(−i)(1 −
298  7 Solutions to the chapterwise exercises where log(1 −
z + i = 0. ) z=−i i
Using the formula, ∞
log(1 + ζ ) = ∑ (−1)n+1 n=1
ζn , n
ζ ∈ U(0; 1),
(7.40)
we conclude that f (z) = −
5πi ∞ (z + i)n −∑ n , 2 i n n=1
14. The given function is f (z) =
log(1+z) , 1+z
z + 1 = reiθ ,
z + i < 1.
where f (0) = −4πi and z0 = 0. Letting
r > 0,
θ ∈ (−π, π),
then fk (z) =
ln r + i(θ + 2kπ) , reiθ
k ∈ ℤ.
Using the given condition, we determine the required branch. If z = 0 ⇔ 1 = ei⋅0 ⇒ fk (0) = 2kπi, then f (0) = −4πi ⇒ k = −2, i. e., f (z) = f−2 (z) =
ln r + i(θ − 4π) . reiθ
The disc U(0; 1) is the maximal disc with the center at the origin where the function f is welldefined and it is holomorphic. Since f (z) =
1 log(1 + z) = [−4πi + log(1 + z)], 1+z z+1
where log(1 + z)z=0 = 1, from the relation (7.40) used to point 13, we get ∞
∞
n=0
n=1
f (z) = ∑ (−1)n z n (−4πi + ∑ (−1)n+1
zn ), n
z < 1.
7.4 Solutions to the exercises of Chapter 4  299
Solution of Exercise 4.10.6 Since we need to expand all these functions in Laurent series about the point z0 = 0, the expansions will have the form ∞
f (z) = ∑ an z n . n=−∞
1 1. The function f (z) = z 2 −5z+6 is not defined in the points z1 = 2 and z2 = 3. (a) If z < 2, from the relation (7.38) used in Exercise 4.10.5 we get
f (z) =
1 1 1 1 1 1 = = − = z 2 − 5z + 6 (z − 3)(z − 2) z − 3 z − 2 2 1 − n
=
n
∞ 1 1 ∞ z 1 ∞ z 1 ∑ ( ) − ∑ ( ) = ∑ ( n+1 − n+1 )z n , 2 n=0 2 3 n=0 3 3 n=0 2
z 2
−
1 1 31−
z 3
where z < 2.
(b) If 2 < z < 3, from the relation (7.38) used in Exercise 4.10.5 we get f (z) =
1 1 1 1 1 = − =− z 1− z 2 − 5z + 6 z − 3 z − 2 n
=−
n
1 ∞ z 1 ∞ 2 ∑( ) − ∑( ) , z n=0 z 3 n=0 3
2 z
−
1 1 31−
z 3
where 2 < z < 3,
and where we used the fact that 2 z > 2 ⇔ < 1 z
z and z < 3 ⇔ < 1. 3
(c) If 3 < z, using the relation (7.38) we get f (z) =
1 1 1 1 1 = − =− z 1− z 2 − 5z + 6 z − 3 z − 2 n
=−
n
2 z
+
1 1 z 1−
∞ 1 ∞ 3 1 1 ∞ 2 ∑ ( ) + ∑ ( ) = ∑ (3n − 2n ) n+1 , z n=0 z z n=0 z z n=0
3 z
where 3 < z.
1 2. The function f (z) = z(z+1)(z+2) is not defined in the points z1 = 0, z2 = −1 and z3 = −2. (a) If 0 < z < 1, from the relation (7.36) used in Exercise 4.10.5, we get
f (z) =
1 1 1 1 1 1 1 = − + = − + z(z + 1)(z + 2) 2z z + 1 2(z + 2) 2z z + 1 4(1 + z2 ) n
= =
∞ 1 ∞ z 1 − ∑ (−1)n z n + ∑ (−1)n ( ) 2z n=0 4 n=0 2 ∞ 1 1 − ∑ (−1)n (1 − n+2 )z n , 2z n=0 2
where 0 < z < 1.
300  7 Solutions to the chapterwise exercises (b) If 1 < z < 2, from the relation (7.36) used in Exercise 4.10.5, we get f (z) = =
1 1 1 1 1 1 1 = − + = − z(z + 1)(z + 2) 2z z + 1 2(z + 2) 2z z 1 + ∞ ∞ 1 (−1)n 1 − ∑ (−1)n n+1 + ∑ n+2 z n , 2z n=0 z n=0 2
and where we used the fact that 1 z > 1 ⇔ < 1 z
1 z
1 4(1 + z2 )
+
where 1 < z < 2,
z and z < 2 ⇔ < 1. 2
(c) If 2 < z, from the relation (7.36) we get f (z) = = =
1 1 1 1 1 1 1 = − + = − z(z + 1)(z + 2) 2z z + 1 2(z + 2) 2z z 1 + ∞ ∞ 1 2n−1 1 − ∑ (−1)n n+1 + ∑ (−1)n n+1 2z n=0 z z n=0 ∞ 1 1 − ∑ (−1)n (2n−1 − 1) n+1 , 2z n=1 z
1 z
+
1 1 2z (1 + 2 ) z
where 2 < z.
2
get
3. The function f (z) = z 32z−2z−3z+3 2 +z−2 is not defined in the points z1 = 2 and z2,3 = ±i. (a) If z < 1, from the relation (7.36) used in Exercise 4.10.5 together with (7.38), we 1 1+i 1 1−i 1 2z 2 − 3z + 3 = + + 2 z−i 2 z+i z 3 − 2z 2 + z − 2 z − 2 1 1 1+i 1 1−i 1 =− − + 2 1 − z2 2i 1 − zi 2i 1 + zi
f (z) =
n
n
n
z 1−i ∞ z 1+i ∞ 1 ∞ z = − ∑( ) − ∑( ) − ∑ (−1)n ( ) 2 n=0 2 2 n=0 i 2 n=0 i
1 ∞ 1 1 − i + (1 + i)(−1)n n = − ∑( n + )z 2 n=0 2 in
where z < 1.
(b) If 1 < z < 2, similarly as above, we have 2z 2 − 3z + 3 1 1+i 1 1−i 1 = + + 3 2 2 z−i 2 z+i z − 2z + z − 2 z − 2 1−i 1 1+i 1 1 1 + + =− 2 1 − z2 2z 1 − i 2z 1 + i z z
f (z) =
n
n
n
1 ∞ z 1+i ∞ i 1−i ∞ i = − ∑( ) + ∑( ) + ∑ (−1)n ( ) 2 n=0 2 2z n=0 z 2z n=0 z ∞ 1 + i + (1 − i)(−1)n in zn + , ∑ n+1 2 z n+1 n=0 n=0 2 ∞
=−∑
where 1 < z < 2,
7.4 Solutions to the exercises of Chapter 4  301
where we used the fact that i z > 1 ⇔ < 1 z
z and z < 2 ⇔ < 1. 2
(c) If 2 < z, using the formulas (7.36) and (7.38), we obtain 1+i 1 1−i 1 2z 2 − 3z + 3 1 + + = 2 z−i 2 z+i − 2z 2 + z − 2 z − 2 1 1 1+i 1 1−i 1 = + + z 1− 2 2z 1 − i 2z 1 + i z z z
f (z) =
z3
n
=
n
n
1+i ∞ i 1−i ∞ i 1 ∞ 2 ∑( ) + ∑( ) + ∑ (−1)n ( ) z n=0 z 2z n=0 z 2z n=0 z
∞ 2n 1 + i + (1 − i)(−1)n in + ∑ n+1 2 z n+1 n=0 z n=0 ∞
=∑ ∞
= ∑ (2n + n=0
1 + i + (1 − i)(−1)n n 1 i ) n+1 , 2 z
where 2 < z.
1−z 4. The function f (z) = log 1+z , with f (0) = 0 is not defined in the point z1 = 1. Since f (0) = 0, the logarithmic function that appears is the main branch, i. e., log 1 = 0. (a) If z < 1, from the relation (7.40) used in Exercise 4.10.5 we get
f (z) = log
1−z = log(1 − z) − log(1 + z), 1+z
where log(1 − z)z=0 = 0,
log(1 + z)z=0 = 0,
and it follows that zn ∞ zn − ∑ (−1)n+1 n n=1 n n=1 ∞
f (z) = log(1 − z) − log(1 + z) = − ∑ 1 + (−1)n+1 n z , n n=1 ∞
=−∑
where z < 1.
(b) If 1 < z, from the relation (7.40) used in Exercise 4.10.5 we get f (z) = log
1− 1−z = log(−1) 1+z 1+
1 z 1 z
= log(−1) + log
1−
1+
where 1 log(1 − ) = 0, z z=∞
1 log(1 + ) = 0, z z=∞
1 z 1 z
,
302  7 Solutions to the chapterwise exercises and thus f (z) = log
1− 1−z = log(−1) 1+z 1+ ∞
1 + (−1) n n=1
= iπ − ∑
n+1
1 , zn
1 z 1 z
= log(−1) + log
1−
1+
1 z 1 z
where z > 1.
Solution of Exercise 4.10.7 1. Since f ∈ H(ℂ \ {0, 1, −1±i2 follows that
√3
f (z) =
}), the function is holomorphic in the given domain. It
z5 ∞
1 1 1 1 ∞ 1 = 5 = 5 ∑ 3n 1 2 z 1− 3 z n=0 z −z z
1 , 3n+5 z n=0
=∑
where z > 1,
and where we used the relation (7.38), and the fact that 1 1 z > 1 ⇔ < 1 ⇔ 3 < 1. z z 2. Since f ∈ H(ℂ \ {±i, ±2, ±2i}), the function is holomorphic in the given domain. We have f (z) =
z 15 = (z 2 + 1)5 (z 4 − 16) (1 +
z
1 5 ) (1 z2
−
16 ) z4
= z(1 +
−5
1 1 ) z2 1 − z164 n
∞ 16 (−5)(−5 − 1) ⋅ . . . ⋅ (−5 − n + 1) 1 ) ( 4) ∑ 2n n! z n=1 n=0 z ∞
= z(1 + ∑ =z−
5 31 115 + − 5 + ⋅⋅⋅, z z3 z
where z > 2,
where we used the relations (7.38) and (7.39), and the fact that 1 1 z > 2 ⇔ < 1 ⇔ 2 < 1, z z
2 16 z > 2 ⇔ < 1 ⇔ 4 < 1. z z
3. The given form of the function is the same with the Laurent series expansion about the point z0 = ∞, and thus f (z) = a0 + a1 z + ⋅ ⋅ ⋅ + an z n ,
z ∈ ℂ.
7.4 Solutions to the exercises of Chapter 4  303
4. We have f ∈ H(ℂ \ {0}), and f (z) = z 3 sin
∞ (−1)n 1 1 = z3 ∑ 2n+1 z (2n + 1)! z n=0
(−1)n 1 , 2(n−1) (2n + 1)! z n=0 ∞
where z > 0,
=∑
and where we used the wellknown formula: ∞
sin ζ = ∑ (−1)n n=0
ζ 2n+1 , (2n + 1)!
ζ ∈ ℂ.
(7.41)
5. First, we will expand in Laurent series z0 = ∞, the derivative of the given function, i. e., f (z) =
1 1 a−b = ( (z − a)(z − b) z 1 −
a z n
−
∞ n an − bn a −b =∑ , n+1 n+1 z n=1 z n=0 ∞
1
1−
=∑
b z
)
z > max{a, b},
where we used the relation (7.38). Integrating the obtained relation, we deduce that f (z) = log
∞ n z−a a − bn 1 =−∑ + c, z−b n zn n=1
z > max{a, b}.
The constant c ∈ ℂ will be determined by using the given condition log 1 = 2kπi, i. e., lim f (z) = log 1 = 2kπi ⇒ c = 2kπi.
z→∞
Thus, an − bn 1 , n zn n=1 ∞
where z > max{a, b}.
f (z) = 2kπi − ∑
6. Since f ∈ H(ℂ \ {−1}), the function is holomorphic in the given domain, and 2
f (z) = e z+1 = e 1+ z = e
2(1− z1 +
∞
−
2z
= e2 ∑
1
2n (− z1 +
n=0
= e2 (1 −
1 z2
n!
1 z2
1 z3
−
1 z3
+⋅⋅⋅)
+ ⋅ ⋅ ⋅)n
= e2 e
2 4 14 1 − + ⋅ ⋅ ⋅), + z z2 3 z3
2(− z1 +
1 z2
−
1 z3
+⋅⋅⋅)
where z > 1,
where we used the relation (7.36), and the fact that ζn , n! n=0 ∞
eζ = ∑
ζ ∈ ℂ.
(7.42)
304  7 Solutions to the chapterwise exercises Solution of Exercise 4.10.8 1. We have f ∈ H(ℂ \ {±1}). The expansion regarding point z0 = 1 will be given in the circular ring U(1; 0, 2), i. e., f (z) =
1 1 1 1 = = (z 2 − 1)2 (z − 1)2 (z + 1)2 (z − 1)2 [(z − 1) + 2]2
∞ 1 1 (−1)n−1 n 1 = (z − 1)n−1 ∑ z−1 2 2 2 4(z − 1) (1 + 4(z − 1) n=1 2n−1 ) 2 1 1 3 1 − = + − (z − 1) + ⋅ ⋅ ⋅ , where z − 1 < 2, 4(z − 1)2 4(z − 1) 16 8
=
and where we used that ∞ 1 = ∑ (−1)n−1 nζ n−1 , 2 (1 + ζ ) n=1
ζ ∈ U(0; 1).
(7.43)
It follows that point z0 = 1 is a secondorder pole for f . For the expansion regarding the point z0 = ∞, we will expand the function in Laurent series on the circular ring z > 1. Letting ζ = z1 , then instead of z0 = ∞ we will expand the function in Laurent series regarding the point ζ0 = 0. Thus, the circular ring of convergence z > 1 will be replaced by the unit disc U(0; 1), while the type of the singular point z0 = ∞ will be given by the type of the point ζ0 = 0. Thus, f (z) =
(z 2
∞ 1 1 1 ∞ n n 1 = 4 ∑ 2(n−1) = ∑ 2(n+1) , = 4 1 2 z (1 − 2 )2 z n=1 z − 1) z n=1 z
where z > 1,
hence ∞ 1 f ( ) = ∑ nζ 2(n+1) , ζ n=1
ζ  < 1.
We conclude that z0 = ∞ is a removable singular point. 2. We have f ∈ H(ℂ \ {1}), and we will expand the function in Laurent series about the point z0 = 1 in the circular ring U(0; 1, ∞), i. e., f (z) = sin =−
∞ 1 1 1 = ∑ (−1)n+1 1 − z n=0 (2n + 1)! (z − 1)2n+1
1 1 1 − + ⋅⋅⋅, + (z − 1) 3!(z − 1)3 5!(z − 1)5
where 0 < z − 1 < ∞,
and where we used the relation (7.41). Hence the point z0 = 1 is an essential isolated singular point.
7.4 Solutions to the exercises of Chapter 4  305
For the expansion about the point z0 = ∞ we will expand the function in Laurent series in the circular ring z > 1 as follows: f (z) = sin
1 1 1 1 ∞ 1 = sin(− ) = sin(− ∑ ) 1−z z 1− 1 z n=0 z n z
1 1 1 = − sin( + 2 + 3 + ⋅ ⋅ ⋅) z z z ∞
= ∑ (−1)n+1 n=0
2n+1
1 1 1 1 ( + 2 + 3 + ⋅ ⋅ ⋅) (2n + 1)! z z z
1 1 1 5 1 1 =− − 2 − − + ⋅⋅⋅, 3 z z 6z 2 z4
where z > 1,
and where we used the formulas (7.38) and (7.41). From here, since ∞ 1 1 2n+1 (ζ + ζ 2 + ζ 3 + ⋅ ⋅ ⋅) , f ( ) = ∑ (−1)n+1 ζ (2n + 1)! n=0
ζ  < 1,
we conclude that z0 = ∞ is a removable singular point. 3. We have f ∈ H(U(0; 0, π)), and f (z) =
1 1 z 1 = = g(z), z sin z z 2 sin z z 2
0 < z < π,
where g(z) =
1 z = ∈ H(U(0; π)), sin z ∑∞ (−1)n z 2n n=0 (2n+1)!
n because limz→0 g(z) = 1. If g(z) = ∑∞ n=0 an z , then ∞
∑ (−1)n
n=0
∞ z 2n ∑ an z n = 1, (2n + 1)! n=0
z ∈ U(0; π),
thus we get a0 = 1,
a1 = 0,
a2 =
1 , 6
a3 = 0,
a4 =
7 , 360
....
Hence f (z) =
1 1 7 2 + + z + ⋅⋅⋅, z 2 6 360
where z ∈ U(0; 0, π),
where we used the relation (7.41). It follows that the point z0 = 0 is a second order pole.
306  7 Solutions to the chapterwise exercises For the expansion about the point z0 = π, we will expand the function in Laurent series on the circular ring U(π; 0, π). Substituting z = t + π, the new function will be expanded for the corresponding point t0 = 0, i. e., f (t + π) =
1 1 −1 −1 = = t (t + π) sin(t + π) (t + π) sin t π(1 + ) sin t π
=−
tn t 2 7t 4 1 1 ∞ + ⋅ ⋅ ⋅) ∑ (−1)n n (1 + + π n=0 π 6 360 t
1 t 2 7t 4 t 1 t3 t2 + 2 − 3 + 4 − ⋅ ⋅ ⋅)(1 + + + ⋅ ⋅ ⋅) π π 6 360 t π π π2 + 6 1 11 π2 + 6 2 − + =− t − ⋅ ⋅ ⋅ , t < π, t+ π t π2 6π 4 6π 3
= (−
where we used the formulas (7.36) and (7.41). Returning to the original variable z, we get f (z) = −
π2 + 6 1 1 1 π2 + 6 (z − π)2 − ⋅ ⋅ ⋅ , + 2 − (z − π) + π z−π π 6π 4 6π 3
where z ∈ U(π; 0, π),
and thus the point z0 = π is a firstorder pole. 4. We have f ∈ H(ℂ\{−1}), and the expansion in Laurent series for the point z0 = ∞ will be made in the circular ring z > 1, i. e., 2
f (z) = e 1+ z = e2 ∑n=0 ∞
1
2n (− z1 +
∞
= e2 ∑
(−1)n zn
n=0
= e2 (1 −
1 z2
= e2 e −
n!
1 z3
2(− z1 +
1 z2
−
1 z3
+⋅⋅⋅)
+ ⋅ ⋅ ⋅)n
2 4 + + ⋅ ⋅ ⋅), z z2
where z ∈ U(0; 1, ∞),
and where we used the relations (7.36) and (7.42). From here, since 1 f ( ) = e2 (1 − 2ζ + 4ζ 2 + ⋅ ⋅ ⋅), ζ
ζ  < 1,
it follows that z0 = ∞ is a removable singular point. 5. We have f ∈ H(ℂ \ {i}), and the expansion in Laurent series for the point z0 = i will be made in the circular ring U(i; 0, ∞), i. e., z+i
2π
2i
2π
f (z) = eiπ z−i = eiπ(1+ z−i ) = eiπ e− z−i = −e− z−i n
(−1)n 2π ( ) , n! z−i n=0 ∞
=−∑
where z ∈ ℂ \ {i},
where we used the formula (7.42). Thus, z0 = i is an essential isolated singular point.
7.4 Solutions to the exercises of Chapter 4  307
6. The function is not defined at z0 = 0, and in the points of the form zk = kπ, k ∈ ℤ. We will expand the function in Laurent series for the point z0 = 0 in the disc U(0; π). Using the wellknown relation, cos z 1 1 − f (z) = cot z = = sin z z 1−
z2 2! z2 3!
+ +
z4 4! z4 5!
− ⋅⋅⋅ − ⋅⋅⋅
0 < z < π,
,
and using the undetermined coefficients method, we get f (z) =
1 2 5 1 1 − z − z3 − z − ⋅⋅⋅, z 3 45 945
0 < z < π.
We conclude that z0 = 0 is a firstorder pole. For the points of the form zk = kπ, k ∈ ℤ, using the above results we get f (z) = cot(z − kπ + kπ) = cot(z − kπ) 1 1 2 1 − (z − kπ) − (z − kπ)3 − (z − kπ)5 − ⋅ ⋅ ⋅ , = z − kπ 3 45 945 0 < z − kπ < π. Thus, the points zk = kπ, k ∈ ℤ, are firstorder poles. 7. We have f ∈ H(ℂ \ {0}), and f (z) = =
∞ 2n 2n sin2 z 1 − cos 2z 1 n2 z = = (1 − (−1) ) ∑ (2n)! z5 z5 z5 n=0
4 21 2 + z − ⋅⋅⋅, − z 3 3 z 45
where z ∈ ℂ∗ .
From here, we get that z0 = 0 is a thirdorder pole for f . Solution of Exercise 4.10.9 1. The function is not defined in the points z1 = 1 and z2 = −2. Decomposing the function into the simple fractions, we get 1 1 1 3 1 = ( − + ) (z − 1)2 (z + 2) 9 (z − 1)2 z − 1 z + 2 1 1 1 1 3 1 1 − + = ( 2 ). 9 z (1 − 1 )2 z 1 − 1 2 1 + z2 z z
f (z) =
Using the wellknown Taylor series expansions (7.43), (7.38) and (7.36), we obtain f (z) = =
1 ∞ 1 zn 1 ∞ 1 3 ∞ n ( 2 ∑ n−1 − ∑ n + ∑ (−1)n n ) 9 z n=1 z z n=0 z 2 n=0 2 zn 1 ∞ 3n − 1 1 ∞ ( ∑ n+1 − + ∑ (−1)n n+1 ), 9 n=1 z z n=0 2
1 < z < 2,
308  7 Solutions to the chapterwise exercises where we used that 1 z > 1 ⇔ < 1, z
z z < 2 ⇔ < 1. 2
2. Similarly, for point 1, we have f ∈ H(ℂ \ {−1, −2}) and 2 z − 2 1 2z + 1 2 1 z = + − 2 2 2 2 25 z + 2 + 1) (z + 2) 25 z + 1 5 (z + 1) 1 1 1 2z + 1 2z − 4 1 − + = 5z 4 (1 + 12 )2 25 1 + z2 25z 2 1 + 12 z z
f (z) =
(z 2
Using the wellknown Taylor series expansions (7.43) and (7.36), we get f (z) = =
zn 2z − 4 ∞ 2z + 1 ∞ 1 ∞ n−1 n n 1 (−1) (−1) + − ∑ (−1)n n ∑ ∑ 4 2 2n 2n−2 25 n=0 2 5z n=1 25z n=0 z z n 2z − 4 ∞ 2(n − 5)z + n + 20 (−1)n+1 1 ∞ nz − + (−1) , ∑ ∑ 25 25 n=0 2n 25z 2 z 2(n+1) n=1
1 < z < 2,
where we used that 1 z > 1 ⇔ 2 < 1, z
z z < 2 ⇔ < 1. 2
Solution of Exercise 4.10.10 1. Decomposing the function into simple fractions, and using the wellknown expansion formulas in Laurent series, we get 1 1 1 1 3 = ( − ) + z(z − 3)2 9 z z − 3 (z − 3)2 1 1 3 1 1 1 1 ) + + = ( 1 z−1 9 z−11+ 21− 4 (1 − z−1 )2 z−1 2 2
f (z) =
=
1 1 ∞ 1 ∞ (z − 1)n 3 ∞ (z − 1)n−1 1 ) ( + ∑ + ∑n ∑ (−1)n n 9 z − 1 n=0 (z − 1) 2 n=0 2n 4 n=1 2n−1
=
∞ 1 ∞ 1 3n + 2 ( ∑ (−1)n + ∑ n+1 (z − 1)n−1 ), n+1 9 n=0 (z − 1) n=1 2
1 < z − 1 < 2,
where we used that 1 z − 1 > 1 ⇔ < 1, z − 1
z − 1 z − 1 < 2 ⇒ < 1. 2
7.4 Solutions to the exercises of Chapter 4  309
2. Similarly, for point 1, we obtain 1 1 5 1 1 − − = ( ) (z 2 − 1)(z 2 + 4)2 25 z 2 − 1 z 2 + 4 (z 2 + 4)2 1 1 1 1 1 1 5 − = ( 2 − ) 25 z 1 − 12 z 2 1 + 42 z 4 (1 + 42 )2
f (z) =
z
z
z
4n 1 ∞ 5 ∞ 4n−1 1 1 ∞ 1 = ( 2 ∑ 2n − 2 ∑ (−1)n 2n − 4 ∑ (−1)n−1 n 2n−2 ) 25 z n=0 z z n=1 z n=0 z z 1 − (−1)n 4n − 5n(−1)n−1 4n−1 1 , 25 z 2n+2 n=1 ∞
z > 2,
=∑ where we used that
1 z > 2 ⇔ z > 1 ⇔ 2 < 1, z
4 z > 2 ⇔ 2 < 1. z
3. From the relation (7.42), we have 1
∞
∞ 1 1 , = ∑ n n+3 n!z n!z n=0 n=0
f (z) = z 3 e z = z 3 ∑
z ∈ ℂ∗ .
4. Using the wellknown expansion formula in Taylor series for the cosine function, we obtain f (z) = z 3 cos
∞ 1 1 3 = [(z − 2) + 2] ∑ (−1)n z−2 (2n)!(z − 2)2n n=0 ∞
= [(z − 2)3 + 6(z − 2)2 + 12(z − 2) + 8] ∑ (−1)n z ∈ ℂ \ {2}.
n=0
1 , (2n)!(z − 2)2n
5. We will use the relation (7.36) and the undetermined coefficients method, hence 1 1 1 7 4 = (1 + z 2 + z + ⋅ ⋅ ⋅), sin z z 6 360
0 < z < π.
Thus 1 7 4 1 1 = (1 + z + z 2 + ⋅ ⋅ ⋅)(1 + z 2 + z + ⋅ ⋅ ⋅) 6 360 z 2 (1 − z) sin z z 3 1 7 427 1 7 = 3 + 2 + + + z + ⋅ ⋅ ⋅ , 0 < z < 1. 6z 6 360 z z
f (z) =
6. Similarly, for point 4, we deduce that f (z) = cos2
2 1 1 + cos z 22n−1 1 1 2 1 ∞ , = = + cos = + ∑ (−1)n z 2 2 2 z 2 n=0 (2n)!z 2n
z ∈ ℂ∗ .
310  7 Solutions to the chapterwise exercises Solution of Exercise 4.10.11 In both of these cases, we will use the undetermined coefficient method1 1. Using the wellknown relation, ∞
sin z = ∑ (−1)n n=0
z3 z5 z7 z 2n+1 =z− + − + ⋅⋅⋅, (2n + 1)! 3! 5! 7!
z ∈ ℂ,
we get f (z) =
1 1 z 1 = = sin z z sin z z 1 −
z2 3!
1 +
z4 5!
− ⋅⋅⋅
,
z ∈ ℂ \ {kπ : k ∈ ℤ}.
From the undetermined coefficients method, 1−
z2 3!
1 +
z4 5!
= a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅
− ⋅⋅⋅
we get a0 = 1,
a1 = 0,
a2 =
1 , 6
a3 = 0,
a4 =
7 , 360
...,
and thus f (z) =
1 z 7z 3 1 = + + + ⋅⋅⋅, sin z z 6 360
z ∈ U(0; 0, π).
2. Using the wellknown relation, z2 z3 zn =1+z+ + + ⋅⋅⋅, n! 2! 3! n=0 ∞
ez = ∑
z ∈ ℂ,
we get f (z) =
ez
1 1 z 1 = = z −1 z e −1 z 1+
z 2!
1 +
z2 3!
+ ⋅⋅⋅
,
z ∈ ℂ \ {2kπi : k ∈ ℤ}.
According to the undetermined coefficients method, 1+
z 2!
1 +
z2 3!
+ ⋅⋅⋅
= a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ ⇒ a0 = 1,
1 a1 = − , 2
a2 =
1 , 12
...,
we conclude that f (z) =
ez
1 1 1 z = − + − ⋅⋅⋅, − 1 z 2 12
z ∈ U(0; 0, 2π).
1 The method of an undetermined coefficient is an approach to find a particular solution for certain problems.
7.4 Solutions to the exercises of Chapter 4  311
Solution of Exercise 4.10.12 Let denote z − i = r1 eiθ1 ,
π 3π where θ1 , θ2 ∈ [− , ] 2 2
z + i = r2 eiθ2 ,
and π π (θ1 , θ2 ) ≠ (− , ), 2 2
(θ1 , θ2 ) ≠ (
3π π , ). 2 2
Using the above notation, the function f will have the form f (z) = √r1 r2 ei
θ1 +θ2 +2kπ 2
,
k ∈ {0, 1}.
Using the given assumption, we will determine the branch which satisfies the condition 3 5 f ( ) = ⇒ k = 0. 4 4 Hence we deduce that 1
1 2 1 f (z) = √z 2 (1 + 2 ) = z(1 + 2 ) , z z
1
1 2 where (1 + 2 ) = √2, z=1 z
thus 1
∞ 1 ( 1 − 1) ⋅ . . . ⋅ ( 1 − n + 1) 1 2 2 ) f (z) = z(1 + 2 ) = z(1 + ∑ 2 2 z n!z 2n n=1 ∞ 1(1 2 2
=z+∑
− 1) ⋅ . . . ⋅ ( 21 − n + 1) n!z 2n−1
n=1
,
z > 1,
where we used the formula of the binomial expansion, and the fact that 1 z > 1 ⇔ 2 < 1. z Solution of Exercise 4.10.13 We will determine the required branch similarly as in Exercise 4.10.12 If z + 1 + r1 eiθ1 ,
z − 1 = r2 eiθ2 ,
where θ1 , θ2 ∈ [0, 2π], (θ1 , θ2 ) ≠ (0, π),
312  7 Solutions to the chapterwise exercises then f (z) = √r1 r2 ei
θ1 +θ2 +2kπ 2
k ∈ {0, 1},
,
and f (√5) = 2 ⇒ k = 0. It follows that 1
−1
1
1
2 1 1 = = (1 − 2 ) , 1 1 z z (z 2 − 1) 2 z(1 − z12 ) 2
2 1 2 , = where (1 − 2 ) z=√5 √5 z
hence 1
− 21
1 1 = = (1 − 2 ) 1 1 z z (z 2 − 1) 2 z(1 − z12 ) 2 1
=
− 1 (− 1 − 1) ⋅ . . . ⋅ (− 21 − k + 1) 1 ∞ , + ∑ (−1)k 2 2 z k=1 k!z 2k+1
z > 1,
where we used the formula of the binomial expansion, and the fact that 1 z > 1 ⇒ 2 < 1. z Substituting n = −k, we deduce the required formula 1
√z 2 − 1
=
−1 1 + ∑ z n=−∞
1 2
⋅
3 2
⋅ . . . ⋅ ( 21 − n − 1) (−n)!
z 2n−1 ,
z > 1.
Solution of Exercise 4.10.14 First, we will decompose the function f into the simple fractions, i. e., f (z) =
1 1 1 1 ( = − ). 2 √ √ −1− −1+ 5 √5 z − 1−z−z z− 2 5 2
The expansion in Taylor series is ∞
f (z) = ∑ an z n , n=0
1 + √5 ), 2
z ∈ U(0;
where an =
Since (n)
(
c ) az + b
= (−1)n c
n!an , (az + b)n+1
1 (n) f (0), n ∈ ℕ. n!
7.4 Solutions to the exercises of Chapter 4  313
we get f (n) (z) =
(−1)n n! [ √5 (z −
and thus an =
1
−1−√5 n+1 ) 2 n+1
1 1 + √5 ) [( √5 2
−
1
(z −
−1+√5 n+1 ) 2
],
n+1
−(
1 − √5 ) 2
].
Solution of Exercise 4.10.15 1. We have
2 z z+2−2 1 − = = , z + 2 (z + 2)2 (z + 2)2 (z + 2)2
f (z) =
z ∈ ℂ \ {−2}.
2. The function f has the poles zk = 2kπi, where k ∈ ℤ. For z0 = 0, the circular ring U(0; 0, 2π) is the maximal ring with the center at the origin where the function f is holomorphic. Then
2 ez + 1 ez − 1 + 2 = =1+ z , ez − 1 ez − 1 e −1 and using point 2 of Exercise 4.10.11 we get f (z) =
1 1 z − ⋅ ⋅ ⋅), f (z) = 1 + 2( − + z 2 12
z ∈ U(0; 0, 2π),
z ∈ U(0; 0, 2π).
Thus, the main part of the Laurent series expansion regarding the point z0 = −2 for the function f is z2 . For the points of the form z0 = 2πi, substituting z = t + 2πi we get f (t + 2πi) = 1 + and thus
2
e2πi+t
−1
f (z) = 1 + 2(
=1+
et
1 1 2 t = 1 + 2( − + − ⋅ ⋅ ⋅), t 2 12 −1
1 z − 2πi 1 − + − ⋅ ⋅ ⋅), z − 2πi 2 12
0 < t < 2π,
z ∈ U(2πi; 0, 2π).
Thus, the main part of the Laurent series expansion regarding the point z0 = 2πi for 2 the function f is z−2πi . 3. Using the result of point 1 of Exercise 4.10.11, we have z−1
2
1 ) sin z sin2 z 2 1 1 z 7z 3 1 = (z − 1)( + + + ⋅ ⋅ ⋅) = (z − 1)( 2 + + ⋅ ⋅ ⋅) z 6 360 3 z 1 1 1 = − 2 + − + ⋅ ⋅ ⋅ , z ∈ U(0; 0, π), z 3 z
f (z) =
= (z − 1)(
hence the main part of the Laurent series expansion is − z12 + z1 .
314  7 Solutions to the chapterwise exercises Solution of Exercise 4.10.16 We will use the first removability criterion. 1. Since lim f (z) = lim z→1
z→1
z2 − 1 = 2 ∈ ℂ, z−1
it follows that z0 = 1 is a removable singular point. 2. The point z0 = 0 is a removable singular point, because lim f (z) = lim
z→0
z→0
sin z = 1 ∈ ℂ. z
3. Since lim f (z) = lim
z→0
z→0
z cos z z = lim = 1 ∈ ℂ, = lim tan z z→0 sin z z→0 sin z cos z
z
it follows that z0 = 0 is a removable singular point. 4. To determine the type of the point z0 = π2 , we will expand the given function in Laurent series for the point z0 . Since f (z) =
1 1 1 1 − − = , cos2 z (z − π2 )2 sin2 (z − π ) (z − π2 )2 2
we will use the result obtained in point 1 of Exercise 4.10.11, for t = z − π2 , i. e., 1 1 = sin(z − π2 ) z −
π 2
+
z−
6
π 2
+
Then the main part of the expansion of
7(z − π2 )3 360
1 sin2 (z− π2 )
+ ⋅⋅⋅,
will be
π 0 < z − < π. 2
1 , hence the main part for (z− π2 )2
the Laurent series expansion of the function f about z0 is 0. From the second removability criterion, it follows that z0 = π2 is a removable singular point. 5. In this case, we will also use the second removability criterion. Since f (z) = =
2n 1 1 1 ∞ 1 z2 z4 1 − cos z n z = + − ⋅ ⋅ ⋅) (−1) = − − (1 − ∑ (2n)! z 2 z 2 2! 4! z2 z 2 z 2 n=0
1 z2 − + ⋅⋅⋅, 2 4!
z ∈ ℂ∗ ,
we conclude that z0 = 0 is a removable singular point. 6. Using the results of Exercise 4.10.11, we have 1 1 1 z 1 z 7z 3 1 − =( − + − ⋅ ⋅ ⋅) − ( + + + ⋅ ⋅ ⋅) − 1 sin z z 2 12 z 6 360 1 z − ⋅ ⋅ ⋅ , 0 < z < π. =− − 2 12
f (z) =
ez
7.4 Solutions to the exercises of Chapter 4  315
The main part of the Laurent series expansion of f for z0 is 0; then from the second removability criterion, it follows that z0 = 0 is a removable singular point. 7. Since z2 − 1 2 z−1 = − ∈ ℂ, = lim 2 z→−1 z 3 + 1 z→−1 z − z + 1 3
lim f (z) = lim
z→−1
from the first removability criterion it follows that a z0 = −1 is a removable singular point. Solution of Exercise 4.10.17 We will prove the above results by using the theorem of characterization of the poles. 1. Since 1 = ∞, z→0 z
lim f (z) = lim
z→0
it follows that z0 = 0 is a firstorder pole for f . 2. The point z0 = i is a double pole for f , because lim f (z) = lim
z→i (z 2
z→i
1 = ∞. + 1)2
3. Since z2 + 1 = ∞, z→−1 z + 1
lim f (z) = lim
z→−1
the point z0 = −1 is a firstorder pole for f . 4. We will determine the main part of the Laurent series about the point z0 . Since f (z) =
1 1 = , 1 − cos z 2 sin2 z 2
using point 1 of Exercise 4.10.11 we get 1 t 7t 3 1 = + + + ⋅⋅⋅, sin t t 6 360
0 < t < π,
1 1 t2 + ⋅⋅⋅, + + t 2 3 15
0 < t < π.
hence it follows that 1
sin2 t
=
Substituting t = z2 , we have f (z) =
1
2 sin2 z2
=
2 1 z2 + + + ⋅⋅⋅, 2 6 120 z
and it follows that z0 = 0 is a double pole for f .
0 < z < 2π,
(7.44)
316  7 Solutions to the chapterwise exercises 5. Since lim f (z) = lim
z→−i
z→−i
(z 2
1 = ∞, + 1)2
the point z0 = −i double pole for f . 6. We will determine the main part of the Laurentseries about the point z0 . A simple calculation shows that f (z) =
z − 2π z z − 2π + 2π 2π = = + . 2 z 2 z−2π 1 − cos z 2 sin 2 2 sin 2 2 sin2 z−2π 2
Using the relation (7.44) from the point 4, we get 1
sin2 t Letting t = f (z) =
z−2π , 2
+ π( =
1 1 t2 + + + ⋅⋅⋅, t 2 3 15
0 < t < π,
we have
z − 2π
2 sin2
=
z−2π 2
+
2π
2 sin2
z−2π 2
=
4 z − 2π 1 (z − 2π)2 ( + + ⋅ ⋅ ⋅) + 2 60 (z − 2π)2 3
4 1 (z − 2π)2 + + + ⋅ ⋅ ⋅) 2 3 60 (z − 2π)
π z − 2π π(z − 2π)2 4π 2 + + + + ⋅⋅⋅, + 2 z − 2π 3 6 60 (z − 2π)
0 < z − 2π < 2π,
then it follows that z0 = 2π is a double pole for f . 7. We will determine the main part of the Laurent series about the point z0 . Since f (z) =
z z z − πi + πi z − πi πi = = − z−πi = − z−πi − , ez + 1 eπi ez−πi + 1 e −1 e − 1 ez−πi − 1
using the result from point 2 of Exercise 4.10.11 we have t 1 1 1 − ⋅⋅⋅, = − + et − 1 t 2 12
0 < t < 2πi,
and substituting t = z − πi it follows that z − πi 1 z − πi πi 1 − + − ⋅ ⋅ ⋅) − z−πi = −(z − πi)( z − πi 2 12 −1 e −1 1 1 z − πi − πi( − + − ⋅ ⋅ ⋅) z − πi 2 12 πi − 2 6 − πi −πi + + (z − πi) + ⋅ ⋅ ⋅ , 0 < z − πi < 2π. = z − πi 2 12
f (z) = −
ez−πi
Thus, z0 = πi is a firstorder pole for f .
7.4 Solutions to the exercises of Chapter 4  317
Solution of Exercise 4.10.18 1. We will determine the main part of the Laurent series for the point z0 , by using the formula (7.41), i. e., f (z) = sin
∞ π π 2n+1 = ∑ (−1)n , 2 z (2n + 1)!z 4n+2 n=0
z ∈ ℂ∗ .
Since the main part of the Laurent series expansion has an infinite number of terms, it follows that z0 = 0 is an essential isolated singular for f . 2. The Laurent series expansion of f for z0 is f (z) = (z − 1)2 cos
∞ π π 2n , = ∑ (−1)n z − 1 n=0 (2n)!(z − 1)2(n−1)
z ∈ ℂ \ {1}.
The main part of the Laurent series expansion has an infinite number of terms, hence z0 = 1 is an essential isolated singular for f . 3. The Laurent series expansion of f about z0 is 1
∞
1 , n!(z + 1)n n=0
f (z) = e z+1 = ∑
z ∈ ℂ \ {−1},
thus z0 = −1 is an essential isolated singular for f , because the main part of the Laurent series expansion has an infinite number of terms. 4. The Laurent series expansion for z0 may be determined as follows: f (z) = cos
1 1 1 z = cos(1 − ) = cos 1 cos + sin 1 sin z+1 z+1 z+1 z+1 ∞ (−1)n (−1)n + sin 1 ∑ , 2n 2n+1 n=0 (2n + 1)!(z + 1) n=0 (2n)!(z + 1) ∞
= cos 1 ∑
z ∈ ℂ \ {−1}.
Since the main part of the Laurent series expansion has an infinite number of terms, it follows that z0 = −1 is an essential isolated singular for f . 5. The function is not defined in the points z1,2 = ±i, thus f ∈ H(U(i; 0, 2)). Since m
∞ π iπ 1 z−i iπ = − (−1)m ( ) , = − ∑ 2 z−i 2(z − i) 2(z − i) 2i z +1 1+ m=0
0 < z − i < 2,
2i
where z − i z − i < 2 ⇔ < 1, 2i
318  7 Solutions to the chapterwise exercises by using the expansion of the sine function in Taylor series, it follows that m 2n+1
∞ ∞ π iπ z−i (−1)n f (z) = sin 2 ) ] = ∑ [− ∑ (−1)m ( 2i z + 1 n=0 (2n + 1)! 2(z − i) m=0
,
0 < z − i < 2. For m = 0, the main part of the Laurent series expansion for z0 has an infinite number of terms, hence z0 = i is an essential isolated singular for f . 6. We will consequently use the expansion formulas in the Taylor series of the sine and of the exponential functions, i. e., 1
∞ ∞ (−1)n 2n+1 (−1)n (2n + 1)m e z = ∑ (∑ ), (2n + 1)! (2n + 1)! m=0 m!z m n=0 n=0 ∞
f (z) = sin e z = ∑
z ∈ ℂ∗ .
We deduce that z0 = 0 is an essential isolated singular for f , because the main part of the Laurent series expansion for z0 has an infinite number of terms. Solution of Exercise 4.10.19 Since a ∈ D, g ∈ H(D) and g(a) = 0, according the properties of holomorphic functions ̇ r). Hence the of zeros, it follows that there exists r > 0, such that g(z) ≠ 0, ∀z ∈ U(a; point a ∈ D is an isolated singular point for the function h. Then f (z) = lim z→a g(z) z→a lim
f (z)−f (a) z−a g(z)−g(a) z−a
=
f (a) ∈ ℂ, g (a)
because g (a) ≠ 0, and using the first removability criterion we deduce that the point a ∈ D is a removable isolated singular point for the function h. Solution of Exercise 4.10.20 Since f : U(a; r) \ {a} → ℂ∗ , the function f has no zeros, thus the function h is welldefined, and f ∈ H(U(a; r) \ {a}) ⇒
1 ∈ H(U(a; r) \ {a}) ⇒ h ∈ H(U(a; r) \ {a}). f
We need to prove that the function h is differentiable at the point a, whenever a is a pole for f . Using the theorem of characterization of the poles, the point z0 = a is a pole for f if and only if ∃! n ∈ ℕ∗ , ∃ρ > 0, ∃g ∈ H(U(a; ρ)) : g(a) ≠ 0,
f (z) =
g(z) , (z − a)n
̇ ρ). ∀z ∈ U(a;
7.4 Solutions to the exercises of Chapter 4  319
Using these 1
h(z) − h(a) f (z) = lim z→a z→a z − a z−a lim
= lim
z→a
i. e., ∃ limz→a
h(z)−h(a) z−a
(z−a)n g(z)
z−a
0, (z − a)n−1 ∈{ 1 z→a , g(z) g(a)
= lim
if n > 1 if n = 1,
∈ ℂ, hence the function h is holomorphic in the disc U(a; r).
Solution of Exercise 4.10.21 Since f ∈ H(D), the function φ(z) = ef (z) is holomorphic on the domain D, with φ(z) ≠ 0, ∀z ∈ D. From here, according to the theorem of the module maximum of the holomorphic functions, we have that φ has no locally maximum or minimum points in the domain D. But f (z) Re f (z) , φ(z) = e = e hence the function g = Re f has no locally maximum or minimum points in the domain D. Similarly, if we denote h = Im f , and using the fact Re(−if ) = Im f , it follows that h = Im f has no locally maximum or minimum points in the domain D. Solution of Exercise 4.10.22 From Exercise 4.10.21, it is known that the function Re f has no locally minimum point in the disc U(0; r), whenever f is a nonconstant function. Then Re f (z) < A,
∀z ∈ U(0; r),
(7.45)
hence, from Re f (0) = 0 it follows that A > 0. Let φ : U(0; r) → ℂ,
φ(z) =
f (z) , 2A − f (z)
that is also a holomorphic function on the disc U(0; r), because Re f (z) ≤ A, ∀z ∈ U(0; r). Then φ(0) = 0, and since A > 0, from the inequality (7.45), we deduce that f (z)2 f (z)2 2 = < 1, φ(z) = 2 2A − f (z) 4A[A − Re f (z)] + f (z)2 i. e., φ(z) < 1,
∀z ∈ U(0; r).
∀z ∈ U(0; r),
320  7 Solutions to the chapterwise exercises Using the generalized Schwarz lemma, we get z , φ(z) ≤ r
z ∈ U(0; r).
From here, since φ(z) =
2Aφ(z) f (z) ⇔ f (z) = , 2A − f (z) 1 + φ(z)
we deduce the required inequality 2Az 2Aφ(z) 2Aφ(z) ≤ , ≤ f (z) = 1 + φ(z) 1 − φ(z) r − z
∀z ∈ U(0; r).
Solution of Exercise 4.10.23 Since a < 1, the function h : U(0; 1) → U(0; 1),
h(z) =
z+a 1 + az
is bijective and its inverse will be h−1 : U(0; 1) → U(0; 1),
h−1 (z) =
z−a . 1 − az
Thus g = h−1 ∘ f ∘ h ∈ H(U(0; 1)), g(0) = h−1 (f (a)) = h−1 (a) = 0 and g(z) < 1, ∀z ∈ U(0; 1). From here, according to the Schwarz lemma, we get g(z) ≤ z,
∀z ∈ U(0; 1),
and since g(
b−a b−a ) = h−1 (f (b)) = h−1 (b) = , 1 − ab 1 − ab
using again the Schwarz lemma we deduce that g(z) = cz,
z ∈ U(0; 1) where c = 1.
b−a since there exists z0 = 1−ab ∈ U(0; 1), such that g(z0 ) = z0 , it follows c = 1. Hence g(z) = z, ∀z ∈ U(0; 1). From here, we have f (h(z)) = h(z), ∀z ∈ U(0; 1) and since h : U(0; 1) → U(0; 1) is a bijection we conclude that
f (z) = z,
∀z ∈ U(0; 1).
7.4 Solutions to the exercises of Chapter 4  321
Solution of Exercise 4.10.24 Since α < 1, the function φ : U(0; 1) → U(0; 1),
φ(z) =
z−α 1 − αz
is bijective and its inverse will be φ−1 : U(0; 1) → U(0; 1),
φ−1 (z) =
z+α . 1 + αz
Then g = f ∘ φ−1 ∈ H(U(0; 1)), g(0) = f (α) = 0 and g(z) = f (φ−1 (z)) < 1, ∀z ∈ U(0; 1), because from the theorem of the module maximum of the holomorphic functions there does not exist points z0 ∈ U(0; 1) such that f (z0 ) = 1. Using the Schwarz lemma for the function g, we have −1 f (φ (z)) ≤ z,
∀z ∈ U(0; 1).
From here, since the function φ : U(0; 1) → U(0; 1) is bijective, it follows that z − α , f (z) ≤ 1 − αz
∀z ∈ U(0; 1),
Solution of Exercise 4.10.25 Since 1 1 f( ) = , n n+1
1
1 ∀n ∈ ℕ∗ ⇔ f ( ) = n 1 , n 1+ n
∀n ∈ ℕ∗ ,
it follows that the function f (z) =
z ∈ H(ℂ \ {−1}) z+1
satisfies the assumption of the problem. Since limn→+∞ n1 = 0 ∈ ℂ \ {−1}, according to the theorem of the uniqueness of holomorphic functions there does not exist another function that satisfies these properties. Solution of Exercise 4.10.26 1. Since 0, 1 f( ) = { 2 , n n
if n = 2k + 1, k ∈ ℕ if n = 2k, k ∈ ℕ∗
and
lim
k→∞
1 5 1 = 0 ∈ U( ; ), 2k + 1 2 6
322  7 Solutions to the chapterwise exercises from the theorem of the uniqueness of holomorphic functions, since f ∈ H(U( 21 ; 65 )), it follows that f (z) = 0, ∀z ∈ U( 21 ; 65 ). But this function does not satisfy the assumption f ( n1 ) = n2 , n = 2k and k ∈ ℕ∗ . 2. We will use the same method as in point 1. Since 1 f( ) = { n { { { ={ { { {
1 , k+2 1 , k+1 2 n
if n = 2k + 1, k ∈ ℕ if n = 2k, k ∈ ℕ∗
1+ n3
,
if n = 2k + 1, k ∈ ℕ
1+ n2
,
if n = 2k, k ∈ ℕ∗
2 n
and limk→∞ 2k1 = 0 ∈ U( 21 ; 65 ), from the theorem of the uniqueness of holomorphic func2z tions, since f ∈ H(U( 21 ; 65 )), it follows that f (z) = 2z+1 , ∀z ∈ U( 21 ; 65 ). But this function 1 2 does not satisfy the assumptions f ( n ) = n+3 , n = 2k + 1 and k ∈ ℕ. Solution of Exercise 4.10.27 1. The function f (z) = z 2 , ∀z ∈ U(0; 1), is holomorphic in the disc U(0; 1) and it satisfies the assumption. From the theorem of the uniqueness of holomorphic functions, this function is the only one from H(U(0; 1)) that satisfies the given assumptions. 2. Since 1 1 f( ) = 3, n n
n ∈ ℕ∗ and lim
n→+∞
1 = 0 ∈ U(0; 1), n
from the theorem of the uniqueness of holomorphic functions, since f ∈ H(U(0; 1)), it follows that f (z) = z 3 , ∀z ∈ U(0; 1). But this function does not satisfy the assumption f (− n1 ) = n13 , n ∈ ℕ∗ .
7.5 Solutions to the exercises of Chapter 5 Solution of Exercise 5.4.1 1. From the formula of the residues value at the pole, we get Res(f ; 2) = lim f (z)(z − 2) = lim z→2
z→2
sin z (z − 2) = sin 2, z−2
because z0 = 2 is a simple pole for the function f . 2. The point z0 = 1 is a simple pole for the function, hence Res(f ; 1) = lim f (z)(z − 1) = lim z→1
z→1
z (z − 1) = 1. (z − 1)(z − 2)2
7.5 Solutions to the exercises of Chapter 5  323
Since the point z1 = 2 is a secondorder pole, we have Res(f ; 2) =
1 z −1 = −1. lim[f (z)(z − 2)2 ] = lim( ) = lim z→2 z − 1 z→2 (z − 1)2 1! z→2
3. The residue of the function in the point z1 is Res(f ; z1 ) =
(p−1)
1 z (z − z1 )p ] lim[ (p − 1)! z→1 (z − z1 )p (z − z2 ) (p−1)
z 1 ) lim( z→1 (p − 1)! z − z2 z2 = (−1)p−1 , (z1 − z2 )p =
=
(p − 1)!z2 1 lim(−1)p−1 z→1 (p − 1)! (z − z2 )p
because z1 is a pth order pole. The point z2 is a simple pole for the function, hence z2 z (z − z2 ) = . (z − z1 )p (z − z2 ) (z2 − z1 )p
Res(f ; z2 ) = lim
z→z2
4. The point z0 = −1 is a firstorder pole, hence Res(f ; −1) = lim
z→−1
1 1 1 (z + 1) = lim 3 =− . z→−1 z (1 − z) 2 z3 − z5
The point z1 = 0 is a thirdorder pole, thus Res(f ; 0) = =
1 1 1 1 lim ( 3 z 3 ) = lim ( ) 5 2! z→0 z − z 2! z→0 1 − z 2 2 + 6z 2 1 lim = 1. 2 z→0 (1 − z 2 )3
The residue in the point z2 = 1 is Res(f ; 1) = lim z→1
−1 1 1 (z − 1) = lim 3 =− , z→1 z (1 + z) 2 z3 − z5
since z2 is a simple pole. 5. Both of the points are simple poles, hence the residues are Res(f ; −i) = lim
z→−i
z2 i z2 (z + i) = lim =− 2 z→−i z − i 2 z +1
and Res(f ; i) = lim
i z2 z2 = . (z − i) = lim z→i z + i 2 +1
z→i z 2
324  7 Solutions to the chapterwise exercises 6. The point z0 = −1 is an nth order pole, hence (n−1)
1 z 2n 1 (n−1) lim [ lim (z 2n ) (z + 1)n ] = (n − 1)! z→−1 (z + 1)n (n − 1)! z→−1 (2n)! 1 (−1)n+1 = (−1)n+1 Cn−1 = 2n . (n − 1)! (n + 1)!
Res(f ; −1) =
Since the sum of the residues of all the poles is zero, we deduce that the residue in z1 = ∞ is Res(f ; ∞) = − Res(f ; −1) = (−1)n Cn−1 2n . 7. The points zk = 2kπi are firstorder poles, for all k ∈ ℤ, hence ζ ζ z − 2kπi = lim = lim 1 − ez ζ →0 1 − eζ +2kπi ζ →0 1 − eζ ζ 1 = −1. = lim n = lim n−1 ζ →0 − ∑∞ ζ ζ →0 1 − ∑∞ ζ n=0 n! n=1 n!
Res(f ; zk ) = lim
z→2kπi
Solution of Exercise 5.4.2 1. The function f has the simple poles zk = ei zn = ∞ is an nth order pole. The residues in the points zk = ei Res(f ; zk ) =
(2k+1)π n
(2k+1)π n
, where k ∈ {0, 1, . . . , n − 1}, while
are
(2k+1)π zk z n z ei n z 2n = = − = − . n z=zk n n nz n−1 z=zk
To calculate the residue in the pole zn = ∞, we will use the definition of the residue at the infinity, i. e., Res(f ; ∞) = −a−1 , where ∞
f (z) = ∑ an z n , n=−∞
z ∈ U(0; r, ∞).
In a neighborhood of the infinity, the expansion of the function in Laurent series is f (z) =
z 2n 1 = zn 1 + zn 1+
∞
1 zn
= z n ∑ (−1)k ( k=0
k
1 ) zn
1 1 1 1 1 − + ⋅ ⋅ ⋅) = z n − 1 + n − 2n + ⋅ ⋅ ⋅ , + z n z 2n z 3n z z where z > 1,
= z n (1 −
7.5 Solutions to the exercises of Chapter 5  325
hence 0, −1,
Res(f ; ∞) = {
if n ≠ 1 if n = 1.
2. The points of the form zk = i(2k + 1)2π, k ∈ ℤ, are secondorder poles. The residues in the points zk are Res(f ; zk ) = lim [ z→zk
eaz
z
(1 + e 2 )2
(z − zk )2 ] = eazk lim[ t→0
t 2 eat t
(1 − e 2 )2
],
where we used the substitution z = t + zk . The expansion in Taylor series of the function ez is well known, hence 2
∞ tn tn (1 − e ) = (1 − ∑ n ) = ( ∑ n ) 2 n! n=1 2 n! n=0 t 2
∞
2
2
=
t2 t t2 (1 + + + ⋅ ⋅ ⋅) , 4 4 24
2
where t ∈ ℂ,
and thus Res(f ; zk ) = 4eazk lim[ t→0
= 4e
azk
(1 +
lim[e
t→0
at a
eat
t 4
+
−
1 2
t2 24
+ ⋅ ⋅ ⋅)2
t 4
+
]
+ ( a4 − 61 )t + ⋅ ⋅ ⋅
(1 +
t2 + ⋅ ⋅ ⋅)3 24 i(2k+1)2πa
= 2(2a − 1)eazk = 2(2a − 1)e
]
,
k ∈ ℤ.
3. The function has the essential singular points z0 = 0 and z1 = ∞. To determine the residue in the point z0 = 0, we will expand the function in Taylor series, i. e., 1
1 zk ∞ ⋅ ∑ (−1)j j k! j!z j=0 k=0 ∞
1
f (z) = ez− z = ez e− z = ∑ ∞ ∞
= ∑ ∑ (−1)j k=0 j=0
1 k−j z , k!j!
where z ∈ ℂ∗ ,
hence ∞
∞
Res(f ; 0) = − Res(f ; ∞) = ∑ ∑ (−1)j k=0 j=0 k−j=−1
∞ 1 (−1)k+1 = ∑ . k!j! k=0 k!(k + 1)!
4. The point z0 = 0 is an essential singular point, z1 = −1 is a simple pole while z2 = ∞ is a n − 1th order pole.
326  7 Solutions to the chapterwise exercises To determine the residue of f in the point z0 = 0, we will expand the function in Laurent series. Using the following formulas, ∞ 1 = ∑ (−1)n z n , 1 + z n=0
1
if z < 1,
∞
1 , n n!z n=0
ez = ∑
if z ∈ ℂ∗ ,
we get 1
1 1 1 zne z + = z n (1 − z + z 2 − z 3 + ⋅ ⋅ ⋅)(1 + + f (z) = + ⋅ ⋅ ⋅), 1+z z 2!z 2 3!z 3 0 < z < 1,
hence Res(f ; 0) = a−1 =
1 1 1 1 − + − + ⋅⋅⋅ (n + 1)! (n + 2)! (n + 3)! (n + 4)!
− e1 +
={
1 e
−
1 2!
1 2!
−
+
1 3!
1 3!
+ ⋅⋅⋅ +
− ⋅⋅⋅ +
1 , n!
1 , n!
if n is even
if n is odd.
From the formulas of the residues value at the pole, for the point z1 = −1 we have 1
Res(f ; −1) = lim [ z→−1
1 zne z (−1)n (1 + z)] = lim z n e z = . z→−1 1+z e
To calculate the residue at the infinity, we will use the definition of the residue at the infinity. Case 1. If n = 1, we will use the wellknown Taylor series expansion formula, 1
f (z) =
ez
1+
1 ∞ (−1)j ∑ j k k=0 k!z j=0 z ∞
1 z
= (1 +
= ∑
1 1 1 1 1 1 + + + ⋅ ⋅ ⋅)(1 − + 2 − 3 + ⋅ ⋅ ⋅), 1!z 2!z 2 3!z 3 z z z
z > 1,
hence Res(f ; ∞) = −a−1 = 0. Case 2. If n > 1, we get similarly 1
f (z) =
z n−1 e z 1+
1 z
= z n−1 (1 + z > 1,
1 ∞ (−1)j ∑ j k k=0 k!z j=0 z ∞
= z n−1 ∑
1 1 1 1 1 1 + ⋅ ⋅ ⋅)(1 − + 2 − 3 + ⋅ ⋅ ⋅), + + 1!z 2!z 2 3!z 3 z z z
and thus Res(f ; ∞) = −a−1 = −
1 1 1 1 + − + ⋅ ⋅ ⋅ + (−1)n−1 . n! (n − 1)! (n − 2)! 2!
7.5 Solutions to the exercises of Chapter 5  327
Solution of Exercise 5.4.3 1. The function has the simple poles zk = kπi, where k ∈ ℤ, hence Res(f ; zk ) = lim [ z→zk
eiaz teia(zk +t) teiat (z − zk )] = lim = eiazk lim t→0 sinh(zk + t) t→0 (−1)k sinh t sinh z
= (−1)k eiazk lim
t→0
= (−1)k e−akπ .
teiat
∑∞ k=0
t 2k+1 (2k+1)!
= (−1)k eiazk lim
t→0
eiat
∑∞ k=0
t 2k (2k+1)!
2. The function f has the secondorder pole z0 = 1, hence 1
1
−e z ez (z − 1)2 ] = lim 2 = −e. Res(f ; 1) = lim[ z→1 z z→1 (z − 1)2 The point z1 = 0 is an isolated essential singular point, and 1
f (z) = e z
∞ 1 ∞ j−1 1 = ∑ ∑ jz (1 − z)2 k=0 k!z k j=1
= (1 +
1 1 1 + ⋅ ⋅ ⋅)(1 + 2z + 3z 2 + 4z 3 + ⋅ ⋅ ⋅), + + 1!z 2!z 2 3!z 3
0 < z < 1,
hence Res(f ; 0) = 1 +
1 1 1 + + + ⋅ ⋅ ⋅ = e. 1! 2! 3!
3. For the given function, the points z0 = i and z1 = −i are nth order poles, hence Res(f ; i) =
(n−1)
1 1 (z − i)n ] lim[ 2 z→i (n − 1)! (z + 1)n (n−1)
=
1 1 lim[ ] (n − 1)! z→i (z + i)n
=−
n(n + 1) ⋅ . . . ⋅ (2n − 2) i (n − 1)! 22n−1
and Res(f ; −i) =
(n−1)
1 1 (z + i)n ] lim [ (n − 1)! z→−i (z 2 + 1)n (n−1)
=
1 1 lim[ ] (n − 1)! z→i (z − i)n
=
n(n + 1) ⋅ . . . ⋅ (2n − 2) i . (n − 1)! 22n−1
4. The function has the singular points zk = kπ, where k ∈ ℤ∗ that are simple poles, and z0 = 0 is a thirdorder pole. The point z∗ = ∞ is an accumulation point for the poles of the function f , hence it is not an isolated singular point of the function f .
328  7 Solutions to the chapterwise exercises We have Res(f ; kπ) =
1
(z 2 sin z) z=zk
=
(−1)k 1 = 2 2, 2 2z sin z + z cos z z=zk k π
where k ∈ ℤ∗ .
Since Res(f ; 0) =
1 1 1 z z 3 ] = lim [ lim [ ] , 2! z→0 z 2 sin z 2! z→0 sin z
we define the function g(z) =
1 z . = ∞ sin z ∑ (−1)n z 2n n=0 (2n+1)!
n Then g ∈ H(U(0; π)), because limz→0 g(z) = 1. If g(z) = ∑∞ n=0 an z , then ∞
∑ (−1)n
n=0
∞ z 2n ∑ an z n = 1, (2n + 1)! n=0
z ∈ U(0; π),
hence it follows that a0 = 1,
a1 = 0,
a2 =
1 , 6
a3 = 0,
a4 =
7 , 360
...
and thus 7 4 1 z + ⋅⋅⋅, g(z) = 1 + z 2 + 6 360
z ∈ U(0; π).
From here, we deduce that Res(f ; 0) =
1 11 1 lim g (z) = = , 2! z→0 23 6
since
g (z) = (
z 7z 3 z ) = + + ⋅⋅⋅, sin z 3 90
g (z) = (
z 1 7z 2 ) = + + ⋅⋅⋅, sin z 3 30
0 < z < π,
and
0 < z < π.
5. The point z0 = 1 is an essential singular point for the function, while z1 = ∞ is a thirdorder pole. The function has the expansion about the point z0 = 1 of the form 1
(−1)n , n!(z − 1)n n=0 ∞
f (z) = z 3 e 1−z = [(z − 1)3 + 3(z − 1)2 + 3(z − 1) + 1] ∑ z ∈ U(1; 0, ∞).
7.5 Solutions to the exercises of Chapter 5  329
Hence 1 3 3 1 − + −1= , 4! 3! 2! 24
Res(f ; 1) = a−1 =
and from the sum of the residues it follows that Res(f ; ∞) = − Res(f ; 1) = −
1 . 24
6. The function has only the secondorder poles zk = i(2k + 1) π2 , where k ∈ ℤ. From here, we get Res(f ; zk ) = lim [ z→zk
eiaz
cosh2 z
(z − zk )2 ] = −eiazk lim[
t 2 eiat
sinh2 t t cosh t − sinh t
t→0
= −eiazk lim[ieiat t→0
t t (a + 2i sinh t sinh t
]
2
sinh t
π
)] = −iae−a(2k+1) 2 ,
where we used the substitution z = t + zk and the following relations: cosh(t + zk ) = i(−1)k sinh t,
lim
t→0
sinh t = 1, t
lim
t→0
t cosh t − sinh t sinh2 t
= 0.
Solution of Exercise 5.4.4 1. The function has singular points z0 = 0 that is a firstorder pole, and z1 = 2i, z2 = −2i that are secondorder poles. Then Res(f ; 0) = lim
z→0
1 z+2 = , (z 2 + 4)2 8
Res(f ; 2i) = lim [ z→2i
z+2 1 1 z 2 + 3z + 2i ] = lim −2 2 =− − i 2 z→2i 16 32 z(z + 2i) z (z + 2i)3
and Res(f ; −2i) = lim [ z→−2i
z+2 z 2 + 3z − 2i 1 1 ] = lim −2 2 = − + i. 2 z→−2i 16 32 z(z − 2i) z (z − 2i)3
(a) The path {γ} = 𝜕U(0; 1) that is directly oriented, turns around only the point z0 , hence from the residues theorem we have ∫ γ
πi z+2 dz = 2πi Res(f ; 0) = . 2 2 4 z(z + 4)
(b) The path {γ} = 𝜕U(0; 5) turns around all of the singular points of f , and from the residues theorem it follows that ∫ γ
z+2 dz = 2πi[Res(f ; 0) + Res(f ; 2i) + Res(f ; −2i)] = 0. z(z 2 + 4)2
330  7 Solutions to the chapterwise exercises 2. The point z0 = 1 is a secondorder pole for the function, while z1 = i and z2 = −i are simple poles. The disc U(1 + i; 2) contains the points z0 and z1 . Since Res(f ; 1) = lim[ z→1
1 −2z 1 ] = lim 2 =− 2 2 z→1 (z + 1) 2 z +1
and Res(f ; i) = lim z→i
1 1 = , (z − 1)2 (z + i) 4
from the residues theorem, it follows that ∫ γ
1 πi dz = 2πi[Res(f ; 1) + Res(f ; i)] = − . 2 (z − 1)2 (z 2 + 1)
3. Case 1. If n = 1, then f (z) = follows that ∫ γ
1 z+2
∈ H(ℂ \ {−2}). From the residues theorem, it
1 dz = 2πi Res(f ; −2) = 2πi. z+2
Case 2. If n > 1, then the function f has the singular points of the form zk = √2e , k ∈ {0, 1, . . . , n − 1}, that are simple poles and zk  = √n 2 for all k ∈ {0, 1, . . . , n − 1}. The path γ(t) = (3 + ε)e2πit + 1, t ∈ [0, 1], with ε > 0 turns around all of these poles, hence n
i π+2kπ n
∫ γ
zn
n−1 1 dz = 2πi ∑ Res(f ; zk ) = −2πi Res(f ; ∞). +2 k=0
Since k 1 ∞ 1 1 1 2 22 1 23 k 2 = = + ⋅ ⋅ ⋅), = (−1) (1 − + − ∑ z n + 1 z n 1 + 2n z n k=0 z n z 2n z 3n z kn z n z n
z > √2,
we have Res(f ; ∞) = 0, hence ∫ γ
1 dz = 0. zn + 2
4. The function has singular points all the nth order roots of the unity, and also all of the thirdorder roots of the unity. But all of these points do not belong to the U(0; r)
7.5 Solutions to the exercises of Chapter 5  331
disc, whenever r < 1, and from the Cauchy integral theorem the value of the integral will be zero, i. e., ∫ γ
1 dz = 0. (z n − 1)(z 3 − 1)
5. The points z0 = 1, z1 = i and z2 = −i are simple poles for the function, and the path γ(t) = √2e2πit + (1 + i), t ∈ [0, 1], turns around only the points z0 and z1 . Since Res(f ; 1) = lim z→1
1 1 = z2 + 1 2
and Res(f ; i) = lim z→i
1−i 1 =− , (z − 1)(z + i) 4
it follows that ∫ γ
1 π(i − 1) dz = 2πi[Res(f ; 1) + Res(f ; i)] = . 2 (z − 1)(z 2 + 1) kπi
6. The points zk = e 5 , k ∈ {0, 1, . . . , 9}, are simple poles, while z10 = 0 is a thirdorder pole for f . All these singular points belong to the disc U(0; 2), because zk  = 1, for all k ∈ {0, 1, . . . , 9}, and z10  = 0. Using the theorem related to the sum of the residues in all of the singular points, we get ∫ γ
1
z 3 (z 10 − 1)
10
dz = 2πi ∑ Res(f ; zk ) = −2πi Res(f ; ∞). k=0
Since 1
z 3 (z 10 − 1)
=
1 1 1 ∞ 1 , = ∑ z 13 1 − 110 z 13 n=0 z 10n z
z > 1,
it follows that Res(f ; ∞) = 0, hence ∫ γ
1 dz = 0. z 3 (z 10 − 1)
7. The point z0 = 1 is an essential singular point for the function. Since 1
(−1)n 1 1 1 − + ⋅⋅⋅, + =1− n 2 n!(z − 1) z − 1 3!(z − 1)3 2!(z − 1) n=0 ∞
e 1−z = ∑
where z ∈ ℂ \ {1},
332  7 Solutions to the chapterwise exercises it follows that Res(f ; 1) = a−1 = −1, hence 1
∫ e 1−z dz = 2πi Res(f ; 1) = −2πi. γ
8. The function f has the firstorder poles of the form zk = Res(f ; zk ) =
1 2
+ k, where k ∈ ℤ, and
sin πz sin πz 1 = =− , (cos πz) z=zk −π sin πz z=zk π
∀k ∈ ℤ.
The disc U(0; n), n ∈ ℕ∗ , contains the point zk if and only if 1 zk  < n ⇔ + k < n. 2
Thus,
k=n−1
∫ tan πzdz = 2πi ∑ Res(f ; zk ) = −4ni. γ
k=−n
9. The function has singular points zk = π2 + kπ, where k ∈ ℤ, which are simple poles, and z∗ = 0 is also a simple pole. The disc U(0; 2) contains only the poles z∗ = 0, z1 = π2 and z−1 = − π2 . Since tan z tan z = 1, z] = lim z→0 z z2 π sin z sin z 1 4 Res(f ; ) = 2 = = =− 2 2 2 π π π 2 (z cos z) z= 2 2z cos z − z sin z z= 2 − π 4
Res(f ; 0) = lim [ z→0
and sin z π sin z −1 4 Res(f ; − ) = 2 = = 2 = − 2, 2 π 2 (z cos z) z=− π2 2z cos z − z sin z z=− π2 π 4
it follows that ∫ γ
π 8 tan z π dz = 2πi[Res(f ; 0) + Res(f ; ) + Res(f ; − )] = 2πi(1 − 2 ). 2 2 2 z π π+2kπ
10. The function f has the simple poles of the form zk = ei 4 , where k ∈ {0, 1, 2, 3}. Since the disc U(1; 1) contains only the poles z0 and z3 , we have Res(f ; z0 ) =
√2 z z 1 = 0 = − 0 = − (1 + i) 4 8 (z 4 + 1) z=z0 4z04
7.5 Solutions to the exercises of Chapter 5  333
and similarly Res(f , z3 ) =
(z 4
√2 z z 1 = 34 = − 3 = − (1 − i). 4 8 + 1) z=z3 4z3
Thus, 1 πi√2 . dz = 2πi[Res(f ; z0 ) + Res(f ; z3 )] = − 2 z4 + 1
∫ γ
11. The point z0 = 0 is a simple pole, while z1 = 1 is an essential isolated singular point. The directly oriented path {γ} = 𝜕U(0; 2) turns around both of these points, hence 1 1 Res(f ; 0) = lim [z sin ] = sin 1, z→0 z (z − 1)2 to determine the residue in z1 = 1 we will expand the function in Laurent series about this point. Using the relations, ∞ 1 1 = = ∑ (−1)n (z − 1)n , z (z − 1) + 1 n=0
sin
z − 1 < 1,
∞ 1 1 , = ∑ (−1)n 2 (z − 1) (2n + 1)!(z − 1)2(2n+1) n=0
z ∈ ℂ \ {1},
it follows that f (z) = (1 − (z − 1) + (z − 1)2 − (z − 1)3 + ⋅ ⋅ ⋅ ) ⋅(
1 1 1 − − ⋅ ⋅ ⋅), + (z − 1)2 3!(z − 1)6 5!(z − 1)10
0 < z − 1 < 1,
hence the residue in z1 = 1 is Res(f ; 1) = a−1 = −1 +
1 1 1 − + − ⋅ ⋅ ⋅ = sin(−1) = − sin 1. 3! 5! 7!
Thus, ∫ γ
1 1 dz = 2πi[Res(f ; 0) + Res(f ; 1)] = 0. sin z (z − 1)2
12. The directly oriented path {γ} = 𝜕U(0; 2) turns around the simple pole z0 = 0 and the essential isolated singular point z1 = 1. Then 1 1 Res(f ; 0) = lim (z − 1)3 e z−1 = − . z→0 e
334  7 Solutions to the chapterwise exercises We will expand the function in Laurent series for the point z1 = 1 as follows: ∞ 1 1 = = ∑ (−1)n (z − 1)n , z (z − 1) + 1 n=0 ∞
1 , n!(z − 1)n n=0
1
e z−1 = ∑
z − 1 < 1,
z ∈ ℂ \ {1},
and we deduce that f (z) = (1 − (z − 1) + (z − 1)2 − (z − 1)3 + ⋅ ⋅ ⋅)(z − 1)3 ⋅ (1 +
1 1 1 + + ⋅ ⋅ ⋅), + z − 1 2!(z − 1)2 3!(z − 1)3
0 < z − 1 < 1.
From here, the residue in the point z1 = 1 will be Res(f ; 1) = a−1 =
1 1 1 1 1 1 1 1 1 − + − + ⋅ ⋅ ⋅ = e−1 − (1 − + − ) = − , 4! 5! 6! 7! 1! 2! 3! e 3
hence ∫ γ
1 (z − 1)3 z−1 2πi e dz = 2πi[Res(f ; 0) + Res(f ; 1)] = − . z 3
13. Since the directly oriented path {γ} = 𝜕U(0; r) has the radius r > 0 enough big, such that all the zeros of the function z 111 + z 11 + z + 1 belong to U(0; r), using the theorem related to the sum of the residues in all of the singular points we get ∫ γ
1 dz = −2πi Res(f ; ∞). z 111 + z 11 + z + 1
If f (z) =
1 , z 111 + z 11 + z + 1
z > r,
letting φ(ζ ) = f ( ζ1 ), we have φ(ζ ) =
1+
ζ 100
ζ 111 = ζ 111 (1 − ζ 100 + ⋅ ⋅ ⋅) = ζ 111 − ζ 211 + ⋅ ⋅ ⋅ , + ζ 110 + ζ 111
hence f (z) =
1 1 + ⋅⋅⋅, − z 111 z 211
z ∈ U(0; r, ∞).
From here, it follows that Res(f ; ∞) = 0, and then ∫ γ
1 dz = 0. z 111 + z 11 + z + 1
1 ζ  < , r
7.5 Solutions to the exercises of Chapter 5  335
14. The point z0 = 1 is an essential isolated singular point. Case 1. If r < 1, then the function f is holomorphic in the disc U(0; r). According to the Cauchy integral theorem, we have ∫ γ
1 1 z−1 e dz = 0. z−1
Case 2. If r > 1, then z1 ∈ U(0; r). To determine the residue in the point z1 , we will expand the function in Laurent series about z1 , i. e., f (z) = =
1 1 z−1 1 1 ∞ e = ∑ z−1 z − 1 n=0 n!(z − 1)n
1 1 1 1 + ⋅ ⋅ ⋅), + (1 + + z−1 z − 1 2!(z − 1)2 3!(z − 1)3
z ∈ ℂ \ {1},
hence Res(f ; 1) = a−1 = 1. We conclude that ∫ γ
1 1 z−1 e dz = 2πi Res(f ; 1) = 2πi. z−1
15. The disc U(0; 3) contains the next singular points of f : the point z0 = 0 that is an essential isolated singular point, and the simple poles z1 = 1 and z2 = −1. Then 1
Res(f ; 1) = lim z→1
e zne z = z+1 2
and 1
zne z (−1)n−1 = . z→−1 z − 1 2e
Res(f ; −1) = lim
To obtain the residue in the point z0 , we will expand the function in Laurent series about z0 . If we use the relations, ∞ 1 1 1 ∞ n ∞ 1 1 n n ( − ) = − ( z + (−1) z ) = − z 2n , = ∑ ∑ ∑ 2 n=0 z2 − 1 2 z − 1 z + 1 n=0 n=0 1
∞
1 , n n!z n=0
ez = ∑
z < 1,
z ∈ ℂ∗ ,
then f (z) = −z n (1 + z 2 + z 4 + z 6 + ⋅ ⋅ ⋅)(1 +
1 1 1 + ⋅ ⋅ ⋅), + + z 2!z 2 3!z 3
0 < z < 1.
336  7 Solutions to the chapterwise exercises Since the residue in the origin is Res(f ; 0) = a−1 = −
1 1 1 − − − ⋅⋅⋅, (n + 1)! (n + 3)! (n + 5)!
we get 1
zne z dz = 2πi[Res(f ; 0) + Res(f ; 1) + Res(f ; −1)] ∫ 2 z −1 γ
e (−1)n−1 + Res(f ; 0)] = 2πi[ + 2 2e ={
2πi[ 1!1 +
if n is even
2πi[1 +
if n is odd,
1 1 + 5!1 + ⋅ ⋅ ⋅ + (n−1)! ], 3! 1 1 + 4!1 + ⋅ ⋅ ⋅ + (n−1)! ], 2!
where we used the expansion in Taylor series of the function ez , and we replaced the variable z by 1, and respectively by −1. 16. The function has an essential isolated singular point z0 = 0, and the directly oriented path {γ} = 𝜕U(0; r), r > 0, turns around the point z0 . We will expand the function in Laurent series for the point z0 . Since we know that sin
1 1 1 1 = − + − ⋅⋅⋅, z z 3!z 3 5!z 5
z ∈ ℂ∗ ,
it follows f (z) = sinn
n
1 1 1 1 =( − + − ⋅ ⋅ ⋅) , 3 z z 3!z 5!z 5
z ∈ ℂ∗ .
Case 1. If n = 1, then Res(f ; 0) = 1, hence we get 1 ∫ sinn dz = 2πi Res(f ; 0) = 2πi. z γ
Case 2. If n > 1, then Res(f ; 0) = 0, hence we get 1 ∫ sinn dz = 2πi Res(f ; 0) = 0. z γ
17. The directly oriented path {γ} = 𝜕U(0; 3) turns around the points z0 = 0, z1 = 1 and z2 = 2, that are essential singular points. Let consider the directly oriented paths γ0 , γ1 and γ2 , where {γk } = 𝜕U(zk ; 21 ), with k ∈ {0, 1, 2}. From a consequence of the Cauchy integral theorem, it is well known that ∫ f (z)dz = ∫ f (z)dz + ∫ f (z)dz + ∫ f (z)dz. γ
γ0
γ1
γ2
7.5 Solutions to the exercises of Chapter 5  337
The integral on the path γ0 is given by 1
1
1
∫ f (z)dz = ∫(1 + z + z 2 )e z dz + ∫(1 + z + z 2 )(e z−1 + e z−2 )dz, γ0
γ0
γ0
1
1
and since (1 + z + z 2 )(e z−1 + e z−2 ) is a holomorphic function in the disc U(0; 21 ) we get 1
1
∫(1 + z + z 2 )(e z−1 + e z−2 )dz = 0.
γ0
1
To calculate the integral ∫γ (1 + z + z 2 )e z dz, we need to expand in Laurent series for 0
1
z0 = 0 the function f0 (z) = (1 + z + z 2 )e z , i. e.,
∞
1 n n!z n=0
1
f0 (z) = (1 + z + z 2 )e z = (1 + z + z 2 ) ∑ = (1 + z + z 2 )(1 +
1 1 1 + + ⋅ ⋅ ⋅), + z 2!z 2 3!z 3
z ∈ ℂ∗ .
It follows that 1 5 1 + = , 2! 3! 3
Res(f0 ; 0) = a−1 = 1 + hence
10πi . 3
∫ f0 (z)dz =
γ0
(7.46)
The other integrals can be calculated similarly, i. e., 1
1
1
∫ f (z)dz = ∫(1 + z + z 2 )e z−1 dz + ∫(1 + z + z 2 )(e z + e z−2 )dz, γ1
γ1
γ1
where 1
1
∫(1 + z + z 2 )(e z + e z−2 )dz = 0.
γ1
Since 1
f1 (z) = (1 + z + z 2 )e z−1 = ((z − 1)2 + 3(z − 1) + 3) ⋅ (1 +
1 1 1 + ⋅ ⋅ ⋅), + + 1!(z − 1) 2!(z − 1)2 3!(z − 1)3
it follows that Res(f1 , 1) = a−1 = 3 +
3 1 14 + = , 2! 3! 3
z ∈ ℂ \ {1},
338  7 Solutions to the chapterwise exercises hence 28πi . 3
∫ f1 (z)dz =
γ1
(7.47)
The value of the integral on the path γ2 is given by 1
1
1
∫ f (z)dz = ∫(1 + z + z 2 )e z−2 dz + ∫(1 + z + z 2 )(e z + e z−1 )dz,
γ2
γ2
γ2
where 1
1
∫(1 + z + z 2 )(e z + e z−1 )dz = 0.
γ2
Since 1
f2 (z) = (1 + z + z 2 )e z−2 ((z − 2)2 + 5(z − 2) + 7) ⋅ (1 +
1 1 1 + + + ⋅ ⋅ ⋅), 1!(z − 2) 2!(z − 2)2 3!(z − 2)3
z ∈ ℂ \ {2},
it follows that Res(f2 , 2) = a−1 = 7 +
1 29 5 + = , 2! 3! 3
hence ∫ f2 (z)dz =
γ2
58πi . 3
(7.48)
Using the relations (7.46), (7.47) and (7.48), we conclude that 1
1
1
∫(1 + z + z 2 )(e z + e z−1 + e z−2 )dz = 32πi. γ
18. The points zk = (2k + 1)πi, with k ∈ ℤ are secondorder poles for the function. The directly oriented path {γ} = 𝜕U(0; r), with (2n − 1)π < r < (2n + 1)π and n ∈ ℕ∗ , turns around the point zk , whenever k ∈ {−n, −n + 1, −n + 2, . . . , n − 2, n − 1}. The residues in the points zk are Res(f ; zk ) = lim [ z→zk
1 t2 2 (z − z ) ] = lim [ ], k t→0 (1 − et )2 (1 + ez )2
where we used the substitution z = t + zk . Using the expansion in Taylor series of the function ez , we deduce that 2
2
2
∞ n t t t2 tn (1 − e ) = (1 − ∑ ) = ( ∑ ) = t 2 (1 + + + ⋅ ⋅ ⋅) , n! 2 6 n=1 n! n=0 t 2
∞
z ∈ ℂ.
7.5 Solutions to the exercises of Chapter 5  339
Hence Res(f ; zk ) = lim[ t→0
t 2 (1 +
t2
t 2
+
−1 − 67 t + ⋅ ⋅ ⋅
t2 6
+ ⋅ ⋅ ⋅)2
] = lim
t→0
(1 +
t 2
+
t2 6
+ ⋅ ⋅ ⋅)3
= −1,
and the value of the given integral is ∫ γ
n−1 1 dz = 2πi ∑ Res(f ; zk ) = −4nπi. z 2 (1 + e ) k=−n
Solution of Exercise 5.4.5 1. The point z1 = −i is a fourthorder pole, and the path {γ} = {z ∈ ℂ : z + 2i = 2} turns around this pole. The residue of the function f in this pole is Res(f ; −i) =
1 πz lim [cosh ] 3! z→−i 2
=−
π3i , 48
where we used that πz π sinh , 2 2 2 π πz f (z) = cosh , 4 2 3 πz π sinh . f (z) = 8 2 f (z) =
From the residues theorem, it follows that ∫ γ
cosh πz 2 (z + i)4
dz = 2πi Res(f ; −i) =
π4 . 4!
100 iπz
e has the simple poles z1 = i and z2 = −i, and the ellipse 2. The function f (z) = z z 2 +1 2 2 {γ} = {z = x + iy ∈ ℂ : 4x + y − 4 = 0} turns around these poles. The residues of the given function in these poles are
z 100 eiπz e−π = , z→i z + i 2i z 100 eiπz eπ Res(f ; −i) = lim =− . z→−i z − i 2i Res(f ; i) = lim
The value of the integral is ∫ γ
z 100 eiπz dz = 2πi[Res(f ; i) + Res(f ; −i)] = −2π sinh π. z2 + 1
340  7 Solutions to the chapterwise exercises 5 z+3i √
3. For the function f (z) = z3−z 3 , the points z1 = −3i and z2 = 3 are essential isolated singular points, while z3 = 0 is a thirdorder pole. We will determine the branch of the function involved in this problem. Denoting z + 3i = r1 eiθ1 ,
3 − z = r2 eiθ2 ,
it follows that gk (z) = √ 5
r θ1 −θ2 +2kπ z + 3i = √5 1 ei 5 , 3−z r2
k ∈ {0, 1, 2, 3, 4},
and for k = 2, we get g(z) = g2 (z) = √ 5
19π z + 3i 10 = √2ei 20 . 3 − z z=3i
̇ r), r < 3. From the Thus, the given function f is holomorphic on the circular ring U(0; residues theorem, we need to compute only the residue in the origin, i. e., Res(f ; 0) = because
1 1 5 − i i 9π 5 z + 3i lim [√ ] = lim g (z) = e 10 , 2! z→0 3−z 2! z→0 225
√5 z+3i 3 3 3−z g (z) = ( + i) , 5 5 (3 − z)(z + 3i) g (z) = ( and then
√5 z+3i 6 6 3−z + i)(−6 + 9i + 5z) , 25 25 (z + 3i)2 (3 − z)2 r θ1 −θ2 +4π g(z) = g2 (z) = √5 1 ei 5 . r2
From here, it follows ∫ γ
√5 z+3i 3−z z3
dz = 2πi Res(f ; 0) =
2π(1 + 5i) i 9π e 10 . 225
4. The function has an infinite number of singular points, and the path {γ} = {z = x + iy ∈ ℂ : x2 + y2 − 2y − 3 = 0} turns around only the points z1 = 0, z2 = −√ π2 , z3 = √ π2 , z4 = i√ π2 , z5 = i√ 3π , and z6 = i√ 5π . The residues in these points are 2 2 Res(f ; zk ) = lim
z→zk
1 1 = (z cos z 2 ) cos zk2 − 2zk2 sin zk2
1, { { { 1 − = { 2zk2 , { 1 { , { 2zk2
if k = 1, if k ∈ {2, 3, 5},
if k ∈ {4, 6}.
7.5 Solutions to the exercises of Chapter 5  341
Using the residues theorem, it follows that ∫ γ
6 1 58 ). dz = 2πi Res(f ; zk ) = 2πi(1 − ∑ 15π z cos z 2 k=1 2z
5. The function f (z) = z 2 e z+1 has an essential singular point z0 = −1, and the path {γ} = {z = x + iy ∈ ℂ : x2 + y2 + 2x = 0} turns around this point. Expanding the function in Laurent series for the point z0 = −1, we have 2z
2
f (z) = z 2 e z+1 = (z + 1 − 1)2 e2− z+1
(−2)n n!(z + 1)n n=0 ∞
= [(z + 1)2 − 2(z + 1) + 1]e2 ∑ = ⋅⋅⋅ − hence
22e2 1 + ⋅⋅⋅, 3 z+1 Res(f ; −1) = −
z ∈ ℂ \ {−1},
22 2 e. 3
Thus, 2z
∫ z 2 e z+1 dz = − γ
44i 2 πe . 3
6. The path {γ} = {z = x + iy ∈ ℂ : 4x2 + 9y2 − 36 = 0} turns around to all of the 13 isolated singular points of the function f (z) = (z−2)4z(z 5 +3)2 , that are: the fourthorder
pole z5 = 2, and secondorder poles zk = √5 3ei 5 , with k ∈ {0, 1, 2, 3, 4}. Since the sum of all of the residues is zero, it is easier to calculate the integral by computing the residue at the infinity. Thus, we will expand the function in Laurent series for the infinity, i. e., (2k+1)π
f (z) =
1 1 z 13 = (z − 2)4 (z 5 + 3)2 z (1 − 2 )4 (1 + z
3 2 ) z5
3 32 1 2 4 ⋅ 5 22 + ⋅ ⋅ ⋅)(1 − 2 5 + 3 10 + ⋅ ⋅ ⋅) = (1 + 4 + 2 z z 2 z z z 1 = ⋅ ⋅ ⋅ + + ⋅ ⋅ ⋅ , z > 2. z Thus Res(f ; ∞) = −1, hence it follows that ∫ γ
z 13 dz = −2πi Res(f ; ∞) = 2πi. (z − 2)4 (z 5 + 3)2
342  7 Solutions to the chapterwise exercises 1 z √4z 2 +12z+13
7. The function f (z) =
has isolated essential singular points z1 = − 32 + i
and z2 = − 32 − i, and the firstorder pole z3 = 0. If 0 < r < ∫ γ
√13 , 2
from the residues theorem we have
1 2πi 1 dz = 2πi Res(f ; 0) = 2πi lim = . z→0 √4z 2 + 12z + 13 √13 √4z 2 + 12z + 13
8. Since the function f (z) =
log(z−a) z2
is not defined in the point z = a, we will
replace the path {γ} = {z ∈ ℂ : z = Reiθ , 0 ≤ θ < 2π}, R > a > 0, with the path Γ = γR ∪ γ[R,a+ε] ∪ γε− ∪ γ[a+ε,R] , where γR (t) = Re2πit ,
t ∈ [0, 1],
γ[R,a+ε] (t) = (1 − t)R + t(a + ε), γa,ε (t) = εe
2πit
,
t ∈ [0, 1],
γ[a+ε,R] (t) = (1 − t)(a + ε) + tR,
t ∈ [0, 1], t ∈ [0, 1].
The point z = 0 is a secondorder pole for the function, the residue in this point is 1 Res(f ; 0) = lim [log(z − a)] = − , z→0 a hence ∫ Γ
log(z − a) 2πi . dz = − 2 a z
Also, we have a+ε
∫
γR
log(z − a) log(x − a) + 2πi dz + ∫ dx z2 x2 R
R
+ ∫
γa,ε
log(x − a) 2πi log(z − a) dz + ∫ dx = − . a z2 x2 a+ε
If z − a = εeiθ , then dz = εdθ and log(z − a) = √(ln ε)2 + θ ≤ √(ln ε)2 + 2π, 1 1 1 1 = ≤ = . z2 z − a + a2 (z − a − a)2 a2 − 2aε + ε2 Hence, if z − a = ε → 0 then the integral ∫γ
a,ε
log(z−a) dz z2
tends to zero, because
log(z − a) 2π √(ln ε)2 + 2π 2πε√(ln ε)2 + 2π dz ≤ ∫ 2 εdθ = → 0, ∫ 2 2 z a − 2aε + ε a2 − 2aε + ε2 γ a,ε
0
if ε → 0.
7.5 Solutions to the exercises of Chapter 5  343
Thus, if ε → 0 it follows that a
log(z − a) 2πi 1 dz + 2πi ∫ 2 dx = − , ∫ 2 a z x
γR
R
hence ∫ γ
2πi log(z − a) . dz = − R z2
Solution of Exercise 5.4.6 1 1. The function f (z) = z sin has the following isolated singular points: the secondz order pole z0 = 0, and the simple poles zk = kπ, where k ∈ ℤ∗ . The residues of the function in these points are
Res(f ; 0) = lim [ z→0
sin z − z cos z z2 = 0, ] = lim z→0 z sin z sin2 z
and Res(f ; kπ) = lim
z→kπ
z − kπ z sin z
ε (−1)k ε (−1)k ε = lim = , ε→0 (kπ + ε) sin(kπ + ε) ε→0 (kπ + ε) sin ε kπ
= lim
where k ∈ ℤ∗ . From here, we see that Res(f ; kπ) + Res(f ; −kπ) = 0. If the path {γ} = {z ∈ ℂ : z = r}, r ≠ kπ, k ∈ ℤ, turns around the points zk , then it turns around the points z−k , hence ∫ γ
n 1 dz = 2πi[Res(f ; 0) + ∑ [Res(f ; kπ) + Res(f ; −kπ)]] = 0, z sin z k=1
where nπ < r < (n + 1)π. 2. The function f (z) =
1 3 sin z−sin 3z
may be written in the form
f (z) =
1
4 sin3 z
,
where we used that sin 3z = 3 sin z − 4 sin3 z. Hence the function f has the third order poles zk = kπ, with k ∈ ℤ. The circle {γ} = {z ∈ ℂ : z = 4} turns around the points
344  7 Solutions to the chapterwise exercises z0 = 0, z1 = π and z−1 = −π. To determine the residues of f , we will expand the function in Laurent series for these points. Using the relation, 1 1 z 2 7z 4 = (1 + + + ⋅ ⋅ ⋅), sin z z 6 360
z ∈ U(0; 0, π),
we get f (z) =
1
=
4 sin3 z
z 2 17z 4 1 + + ⋅ ⋅ ⋅), (1 + 2 120 4z 3
z ∈ U(0; 0, π).
From here, it follows that Res(f ; 0) =
1 . 8
For the point z1 = π, substituting z = t + π and using the expansion about the point z∗ = 0 we get f (t + π) = −
1
3
4 sin t
=−
t 2 17t 4 1 (1 + + + ⋅ ⋅ ⋅), 3 2 120 4t
t ∈ U(0; 0, π),
or f (z) = −
(z − π)2 17(z − π)4 1 + + ⋅ ⋅ ⋅), (1 + 2 120 4(z − π)3
z ∈ U(π; 0, π),
hence 1 Res(f ; π) = − . 8 For the point z−1 = −π, substituting z = t − π and using the expansion for the point z∗ = 0 we get f (z) = −
1 (z + π)2 17(z + π)4 (1 + + + ⋅ ⋅ ⋅), 2 120 4(z + π)3
z ∈ U(−π; 0, π),
hence 1 Res(f ; −π) = − . 8 Combining the above results, it follows that ∫ γ
1 πi dz = 2πi[Res(f ; 0) + Res(f ; π) + Res(f ; −π)] = − . 3 sin z − sin 3z 4 1 z
e 3. For the function f (z) = (z−1) 2 , the point z0 = 0 is an isolated essential singular point, while z1 = 1 is a secondorder pole. Since the origin is an isolated essential
7.5 Solutions to the exercises of Chapter 5  345
singular point, we will expand the function in Laurent series for the origin, using also the relations 1
∞
1 , n n!z n=0
ez = ∑
z ∈ ℂ∗ ,
and ∞ 1 = nz n−1 , ∑ (z − 1)2 n=1
z < 1.
Thus, 1 1 1 + + ⋅ ⋅ ⋅), + 2 z 2!z 3!z 3
f (z) = (1 + 2z + 3z 2 + ⋅ ⋅ ⋅)(1 +
0 < z < 1,
hence it follows that Res(f ; 0) = 1 +
∞ 1 1 1 1 1 + + + + ⋅⋅⋅ = ∑ = e. 1! 2! 3! 4! n! n=0
The residue in z1 = 1 is 1
Res(f ; 1) = lim(e z ) = lim(− z→1
z→1
(a) If {γ} = {z ∈ ℂ : z =
√2 }, 2
1 z1 e ) = −e. z2
using the residues theorem it follows that
1
ez dz = 2πi Res(f ; 0) = 2πie. ∫ (z − 1)2 γ
(b) If {γ} = {z ∈ ℂ : z = 2}, we have 1
∫ γ
ez dz = 2πi[Res(f ; 0) + Res(f ; 1)] = 0, (z − 1)2
because the path of integration turns around all the singular points of f . 4. The function f (z) =
log z−1 z+1 (z 2 +1)(z 2 −4)
has the simple poles z1,2 = ±i, z3,4 = ±2, and the
z−1 z=0 = πi, isolated essential singular points z5,6 = ±1. Using the given relation log z+1 we deduce that k = 0 because z − 1 log = πi ⇔ log(−1) = i(π + 2kπ) = πi ⇔ k = 0. z + 1 z=0
(a) If {γ} = {z ∈ ℂ : z = 21 }, then the path does not turn around any singular points, and since f ∈ H(D) where D = ℂ \ ((−∞, −1] ∪ [1, +∞)), it follows that ∫ γ
z−1 log z+1
(z 2 + 1)(z 2 − 4)
dz = 0.
346  7 Solutions to the chapterwise exercises (b) If {γ} = {z ∈ ℂ : z = √2}, then the path turns around the points z1,2 . Since f ∈ H(D), where D = ℂ \ [−1, 1], we need to calculate the next residues: Res(f ; i) = lim z→i
z−1 log z+1
=−
(z + i)(z 2 − 4)
π , 20
and Res(f ; −i) = lim
z→−i
log z−1 z+1 (z −
i)(z 2
− 4)
=−
π . 20
From here, it follows that ∫ γ
z−1 log z+1
(z 2 + 1)(z 2 − 4)
dz = 2πi(Res(f ; i) + Res(f ; −i)) = −
π2i . 5
(c) The path {γ} = {z = x + iy ∈ ℂ : 3x2 + y2 − 2 = 0} turns around only the simple poles z1,2 = ±i, and f ∈ H(D), where D = ℂ \ ((−∞, −1] ∪ [1, +∞)). From the residues theorem, it follows that ∫ γ
z3,4
(z 2
z−1 log z+1
+
1)(z 2
− 4)
dz = 2πi(Res(f ; i) + Res(f ; −i)) = −
π2i . 5
(d) The path {γ} = {z ∈ ℂ : z = 3} turns around the simple poles z1,2 = ±i and a = ±2, and f ∈ H(D) where D = ℂ \ [−1, 1]. Since Res(f ; 2) = lim
z→2
log z−1 z+1
(z 2 + 1)(z + 2)
=−
ln 3 , 20
and Res(f ; −2) = lim
z→−2
(z 2
z−1 log z+1
+ 1)(z − 2)
=−
ln 3 , 20
we have ∫ γ
log z−1 z+1
(z 2 + 1)(z 2 − 4)
dz = 2πi(Res(f ; i) + Res(f ; −i) + Res(f ; 2) + Res(f ; −2)) =−
5. The function f (z) = z1 =
√2 (1 + i), z2 2
=
πi (π + ln 3). 5
sin z z 2 (z 4 +1)
√2 (1 − i), z3 2
=
has the following isolated singular points: z0 = 0,
√ − 22 (1 + i) and z4
sin z = z −
=−
z3 z5 + − ⋅⋅⋅, 3! 5!
√2 (1 − i). Since sin z 2
z ∈ ℂ,
∈ H(ℂ) and
7.5 Solutions to the exercises of Chapter 5  347
all of these singular points are simple poles. The circle {γ} = {z ∈ ℂ : z = 2} turns around all these points, and the residues are Res(f ; 0) = lim
z→0
sin z sin z 1 = lim = 1, 4 z→0 z z4 + 1 z(z + 1)
and Res(f ; zk ) = lim
z→zk
z 2 (z
(z − zk ) sin z . − z1 )(z − z2 )(z − z3 )(z − z4 )
It follows that Res(f ; z1 ) = Res(f ; z3 ) =−
√2 √2 √2 √2 √2 √2 1 + (1 + i)(e 2 − e− 2 ) cos ], [(1 − i)(e 2 + e− 2 ) sin √ 2 2 8 2
and Res(f ; z2 ) = Res(f ; z4 ) =−
√2 √2 √2 √2 √2 √2 1 [(1 + i)(e 2 + e− 2 ) sin + (1 − i)(e 2 − e− 2 ) cos ], 2 2 8√2
where we used that sin[
√2 √2 √2 i √2 i√2 (1 + i)] = sin cos + sin cos 2 2 2 2 2 √ √ √2 2 2 √2 √2 √2 1 (e 2 + e− 2 ) + i cos (e 2 − e− 2 )]. = [sin 2 2 2
Using the residues theorem, we conclude ∫ γ
sin z dz = 2πi(Res(f ; 0) + 2 Res(f ; z1 ) + 2 Res(f ; z2 )) + 1)
z 2 (z 4
= 2πi −
√2 √2 √2 √2 √2 √2 π √2 [(e 2 + e− 2 ) sin + (e 2 − e− 2 ) cos ]i. 2 2 2
1 6. The function f (z) = z 2 (z−1) has the thirdorder pole z0 = 0, and the simple sin z poles z∗ = 1 and zk = kπ, with k ∈ ℤ∗ . The path {γ} = {z ∈ ℂ : z = r}, nπ < r < (n + 1)π, n ∈ ℕ, turns around the points z0 , z∗ and zk , with k ∈ {±1, ±2, . . . , ±n}. The residue of the function in these points is
Res(f ; 1) = lim z→1
1 1 = , z 2 sin z sin 1
and Res(f ; kπ) = lim
z − kπ − 1) sin z
z→kπ z 2 (z
=
(−1)k 1 ε = lim , k 2 π 2 (kπ − 1) ε→0 sin(kπ + ε) k 2 π 2 (kπ − 1)
348  7 Solutions to the chapterwise exercises whenever k ∈ {±1, ±2, . . . , ±n}. To determine the residue in the origin, we will expand the function in Laurent series for the origin, i. e., 1 1 1 1 z 2 7z 4 = − 3 (1 + z + z 2 + z 3 + ⋅ ⋅ ⋅)(1 + + + ⋅ ⋅ ⋅) 2 6 360 z z z − 1 sin z 71 7 1 1 + + ⋅ ⋅ ⋅), z ∈ U(0; 0, 1), = −( 3 + 2 + 6z 6 z z
f (z) =
hence 7 Res(f ; 0) = − . 6 Combining the above results, by using the residues theorem, we get ∫ γ
n 1 dz = 2πi(Res(f ; 0) + Res(f ; 1) + Res(f ; kπ)) ∑ z 2 (z − 1) sin z k=−n
= 2πi( 7. The function f (z) =
z sin z(1−cos z)
n (−1)k 7 1 ). − + ∑ 2 2 sin 1 6 k=−n k π (kπ − 1)
has the isolated singular points zk = kπ, where
k ∈ ℤ. The point z0 = 0 is a secondorder pole, while zk = kπ, k = 2k + 1, k ∈ ℤ, are simple poles, and zk = kπ, k = 2k , k ∈ ℤ∗ , are thirdorder poles. The circle {γ} = {z ∈ ℂ : z = 4} turns around the points z0 = 0, z−1 = −π and z1 = π. The residues in z1 and z−1 are given by Res(f ; π) = lim
z→π
z(z − π) z z−π π = lim lim =− , z→π z→π sin z(1 − cos z) 1 − cos z sin z 2
and similarly for the point z−1 = −π, we have Res(f ; −π) = lim
z→−π
π z(z + π) = . sin z(1 − cos z) 2
To determine the residue in z0 = 0, we will expand the function in Laurent series for the origin. Using the facts that z3 z5 + − ⋅ ⋅ ⋅ , z ∈ ℂ, 3! 5! 2 4 z z 1 − cos z = − + ⋅ ⋅ ⋅ , z ∈ ℂ, 2! 4! sin z = z −
we deduce f (z) =
1 z2 1 −
z2 3!
1 +
z4 5!
− ⋅⋅⋅
1 2!
1
−
z2 4!
+ ⋅⋅⋅
=
1 1 1 ( − z 2 + ⋅ ⋅ ⋅), z 2 2 12
z ∈ U(0; 0, π),
7.5 Solutions to the exercises of Chapter 5  349
hence Res(f ; 0) = 0. Finally, we conclude that ∫ γ
z dz = 2πi(Res(f ; π) + Res(f ; −π) + Res(f ; 0)) = 0. sin z(1 − cos z) 3
1 z
e 8. The function f (z) = zz+1 has the isolated essential singular point z0 = 0, and the simple pole z1 = −1. √ (a) The paths {γ} = {z ∈ ℂ : z = 22 } turns around only the point z0 = 0. Since the origin is an isolated essential singular point, we will expand the function in Laurent series for the origin. Using the relations,
1 = 1 − z + z 2 − ⋅ ⋅ ⋅ , z ∈ U(0; 1), z+1 1 1 1 ez = 1 + + + ⋅ ⋅ ⋅ , z ∈ ℂ∗ , z 2!z 2 it follows that f (z) = z 3 (1 − z + z 2 − ⋅ ⋅ ⋅)(1 + hence the coefficient of the term
1 z
Res(f ; 0) =
1 1 + ⋅ ⋅ ⋅), + 1!z 2!z 2
z ∈ U(0; 0, 1),
is given by 1 1 1 1 1 − + − ⋅⋅⋅ = − . 4! 5! 6! e 3
Thus, 1
z3e z 2πi(3 − e) dz = 2πi Res(f ; 0) = . z+1 3e
∫ γ
(b) The path {γ} = {z ∈ ℂ : z = √2} turns around both of the singular points of f . The residue in the simple pole z1 = −1 is 1 1 Res(f ; −1) = lim z 3 e z = − , z→−1 e
and using the residues theorem, it follows that 1
z3e z 2πi dz = 2πi(Res(f ; 0) + Res(f ; −1)) = − . ∫ z+1 3 γ
350  7 Solutions to the chapterwise exercises n
1 z
9. The function f (z) = zz 2e−1 has the following isolated singular points: the isolated essential singular point z0 = 0, and the simple poles z1 = 1 and z2 = −1. The residues in these points are 1
Res(f ; 1) = lim z→1
zne z e = z+1 2
and 1
(−1)n+1 zne z = . Res(f ; −1) = lim z→−1 z − 1 2e The expansion of the function in Laurent series about the origin is f (z) = −z n
1
ez 1 − z2
= −z n (1 + z 2 + z 4 + ⋅ ⋅ ⋅)(1 +
1 1 + ⋅ ⋅ ⋅), + 1!z 2!z 2
z ∈ U(0; 0, 1),
hence it follows that Res(f ; 0) = −[
1 1 1 + + + ⋅ ⋅ ⋅]. (n + 1)! (n + 3)! (n + 5)!
Using the residues theorem, it follows that: (a) If {γ} = {z ∈ ℂ : z = 21 }, then 1
1 zne z 1 1 dz = 2πi Res(f ; 0) = −2πi[ + + + ⋅ ⋅ ⋅]. ∫ 2 (n + 1)! (n + 3)! (n + 5)! z −1 γ
(b) If {γ} = {z ∈ ℂ : z = 2}, then 1
∫ γ
zne z dz = 2πi(Res(f ; 0) + Res(f ; 1) + Res(f ; −1)) z2 − 1 = −2πi[
1 1 1 πi(e2 − (−1)n ) + + + ⋅ ⋅ ⋅] + . (n + 1)! (n + 3)! (n + 5)! e
(c) If {γ} = {z = x + iy ∈ ℂ : x2 + y2 − 2x −
5 4
= 0}, then
1
∫ γ
zne z dz = 2πi(Res(f ; 0) + Res(f ; 1)) z2 − 1 = −2πi[
1 1 1 + + + ⋅ ⋅ ⋅] + πie. (n + 1)! (n + 3)! (n + 5)!
7.5 Solutions to the exercises of Chapter 5  351 π z−i
10. The function f (z) = ez 2 +1 has the essential singular point z1 = i, and the simple pole z2 = −i. The given integration paths {γ} = {z ∈ ℂ : z = 2} turns around both of these singular points, and the first residue is π
Res(f ; −i) = lim
z→−i
π
πi
e z−i e− 2i e2 1 = = =− . z−i −2i −2i 2
To determine the residue in z1 = i, we will expand the function in Laurent series for this point, using the substitution z = t + i, i. e., π
π 1 1 et = et f (t + i) = 2 t + 2it 2it 1 − ti 2
=
π3 ti (ti)2 (ti)3 π2 1 π (1 + + 2 + 3 + ⋅ ⋅ ⋅)(1 + + 2 + 3 + ⋅ ⋅ ⋅), 2it 2 1!t 2!t 2 3!t 2 t ∈ U(0; 0, 2),
hence f (z) =
(z − i)i ((z − i)i)2 ((z − i)i)3 1 + (1 + + + ⋅ ⋅ ⋅) 2i(z − i) 2 23 22 π3 π2 π + ⋅ ⋅ ⋅), z ∈ U(i; 0, 2). + + ⋅(1 + 1!(z − i) 2!(z − i)2 3!(z − i)3
It follows that Res(f ; i) =
πi 2 ( πi )3 1 πi ( 2 ) 1 πi 1 (1 + + + 2 + ⋅ ⋅ ⋅) = e 2 = . 2i 1!2 2! 3! 2i 2
From here, we deduce that π
∫ γ
e z−i dz = 2πi(Res(f ; i) + Res(f ; −i)) = 0. z2 + 1
1 11. The function f (z) = z(z 2 +a 2 )2 has the simple pole z0 = 0, while z1 = ai and z2 = −ai are secondorder poles. Thus,
Res(f ; 0) = lim
z→0 (z 2
Res(f ; ai) = lim ( z→ai
1 1 = 4, 2 2 a +a )
1 −3z − ai 1 ) = lim 2 = − 4, z→ai z (z + ai)3 2a z(z + ai)2
and Res(f ; −ai) = lim ( z→−ai
1 −3z + ai 1 ) = lim 2 = − 4. z→−ai z (z − ai)3 2a z(z − ai)2
352  7 Solutions to the chapterwise exercises From the residues theorem, it follows that ∫ γ
1 dz = 2πi(Res(f ; 0) + Res(f ; ai) + Res(f ; −ai)) = 0. z(z 2 + a2 )2 π+2kπ
12. The function f (z) = z n1+1 has the nth order roots of the unit zk = ei n , with k ∈ {0, 1, 2, . . . , n − 1}, as simple poles. We have the following discussion with respect to the radius r > 0, and r ≠ 1 of the integration circle {γ} = {z ∈ ℂ : z = r}. Case 1. If r < 1, the circle does not turn around any of the above poles. Using the Cauchy integral theorem, the value of the integral is zero, i. e., ∫ γ
zn
1 dz = 0. +1
Case 2. If r > 1, the circle turns around all the above poles. Using the theorem related to the sum of the residues in all of the singular points, we need to determine the residue of the function in the point z = ∞. Hence we will expand the function in Laurent series for this point, i. e., f (z) =
1 1 1 1 1 1 1 1 = = n (1 − n + 2n − ⋅ ⋅ ⋅) = n − 2n + ⋅ ⋅ ⋅ , z n + 1 z n 1 + 1n z z z z z z
1 < 1, z
and it follows the two cases: 2. Subcase 1. If n = 1, then Res(f ; ∞) = −1, hence ∫ γ
1 dz = −2πi Res(f ; ∞) = 2πi. zn + 1
2. Subcase 2. If n > 1, then Res(f ; ∞) = 0, hence we have ∫ γ
1 dz = −2πi Res(f ; ∞) = 0. zn + 1
1 13. The singular points of the functions f (z) = z 2 sin have been determined to z point 4 of Exercise 5.4.3. Thus, the points of the form zk = kπ, with k ∈ ℤ∗ , are simple
7.5 Solutions to the exercises of Chapter 5  353
poles, while z0 = 0 is a thirdorder pole. The circle {γ} = {z ∈ ℂ : z = r}, with nπ < r < (n + 1)π, and n ∈ ℕ∗ , turns around all these zk points, whenever k < n + 1. We easily deduce that Res(f ; 0) =
1 , 6
and Res(f ; kπ) =
(−1)k , k2 π2
k ∈ ℤ∗ .
Thus, ∫ γ
1 dz z 2 sin z n πi 4i n (−1)k 2 n (−1)k 1 . + = 2πi ∑ Res(f ; zk ) = 2πi( + 2 ∑ 2 ) = ∑ 6 π k=1 k 3 π k=1 k 2 k=−n
14. The function f (z) = z4 = −2. If z − i = r1 eiθ1 ,
√1+z 2 (1−z 2 )(z 2 −4)
has the simple poles z1 = 1, z2 = −1, z3 = 2 and
π 3π where r1 > 0, r2 > 0, θ1 , θ2 ∈ [− , ], 2 2
z + i = r2 eiθ2 ,
then √z 2 + 1 = √r1 r2 ei
θ1 +θ2 +2kπ 2
,
k ∈ {0, 1}.
The branch of the given function is obtained for k = 0, because of the given condition √1 + z 2 z=0 = 1. Then f ∈ H(ℂ \ (T1 ∪ T2 )), where T1 = {z = x + iy ∈ ℂ : x = 0, y ≤ −1},
T2 = {z = x + iy ∈ ℂ : x = 0, y ≥ 1}.
(a) The circle {γ} = {z ∈ ℂ : z = 22 } does not turn around any of the above poles, and from the Cauchy integral theorem we have √
∫ γ
√1 + z 2 dz = 0. (1 − z 2 )(z 2 − 4) 2
(b) The path {γ} = {z = x + iy ∈ ℂ : x2 + 4y2 − 1 = 0, x ≥ 0} ∪ {z = x + iy ∈ ℂ : 2 x + y2 − 41 = 0, x < 0} turns around only the point z1 = 1. The residue of the function at this point is Res(f ; 1) = lim z→1
√1 + z 2 √2 = , 2 6 −(z + 1)(z − 4)
354  7 Solutions to the chapterwise exercises hence ∫ γ
√1 + z 2
(1 − z 2 )(z 2 − 4)
15. The function f (z) = z4 = −2i. If
z−1 z+1 2 2 (z +1)(z +4)
√z 2 −1 log
dz = 2πi Res(f ; 1) =
has the simple poles z1 = i, z2 = −i, z3 = 2i and
z + 1 = r2 eiθ2 ,
z − 1 = r1 eiθ1 ,
√2πi . 3
where r1 , r2 ∈ [0, 2π],
then θ1 +θ2 +2kπ 2
√z 2 − 1 = √r1 r2 ei
,
k ∈ {0, 1}
and log
r z−1 = ln 1 + i(θ1 − θ2 + 2lπ), z+1 r2
l ∈ ℤ.
From the conditions √z 2 − 1z=0 = i and log z−1  = πi, we obtain the values k = 0 z+1 z=0 and l = 0. Then f ∈ H(ℂ \ (T1 ∪ T2 )), where T1 = {z = x + iy ∈ ℂ : x ≤ −1, y = 0},
T2 = {z = x + iy ∈ ℂ : x ≥ 1, y = 0}.
The residues in the simple poles are: Res(f ; i) = lim
√z 2 − 1 log z−1 z+1 i)(z 2
πi√2 , 12
=
(z + + 4) √z 2 − 1 log z−1 πi√2 z+1 =− , Res(f ; −i) = lim z→−i (z − i)(z 2 + 4) 4 √z 2 − 1 log z−1 i √5 z+1 =− (π − 2 arctan 2), Res(f ; 2i) = lim 2 z→2i (z + 1)(z + 2i) 12 z→i
and Res(f ; −2i) = lim
√z 2 − 1 log z−1
z→−2i (z 2
(a) The path {γ} = {z ∈ ℂ : z = points, hence ∫ γ
z+1
+ 1)(z − 2i) √2 } 2
=
i√5 (π + 2 arctan 2). 12
does not turn around any of these singular
√z 2 − 1 log z−1 z+1
(z 2 + 1)(z 2 + 4)
dz = 0.
7.5 Solutions to the exercises of Chapter 5  355
(b) The ellipse {γ} = {z = x + iy ∈ ℂ : 36x 2 + 4y2 − 9 = 0} turns around only the points z1 = i and z2 = −i. From the residues theorem, it follows that ∫ γ
√z 2 − 1 log z−1
(z 2
+
1)(z 2
z+1
+ 4)
dz = 2πi(Res(f ; i) + Res(f ; −i)) =
π 2 √2 . 3
(c) The ellipse {γ} = {z = x + iy ∈ ℂ : 10x2 + y2 − 5 = 0} turns around to all of the simple poles, and thus ∫ γ
√z 2 − 1 log z−1 z+1
(z 2 + 1)(z 2 + 4)
dz = 2πi(Res(f ; i) + Res(f ; −i) + Res(f ; 2i) + Res(f ; −2i)) = 2π(
π √2 √ 5 − arctan 2). 6 3
(d) The circle {γ} = {z = x + iy ∈ ℂ : x2 + y2 − 5y = 0} turns around only the simple poles z1 = i and z3 = 2i, and thus ∫ γ
√z 2 − 1 log z−1 z+1
(z 2 + 1)(z 2 + 4)
dz = 2πi(Res(f ; i) + Res(f ; 2i)) =
π √ ( 5(π − arctan 2) − π √2). 6
Solution of Exercise 5.4.7 All the integrals are of the form 2π
I = ∫ R(sin θ, cos θ)dθ, 0
that can be computed as follows: I = 2π ∑ Res(g; z), z 1. From Res(g; z1 ) = lim (z − z1 )g(z) = z→z1
√2 4
356  7 Solutions to the chapterwise exercises and Res(g; z2 ) = lim (z − z2 )g(z) = z→z2
√2 , 4
we have that 2π
∫ 0
1
2 − sin2 θ
2. Since f (θ) =
dθ = 2π ∑ Res(g; z) = 2π(Res(g; z1 ) + Res(g; z2 )) = π √2. z 1, the unit disc contains only the point z1 , with the residue Res(g; −a + √a2 − 1) =
2
lim
z→−a+√a2 −1
z + a + √a2 − 1
=
1 . √a2 − 1
It follows that 2π
∫ 0
2π 1 dθ = 2π Res(g; −a + √a2 − 1) = . √a2 − 1 a + cos θ
5. We have f (θ) =
1 , 1+a sin θ
and g will be
g(z) =
1 2i 1 = 2 . z 1 + a z−z −1 az + 2iz − a 2i
2
2
The function g has the simple poles z1 = i −1+ a1−a and z2 = i −1− a1−a . From a < 1, the unit circle turns around only the point z1 . The residue is √
Res(g; i
−1 + √1 − a2 )= a
lim
−1+√1−a2 z→i a
√
2i a(z +
2 √ i 1+ a1−a )
=
1 , √1 − a2
hence 2π
∫ 0
1 −1 + √1 − a2 2π dθ = 2π Res(g; i )= . √1 − a2 1 + a sin θ a
6. Since f (θ) =
1 , (a+b cos θ)2
g(z) =
the correspondent function g is 4z 1 1 . = z (a + b z+z −1 )2 (bz 2 + 2az + b)2 2
358  7 Solutions to the chapterwise exercises 2
2
2
2
The function g has the singular points z1 = −a+ ba −b and z2 = −a− ba −b that are secondorder poles. The unit disc contains only the point z1 , whenever a > b > 0. The residue is √
Res(g; z1 ) = lim ( z→z1
4z b2 (z +
√
a+√a2 −b2 2 ) b
) =
a (a2
3
− b2 ) 2
,
hence we deduce that 2π
∫ 0
7. Since f (θ) =
1 2πa . dθ = 2π Res(g; z1 ) = 3 2 (a + b cos θ)2 (a − b2 ) 2
1+cos θ , (13−5 cos θ)2
we have that g will be
z+z 1 1+ 2 2(z + 1)2 g(z) = = . −1 z (13 − 5 z+z )2 25(z − 5)2 (z − 1 )2 5 −1
2
The function g has the secondorder poles z1 = 5 and z2 = 51 , and the unit disc contains only the pole z2 . The residue in this point is
1 1 −24(z + 1) 2(z + 1)2 = ) = lim Res(g; ) = lim ( , 3 2 1 1 5 96 z→ 5 25(z − 5) z→ 5 25(z − 5) hence we get 2π
∫ 0
1 π 1 + cos θ dθ = 2π Res(g; ) = . 5 48 (13 − 5 cos θ)2
8. The function f (θ) = b = 8. Hence 2π
∫ 0
1 (17+8 cos θ)2
is a special case of the point 6, where a = 17 and
34π 1 2π ⋅ 17 = 3. dθ = 3 2 2 15 (17 + 8 cos θ)2 2 (17 − 8 )
9. First, we will use the exponential form of the function cos θ, and then we will use the following substitutions: cos θ =
eiθ + e−iθ , 2
θ = 2πt,
z = ei2πt .
7.5 Solutions to the exercises of Chapter 5  359
Thus 2π
∫ 0
2π
1 1 dθ dθ = ∫ 1 − 2p cos θ + p2 1 − p(eiθ + e−iθ ) + p2 0
1
= 2π ∫ 0
1 0
1 i
1 dt + e−i2πt ) + p2
1 2πiei2πt dt (1 + p2 )ei2πt − p(ei4πt + 1)
=∫ =
1−
p(ei2πt
∫ 𝜕U(0;1)
1 1 dz = i (1 + p2 )z − p(z 2 + 1)
∫ 𝜕U(0;1)
1 dz. (z − p)(1 − pz)
1 , where 0 < p < 1, has the only one singular point z0 = p The function g(z) = (z−p)(1−pz) with the module less than 1, and then
Res(g; p) = lim
z→p
1 1 . = 1 − pz 1 − p2
It follows that 2π
∫ 0
get 2π
∫ 0
1 1 dθ = 2 i 1 − 2p cos θ + p
∫ 𝜕U(0;1)
1 2π . dz = 2π Res(g; p) = (z − p)(1 − pz) 1 − p2
10. We will use the same method as in point 9. Using the substitution z = eiθ , we 2π
1 + cos 4θ 1 cos2 2θ dθ = ∫ dθ 2 1 − 2p cos θ + p2 1 − 2p cos θ + p2 0
2π
1 1 2 + ei4θ + e−i4θ = ∫ dθ = iθ −iθ 2 4 1 − p(e + e ) + p 4 0
= = where g(z) =
1 4i 1 4i
8
∫ 𝜕U(0;1)
∫ 𝜕U(0;1)
(z 4 +1)2 . −pz 4 (z−p)(z− p1 )
∫ 𝜕U(0;1)
1 2 + z 4 + z −4 dz −1 2 1 − p(z + z ) + p iz
4
z + 2z + 1 dz (p2 z − p − pz 2 + z)z 4 (z 4 + 1)2
−pz 4 (z
− p)(z −
1 ) p
dz =
π ∑ Res(g; z), 2 z 1, the unit disc contains only the simple pole z1 = a . From the −a(z−a)(z− a )
above result,
2π
I=∫ 0
cos nθ 2π , dθ = n 2 1 − 2a cos θ + a2 a (a − 1)
2π
J=∫ 0
sin nθ dθ = 0. 1 − 2a cos θ + a2
If we use the notation, 2π
I=∫ 0
cos nθ dθ, 1 − 2a sin θ + a2
2π
J=∫ 0
sin nθ dθ, 1 − 2a sin θ + a2
then 2π
I + iJ = ∫ 0
=
einθ dθ = 1 − 2a sin θ + a2
∫ 𝜕U(0;1)
∫ 𝜕U(0;1)
zn
1−
−1 a z−zi
1 dz + a2 iz
2πin i , ) = dz = 2πi Res(g; a an (a2 − 1) −a(z − ia)(z − ai ) zn
where we used the substitution z = eiθ , and we denoted by g the function g(z) = zn i i . Since a > 1, the unit disc contains only the simple pole z1 = a . From here, −a(z−ia)(z− a )
we get
2π
I=∫ 0
2π
J=∫ 0
cos nθ 2π Re in , dθ = n 2 2 1 − 2a sin θ + a a (a − 1) 2π sin nθ Im in . dθ = n 2 1 − 2a sin θ + a2 a (a − 1)
362  7 Solutions to the chapterwise exercises 2. We will use a similar method with in point 1. Let 2π
2π
sin θ sin nθ I=∫ dθ, 5 − 4 cos θ
J=∫
0
0
sin θ cos nθ dθ. 5 − 4 cos θ
Then 2π
J + iI = ∫ 0
=
1 2
1 sin θeinθ dθ = − 5 − 4 cos θ 2 2
∫ 𝜕U(0;1)
∫ 𝜕U(0;1)
(z − z −1 )z n 1 dz 5 − 2(z + z −1 ) z
n−1
1 (z − 1)z dz = 2 2z 2 − 5z + 2
∫ 𝜕U(0;1)
(z 2 − 1)z n−1
2(z − 2)(z − 21 )
dz
πi 1 = πi Res(g; ) = n+1 , 2 2 where g(z) =
this point is
(z 2 −1)z n−1 . 2(z−2)(z− 21 )
Only the pole z1 =
1 2
lies in the unit disc, and the residue in
1 1 (z 2 − 1)z n−1 = n+1 . Res(g; ) = lim 2 2(z − 2) 2 z→ 21 The integrals are 2π
2π
sin θ sin nθ π I=∫ dθ = n+1 , 5 − 4 cos θ 2
J=∫ 0
0
sin θ cos nθ dθ = 0. 5 − 4 cos θ
Solution of Exercise 5.4.9 1. Since f (θ) =
cos nθ b−ia cos θ
is an even function, we have π
π
0
−π
cos nθ cos nθ 1 I=∫ dθ = ∫ dθ. b − ia cos θ 2 b − ia cos θ Using this property, we will define the integral π
J=
sin nθ 1 dθ, ∫ 2 b − ia cos θ −π
and from the fact that h(z) =
sin nθ b−ia cos θ
is an odd function, it follows that J = 0.
7.5 Solutions to the exercises of Chapter 5  363
Using the substitution z = eiθ , we get dθ =
1 dz, iz
π
einθ 1 dθ = I + iJ = ∫ 2 b − ia cos θ −π
n
=
∫ 𝜕U(0;1)
∫ 𝜕U(0;1)
hence
1 zn dz 2b − ia(z + z −1 ) iz
z dz. az 2 + 2ibz + a
n
2
2
z has the simple poles z1 = i −b+ aa +b and z2 = i −b− The function g(z) = az 2 +2ibz+a and only z1 lies in the unit disc, whenever a > 0, b > 0. The residue will be √
Res(g; z1 ) = lim
z→z1
√a2 +b2 , a
in an zn = , 2az + 2bi 2i√a2 + b2 (√a2 + b2 + b)n
and since I + iJ = 2πi Res(g; z1 ) we obtain that π
∫ 0
cos nθ πin an dθ = . √a2 + b2 (√a2 + b2 + b)n b − ia cos θ
2. We will substitute z = eiθ , hence 2π
I=∫ 0
4 1 dθ = i a + b cos2 θ
∫ 𝜕U(0;1)
z dz. bz 4 + 2(2a + b)z 2 + b
If we denote ζ = z 2 , then we get I=
2 i
∫ 2×𝜕U(0;1)
The function g(z) = ζ1 =
4 1 dζ = i bζ 2 + 2(2a + b)ζ + b
1 bζ 2 +2(2a+b)ζ +b
∫ 𝜕U(0;1)
1 dζ . bζ 2 + 2(2a + b)ζ + b
has the simple poles
−(2a + b) + 2√a(a + b) b
and ζ2 =
−(2a + b) − 2√a(a + b) . b
Only the point ζ1 lies in the unit disc, whenever a > 0, b > 0, and the residue will be Res(g; ζ1 ) = lim
z→ζ1
1 1 . = b(ζ − ζ2 ) 4√a(a + b)
From the residues theorem, we have 2π
∫ 0
2π 1 dθ = 8π Res(g; ζ1 ) = . a + b cos2 θ √a(a + b)
364  7 Solutions to the chapterwise exercises Solution of Exercise 5.4.10 1. To calculate this integral, we will use the next wellknown result: if P and Q are real polynomials, such that Q has no real roots, and gr Q ≥ gr P, then +∞
∫[ 0
P(x) ix P(−x) −ix dx P(z) iz 1 P(0) e − e ] = 2πi[ + ∑ Res( e ; zk )]. Q(x) Q(−x) x 2 Q(0) Im z >0 zQ(z) k
We will write the required integral in a such form, to be able to use the above formula, i. e., +∞
+∞
0
0
eix − e−ix 1 2πi 1 P(0) π sin x dx = dx = = , ∫ ∫ x 2i x 2i 2 Q(0) 2
where P(x) = 1 and Q(x) = 1. Hence +∞
∫ 0
sin x π dx = . x 2
2. In this case, we will use another wellknown result: if P and Q are real polynomials, such that Q has no real roots, and gr Q ≥ gr P + 2, then +∞
∫ −∞
The function f (z) =
1 z 4 +1
P P(x) dx = 2πi ∑ Res( ; zk ). Q(x) Q Im z >0
(7.49)
k
has the simple poles zk = ei
condition Im zk > 0 holds only for the points z0 = e residues are
i π4
π+2kπ 4
3π
and z1 = ei 4 . The corresponding
Res(f ; z0 ) = lim
z 1 = − 0, 3 4 4z
Res(f ; z1 ) = lim
z 1 = − 1, 4 4z 3
z→z0
, where k ∈ {0, 1, 2, 3}. The
and z→z1
hence +∞
∫ −∞
x4
1 π √2 πi . dx = 2πi(Res(f ; z0 ) + Res(f ; z1 )) = − (z0 + z1 ) = 2 2 +1
7.5 Solutions to the exercises of Chapter 5  365
3. As in point 2, we will use the relation (7.49). To change the integration limits, we have +∞
0
+∞
−∞
−∞
0
1 1 1 dx = ∫ dx + ∫ dx ∫ 2 n 2 n (a + bx ) (a + bx ) (a + bx2 )n 0
+∞
1 1 (−dy) + ∫ dx = ∫ (a + by2 )n (a + bx2 )n 0
+∞
+∞
=2 ∫ 0
1 dx, (a + bx2 )n
(7.50)
where we used the y = −x variable changes. Now we will apply the above mentioned 1 a a relation. The function f (z) = (a+bz 2 )n has the nth order poles z1 = i√ b and z2 = −i√ b . The condition Im zk > 0 holds only for the point z1 = i√ ba , and the residue is Res(f ; z1 ) =
(n−1)
1 1 lim ( ) (n − 1)! z→z1 bn (z − z2 )n
(−1)n−1 (2n − 2)!√a 1 (−1)n−1 (2n − 2)! = i lim n 2n−1 (n − 1)!b z→z1 (n − 1)!(z − z2 ) (−1)n [(n − 1)!]2 22n−1 an √b (2n − 2)!√a =− i. [(n − 1)!]2 22n−1 an √b =
Using the relation (7.50), we get +∞
∫ 0
(2n − 2)!π√a 1 1 . dx = 2πi Res(f ; z1 ) = 2 (a + bx2 )n [(n − 1)!]2 22n−1 an √b
4. Similarly, as in the previous point, we will use the relation (7.49). The function π+2kπ f (z) = 1+z1 2n has the simple poles zk = ei 2n , where k ∈ {0, 1, 2, . . . , 2n−1}. The condition Im zk > 0 holds only for the points zk = ei these points are
π+2kπ 2n
Res(f ; zk ) = lim
z→zk
hence +∞
∫ −∞
, k ∈ {0, 1, . . . , n − 1}, and the residues in
z 1 =− k, 2n 2nz 2n−1
n−1 πi n−1 1 πi i π dx = 2πi Res(f ; z ) = − ∑ ∑ zk = − k π = π . n n 1 + x2n sin n sin k=0 k=0 2n 2n
5. Here, we will use another wellknown result: if P and Q are real polynomials, such that Q has no real nonnegative roots, and gr Q ≥ gr P + 2, then +∞
∫ 0
1 P(z) P(x) ln xdx = − Re[ ∑ Res( (log z)2 ; zk )], Q(x) 2 Q(z) z ∈ℂ∗ k
366  7 Solutions to the chapterwise exercises 2
(log z) where log 1 = 0. It follows that we need to study the function f (z) = (z+a) 2 +b2 . The numbers z1 = −a + bi and z2 = −a − bi are simple poles, and both satisfy the condition zk ∈ ℂ∗ , because a, b > 0. The residues in these points are
Res(f ; z1 ) = lim
(log z)2 (log(−a + bi))2 = , z + a + bi 2bi
Res(f ; z2 ) = lim
(log(−a − bi))2 (log z)2 = . z + a − bi −2bi
z→z1
and z→z2
Thus +∞
∫ 0
ln x dx (x + a)2 + b2
1 = − Re(Res(f ; z1 ) + Res(f ; z2 )) 2 1 (log(−a + bi))2 − (log(−a − bi))2 = − Re 2 2bi 2 + b2 + i(π − arctan b ))2 − (ln √a2 + b2 + i(π + arctan b ))2 √ (ln a 1 a a = − Re 2 2bi 2π arctan ba − 2i ln √a2 + b2 arctan ba 1 ln √a2 + b2 b = − Re = arctan . 2 bi b a
6. We will change the limits of integration as follows: +∞
0
−∞
−∞
+∞
cos ax cos ax cos ax I= ∫ 2 dx = ∫ 2 dx + ∫ 2 dx 2 2 2 2 (x + b ) (x + b ) (x + b2 )2 0
0
+∞
+∞
+∞
0
0
cos ax cos ax cos ay (−dy) + ∫ 2 dx = 2 ∫ 2 dx. = ∫ 2 (y + b2 )2 (x + b2 )2 (x + b2 )2 Let +∞
0
−∞
−∞
+∞
sin ax sin ax sin ax J= ∫ 2 dx = ∫ 2 dx + ∫ 2 dx 2 2 2 2 (x + b ) (x + b ) (x + b2 )2 0
= ∫ +∞
0
+∞
sin ax − sin ay (−dy) + ∫ 2 dx = 0. (y2 + b2 )2 (x + b2 )2 0
We get +∞
I = I + iJ = ∫ −∞
eixa dx. (x2 + b2 )2
7.5 Solutions to the exercises of Chapter 5  367
To compute this last integral, we will use the following wellknown result: if P and Q are real polynomials, such that Q has no real roots, and gr Q ≥ gr P + 1, then +∞
∫ −∞
The function f (z) = residue
P(x) ix P(z) iz e dx = 2πi ∑ Res( e ; zk ). Q(x) Q(z) Im z >0
(7.51)
k
eiaz (z 2 +b2 )2
has the secondorder pole z1 = bi, with the corresponding
Res(f ; z1 ) = lim ( z→bi
eiaz ab + 1 ) = − ab 3 i, (z + bi)2 4e b
hence +∞
I= ∫ 0
cos ax 1 (ab + 1)π . dx = 2πi Res(f ; z1 ) = 2 (x2 + b2 )2 4eab b3
7. We will integrate the function f (z) = γ[ε,r] ∪ γr , where
eiaz −eibz z2
along the curve γ = γ[−r,−ε] ∪ γε− ∪
γ[−r,−ε] (t) = (1 − t)(−r) + t(−ε),
γε (t) = εeiπt ,
γ[ε,r] (t) = (1 − t)ε + tr,
γr (t) = reiπt ,
t ∈ [0, 1],
and 0 < ε < r. Since ∫γ f (z)dz = 0, then r
−ε
∫ f (x)dx − ∫ f (z)dz + ∫ f (x)dx + ∫ f (z)dz = 0, γε
−r
γr
ε
and we deduce that 0
+∞
−∞
0
eiax − eibx eiax − eibx dx + dx = lim ∫ f (z)dz − lim ∫ f (z)dz. ∫ ∫ r→+∞ ε→0 x2 x2 γε
γr
But 0
+∞
∫ −∞
eiax − eibx eiax − eibx dx + dx ∫ x2 x2 0
0
+∞
+∞
0
0
e−iat − e−ibt eiax − eibx cos ax − cos bx = ∫ dt + dx = 2 ∫ dx, ∫ t2 x2 x2 +∞
368  7 Solutions to the chapterwise exercises and thus +∞
∫ 0
cos ax − cos bx 1 dx = [lim ∫ f (z)dz − lim ∫ f (z)dz]. 2 r→+∞ ε→0 2 x γε
(7.52)
γr
Since f ∈ H(S0 [0, π] \ {0}) and z0 = 0 is a simple pole for f , using the fact that Res(f ; 0) = lim (eiaz − eibz ) = i(a − b),
z→0
it follows lim ∫ f (z)dz = πi Res(f ; 0) = π(b − a).
ε→0
(7.53)
γε
Substituting az = ζ , we have ∫
γr
eiaz 1 dz = ∫ φ(ζ )eiζ dζ , a z2
where φ(ζ ) =
γr
a2 , ζ2
hence we get lim φ(ζ ) = 0 ⇒ lim ∫ φ(ζ )eiζ dζ = 0.
r→+∞
r→+∞
γr
Thus lim ∫
eiaz dz = 0, z2
lim ∫
eibz dz = 0, z2
r→+∞
γr
and similarly we obtain r→+∞
γr
thus lim ∫ f (z)dz = lim ∫
r→+∞
γr
r→+∞
γr
eiaz − eibz dz = 0. z2
Combining the relations (7.52), (7.53) and (7.54), we have +∞
∫ 0
cos ax − cos bx π(b − a) dx = . 2 x2
(7.54)
7.5 Solutions to the exercises of Chapter 5  369
8. Define the integral 2
I = ∫ e−az dz, 𝜕𝒞 b where 𝒞 is the boundary of the rectangle determined by the points A(R), B(R + i 2a ), 2
b C(−R + i 2a ) and D(−R). The function f (z) = e−az is holomorphic in the whole complex plane, thus 2
I = ∫ e−az dz = 0, 𝒞
i. e., ∫ f (z)dz + ∫ f (z)dz + ∫ f (z)dz + ∫ f (z)dz = 0. AB
BC
DA
CD
On the real axis, we have R
R
2
2
∫ f (z)dz = ∫ e−ax dx = 2 ∫ e−ax dx, 0
−R
DA
and using the substitution x =
t √a
we get R√a
2 2 ∫ e−t dt. ∫ f (z)dz = √a
0
DA
b On the BC segment, where z = x + i 2a , we have b2
−R
2
∫ f (z)dz = e 4a ∫ e−ax e−ibx dx. R
BC
Since 0
∫e
−ax2 −ibx
e
R
2
dx = ∫ e−ax eibx dx, 0
−R
we obtain that b2
R
2
∫ f (z)dz = −e 4a ∫ e−ax (eibx + e−ibx )dx, BC
0
(7.55)
370  7 Solutions to the chapterwise exercises or b2
R
2
∫ f (z)dz = −2e 4a ∫ e−ax cos bxdx. 0
BC
Replacing the above results in (7.55), it follows that R√a
R
0
0
b2 2 2 2 ∫ e−t dt − 2e 4a ∫ e−ax cos bxdx = − ∫ f (z)dz − ∫ f (z)dz. √a
AB
CD
Since +∞
2
∫ e−t dt = 0
√π , 2
for R → +∞, we have +∞
√
b2 2 π − 2e 4a ∫ e−ax cos bxdx = − lim [ ∫ f (z)dz + ∫ f (z)dz]. R→∞ a
0
AB
CD
b ] and, respectively, The equations of the segments AB and CD are z = R + iy, y ∈ [0, 2a b z = −R + iy, y ∈ [0, 2a ], thus 2
b 2 −a(R2 ±2Riy−y2 ) −a(R2 −y2 ) ≤ e 4a e−aR , f (z) = e = e
z ∈ AB ∪ CD.
From here, we get b2 −aR2 b , ∫ f (z)dz ≤ max{f (z) : z ∈ AB}V(AB) ≤ e 4a e 2a AB
hence lim ∫ f (z)dz = 0.
R→∞
AB
We will prove similarly that lim ∫ f (z)dz = 0,
R→∞
CD
and thus it follows that +∞
2 1 π b2 ∫ e−ax cos bxdx = √ e− 4a . 2 a
0
7.5 Solutions to the exercises of Chapter 5  371
Solution of Exercise 5.4.11 We will determine the number of the zeros that lies in the disc U(0; 1) by using the Rouché theorem. 1. The equation z 8 − 4z 5 + z 2 − 1 = 0 may be written as f (z) + g(z) = 0, where f (z) = −4z 5 and g(z) = z 8 + z 2 − 1. If ζ ∈ {γ} = 𝜕U(0; 1), then 5 f (ζ ) = −4ζ = 4, and 8 2 8 2 g(ζ ) = ζ + ζ − 1 ≤ ζ + ζ + 1 = 3, i. e., g(ζ ) ≤ 3 < 4 = f (ζ ),
∀ζ ∈ {γ}.
From the Rouché theorem, we obtain that the given equation has five zeros in the unit disc, because Θ(f + g, U(0; 1)) = Θ(f , U(0; 1)) = 5. 2. The equation z 8 − 11z + 1 = 0 may be written as f (z) + g(z) = 0, where f (z) = −11z and g(z) = z 8 + 1. If ζ ∈ {γ} = 𝜕U(0; 1), then f (ζ ) = −11ζ  = 11, and 8 8 g(ζ ) = ζ + 1 ≤ ζ + 1 = 2,
372  7 Solutions to the chapterwise exercises i. e., g(ζ ) ≤ 2 < 11 = f (ζ ),
∀ζ ∈ {γ}.
From the Rouché theorem, we obtain that the given equation has one zero in the unit disc, because Θ(f + g, U(0; 1)) = Θ(f , U(0; 1)) = 1. 3. The equation 23z 5 − 10z 3 + 6z − 5 = 0 may be written as f (z) + g(z) = 0, where f (z) = 23z 5 and g(z) = −10z 3 + 6z − 5. If ζ ∈ {γ} = 𝜕U(0; 1), then 5 f (ζ ) = 23ζ = 23, and 3 g(ζ ) = −10ζ + 6ζ − 5 ≤ 21, i. e., g(ζ ) ≤ 21 < 23 = f (ζ ),
∀ζ ∈ {γ}.
From the Rouché theorem, we obtain that the given equation has five zeros in the unit disc. 4. For the equation ez − 3iz = 0, let f (z) = −3iz,
g(z) = ez ,
where g(z) + f (z) = 0. If ζ ∈ {γ} = 𝜕U(0; 1), then f (ζ ) =  − 3iζ  = 3, and ζ g(ζ ) = e ≤ e. We deduce that g(ζ ) ≤ e < 3 = f (ζ ),
∀ζ ∈ {γ}.
7.5 Solutions to the exercises of Chapter 5  373
From the Rouché theorem, we obtain that the given equation has one zero in the unit disc, because Θ(f + g, U(0; 1)) = Θ(f , U(0; 1)) = 1. 5. The equation z n + 8z 2 + 1 = 0, n ∈ ℕ, n ≥ 3, may be written as f (z) + g(z) = 0, where f (z) = 8z 2 ,
g(z) = z n + 1.
If ζ ∈ {γ} = 𝜕U(0; 1), then 2 f (ζ ) = 8ζ = 8, and n n g(ζ ) = ζ + 1 ≤ ζ + 1 = 2, i. e., g(ζ ) ≤ 2 < 8 = f (ζ ),
∀ζ ∈ {γ}.
From the Rouché theorem, we obtain that the given equation has two zeros in the unit disc. 6. The equation z n + 3z 2 + 1 = 0, n ∈ ℕ, n ≥ 3, may be written as f (z) + g(z) = 0, where f (z) = 3z 2 and g(z) = z n + 1. If ζ ∈ {γ} = 𝜕U(0; 1), then 2 f (ζ ) = 3ζ = 3, and n n g(ζ ) = ζ + 1 ≤ ζ + 1 = 2, i. e., g(ζ ) ≤ 2 < 3 = f (ζ ),
∀ζ ∈ {γ}.
374  7 Solutions to the chapterwise exercises From the Rouché theorem, we obtain that the given equation has two zeros in the unit disc, because Θ(f + g, U(0; 1)) = Θ(f , U(0; 1)) = 2. 7. Similar to point 4, the equation has the form f (z) + g(z) = 0, where f (z) = −4z n , and g(z) = ez + 1. If ζ ∈ {γ} = 𝜕U(0; 1), then n f (ζ ) = −4ζ = 4, ζ g(ζ ) = e + 1 ≤ e + 1, i. e., g(ζ ) ≤ e + 1 < 4 = f (ζ ),
∀ζ ∈ {γ}.
From the Rouché theorem, the given equation has n zeros in the unit disc. 8. If z n + pz 2 + qz + r = 0, n ∈ ℕ, n ≥ 3, where p > q + r + 1, let f (z) = pz 2 and g(z) = z n + qz + r. If ζ ∈ {γ} = 𝜕U(0; 1), we deduce that 2 f (ζ ) = pζ = p, and n g(ζ ) = ζ + qζ + r ≤ 1 + q + r. Because g(ζ ) ≤ 1 + q + r < p = f (ζ ),
∀ζ ∈ {γ},
according to the Rouché theorem, the given equation has two zeros in the unit disc.
7.5 Solutions to the exercises of Chapter 5  375
9. The equation z 10 − 9z 6 + 3z 3 + z 2 − 2 = 0 has the form f (z) + g(z) = 0, where f (z) = −9z 6 and g(z) = z 10 + 3z 3 + z 2 − 2. If ζ ∈ {γ} = 𝜕U(0; 1), we get 6 f (ζ ) = −9ζ = 9, and 10 3 2 g(ζ ) = ζ + 3ζ + ζ − 2 ≤ 7. From here, it follows that g(ζ ) ≤ 7 < 9 = f (ζ ),
∀ζ ∈ {γ},
according to the Rouché theorem, the given equation has six zeros in the unit disc, because Θ(f + g, U(0; 1)) = Θ(f , U(0; 1)) = 6. Solution of Exercise 5.4.12 To solve this problem, we will use the Rouché theorem. Let f (z) = az n , and g(z) = −ez . Since we need to determine the number of the zeros from the disc U(0; r), if ζ ∈ {γ} = 𝜕U(0; r), then n n f (ζ ) = aζ = ar , and ζ r g(ζ ) = −e ≤ e . From the assumption ar n > er , we get r n g(ζ ) ≤ e < ar = f (ζ ),
∀ζ ∈ {γ},
and according to the Rouché theorem, the equation has n zeros in the disc U(0; r).
376  7 Solutions to the chapterwise exercises Solution of Exercise 5.4.13 We will again use the Rouché theorem. The solution consists of the following two steps: we will determine the number of the roots in the bigger circular ring, then in the smaller circular ring, and the result will be the difference between these numbers. 1. (i) First, we will determine the number of the roots in the disc U(0; 1). The lefthand side of the equation may be written like the sum of the next functions: f (z) = −9z,
g(z) = z 4 + 1.
If ζ ∈ {γ1 } = 𝜕U(0; 1), then f (ζ ) =  − 9ζ  = 9, and 4 g(ζ ) = ζ + 1 ≤ 2. It follows that g(ζ ) ≤ 2 < 9 = f (ζ ),
∀ζ ∈ {γ1 },
and using the Rouché theorem we obtain that the equation has one root in the disc U(0; 1). (ii) Now we will determine the number of the roots in the disc U(0; 3). The lefthand side of the equation may be written like the sum of the next functions: f (z) = z 4 ,
g(z) = −9z + 1.
If ζ ∈ {γ2 } = 𝜕U(0; 3), then 4 4 f (ζ ) = ζ = 3 = 81, and g(ζ ) =  − 9ζ + 1 ≤  − 9ζ  + 1 = 28, hence g(ζ ) ≤ 28 < 81 = f (ζ ),
∀ζ ∈ {γ2 }.
7.5 Solutions to the exercises of Chapter 5
 377
Using the Rouché theorem, we obtain that the equation has four roots in the disc U(0; 3). In conclusion, the equation has three roots in the circular ring U(0; 1, 3). 2. (i) We will determine the number of the roots in the disc U(0; 1). The lefthand side of the equation may be written like the sum of the next functions: f (z) = −7z 3 ,
g(z) = 3z 5 + z 2 − 2. If ζ ∈ {γ1 } = 𝜕U(0; 1), then 3 f (ζ ) = −7ζ = 7, g(ζ ) = 3ζ 5 + ζ 2 − 2 ≤ 6. From here, we get g(ζ ) ≤ 6 < 7 = f (ζ ),
∀ζ ∈ {γ1 },
and from the Rouché theorem we obtain Θ(f + g, U(0; 1)) = Θ(f , U(0; 1)) = 3. (ii) Next, we will determine the number of the roots in the disc U(0; 2). The lefthand side of the equation may be written like the sum of the next functions: f (z) = 3z 5 ,
g(z) = −7z 3 + z 2 − 2. If ζ ∈ {γ2 } = 𝜕U(0; 2), then 5 5 f (ζ ) = 3ζ = 3 ⋅ 2 = 96, and 3 2 3 2 g(ζ ) = −7ζ + ζ − 2 ≤ −7ζ + ζ + 2 = 62, hence g(ζ ) ≤ 62 < 96 = f (ζ ),
∀ζ ∈ {γ2 }.
Using the Rouché theorem, we obtain Θ(f + g, U(0; 2)) = Θ(f , U(0; 2)) = 5.
378  7 Solutions to the chapterwise exercises We conclude that Θ(f + g, U(0; 1, 2)) = 5 − 3 = 2, hence the equation has two roots in the circular ring U(0; 1, 2). 3. (i) We will determine the number of the roots in the disc U(0; 1). The lefthand side of the equation may be written like the sum of the next functions: f (z) = −4z,
g(z) = z 4 + z 3 + 1.
If ζ ∈ {γ1 } = 𝜕U(0; 1), we have f (ζ ) =  − 4ζ  = 4, and 4 3 g(ζ ) = ζ + ζ + 1 ≤ 3. It follows that g(ζ ) ≤ 3 < 4 = f (ζ ),
∀ζ ∈ {γ1 },
and using the Rouché theorem we obtain Θ(f + g, U(0; 1)) = Θ(f , U(0; 1)) = 1. (ii) Next, we will determine the number of the roots in the disc U(0; 3). The lefthand side of the equation may be written like the sum of the next functions: f (z) = z 4 ,
g(z) = z 3 − 4z + 1.
If ζ ∈ {γ2 } = 𝜕U(0; 3), we get 4 4 f (ζ ) = ζ = 3 = 81, and 3 3 2 g(ζ ) = ζ − 4ζ + 1 ≤ ζ + −4ζ + 1 = 40. Thus g(ζ ) ≤ 40 < 81 = f (ζ ),
∀ζ ∈ {γ2 },
7.5 Solutions to the exercises of Chapter 5  379
and using the Rouché theorem we obtain Θ(f + g, U(0; 2)) = Θ(f , U(0; 2)) = 4. We conclude that Θ(f + g, U(0; 1, 3)) = 4 − 1 = 3, hence the equation has three roots in the circular ring U(0; 1, 3). Solution of Exercise 5.4.14 1. We will use the Rouché theorem for the functions f (z) = z, and g(z) = −aez . Since we need to determine the number of the roots in the disc U(0; 1), if ζ ∈ {γ} = 𝜕U(0; 1) then f (ζ ) = ζ  = 1, ζ g(ζ ) = −ae ≤ ae.
From the assumption ae < 1, we get
g(ζ ) ≤ ae < 1 = f (ζ ),
∀ζ ∈ {γ}.
It follows that the equation has only one root in the disc U(0; 1). 2. The proof is identical to the previous one. The lefthand side of the equation z 2 = aez may be written as the sum of f (z) = z 2 , and g(z) = −aez . Since we need to determine the number of the roots in the disc U(0; 1), if ζ ∈ {γ} = 𝜕U(0; 1) then 2 f (ζ ) = ζ = 1, ζ g(ζ ) = −ae ≤ ae.
From the assumption ae < 1, we get
g(ζ ) ≤ ae < 1 = f (ζ ),
∀ζ ∈ {γ}.
It follows that the equation has two roots in the disc U(0; 1).
380  7 Solutions to the chapterwise exercises
7.6 Solutions to the exercises of Chapter 6 Solution of Exercise 6.5.1 1. It is sufficient to prove that the function f is bijective, f ∈ H(D) and f (z) ≠ 0, ∀z ∈ D. If ∀ζ = u + iv ∈ Ω, then v > 0 and f (z) = ζ ⇔ iz = ζ ⇔ ∃!z = −iζ = v − iu ∈ D. It follows that f is bijective. But f ∈ H(D) and f (z) = i ≠ 0, ∀z ∈ D, thus the conclusion of the problem holds, i. e., the function f maps conformally the halfplane D onto the halfplane Ω. 2. Let now g : Ω → D, g(z) = −iz. For ∀ζ = u + iv ∈ D, then u > 0 and g(z) = ζ ⇔ −iz = ζ ⇔ ∃!z = iζ = −v + iu ∈ Ω. It follows that g is bijective. But g ∈ H(Ω) and g (z) = −i ≠ 0, ∀z ∈ Ω, thus the conclusion of the problem holds, i. e., the function g maps conformally the halfplane Ω onto the halfplane D. Solution of Exercise 6.5.2 We know that conformal mappings maps the domains onto the domains, and maps the boundary of the domain onto the boundary of the image domain. These properties will ⌢
be used in the following solution. The boundary of the domain D is 𝜕D = AB ∪ [BA], where ⌢
AB : γ ⌢ (t) = eiπ AB
2t−1 2
,
t ∈ [0, 1],
and [BA] : γ[BA] (t) = i(1 − t) + t(−i) = i(1 − 2t),
t ∈ [0, 1].
The image of 𝜕D will be ⌢
f (AB) = {f (eiπ
2t−1 2
) : t ∈ [0, 1]} = {eiπ(2t−1) : t ∈ [0, 1]} = 𝜕U(0; 1),
that represents the unit circle, and f ([BA]) = {f (i − 2ti) : t ∈ [0, 1]} = {−(2t − 1)2 : t ∈ [0, 1]}. Thus, if 1 t ∈ [0, ] ⇒ f (i − 2ti) = [−1, 0], 2 1 t ∈ [ , 1] ⇒ f (i − 2ti) = [0, −1], 2
7.6 Solutions to the exercises of Chapter 6  381
hence f ([BA]) = [−1, 0] ∪ [0, −1]. We need to determine in which of the domains bounded by 𝜕Ω, the function f maps the domain D. For this, let us choose an arbitrary point in D, and we will check if its image belongs to Ω: 1 1 1 ∈ D ⇒ f ( ) = ∈ Ω. 2 2 4 Thus, the function f (z) = z 2 conformally maps the domain D onto the domain Ω. Solution of Exercise 6.5.3 We will try to determine a circular transform f that conformally maps the halfplane D onto the disc Ω. It is well known that a circular transform maps the circles from ℂ∞ onto the circles from ℂ∞ , hence it is sufficient to determine the image of the three points from the boundary of the domain of definition. Thus {−1, 0, 1} ⊂ 𝜕D ⇒ {g(−1) = i, g(−0) = −1, g(1) = −i} ⊂ 𝜕Ω, and i ∈ D ⇒ g(i) = 0 ∈ Ω. It follows that the function f = gD conformally maps the halfplane D onto the disc Ω. Solution of Exercise 6.5.4 Similarly, we will determine a circular transform f that conformally maps the domain D onto the domain Ω. Thus {−i, −1, i} ⊂ 𝜕D ⇒ {f (−i) = 1, f (−1) = 0, f (i) = −1} ⊂ 𝜕Ω, where 𝜕D = 𝜕U(0; 1),
𝜕Ω = {z ∈ ℂ : Im z = 0},
and let consider a point from D: 0 ∈ D ⇒ f (−0) = i ∈ Ω. It follows that the function f conformally maps the disc D onto the halfplane Ω.
382  7 Solutions to the chapterwise exercises Solution of Exercise 6.5.5 We will determine a circular transform f that conformally maps the domain D onto the domain, but we cannot use the previous method because the boundary of the domain of the definition is not a circle from ℂ∞ . We will follow the solution of Exercise 6.5.2, since a conformal mappings maps the domains onto the domains, and maps the boundary of the domain onto the boundary of the image domain. Thus 𝜕D = {z ∈ ℂ : Re z = 0, Im z ≥ 0} ∪ {z ∈ ℂ : Re z ≤ 0, Im z = 0} = {x + iy ∈ ℂ : x = 0, y ≥ 0} ∪ {x + iy ∈ ℂ : x ≤ 0, y = 0}.
Case 1. If z ∈ {x + iy ∈ ℂ : x = 0, y ≥ 0}, then f (z) = f (iy) =
yi − i y − 1 = , yi + i y + 1
y ≥ 0.
Let f1 : [0, +∞) → ℝ,
f1 (y) =
y−1 , y+1
and we need to determine the image of f1 . Since f1 (y) =
2 > 0, (y + 1)2
∀y ∈ [0, +∞)
and lim f (y) y→+∞ 1
= 1,
lim f1 (y) = −1, y↓0
we get f ({x + iy ∈ ℂ : x = 0, y ≥ 0}) = [−1, 1]. Case 2. If z ∈ {x + iy ∈ ℂ : x ≤ 0, y = 0}, then the function will be f (z) = f (x) =
x − i x2 − 1 2x − i, = x + i x2 + 1 x2 + 1
x ≤ 0,
hence 2
2
x2 − 1 2x ) +( 2 ) = 1. f (z) = √( 2 x +1 x +1 We have obtained that the images f (z) that belong to the unit circle. We need to prove that these images are only on the required halfcircle.
7.6 Solutions to the exercises of Chapter 6  383
Since Im f (z) = − x2x 2 +1 , let f2 : (−∞, 0] → ℝ,
f2 (x) = −
2x . x2 − 1
Then f2 (x) = 2
x2 − 1 , (x2 + 1)2
x ∈ (−∞, 0]
and lim f (x) x→−∞ 2
= 0,
lim f2 (x) = 0, x↑0
f2 (−1) = 1,
hence we have that f ({x + iy ∈ ℂ : x ≤ 0, y = 0}) = {z ∈ ℂ : z = 1, Im z ≥ 0}. From both of these cases, we deduce that f (𝜕D) = 𝜕Ω. We will choose a point from the domain D, i. e., −1 + i ∈ D ⇒ f (−1 + i) =
1 2 + i ∈ Ω, 5 5
which completes our proof. Solution of Exercise 6.5.6 We will use a similar method as in Exercise 6.5.5. We will determine the image of the boundary of the domain a D, i. e., 𝜕D = {eiθ ∈ ℂ : θ ∈ [0, π]} ∪ {x ∈ ℝ : x ∈ [−1, 1]}. Case 1. If z ∈ {eiθ ∈ ℂ : θ ∈ [0, π]}, the image is 1 1 f (z) = f (eiθ ) = (eiθ + iθ ) = cos θ, 2 e
θ ∈ [0, π],
i. e., {f (eiθ ) : θ ∈ [0, π]} = [−1, 1]. Case 2. if z ∈ {x ∈ ℝ : x ∈ [−1, 1]}, then 1 1 1 x2 + 1 f (z) = f (x) = (x + ) = , 2 x 2 x
x ∈ [−1, 1].
384  7 Solutions to the chapterwise exercises Let f1 : [−1, 1] \ {0} → ℝ,
f1 (x) =
1 x2 + 1 , 2 x
hence we need to find the image of f1 . But f1 (x) =
1 x2 − 1 ⇒ f1 2 x2
is decreasing on [−1, 0) and (0, 1],
and lim f1 (x) = −1,
lim f1 (x) = −∞,
lim f1 (x) = +∞,
lim f1 (x) = 1,
x↓−1 x↓0
x↑0
x↑1
thus f ([−1, 1] \ {0}) = ℝ \ [−1, 1]. From the above results, we get f (𝜕D) = 𝜕Ω. The image of an arbitrary point from D is i i 3i ∈ D ⇒ f ( ) = − ∈ Ω, 2 2 4 hence the Joukowski function conformally maps the domain D onto the halfplane Ω.
Solution of Exercise 6.5.7 We will use a similar proof as in Exercise 6.5.6, by determining the image of the boundary of the domain D, where 𝜕D = (ℝ \ [−1, 1]) ∪ {eiθ ∈ ℂ : θ ∈ [0, π]}. Case 1. If z ∈ ℝ \ [−1, 1], then g(z) = g(x) = x +
1 x2 + 1 = , x x
x ∈ ℝ \ [−1, 1].
If g1 : ℝ \ [−1, 1] → ℝ,
g1 (x) =
x2 + 1 , x
7.6 Solutions to the exercises of Chapter 6  385
we will determine the image of the function g1 . Since g1 (x) =
x2 − 1 ⇒ g1 x2
is increasing in (−∞, −1) and (1, +∞),
and lim g (x) x→−∞ 1
= −∞,
lim g1 (x) = 2, x→1
lim g1 (x) = −2,
x→−1
lim g (x) x→+∞ 1
= +∞,
we get g(ℝ \ [−1, 1]) = ℝ \ [−2, 2]. Case 2. If z ∈ {eiθ ∈ ℂ : θ ∈ [0, π]}, the image of the function will be g(z) = g(eiθ ) = eiθ +
1 = 2 cos θ, eiθ
θ ∈ [0, π],
i. e., g({eiθ ∈ ℂ : θ ∈ [0, π]}) = [−2, 2]. Combining the results of the both cases, we deduce that the function g maps the boundary of the domain D onto the boundary of the domain Ω. The image of a point that belongs to the domain D is 2i ∈ D ⇒ f (2i) =
3i ∈ Ω, 2
thus the function f = gD conformally maps the domain D onto the halfplane Ω. Solution of Exercise 6.5.8 We will determine the required function by using the images of some wellknown elementary functions. It is easy to prove that g(D) = {z ∈ ℂ : Im z > 0},
where g(z) = (
2
z−1 ). z+1
According to Exercise 6.5.3, we have h({z ∈ ℂ : Im z > 0}) = Ω,
where h(z) =
z−i , z+i
and composing both of the above functions we get f = h ∘ g, i. e., f (z) =
(z − 1)2 − i(z + 1)2 . (z − 1)2 + i(z + 1)2
386  7 Solutions to the chapterwise exercises Solution of Exercise 6.5.9 We will use a similar proof as in Exercise 6.5.6. If 𝜕D = {z = eiθ ∈ ℂ : θ ∈ [0, 2π]}, then z ∈ 𝜕D ⇒ f (z) = f (eiθ ) = cos θ,
θ ∈ [0, 2π].
For the function f1 (θ) = cos θ, θ ∈ [0, 2π], we have lim f1 (θ) = 1,
lim f1 (θ) = −1,
θ↓0
θ→π
lim f1 (θ) = 1,
θ→2π
thus f (𝜕U(0; 1)) = [−1, 1]. The image of a point that belongs to the domain D is 2 ∈ D ⇒ f (2) =
5 ∈ Ω. 4
But f (z) =
z2 − 1 ≠ 0, 2z 2
∀z ∈ D,
hence the Joukowski function conformally maps the domain D onto the domain Ω.
Solution of Exercise 6.5.10 It is well known that a conformal mappings maps the domains onto the domains, and maps the boundary of the domain onto the boundary of the image domain. Thus, the boundary of the domain D is 𝜕D = {z ∈ ℂ : Re z = 0}. The image of a point that belongs to this boundary is z = iy ∈ 𝜕D ⇒ f (iy) = (iy)2 = −y2 ,
y ∈ ℝ.
If we let f1 (y) = −y2 , then lim f (y) y→−∞ 1
= −∞,
lim f1 (y) = 0,
y→0
lim f (y) y→+∞ 1
= +∞.
7.6 Solutions to the exercises of Chapter 6  387
From here, we get f (𝜕D) = (−∞, 0] = 𝜕Ω, and the image of a point that belongs to the domain D will be 1 ∈ D ⇒ f (1) = 1 ∈ Ω. Hence the function f conformally maps the halfplane D onto the domain Ω. Solution of Exercise 6.5.11 Similar to the proof of Exercise 6.5.8, the required function will be determined by using the images of some known functions. Using the result of Exercise 6.5.6, we know that f1 (D) = {z ∈ ℂ : Im z < 0} = Δ1 ,
1 1 where f1 (z) = (z + ). 2 z
Rotating the domain Δ1 with the angle π we get f2 (Δ1 ) = {z ∈ ℂ : Im z > 0} = Δ2 ,
where f2 (z) = eiπ z = −z.
Finally, we will use the result of Exercise 6.5.3, i. e., f3 (Δ2 ) = Ω,
where f3 (z) =
z−i . z+i
Composing the above functions, we get f = f3 ∘ f2 ∘ f1 , thus f (z) =
z 2 + 2iz + 1 . z 2 − 2iz + 1
Solution of Exercise 6.5.12 We will use a similar solution as in the previous problem. Multiplying the argument by 3, we have f1 (D) = {z ∈ ℂ : z < 2, Im z > 0} = Δ1 ,
where f1 (z) = z 3 .
The function f2 : Δ1 → {z ∈ ℂ : z < 1, Im z > 0} = Δ2 ,
f2 (z) =
z 2
divide the modules by 2 and f2 (Δ1 ) = Δ2 . According to Exercise 6.5.6 we get f3 (Δ2 ) = {z ∈ ℂ : Im z < 0} = Δ3 ,
1 1 f3 (z) = (z + ). 2 z
388  7 Solutions to the chapterwise exercises Rotating the domain Δ3 with the angle π, we have f4 (Δ3 ) = Ω,
f4 (z) = eiπ z = −z.
From the above results, we conclude that f = f4 ∘ f3 ∘ f2 ∘ f1 , hence f (z) = −
z6 + 4 . 4z 3
Solution of Exercise 6.5.13 The function will be determined by using the images of some known functions. First, if f1 : D → {z ∈ ℂ : 0 < Re z < 1} = Δ1 ,
f1 (z) =
z−a , b−a
we easily prove that f1 (D) = Δ1 . The domain Δ1 is independent on the parameters a and b. Rotating the domain Δ1 with the angle π2 and multiplying the modules by π, we get f2 : Δ1 → {z ∈ ℂ : 0 < Im z < π} = Δ2 ,
π
f2 (z) = πei 2 z = πiz,
and f2 (Δ1 ) = Δ2 . Finally, using the function f3 : Δ2 → Ω,
where f3 (z) = ez ,
we get the required domain, because f3 (Δ2 ) = Ω. The function will be f = f3 ∘ f2 ∘ f1 , i. e., πi
f (z) = e b−a (z−a) .
Solution of Exercise 6.5.14 We will determine a circular transform f that conformally maps the halfplane D onto the disc Ω, using the fact that a circular transform maps the circles from ℂ∞ onto the circles from ℂ∞ . It is sufficient to determine the image of the three points from the boundary of the domain of definition. The boundary of the domain D consists in the union of two circles, i. e., 𝜕D = 𝜕U(0; 2r) ∪ 𝜕U(a; r).
7.6 Solutions to the exercises of Chapter 6  389
To find the image of one of these circles, it is sufficient to calculate the images of the three points from the circle: {2r, 2ri, −2r} ⊂ 𝜕U(0; 2r) ⇒ f (2r) =
r , r−a
where a = r, i. e., a = reiθ . Thus
f (2ri) =
ar r2 −i 2 2, a2 + r 2 a +r
f (−2r) =
r , r+a
1 sin θ 1 r = +i ⇒ Re f (2ri) = . r−a 2 2 − 2 cos θ 2
f (2r) = We get similarly
1 Re f (2ri) = , 2
and thus, we deduce
1 Re f (−2r) = , 2
1 f (𝜕U(0; 2r)) = {z ∈ ℂ : Re z = }. 2 Now we will determine the image of the circle 𝜕U(a; r): r−a , {0, r − a, −r − a} ⊂ 𝜕U(a; r) ⇒ f (0) = 0, f (r − a) = r − 3a
f (−r − a) =
where a = reiθ , hence
Re f (r − a) = 0,
r+a , r + 3a
Re f (−r − a) = 0.
Thus f (𝜕U(a; r)) = {z ∈ ℂ : Re z = 0}. Combining the above results, we have f (𝜕D) = 𝜕Ω. The image of a point from D will be 1 −a ∈ D ⇒ f (−a) = ∈ Ω, 3 hence the function f (z) =
z z−2a
conformally maps the domain D onto the domain Ω.
Solution of Exercise 6.5.15 As we have previously mentioned, the required function will be determined by using the images of some known functions. Using the result of Exercise 6.5.14, consider the special case 1 f1 : D → {z ∈ ℂ : 0 < Re z < }, 2 obtained for r = 1 and a = i.
with f1 (z) =
z , z − 2i
390  7 Solutions to the chapterwise exercises Considering the special case of Exercise 6.5.13, 1 f2 : {z ∈ ℂ : 0 < Re z < } → Ω, 2
with f2 (z) = e2πiz ,
obtained for a = 0 and b = 21 , from these two results we conclude that the function will be f = f2 ∘ f1 , i. e., 2πiz
f (z) = e z−2i . Solution of Exercise 6.5.16 The function will be determined by using the images of some known functions. Using the result of Exercise 6.5.14, we have 1 f1 (D) = {z ∈ ℂ : 0 < Re z < , Im z > 0} = Δ1 , 2
f1 (z) =
z , z+r
where a = − 2r . Multiplying the modules of the numbers of the domain Δ1 by 2π, and translating by − π2 , i. e., f2 : Δ1 → {z ∈ ℂ : −
π π < Re z < , Im z > 0} = Δ2 , 2 2
f2 (z) = 2πz −
π , 2
we get f2 (Δ1 ) = Δ2 . The function f3 : Δ2 → Ω,
f3 (z) = sin z
has the property that f3 (Δ2 ) = Ω, and the result will be given by f = f3 ∘ f2 ∘ f1 , i. e., f (z) = sin(
2πz π − ). z+r 2
Solution of Exercise 6.5.17 Similar to the previous problem, according to Exercise 6.5.14 we have 1 f1 (D) = {z ∈ ℂ : 0 < Re z < } = Δ1 , 2 where a = 2r .
f1 (z) =
z , z−r
7.6 Solutions to the exercises of Chapter 6  391
Multiplying the modules of the numbers of the domain Δ1 by 2π, and rotating with the angle π, we get f2 : Δ1 → {z ∈ ℂ : 0 < Im z < π} = Δ2 ,
f2 (z) = 2πiz,
hence f2 (Δ1 ) = Δ2 . Similar to Exercise 6.5.13, the function f3 : Δ2 → Ω,
f3 (z) = ez
has the property f3 (Δ2 ) = Ω. Thus, the required function will be f = f3 ∘ f2 ∘ f1 , i. e., 2πiz
f (z) = e z−r . Solution of Exercise 6.5.18 The function f1 : D → {z ∈ ℂ : z < 1, 0 < arg z < π} = Δ1 ,
f1 (z) = z 2
doubles the arguments, hence f1 (D) = Δ1 . Using now the result obtained to Exercise 6.5.11, we have that f2 : Δ1 → Ω,
f2 (z) =
z 2 + 2iz + 1 z 2 − 2iz + 1
satisfies f2 (Δ1 ) = Ω. From the above results, we conclude that f = f2 ∘ f1 , i. e., f (z) = z 4 +2iz 2 +1 . z 4 −2iz 2 +1 Solution of Exercise 6.5.19 We will prove that the function f1 : D → ℂ \ {z ∈ ℂ : Im z = 0, Re z ≥ 0} = Δ1 ,
f1 (z) =
z−a , z−b
conformally maps the domain D onto the domain Ω. The boundary of the domain D is 𝜕D = (−∞, a] ∪ [b, +∞). The function f1∗ : (−∞, a] ∪ (b, +∞) → ℝ,
f1∗ (x) =
is decreasing on (−∞, a] and (b, +∞). We also have lim f ∗ (x) x→−∞ 1
= 1,
lim f ∗ (x) x→a 1
= 0,
x−a x−b
392  7 Solutions to the chapterwise exercises and lim f1∗ (x) = +∞,
lim f ∗ (x) x→+∞ 1
x↓b
= 1,
because b > a. From here, it follows that f1∗ ((−∞, a] ∪ (b, +∞)) = [0, +∞), thus we proved that f1 (D) = ℂ \ [0, +∞) = Δ1 . Now we will return to our problem, where we divide the argument by 21 . If f2 : Δ1 → Ω,
f2 (z) = √z,
where f2 (−1) = √−1 = i, then f2 (Δ1 ) = Ω. Composing both of these functions, we have f = f2 ∘ f1 , i. e., f (z) = √
z−a , z−b
where f (
a+b ) = √−1 = i. 2
Solution of Exercise 6.5.20 First, we will make a translation with 1, then we will divide the argument by 21 , i. e., f1 (D) = {z ∈ ℂ : Im z > 0} = Δ1 ,
f1 (z) = √z − 1,
where f1 (0) = √−1 = i. Using now the result of Exercise 6.5.3, we get f2 (Δ1 ) = Ω,
f2 (z) =
z−i . z+i
The required function will be f = f2 ∘ f1 , i. e., f (z) =
√z − 1 − i , √z − 1 + i
where f (0) = 0.
Solution of Exercise 6.5.21 The proof of the first point will be used for all the points of the problem. 1. We will start similarly as to the solution of Exercise 6.5.19. Consider the circular transform f1 : D → Δ1 ,
f1 (z) =
z−a , z−b
where Δ1 = ℂ \ ({z ∈ ℂ : Im z = 0, Re z ≤ 0} ∪ {1}).
7.6 Solutions to the exercises of Chapter 6  393
The boundary of the domain D is 𝜕D = [a, b] = {z ∈ ℂ : z = z(t) = (1 − t)a + tb, t ∈ [0, 1]}, and the values of f1 in these boundary points are f1 (z(t)) = f1 ((1 − t)a + tb) =
t , t−1
t ∈ [0, 1).
Let f1∗ : [0, 1) → ℝ,
f1∗ (t) =
t , t−1
a decreasing function on the domain of definition, and lim f1∗ (t) = −∞.
lim f1∗ (t) = 0,
t↑1
t↓0
From here, we get f1∗ ([0, 1)) = (−∞, 0], and since limz→∞ f1 (z) = 1 it follows that f1 (D) = Δ1 . Thus, the function f1 conformally maps the domain D onto the domain Δ1 . Now we will divide the arguments of the numbers of Δ1 by 2, and we will rotate these with the angle π2 , i. e., f2 : Δ1 → Ω,
f2 (z) = i√z,
where f2 (2) = i√2. Since f2 (Δ1 ) = Ω, composing both of these functions we deduce that f = f2 ∘ f1 , i. e., f (z) = i√
z−a , z−b
where f (2b − a) = i√2.
2. We will use the result of the point 1, where T = [1 + i, −3 − i], i. e., a = 1 + i and b = −3 − i. We obtain f (z) = i√
z−1−i , z+3+i
where f (−7 − 3i) = i√2.
3. In this case, the function will be f (z) = i√
z+i , z−i
where f (3i) = i√2.
Here, we used the result obtained in the point 1, for the special case a = −i and b = i. 4. Considering in the point 1 the set T = [−1, 1], i. e., a = −1 and b = 1, then we deduce that f (z) = i√
z+1 , z−1
where f (3) = i√2.
394  7 Solutions to the chapterwise exercises Solution of Exercise 6.5.22 The function will be determined by using the images of some known functions. Let f1 : D → {z ∈ ℂ : −1 < Im z < 0} = Δ1 ,
f1 (z) =
2 . z
The boundary of the domain D is 𝜕D = 𝜕U(i; 1) ∪ {z ∈ ℂ : Im z = 0}. Case 1. If z ∈ {z ∈ ℂ : Im z = 0}, then ∀z ∈ {z ∈ ℂ : Im z = 0} \ {0} ⇒ f1 (z) =
2 ∈ ℝ, z
hence f1 ({z ∈ ℂ : Im z = 0} \ {0}) = ℝ. Case 2. If z ∈ 𝜕U(i; 1), then z = i + eiθ , θ ∈ [0, 2π]. The images of these points by the function f1 are f1 (z) = f1 (i + eiθ ) =
cos θ 2 = − i, i + eiθ 1 + sin θ
θ ∈ [0, 2π] \ {
3π }, 2
hence f1 (𝜕U(i; 1)) = {z ∈ ℂ : Im z = −1}. The image of a point that lies in D is 3i ∈ D ⇒ f (3i) = −
2i ∈ Δ1 , 3
thus the function f1 conformally maps the domain D onto the domain Δ1 . Now we will multiply the modules of the numbers of Δ1 by π, and then we will make a translation with πi, i. e., f2 : Δ1 → {z ∈ ℂ : 0 < Im z < π} = Δ2 ,
f2 (z) = πz + πi,
hence f2 (Δ1 ) = Δ2 . Finally, the function f3 : Δ2 → Ω,
f3 (z) = ez
satisfies f3 (Δ2 ) = Ω, and composing all the above functions we conclude that f = f3 ∘ f2 ∘ f1 , i. e., f (z) = e
π(iz+2) z
.
7.6 Solutions to the exercises of Chapter 6  395
Solution of Exercise 6.5.23 We will multiply by 3 the arguments of all the complex numbers of D, i. e., f1 (z) = z 3 ,
f1 : D → {z ∈ ℂ : Im z > 0} = Δ1 , hence f1 (D) = Δ1 . If f2 : Δ1 → {z ∈ ℂ : 0 < Im z < π} = Δ2 , where log i = Letting
iπ , 2
f2 (z) = log z,
then f2 (Δ1 ) = Δ2 . f3 : Δ2 → Ω,
f3 (z) =
z , 2
we have f3 (Δ2 ) = Ω. Composing both of the previous functions, we get f = f2 ∘ f1 , i. e., f (z) =
1 log z 3 , 2
where f (1 + i) =
3 3πi ln 2 + . 4 8
Solution of Exercise 6.5.24 We will determine the images of the boundaries of the definition domains, and then we will calculate the image of an interior point. We have 𝜕D = 𝜕U(a; r) = {a+reiθ : θ ∈ [0, 2π]}. Since the given function is a circular transform, and it is well known that it maps any circles of ℂ∞ onto the circles of ℂ∞ , it is sufficient to determine the images of three points of the boundary. Thus, let {a + r, a − r, a + ir} ⊂ 𝜕D. that
Since α∗ is the inverse of the point α with respect to the circle 𝜕U(a; r), it follows (α∗ − a)(α − a) = r 2 ⇔ α∗ = a +
r2 . α−a
From here, we get α−aa−α+r α − a ⇒ f (a + r) = , r a−α+r r α−aa−α−r α − a f (a − r) = ⇒ f (a − r) = r a−α−r r f (a + r) = −
and f (a + ir) = i
α − a a − α + ir α − a ⇒ f (a + ir) = , r a − α − ir r
396  7 Solutions to the chapterwise exercises hence {f (a + r), f (a − r), f (a + ir)} ⊂ 𝜕Ω. Now we will study both of the points of the problem. 1. The image of the domain of the definition will be given by 2 a − α a − α a − α z = a ∈ D ⇒ f (a) = ) < , = ( a − α∗ r r
because a − α < r,
i. e., f (a) ∈ Ω. From here, it follows that f (D) = Ω, which completes our proof. 2. If α and α∗ are two inverse points with respect to the circle 𝜕U(a; r), then α = a + r1 eiφ , α∗ = a + r2 eiφ , where r1 r2 = r 2 , hence α∗ = a +
r 2 iφ e . r1
The image of the domain of the definition will be given by z = a + λreiφ ∈ D,
r λr − r1 r λ ∈ (1, +∞) \ { } ⇒ f (a + λreiφ ) = 1 r1 r λr1 − r
r λr − r1 > 1, ⇒ f (a + λreiφ ) = 1 r λr1 − r
r ∀λ ∈ (1, +∞) \ { } ⇒ f (a + λreiφ ) ∈ Ω, r1
and limz→∞ f (z) = 1. We obtained that the function f (z) = set D onto the set Ω.
z−α z−α∗
conformally maps the
Solution of Exercise 6.5.25 We will use the result obtained in Exercise 6.5.9, i. e., z ∈ 𝜕D ⇒ f (z) = f (eiθ ) = cos θ,
θ ∈ [0, 2π],
hence f (𝜕D) = [−1, 1]. The image of an interior point will be 1 5 1 ∈ D ⇒ f ( ) = ∈ Ω, 2 2 4 hence the function f conformally maps the domain D onto the domain Ω. Solution of Exercise 6.5.26 Considering the Joukowski function 1 1 f (z) = (z + ), 2 z
7.6 Solutions to the exercises of Chapter 6  397
we have f ∈ H(D). To Exercise 6.5.25, we proved that f (𝜕U(0; 1)) = [−1, 1], hence it is sufficient to show that 1 5 f ([ , 1)) = (1, ]. 2 4 This last result holds, since 1 1 1 z = x ∈ [ , 1) ⇒ f (z) = f (x) = (x + ), 2 2 x and from 1 1 f (x) = (1 − 2 ) < 0, 2 x
1 x ∈ [ , 1) ⇒ f is a decreasing function, 2
with 1 5 f( ) = , 2 4
lim f (x) = 1. x↑1
It follows that the Joukowski function conformally maps the domain D onto the domain Ω. Solution of Exercise 6.5.27 Similar to the previous problem, we define the Joukowski function f1 : D → Δ1 ,
1 1 f1 (z) = (z + ), 2 z
where the boundary of the domain D is 𝜕D = 𝜕U(0; 1) ∪ (−1, 0] ∪ [a, 1). To Exercise 6.5.25, we proved that f1 (𝜕U(0; 1)) = [−1, 1].
(7.56)
Now we will determine the images of the intervals (−1, 0] and [a, 1) by the function f1 , i. e., 1 1 z = x ∈ (−1, 0] ∪ [a, 1) ⇒ f1 (z) = f1 (x) = (x + ), 2 x where we know that f1 is a decreasing function (see Exercise 6.5.26), and lim f1 (x) = −1,
x↓−1
lim f1 (x) = −∞, x↑0
1 1 f1 (a) = (a + ), 2 a
lim f1 (x) = 1. x↑1
398  7 Solutions to the chapterwise exercises Thus 1 1 f1 ((−1, 0] ∪ [a, 1)) = (−∞, −1) ∪ (1, (a + )]. 2 a
(7.57)
From the relations (7.56) and (7.57), we get 1 1 f1 (D) = Δ1 = ℂ \ (−∞, (a + )]. 2 a Translating all the numbers of the domain Δ1 with 21 (a + a1 ), then we will divide the arguments by 2, and finally we will rotate by the angle π2 , i. e., f2 : Δ1 → Ω,
1 1 f2 (z) = i√z − (a + ), 2 a
1 where f2 (a + 2a ) = i√ a2 . Since f2 (Δ1 ) = Ω, composing both of these functions we get f = f2 ∘ f1 , i. e.,
1 1 1 1 f (z) = i√ (z + ) − (a + ), 2 z 2 a
a i√a where f ( ) = . 2 2
Solution of Exercise 6.5.28 The function will be determined by using the images of some known functions. Using the result of Exercise 6.5.26, we have 5 f1 : D → ℂ \ [−1, ] = Δ1 , 4
1 1 f1 (z) = (z + ), 2 z
and moreover, f1 (D) = Δ1 . Now we will use the result of the point 1 of Exercise 6.5.21, for the special case a = −1 and b = 45 : f2 : Δ1 → Ω,
f2 (z) = i√
z+1
z−
5 4
,
where f2 ( 72 ) = i√2. The required function will be f = f2 ∘ f1 , i. e., f (z) = i√
2(z + 1)2 , 2z 2 − 5z + 2
1 5i√7 where f ( ) = . 4 7
7.6 Solutions to the exercises of Chapter 6  399
Solution of Exercise 6.5.29 We will determine a circular function with the above property. The points w1 = 0 and w2 = ∞ are inverse points with respect to the circles w = 32 and w = 1. Thus ∃z1 , z2 ∈ ℂ : w(z1 ) = 0, w(z2 ) = ∞ and z1 , z2 are inverse points with respect to the circles z − 3 = 9 and z − 8 = 16. Then z1 − 3z2 − 3 = 81
(7.58)
arg(z1 − 3) = arg(z2 − 3),
(7.59)
z1 − 8z2 − 8 = 256
(7.60)
and respectively, arg(z1 − 8) = arg(z2 − 8).
(7.61)
From (7.59) and (7.61), it follows that arg z1 = arg z2 ∈ {0, π}, i. e., z1 , z2 ∈ ℝ. Now, from (7.58) and (7.60) we deduce that z1 = 0 and z2 = −24 or conversely, z1 = −24 and z2 = 0. Case 1. If z1 = 0, z2 = −24, then {
w(0) = 0 z , ⇔ w(z) = k w(−24) = ∞ z + 24
k ∈ ℂ.
Let denote by 𝒞1 = {z ∈ ℂ : z − 3 > 9}, 𝒞2 = {z ∈ ℂ : z − 8 < 16} and 𝒞1 = {w ∈ ℂ : w = 32 }, 𝒞2 = {w ∈ ℂ : w = 1}. From 2 0 ∈ U(3; 9) ∩ U(8; 16) ⇒ 0 = w(0) ∈ U(0; ) ∩ U(0; 1) 3 ⇒ 𝒞1 = w(𝒞1 ),
C2 = w(𝒞2 ),
we have
and thus
k 24 ∈ 𝒞2 ⇒ w(24) ∈ 𝒞2 ⇔ w(24) = 1 ⇔ = 1 ⇔ k = 2eiθ , 2 w(z) = 2eiθ
Case 2. If z1 = −24, z2 = 0, then {
z , z + 24
θ ∈ ℝ,
θ ∈ ℝ.
w(−24) = 0 z + 24 , ⇔ w(z) = k w(0) = ∞ z
k ∈ ℂ.
From 2 0 ∈ U(3; 9) ∩ U(8; 16) ⇒ ∞ = w(0) ∈ (ℂ \ U(0; )) ∩ (ℂ \ U(0; 1)), 3 ⇒ 𝒞1 = w(𝒞2 ),
C2 = w(𝒞1 ),
400  7 Solutions to the chapterwise exercises we get 2 eiθ 2 24 ∈ 𝒞2 ⇒ w(24) ∈ 𝒞1 ⇔ w(24) = ⇔ 2k = ⇔ k = , 3 3 3
θ ∈ ℝ,
and thus w(z) =
eiθ z + 24 , 3 z
θ ∈ ℝ.
Solution of Exercise 6.5.30 Similar to the previous problem, we will determine a circular function with the above property. The points w1 = 0 and w2 = ∞ are inverse points with respect to the circles w = 1 and w = 45 √ 52 . Then ∃z1 , z2 ∈ ℂ : w(z1 ) = 0, w(z2 ) = ∞ and z1 , z2 are inverse
points with respect to the circles z = 1 and z − 21  = √ 52 . Then z1 z2  = 1
(7.62)
arg z1 = arg z2 ,
(7.63)
and respectively, 1 1 5 z2 + = 2 2 2 1 1 arg(z1 + ) = arg(z2 + ). 2 2 z1 +
(7.64) (7.65)
From (7.63) and (7.65), it follows that arg z1 = arg z2 ∈ {0, π}, i. e., z1 , z2 ∈ ℝ. Now, from (7.62) and (7.64) we deduce that z1 = − 21 and z2 = −2 or conversely, z1 = −2 and z2 = − 21 . Case 1. if z1 = 21 , z2 = 2, then {
z − 21 w( 21 ) = 0 , ⇔ w(z) = k w(2) = ∞ z−2
k ∈ ℂ.
Denote by 𝒞1 = {z ∈ ℂ : z > 1}, 𝒞2 = {z ∈ ℂ : z + 21  < √ 52 } and 𝒞1 = {w ∈ ℂ :
w = 1}, 𝒞2 = {w ∈ ℂ : w = 45 √ 52 }. From
1 1 5 1 4 5 ∈ U(0; 1) ∩ U( ; √ ) ⇒ 0 = w( ) ∈ U(0; 1) ∩ U(0; √ ) 2 2 2 2 5 2 ⇒ 𝒞1 = w(𝒞1 ),
𝒞2 = w(𝒞2 ),
we have k 1 ∈ 𝒞1 ⇒ w(1) ∈ 𝒞1 ⇔ w(1) = 1 ⇔ = 1 ⇔ k = 2eiθ , 2
θ ∈ ℝ,
7.6 Solutions to the exercises of Chapter 6  401
and thus w(z) = 2eiθ
z−
Case 2. If z1 = 2, z2 = 21 , then {
1 2
z−2
θ ∈ ℝ.
,
w(2) = 0 z−2 ⇔ w(z) = k , 1 w( 2 ) = ∞ z− 1
k ∈ ℂ.
2
From 1 5 1 4 5 1 ∈ U(0; 1) ∩ U( ; √ ) ⇒ ∞ = w( ) ∈ (ℂ \ U(0; 1)) ∩ (ℂ \ U(0; √ )) 2 2 2 2 5 2 ⇒ 𝒞1 = w(𝒞2 ),
𝒞2 = w(𝒞1 ),
we get 4 5 2 5 4 5 1 ∈ 𝒞1 ⇒ w(1) ∈ 𝒞2 ⇔ w(1) = √ ⇔ 2k = √ ⇔ k = √ eiθ , 5 2 5 2 5 2
θ ∈ ℝ,
and thus w(z) =
4 5 iθ z − 2 √ e , 5 2 z− 1
θ ∈ ℝ.
2
Solution of Exercise 6.5.31 First, we will rotate the points of the domain D by the angle − π2 , using the function f1 (D) = {z ∈ ℂ : Re z > 0} \ (0, 1] = Δ1 ,
f1 (z) = −iz.
Then we will multiply the arguments of the points of the domain Δ1 by 2, i. e., f2 (z) = z 2 .
f2 (Δ1 ) = ℂ \ (−∞, 1] = Δ2 ,
We will translate the image by −1, with the function f3 : (Δ2 ) = ℂ \ (−∞, 0] = Δ3 ,
f3 (z) = z − 1.
We will divide by 2 the arguments of the points of the domain Δ3 , i. e., f4 (Δ3 ) = {z ∈ ℂ : Re z > 0} = Δ4 , where √1 = 1. Rotating the domain Δ4 with the angle f5 (Δ4 ) = Ω,
f4 (z) = √z, π 2
we get
i π2
f5 (z) = e z = iz.
Composing all the above functions, we conclude that f = f5 ∘ f4 ∘ f3 ∘ f2 ∘ f1 , and f (z) = i√−z 2 − 1,
where f (i√2) = i.
402  7 Solutions to the chapterwise exercises Solution of Exercise 6.5.32 Starting from the Joukowski function, f1 : D → Δ1 ,
1 1 f1 (z) = (z + ), 2 z
then the domain D has the boundary 𝜕D = {eiθ ∈ ℂ : θ ∈ [0, π]} ∪ [−1, 1] ∪ (0, αi]. Using the result of Exercise 6.5.6, we get f1 ({eiθ ∈ ℂ : θ ∈ [0, π]} ∪ [−1, 1]) = {z ∈ ℂ : Im z = 0}. Now we will determine the image of the segment (0, αi] via the function f1 , i. e., 1 1 z ∈ (0, αi] ⇒ f1 (z) = f1 (iy) = (y − )i, 2 y
y ∈ (0, α].
Letting φ1 : (0, α] → ℝ,
1 1 φ1 (y) = (y − ), 2 y
then φ1 is an increasing function because φ1 (y) > 0, ∀y ∈ (0, α], and lim φ1 (y) = −∞, y↓0
1 1 lim φ1 (y) = (α − ). y↑α 2 α
From here, it follows that 1 1 f1 ((0, α]) = {z ∈ ℂ : Re z = 0, Im z ≤ (α − )}. 2 α Since z=
(1 + α)i (1 + α)i (1 + α)2 − 4 ∈ D ⇒ Im f1 ( ) = Im < 0, 2 2 4(1 + α)
∀α ∈ (0, 1),
we get 1 1 Δ1 = {z ∈ ℂ : Im z < 0} \ {z ∈ ℂ : Re z = 0, Im z ≤ (α − )}, 2 α and f1 (D) = Δ1 . Letting f2 (z) = z1 , then f2 (Δ1 ) = {z ∈ ℂ : Im z > 0} \ {z ∈ ℂ : 0 < Im z ≤
2α } = Δ2 . 1 − α2
7.6 Solutions to the exercises of Chapter 6  403
The new problem that we have obtained using this way is similar to Exercise 6.5.31. 2α Similarly, the function f3 will be translated not by −1 but with − 1−α 2 , and thus f3 : {z ∈ ℂ : Im z > 0} \ (0, f3 (z) = i√−z 2 −
2α i] = Δ2 → Ω, 1 − α2
2α , 1 − α2
where f3 ((1 +
4α2 2α 2α )i) = i√1 + + . 2 2 1−α 1−α (1 − α2 )2
The required function will be f = f3 ∘ f2 ∘ f1 , i. e., f (z) = i√−
2α 4z 2 − , + 1)2 1 − α2
(z 2
where the branch of the root function is the principal one.
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Index Abel theorem 108 Analytic function 113 Angle of two lines 5 Angular sector 5 Arcconnected set 55 Argument Principle 161 Axis – imaginary 5 – real 5 Binomial series 114 Bounded functions set 172 Casorati–Weierstrass theorem 131 Cauchy – formula for circular rings 125 – formulas for closed curves 87 – formulas for the disc 80 – inequalities 90 – theorem 74 Cauchy theorem – for triangles (Goursat lemma) 70 – related to the zeros and poles 160 Cauchy–Hadamard theorem or the radius of convergence theorem 109 Characterization – of the zeros of holomorphic functions 115 – theorem for the poles 129 Closed – path 51 – segment 7 – set 7 Compact converging sequence 105 Compactified complex plane 10 Complex number 1 – trigonometric form 5 – algebraic form 2 – conjugate 2 – imaginary part 2 – inverse 2 – module 2 – (principal) argument 3 – real part 2 – set of the arguments 3 Complex plane – line equation 5
Conformal – automorphism 180 – group 180 – isomorphism 180 – radius 184 Conformally equivalent domains 180 Connected set 7 Constant path 54 Continuous function 10 Convex domain 57 Decomposition of a path by a division 53 Differentiable function – complex valued 20 – real valued 17 Disc (open) 6 Domain 7 Essential isolated singular point for a function 129 Existence theorem of the analytical branches of multivalued functions 82 Expandable function in Taylor series 113 Fresneltype integrals 152 Function with bounded variation 58 Fundamental theorem of the algebra 78, 91, 163 Halfplane – left 5 – lower 5 – right 5 – upper 5 Harmonic function 88 Holomorphic (analytic) continuation (extension) of a function 117 Homotopic path 51 Hurwitz theorem 179 Image of a path 51 Inverse of a path 54 Isolated singular point 127 Jordan lemma 147 Kasner theorem 22
408  Index
Lagrangetype theorem 18 Laurent series – 122 – expansion theorem 124 – main part 122 – Taylor part 122 Left halfplane bounded by a line 5 Limit of a function 9 Linear path 55 Liouville theorem 91 Locally constant function 85 Locally finite lines family in a set 82 Meromorphic function 131 Montel theorem 173 Morera theorem 82 Neighborhood system of a point 7 Open set 7 Order of a – pole 130 – rational function 134 Partial sum 107 Path in a set 51 Poisson formula 92 Pole for a function 129 Pompeiu theorem 176 Power series 108 Primitive of a function 66 Properties of the – argument 3 – complex integral 63 – zeros holomorphic functions 116 Radius of convergence of the power series 109 Rectifiable path 56 Regular point 127 Relatively compact functions set 173 Removability criterion – the first 127 – the second 127 – the third, or Cauchy–Riemann 128 Residue of a function at a point 141 Residue’s value in a pole 143 Riemann mapping theorem 182 Riemann–Stieltjes integrability 60 Right halfplane bounded by a line 5
Ring (open) 6 Rouché theorem 162 Schwarz – formula 93 – generalized lemma 121 – lemma 120 Series of functions – compact convergent 107 – convergent 107 – sum 107 – uniformly convergent 107 Set – compact 8 – connected by broken lines 55 – set 7 – starlike with respect to a point 7 – the accumulation points of a 7 – the boundary of a 7 – the closure of a 7 Simply connected domain 57 Special case of the principle of argument variation 162 Starlike domain with respect to a point 57 Taylor series expansion theorem 111 The analyticity of holomorphic functions 113 The complex integral 63 The connection between – the holomorphic functions and the primitives 72 – the primitive and the integral 67 The identity of the Laurent series coefficients 123 The index – of a curve with respect to the point 84 – theorem 85 The maximum principle of the holomorphic functions 117 The order of a function in a – point 158 – set 159 The principle of the – continuation of holomorphic functions 117 – domain conservation 164 The problem of the conformal representation 180 The residue in the infinity 143 The residues theorem 141
Index  409
The set of the zeros of holomorphic functions 115 The theorem of the identity of the power series coefficients 111 Theorem of the annulus of convergence 122 Total variation of a path 56 Two dimensional sphere (ball) 11 Uniformly continuous functions set 171 Union of the paths 53 Uniqueness theorem of holomorphic functions 116 Univalent function 175
Variation – of a function corresponding a the division 58 – of a path corresponding to a division 56 Vitali theorem 175 Weierstrass – lemma 105 – theorem 106 – theorem for holomorphic functions series 108 – uniformly convergence criteria 108 Zero of a function 114