Krishna's Textbook on Trigonometry [26 ed.]

Table of contents :
Trigonometry
Dedication
Preface
Syllabus
Brief Contents
Section- B, Trigonometry
1. Complex Numbers
2. Exponential, Trigonometric and Hyperbolic Functions of a Complex Variable
3. Logarithms of Complex Numbers
4. Inverse Circular and Hyperbolic Functions of Complex Numbers
5. Gregory's Series
6. Summation of Trigonometrical Series

Citation preview

a's n h s Kri

TEXT BOOK on

T rigonometry (For B.A. and B.Sc. Ist year students of All Colleges affiliated to universities in Uttar Pradesh)

As per U.P. UNIFIED Syllabus (w.e.f. 2011-2012)

By

Gopi Ahuja

A. R. Vasishtha

M.Sc., Ph.D., D.Sc.

Retired Head, Dep’t. of Mathematics

Head, Dep’t. of Mathematics

Meerut College, Meerut (U.P.)

Feroze Gandhi (P.G.) College, Raibareilly (U.P.)

Meera Srivastava

S.N. Mishra

M.Sc., Ph.D.

M.Sc., Ph.D.

Head, Dep’t of Mathematics

Dep’t. of Mathematics

D.A.V. (P.G.) College, Kanpur (U.P.)

Brahmanand (P.G.) College, Kanpur (U.P.)

Ashish Kumar

Hemlata Vasishtha

M.Sc., Ph.D.

M.Sc., (Gold Medalist) Ph.D.

Dep’t of Basic Science Dr. B.R.A. College of Agg. & Engg. Etawah (U.P.)

C.C.S. University, Meerut (U.P.)

(Kanpur Edition)

KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

Jai Shri Radhey Shyam

Dedicated to

Lord

Krishna Authors & Publishers

Preface This book on TRIGONOMETRY has been specially written according to the latest Unified Syllabus to meet the requirements of the B.A. and B.Sc. Part-I Students of all Universities in Uttar Pradesh. The subject matter has been discussed in such a simple way that the students will find no difficulty to understand it. The proofs of various theorems and examples have been given with minute details. Each chapter of this book contains complete theory and a fairly large number of solved examples. Sufficient problems have also been selected from various university examination papers. At the end of each chapter an exercise containing objective questions has been given. We have tried our best to keep the book free from misprints. The authors shall be grateful to the readers who point out errors and omissions which, inspite of all care, might have been there. The authors, in general, hope that the present book will be warmly received by the students and teachers. We shall indeed be very thankful to our colleagues for their recommending this book to their students. The authors wish to express their thanks to Mr. S.K. Rastogi, Managing Director, Mr. Sugam Rastogi, Executive Director, Mrs. Kanupriya Rastogi Director and entire team of KRISHNA Prakashan Media (P) Ltd., Meerut for bringing out this book in the present nice form. The authors will feel amply rewarded if the book serves the purpose for which it is meant. Suggestions for the improvement of the book are always welcome.

Preface to the Revised Edition The authors feel great pleasure in presenting the thoroughly revised edition of the book TRIGONOMETRY and wish to record thanks to the teachers and students for their warm reception to the previous edition. The present edition has been specially designed, made up-to-date and well organised in a systematic order according to the latest syllabus. The authors have always endeavoured to keep the text update in the best interests of the students community- a gesture which the authors hope would be appreciated by the students and teachers alike. Suggestions for the improvement of the book will be thankfully received.

June, 2014

— Authors

Syllabus Algebra & Trigonometry U.P. UNIFIED (w.e.f. 2011-12)

B.A./B.Sc. Paper-I

M.M. : 33 / 65

Algebra Unit-1: Sequence and its convergence (basic idea), Convergence of infinite series, Comparison test, ratio test, root test, Raabe’s test, Logarithmic ratio test, Cauchy’s condensation test, DeMorgan and Bertrand test and higher logarithmic ratio test. Alternating series, Leibnitz test, Absolute and conditional convergence, Congruence modulo m relation, Equivalence relations and partitions. Unit-2: Definition of a group with examples and simple properties, Permutation groups, Subgroups, Centre and normalizer, Cyclic groups, Coset decomposition, Lagrange’s theorem and its consequences. Unit-3: Homomorphism and Isomorphism. Cayley’s theorem, Normal subgroups, Quotient group, Fundamental theorem of homomorphism, Conjugacy irelation, Class equation, Direct product. Unit-4: Introduction to rings, subrings, integral domains and fields, Characteristic of a ring, Homomorphism of rings, Ideals, Quotient rings.

Trigonometry Unit-5: Complex functions and separation into real and imaginary parts, Exponential, Direct and inverse trigonometric and hyperbolic functions, Logarithmic functions, Gregory’s series, Summation of series.

B rief C ontents

TRIGONOMETRY.............................................. T-01 — T-124 1.

Complex Numbers ...........................................................................T-03 — T-16

2. Exponential, Trigonometric and Hyperbolic Functions of a Complex Variable (Separation into real & imaginary parts).........T-17 — T-42 3. Logarithms of Complex Numbers ................................................T-43 — T-60 4. Inverse Circular and Hyperbolic Functions of Complex Numbers................................................................................T-61 — T-72 5. Gregory’s Series .................................................................................T-73 — T-82 6. Summation of Trigonometrical Series.......................................T-83 — T-124

SECTION

B TRIGONOMETRY C hapters 1. Complex Numbers 1. 1. Exponential, Trigonometric and Hyperbolic 2. Functions of a Complex Variable (Separation i nto Real a nd Imaginary Parts)

1. Logariths of Complex Numbers 3. 1. Inverse Circular and Hyperbolic 4. Functions of Complex Numbers 1. Gregory's Series 5. 1. Summation of Trigonometrical Series 6.

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1 C omplex N umbers

1.1 Inverse Circular Functions Definitions: The equation sin θ = x means that θ is the angle whose sine is x. To express θ explicitly in terms of x,a convenient notation ‘sin −1 x’ (read as sine inverse x) is introduced. Thus θ = sin −1 x means that θ is the angle whose sine is x. Similarly ‘cos −1 x’ expresses an angle whose cosine is x, tan −1 x denotes an angle whose tangent is x, and so on. The quantities sin −1 x, cos −1 x, tan −1 x etc., are called Inverse Circular Functions. Sometimes sin −1 x is written as ‘arc sine x’ with similar notations for other inverse functions. Note: sin −1 x should not be confused with (sin x) −1 as (sin x) −1 = 1 / sin x.

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1.2 General and Principal Values of Inverse Circular Functions Consider the equation θ = sin −1 x, where −1 ≤ x ≤ 1. In the set of real numbers, there are infinite values of θ which satisfy this equation. All these values of θ taken together form what we call the general value of sin −1 x. Among the values of θ satisfying the equation θ = sin −1 x, the value which is numerically the smallest one is called the principal value of sin −1 x. If θ is the principal value of sin −1 x,obviously we 1 1 must have − π ≤ θ ≤ π. If x is 0, the principal value of sin −1 x is 0; if x is negative, 2 2 1 the principal value of sin −1 x lies between − π and 0; and if x is positive, the 2 1 −1 principal value of sin x lies between 0 and π. Thus the principal value of 2 1 1 sin −1 (1 / √ 2) is π, while the principal value of sin −1 (−1 / √ 2) is − π. 4 4 If θ is the principal value of sin −1 x, then all the angles given by nπ + (−1) n θ, where n is any integer, have their sines equal to x. We call nπ + (−1) n θ as the general value of sin −1 x and denote it by Sin −1 x. Generally if the first letter of an inverse circular function is small, we consider the principal value, while if the first letter is capital, it means the general value. 1 1 1 1 For example, sin −1 = π, while Sin −1 = nπ + (−1) n π, 2 6 2 6 where n is any integer. Thus the inverse sine of x is a many-valued function. Its principal value is denoted by sin −1 x and its general value is denoted by Sin −1 x. Also we have Sin −1 x = nπ + (−1) n sin −1 x, where n is any integer, positive or negative or zero. Similarly we may define the concepts of the principal and general values of the other inverse circular functions. The relations between the general and principal values of various inverse circular functions are as follows : Sin −1 x = nπ + ( − 1) n sin −1 x, Cosec −1 x = nπ + ( − 1) n cosec −1 x Cos −1 x = 2 nπ ± cos −1 x,

Sec −1 x = 2 nπ ± sec −1 x,

Tan −1 x = nπ + tan −1 x,

Cot −1 x = nπ + cot −1 x,

where n is any integer, positive or negative or zero. The principal value of any inverse circular function is the smallest numerical value of that inverse circular function. It may be positive or negative. In case there are two values, one positive and the other negative, which are numerically equal and

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smallest, the principal value is taken as the positive one. For example, the principal 1 1 1 1 value of cos −1 is π, as out of the two numerically smallest values π and − π 2 3 3 3 1 1 of cos −1 , the value π is positive. 2 3 It is to be noted that by the value of an inverse circular function we usually mean its principal value unless the contrary is stated. If x is positive, the principal values 1 of sin −1 x, cosec −1 x, cos −1 x, sec −1 x, tan −1 x and cot −1 x all lie between 0 and π. But 2 if x is negative, the principal values of sin −1 x, cosec −1 x, tan −1 x and cot −1 x lie 1 1 between − π and 0, while those of cos −1 x and sec −1 x lie between π and π. 2 2 Thus the principal values of sin −1 x and tan −1 x (and therefore of cosec −1 x and 1 1 cot −1 x) lie between − π and π, while those of cos −1 x and sec −1 x lie between 0 2 2 and π.

1.3 Relations between Inverse Functions (a) Principle of reciprocity: sin

−1

x = cosec

−1

We have

(1 / x) ; cos −1 x = sec −1 (1 / x); and tan −1 x = cot

−1

(1 / x).

(b) From the definition of an inverse circular function, it is clear that θ = sin −1 (sin θ) and x = sin (sin −1 x). Similarly

θ = cos −1 (cos θ) and x = cos (cos −1 x), θ = tan −1 (tan θ) and x = tan (tan −1 x), etc. sin −1 x = cos −1 √ (1 − x 2 ) = tan –1 { x / √ (1 − x 2 )} cos −1 x = sin −1 √ (1 − x 2 ) = tan −1

and (c)

√ (1 − x 2 ) x

etc.,

tan −1 x = sin −1 {x/ √ (1 + x 2 )} = cos −1 {1 / √ (1 + x 2 )} etc. We have, (i) sin −1 (− x) = − sin −1 x ;

(ii) cos −1 (− x) = π − cos −1 x ;

and (iii) tan −1 (− x) = − tan −1 x.

1.4 Some Important Results about Inverse Functions (a) Complementary inverse functions: We have (i) sin −1 x + cos −1 x = π / 2 ; (iii) sec −1 x + cosec −1 x = π / 2.

(ii) tan −1 x + cot −1 x = π / 2 ;

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(b)

Important formulae: −1

−1

We have,

y = tan −1 {( x + y) / (1 − xy)};

(i)

tan

(ii)

tan −1 x − tan −1 y = tan −1 {( x − y) / (1 + xy)};

x + tan

(iii) 2 tan −1 x = tan −1 {2 x / (1 − x 2 )}; tan −1 x + tan −1 y + tan −1 z = tan −1

(iv)

x + y + z − xyz 1 − yz − zx − xy

⋅

Remark: The formula tan −1 x + tan −1 y = tan −1 {( x + y) / (1 − x y)} does not mean that the sum of the principal values of tan −1 x and tan −1 y will necessarily be equal to the principal value of

tan −1 {( x + y) / (1 − x y)}.

This sum may be equal to the principal value of tan −1 {( x + y) / (1 − x y)} or it may be equal to some other value of tan −1 {( x + y) / (1 − xy)}. (c)

Some more formulae: −1

−1

We have xy − 1

−1

(i)

cot

(ii)

cot −1 x − cot −1 y = cot −1

x + cot

y = cot

x+ y xy + 1 y− x

⋅

1.5 Complex Numbers The equation x 2 = − 1has no solution in the set of real numbers because the square of every real number is either positive or zero. Therefore we feel the necessity to extend the system of real numbers. We all know that this defect is remedied by introducing complex numbers. Complex numbers: Definition: A number of the form x + iy, where i = √ (−1) and x, y are both real numbers, is called a complex number. A complex number is also defined as an ordered pair ( x, y) of real numbers. A complex number x + iy or ( x, y) is usually denoted by the symbol z. If we write z = x + iy or ( x, y) then x is called the real part and y the imaginary part of the complex number z and these are denoted by R (z ) and I (z ) respectively. Thus in the complex number z = √ 3 + 5i, we have R (z ) = the real part of z = √ 3, and I (z ) = the imaginary part of z = 5. A complex number is said to be purely real if its imaginary part is zero, and purely imaginary if its real part is zero. The complex number a + 0 i is simply written as a. We shall denote the set of all complex numbers by C. Equality of two complex numbers: Definition: Two complex numbers z1 = x1 + iy1

or ( x1 , y1 ) and z 2 = x2 + iy2

or ( x2 , y2 )

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are said to be equal if x1 = x2 and y1 = y2 . Thus two complex numbers are equal if and only if the real part of one is equal to the real part of the other and the imaginary part of one is equal to the imaginary part of the other.

1.6 Addition of Complex Numbers If z1 = x1 + iy1 or ( x1 , y1 ) and z 2 = x2 + iy2 or ( x2 , y2 ) are any two complex numbers, then the sum of z1 and z 2 written as z1 + z 2 is defined by z1 + z 2 = ( x1 + iy1 ) + ( x2 + iy2 ) = ( x1 + x2 ) + i ( y1 + y2 ) or

z1 + z 2 = ( x1 , y1 ) + ( x2 , y2 ) = ( x1 + x2 , y1 + y2 ).

Thus

(3 + 5i) + (7 − 8i) = (3 + 7) + (5 − 8) i = 10 − 3i.

Properties of the Addition of Complex Numbers: The addition of complex numbers is commutative, associative, admits of identity element and every complex number possesses additive inverse. Commutativity of addition in C: We have z1 + z 2 = z 2 + z1 where z1 and z 2 are any complex numbers. Associativity of addition in C: We have (z1 + z 2 ) + z 3 = z1 + (z 2 + z 3 ), for all complex numbers z1 , z 2 and z 3 . The complex number (0, 0) or 0 + i0 is the additive identity, since for every complex number ( x, y), we have ( x, y) + (0, 0) = ( x + 0, y + 0) = ( x, y) = (0, 0) + ( x, y). The complex number (0, 0) is called the zero complex number and is simply written as 0. A complex number x + iy is said to be a non-zero complex number if at least one of x and y is not zero. The complex number (− x, − y) is the additive inverse of the complex number ( x, y) since ( x, y) + (− x, − y) = ( x − x, y − y) = (0, 0) = the additive identity and also (− x, − y) + ( x, y) = (0, 0). The complex number (− x, − y) is called the negative of the complex number ( x, y) and we denote (− x, − y) by − ( x, y). Thus if z = ( x, y), then − z = − ( x, y) = (− x, − y). Cancellation law for addition in C: If z1 , z 2 , z 3 are any complex numbers, then z1 + z 3 = z 2 + z 3 ⇒ z1 = z 2 .

1.7 Multiplication of Complex Numbers If z1 = x1 + iy1 or ( x1 , y1 ) and z 2 = x2 + iy2 or ( x2 , y2 ) are any two complex numbers, then the product of z1 and z 2 written as z1 z 2 is defined by

T-8

z1 z 2 = ( x1 + iy1 ) ( x2 + iy2 ) = ( x1 x2 − y1 y2 ) + i ( x1 y2 + y1 x2 ) or

z1 z 2 = ( x1 , y1 ) ( x2 , y2 ) = ( x1 x2 − y1 y2 , x1 y2 + y1 x2 ).

Thus

(3 + 5i) (7 + 6i) = (3 × 7 − 5 × 6) + (3 × 6 + 5 × 7) i = − 9 + 53i,

or using the notation of ordered pairs, we have (3, 5) (7, 6) = (3 × 7 − 5 × 6, 3 × 6 + 5 × 7) = (−9, 53).

Properties of the Multiplication of Complex Numbers The multiplication of complex numbers is commutative, associative, admits of identity element and every non-zero complex number possesses multiplicative inverse. Commutativity of multiplication in C :

We have

z1 z 2 = z 2 z1 , for all complex

numbers z1 and z 2 . Associativity of multiplication in C : We have (z1 z 2 ) z 3 = z1 (z 2 z 3 ), for all complex numbers z1 , z 2 and z 3 . The complex number (1, 0) or 1 + i0 or simply 1 is the multiplicative identity since for every complex number ( x, y), we have ( x, y) (1, 0) = ( x . 1 − y . 0, x . 0 + y . 1) = ( x, y) = (1, 0) ( x, y). Multiplicative inverse: The complex number ( x, y) is called the multiplicative inverse of the complex number (a, b) if ( x, y) (a, b) = (1, 0) or simply 1. We have ( x, y) (a, b) = (1, 0) ⇒ ⇒

( xa − yb, xb + ya) = (1, 0) xa − yb = 1

and

xb + ya = 0.

The equations xa − yb = 1 and xb + ya = 0 give a b x= 2 , y=− 2 , 2 a +b a + b2 provided a2 + b 2 ≠ 0 which implies that a and b are not both zero i.e., (a, b) is a non-zero complex number. Thus every non-zero complex number possesses multiplicative inverse and the multiplicative inverse of the complex number (a, b) ≠ (0, 0) is the complex number  a −b   ⋅ ,  a2 + b 2 a2 + b 2   If z is a non-zero complex number, the multiplicative inverse of z is denoted by1 / z or z −1 . Cancellation law for multiplication in C : If z1 , z 2 , z 3 are complex numbers and z 3 ≠ 0, then z1 z 3 = z 2 z 3 ⇒ z1 = z 2 . Multiplication distributes addition in C : We have z1 (z 2 + z 3 ) = z1 z 2 + z1 z 3 , for all complex numbers z1 , z 2 and z 3 .

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1.8 Difference of Two Complex Numbers If z1 and z 2 are two complex numbers, we define z1 − z 2 = z1 + (− z 2 ). Thus if z1 = ( x1 , y1 ) and z 2 = ( x2 , y2 ), then z1 − z 2 = z1 + (− z 2 ) = ( x1 , y1 ) + (− x2 , − y2 ) = ( x1 − x2 , y1 − y2 ).

1.9 Division in C Definition: A complex number (a, b) is said to be divisible by a complex number (c , d ) if there exists a complex number ( x, y) such that ( x, y) (c , d ) = (a, b). We have ( x, y) (c , d ) = (a, b) ⇒ ( xc − yd, xd + yc ) = (a, b) ⇒ xc − yd = a and xd + yc = b. The equations xc − yd = a and xd + yc = b give ac + bd bc − ad x= 2 , y= 2 2 c +d c + d2 provided c 2 + d2 ≠ 0 which implies that c and d are not both zero. Thus division, except by (0, 0), is always possible in the set of complex numbers. If z1 and z 2 are two complex numbers such that z 2 ≠ 0 then the quotient of the complex numbers z1 and z 2 is defined by the relation z1 1 = z1 ⋅ = z1 (z 2 ) −1 . z2 z2

1.10 The Symbol i and its Powers It is customary to denote the complex number (0, 1) by the symbol i. With this notation i 2 = (0, 1)(0, 1) = (0 . 0 − 1 . 1, 0 . 1 + 1 . 0) = (−1, 0). But we have agreed to write the complex number (a, 0) simply as a. Therefore we have i 2 = − 1. Then i 3 = − i, i 4 = 1, i 5 = i, i 6 = − 1, and so on. Using the symbol i,we may write the complex number ( x, y) in our usual form x + iy. For, we have x + iy = ( x, 0) + (0, 1) ( y, 0) = ( x, 0) + (0 . y − 1 . 0, 0 . 0 + 1 . y) = ( x, 0) + (0, y) = ( x + 0, 0 + y) = ( x, y).

1.11 Conjugate of a Complex Number If z = x + iy is any complex number, then the complex number x − iy is called the conjugate of the complex number z and is written as z. Thus if z = 3 + 4i, then z = 3 − 4i.

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If z = x + iy is any complex number, then we define |z | = √ ( x 2 + y 2 ). Obviously |z | = |z|. The following results are obvious and should be remembered. (i)

Two complex numbers are equal if and only if their conjugates are equal i.e., z1 = z 2 if and only if z1 = z 2 .

(ii) ( z ) = z . (iii) We have z1 + z 2 = z1 + z 2 , z1 − z 2 = z1 − z 2 , z1 z 2 = z1 z 2  z1  z1 , provided z 2 ≠ 0.   =  z2  z2

and

(iv) If z = x + iy, then z + z = ( x + iy) + ( x − iy) = 2 x = 2 R (z ). (v)

A complex number z = x + iy is purely imaginary if and only if z + z = 0.

(vi) If z = x + iy, then z − z = x + iy − ( x − iy) = 2iy = 2i I (z ). (vii) A complex number z is purely real if and only if z − z = 0. (viii) If z = x + iy, then z z = ( x + iy) ( x − iy) = x 2 + y 2 = [√ ( x 2 + y 2 )]2 = |z |2 . Thus the product of two conjugate complex numbers is a purely real number which is always ≥ 0 i.e., which is never negative.

1.12 Modulus of a Complex Number Definition:

If z = ( x, y) or x + iy be any complex number, then the non-negative real

2

+ y 2 ) is called the modulus or absolute value of the complex number z and is

number √ ( x

denoted by |z | or mod z. Thus

|3 + 4i| = √ (32 + 42 ) = 5, |8 − 6i| = √ {82 + (−6)2 } = 10, |cos α + i sin α | = √ (cos 2 α + sin2 α) = 1,

and so on. Remember that the modulus of a complex number is equal to the positive square root of the sum of the squares of the real and imaginary parts of that complex number. The following results about the modulus of a complex number should be remembered : (i)

If z is any complex number, then |z | = |z |. For, if z = x + iy, then |z | = √ ( x 2 + y 2 ).

Also (ii)

|z | = | x − iy | = √ { x 2 + (− y)2 } = √ ( x 2 + y 2 ) = |z |. If z = x + iy be any complex number, then | z | = 0 ⇔ √ ( x 2 + y 2 ) = 0 ⇔ x 2 + y 2 = 0 ⇔ x = 0, y = 0

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⇔ z = 0 + i0 ⇔ z = 0. (iii) If z is any complex number, then z z = |z |2 . For, if z = x + iy, then z z = ( x + iy) ( x − iy) = x 2 + y 2 = { √ ( x 2 + y 2 )}2 = |z |2 . (iv) If z1 , z 2 are any two complex numbers, then |z1 z 2| = |z1||z 2| i. e., the modulus of a product is equal to the product of the moduli. (v) If z1 , z 2 are any two complex numbers and z 2 ≠ 0, then  z1  |z1|  =  z 2  | z 2| i. e., the modulus of a quotient is equal to the quotient of the moduli. (vi) If z = x + iy is any complex number, then R (z ) = x ≤ √ ( x 2 + y 2 ). Thus R (z ) ≤ |z |. Similarly I (z ) ≤ |z |.

1.13 Some Important Results about Complex Numbers (i) The separation of the complex number

a + ib c + id

into real and imaginary

parts i.e., to put it in the form A + iB, where A and B are real numbers. a + ib (a + ib) (c − id ) We have = , c + id (c + id ) (c − id ) multiplying the Nr and the Dr by the conjugate of the Dr =

(ac + bd ) + i (bc − ad ) c

2

+d

2

= A + iB, where

=

ac + bd 2

2

+i

bc − ad

c +d c 2 + d2 ac + bd bc − ad and B = 2 A= 2 ⋅ 2 c +d c + d2

Remember: To put the complex number (a + ib) / (c + id ) in the form A + iB, multiply its numerator and denominator by the conjugate of the denominator. (ii)

If z1 = cos α + i sin α and z 2 = cos β + i sin β, then z1 z 2 = (cos α + i sin α) (cos β + i sin β) = cos α cos β − sin α sin β + i (sin α cos β + cos α sin β) = cos (α + β) + i sin (α + β).

1.14 Modulus-Argument Form or Polar Standard Form or Trigonometric Form of a Complex Number Every non-zero complex number x + iy can always be put in the form r (cos θ + i sin θ), where r and θ are both real numbers.

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Let

x + iy = r (cos θ + i sin θ) = r cos θ + ir sin θ.

Then equating real and imaginary parts on both sides, we get and

x = r cos θ,

…(1)

y = r sin θ.

…(2)

Squaring and adding (1) and (2), we have x2 + y2 = r 2 or

r = + √ ( x 2 + y 2 ), taking the positive sign before the radical sign

or

r = |z |.

Thus r is known and is equal to the modulus of the complex number z. Substituting this value of r in (1) and (2), we have cos θ =

x √ (x

2

2

+ y )

and sin θ =

y √ (x

2

+ y2 )

⋅

…(3)

Whatever be the values of x and y, if they are not both zero, there is one and only one value of θ lying between − π and π which satisfies the equations (3) simultaneously. Thus θ is also determined. If n is any integer, then cos (2nπ + θ) = cos θ and sin (2nπ + θ) = sin θ. So there are an infinite number of values of θ satisfying the equations (3). Any value of θ satisfying the equations (3) is called an argument or amplitude of the complex number z and we write θ = arg z

or

amp z .

Thus every non-zero complex number x + iy can always be put uniquely in the form r (cos θ + i sin θ), where r is positive and − π < θ ≤ π. This trigonometrical form of a complex number is also called its polar standard form or modulus-amplitude form. We have seen that argument of a complex number is not unique. Thus arg z is a many-valued function. The value of argument which satisfies the inequality − π < θ ≤ π is called the principal value of the argument and it is unique. If θ is the principal value of arg z, then 2 nπ + θ, where n is any integer, is the general value of arg z and is represented by Arg z i. e., by writing the first letter of the word arg as capital. Thus Arg z = 2nπ + arg z , where Arg z stands for the general value and arg z for the principal value of the argument of z. Usually by argument we understand its principal value unless stated otherwise. The expression cos θ + i sin θ is sometimes written in short as cis θ. In this notation r (cos θ + i sin θ) is written as r cis θ. Remark 1:

From the equations (1) and (2) to determine r and θ, we also have tan θ = y / x i. e., θ = tan −1 ( y / x).

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But it should be noted that the value of tan −1 ( y / x) lying between −

1 1 π and π 2 2

is not always the principal value of the argument. For example, if z = − 1 + √ 3i = r (cos θ + i sin θ), then

r cos θ = − 1, r sin θ = √ 3.

∴

r 2 = (−1)2 + (√ 3)2 = 4, giving r = 2.

For r = 2, we have cos θ = −

1 1 , sin θ = √ 3. 2 2

The principal value of the argument θ lying between −π and π and found from these equations is 2 π / 3. But if we write tan θ = − √ 3 i. e., θ = tan −1 (− √ 3), then the 1 1 1 value of tan −1 (− √ 3) lying between − π and π is − π which is thus not the 2 2 3 principal value of arg z. So to find the principal value of arg z, we should not simply make use of the equation θ = tan −1 ( y / x), but we must see that the value of θ found from this equation also satisfies the equations (3). Remark 2: The following rule for locating the quadrant in which the principal value of arg z, where z = x + iy, lies should be committed to memory. 1 If x and y are both +ive, the principal value of arg z lies between 0 and π ; 2 1 if x and y are both –ive, the principal value of arg z lies between −π and − π ; 2 1 if x is +ive and y –ive, it lies between − π and 0 ; and if x is –ive and y + ive, it lies 2 1 between π and π . 2 Remark 3:

If a complex number z is given in any of the forms r (cos θ − i sin θ) or

r (sin θ + i cos θ) or r (sin θ − i cos θ), then we cannot have θ = arg z. We can have only if is put exclusively in the form θ = arg z z r (cos θ + i sin θ), where r is positive. Remark 4: Every complex number has a unique modulus and every non-zero complex number has an infinite number of arguments any two of them differing by an integral multiple of 2π. Two complex numbers are equal if and only if their moduli are equal and their arguments differ by integral multiples of 2π. Some particular cases of complex numbers expressed in polar standard form. The following particular cases should be remembered: (i) 1 = 1 + 0 i = cos 0 + i sin 0 , (ii) −1 = − 1 + 0i = cos π + i sin π , 1 1 (iii) i = 0 + 1i = cos π + i sin π , 2 2 1 1 and (iv) − i = 0 − 1i = cos (− π) + i sin (− π) . 2 2

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1.15 De Moivre’s Theorem Whatever be the value of n, positive or negative, integral or fractional, the value, or one of the values, of (cos θ + i sin θ) n is (cos nθ + i sin nθ). (Rohilkhand 2007) Corollary: For all values of n, integral or fractional, positive or negative, the value or one of the values of (cos θ − i sin θ) n is cos nθ − i sin nθ. Note: (i)

The students should note the following very carefully :

(sin θ + i cos θ) n ≠ sin nθ + i cos nθ. 1 1 But (sin θ + i cos θ) n = [cos ( π − θ) + i sin ( π − θ)] n 2 2 1 1 = cos n ( π − θ) + i sin n ( π − θ). 2 2

(ii) (cos θ + i sin φ) ≠ cos nθ + i sin nφ i. e., De Moivre’s theorem is applied only when the real and imaginary parts are cosine and sine of the same angle. (iii) Some authors use the notation cis θ to denote cos θ + i sin θ. In this notation, De Moivre’s theorem would be written as (cis θ) n = cis nθ. (iv) To separate the complex number (a + ib) n into real and imaginary parts, we put a = r cos θ and b = r sin θ so that

r = √ (a2 + b 2 ) and θ = tan −1 (b / a).

Then

(a + ib) n = (r cos θ + i r sin θ) n = r n (cos θ + i sin θ) n = r n (cos nθ + i sin nθ) = A + iB, where

A = r n cos nθ and

B = r n sin nθ.

Results to be remembered: (i) (cos α + i sin α) (cos β + i sin β)(cos γ + i sin γ ) … = cos (α + β + γ + …) + i sin (α + β + γ + …) i. e., the angles are added, (ii) (cos θ + i sin θ) n = cos nθ + i sin nθ,

(Rohilkhand 2006)

(iii) (cos θ − i sin θ) n = cos nθ − i sin nθ, (iv) (cos θ + i sin θ) − n = cos (− nθ) + i sin (− nθ) = cos nθ − i sin nθ, (v) (cos θ − i sin θ) − n = cos (− nθ) − i sin (− nθ) = cos nθ + i sin nθ, 1 (vi) = cos θ − i sin θ, cos θ + i sin θ and (vii)

1 = cos θ + i sin θ. cos θ − i sin θ

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Example 1: Solution:

Find real numbers A and B, if A + iB = We have 3 − 2i 7 + 4i

=

(3 − 2i) (7 − 4i) (7 + 4i) (7 − 4i)

3 − 2i 7 + 4i

⋅

multiplying the Nr. and Dr. by the

,

conjugate complex of the Dr. = =

21 − 12i − 14i + 8i2 2

49 − 16i 13 − 26i 65

=

=

(21 − 8) − 26i 49 + 16

[ ∵ i 2 = − 1]

13 26 1 2 − i = − i. 65 65 5 5

A + iB = (1 / 5) − (2 / 5) i.

∴

Equating real and imaginary parts, we get

A = 1 / 5, B = − 2 / 5.

Example 2: Find the modulus and principal argument of 1 + i. Solution:

Let 1 + i = r (cos θ + i sin θ).

Equating real and imaginary parts, we have and

1 = r cos θ,

…(1)

1 = r sin θ.

…(2)

Squaring and adding (1) and (2), we have r 2 = 1 + 1 = 2. r = + √ 2.

∴

Substituting the value of r in (1) and (2), we have cos θ = 1 / √ 2 and sin θ = 1 / √ 2. ∴

θ = π / 4.

Hence

1 + i = √ 2 [cos (π / 4) + i sin (π / 4)].

∴

modulus of 1 + i = √ 2 and principal argument = π / 4.

Example 3: Express −1 − i in the form r (cos θ + i sin θ). Solution:

Let −1 − i = r (cos θ + i sin θ).

Equating real and imaginary parts, we have and

−1 = r cos θ,

…(1)

−1 = r sin θ.

…(2)

Squaring and adding (1) and (2), we have r 2 = 1 + 1 = 2, or r = √ 2.

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Substituting the value of r in (1) and (2), we have cos θ = − 1 / √ 2 and sin θ = − 1 / √ 2. These give θ = − 3π / 4, choosing the value of θ which lies between − π and π. Hence

−1 − i = √ 2 [cos (− 3π / 4) + i sin (− 3π / 4)]

or

−1 − i = √ 2 [cos (3π / 4) − i sin (3π / 4)].

Example 4:

Express 1 + √ (−3) in the modulus-amplitude form.

Solution: Let

We have 1 + √ (−3) = 1 + i √ 3. 1 + i √ 3 = r (cos θ + i sin θ).

Equating real and imaginary parts, we have and

1 = r cos θ,

…(1)

√ 3 = r sin θ.

…(2)

Squaring and adding (1) and (2), we have r 2 = 1 + 3 = 4, or r = 2. Substituting the value of r in (1) and (2), we have cos θ = 1 / 2 and sin θ = √ 3 / 2. ∴

θ = π / 3.

Hence

1 + √ (−3) = 2 [cos (π / 3) + i sin (π / 3)].

Example 5: Express −1 − √ (−3) in the polar form. Solution: Let

Here

−1 − √ (−3) = − 1 − √ {3 (−1)} = − 1 − i √ 3.

−1 − i √ 3 = r (cos θ + i sin θ).

Equating real and imaginary parts, we have and

−1 = r cos θ,

…(1)

− √ 3 = r sin θ.

…(2)

Squaring and adding (1) and (2), we have r 2 = 1 + 3 = 4 so that r = 2. Dividing (2) by (1), we have tan θ = √ 3. This gives θ = − 2π / 3, choosing the value of θ lying between − π and π for which both cos θ and sin θ are negative. ∴

−1 − √ (−3) = 2 [cos (−2 π / 3) + i sin (−2π / 3)]

or

−1 − √ (−3) = 2 [cos (2π / 3) − i sin (2π / 3)].

¨

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2 E xponential, T rigonometric and H yperbolic F unctions of a C omplex V ariable ( S eparation into R eal and I maginary P arts )

2.1 The Exponential Function of a Complex Variable

W

e know that if x is any real number, then ex =1+

x x2 x3 + + + ……ad. inf. 1! 2 ! 3!

We shall take motivation from this expansion of e x as a power series in x, to define the exponential function e z of a complex variable z = x + iy, where x and y are real. Thus we define ez = 1 +

z z2 z3 + + + …… ad.inf. 1! 2 ! 3!

…(1)

It can be shown by D’ Alembert’s ratio test, that the power series (1) is absolutely convergent for all values of the complex variable z. If we denote the nth term of the series (1) by un , we have

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lim  un +1  lim  z n (n − 1)! lim | z |   = ⋅ = = 0,   n→ ∞ u n → ∞ n ! z n −1 n→ ∞ n    n  which is < 1. Therefore the series (1) is absolutely convergent and hence convergent for all values of the complex variable z. Thus the sum of the series (1) exists for all z and this justifies our definition of e z given in (1). Hence for the complex variable z, we define e z as the sum of the power series (1). The other ways to denote e z are E (z ) and exp (z ). Either of these symbols is read as ‘exponential z’. Putting z = 0 on both sides of (1), we see that e0 = 1. Remark: It should be noted that when z is complex, e z is only a symbol used to represent the sum of the series on the R.H.S. of (1). It does not mean that z 1 1 1 e z =  1 + + + + …… + ∞  ⋅   1! 2 ! 3 !

2.2 Index Law for the Exponential Functions Theorem : To prove that e i. e., Proof:

z1

.e

z2

=e

z1 + z2

E (z1 ) . E (z 2 ) = E (z1 + z 2 ). If z1 and z 2 be two complex numbers, then by the definition of E (z ), we

have E (z1 ) = 1 + and

E (z 2 ) = 1 +

z1 1! z2 1!

+ +

z12 2! z2 2 2!

+

z13

+

3! z2 3 3!

+ …… +

z1 n

+ …… +

n!

+ …… ∞

z2 n n!

+ …… ∞

…(1) …(2 )

Since the series (1) and (2) are absolutely convergent, therefore by Cauchy’s theorem on multiplication of absolutely convergent series, we have     z z 2 z n z z 2 z n E (z1 ) . E (z 2 ) = 1 + 1 + 1 + … + 1 + … 1 + 2 + 2 + … + 2 + … 1! 2! n! 1! 2! n!     =1+

z 2 z z z 2 1 (z1 + z 2 ) +  1 + 1 2 + 2  + … 1! 1! 2!  2!

z n z n −1 z 2 z n −2 z 2 2 z n + 1 + 1 ⋅ + 1 ⋅ +…+ 2 (n − 1)! 1! (n − 2)! 2 ! n!  n! 1 1 = 1 + (z1 + z 2 ) + (z12 + 2 z1 z 2 + z 2 2 ) + … 1! 2! n (n − 1) n −2 1  n + z1 + n z1 n −1 z 2 + z1 z2 2 + … + z2 n  n!  2! =1+

(z1 + z 2 ) 1!

+

(z1 + z 2 )2 2!

+…+

(z1 + z 2 ) n n!

+…

  +… 

  + …

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= E (z1 + z 2 ), by def. of E (z ). The above result may also be written as exp (z1 ) . exp (z 2 ) = exp (z1 + z 2 ). Deductions: (i)

If z1 , z 2 , …, z n be n complex numbers, then from the above law,

we see by induction that E (z1 ) . E (z 2 ) . … . E (z n ) = E (z1 + z 2 + … + z n ). If we put z1 = z 2 = … = z n = z (say), we have [ E (z )] n = E (nz ), n being a positive integer. This result may also be written as [e z ] n = e nz or [exp (z )] n = exp (nz ). (ii) If z be any complex number, then e z . e −z = e

z −z

= e 0 = 1.

From this result, it follows that e z ≠ 0 for all z. For if there were such a value of z, then since e − z would exist for this value of z, we would have 0 = 1, which is absurd.

2.3 Trigonometrical Functions or Circular Functions of a Complex Variable We know that if x is any real number, then

and

cos x = 1 −

x2 x4 x2 n + − … + (−1) n + … ad. inf. 2! 4! (2n)!

sin x = x −

x3 x5 x2 n +1 + − … + (−1) n + … ad.inf. 3! 5! (2n + 1)!

We shall take motivation from these expansions of cos x and sin x as power series in x, to define cos z and sin z, when z is a complex number. So if z = x + iy is a complex variable, we define

and

cos z = 1 −

z2 z4 z 2n + − … + (−1) n + …ad. inf. 2! 4! (2n)!

sin z = z −

z3 z5 z 2 n +1 + − … + (−1) n + … ad. inf. 3! 5! (2n + 1)

Our definitions are sensible because both the series used to define cos z and sin z are absolutely convergent for all z. Replacing z by −z in the above definitions, we find that cos (− z ) = cos z , and sin (− z ) = − sin z . The other circular functions for a complex variable are defined in the same way as those for a real variable. Thus we define

T-20

tan z = Remark:

sin z

, cot z =

cos z

cos z sin z

, sec z =

1 1 and cosec z = ⋅ cos z sin z

For a complex variable z = x + iy, we have defined e z , cos z , sin z etc. in

such a way that if we take z = x + i0 = x

(i. e., real),

then these definitions give results which are in sound agreement with those for a real variable.

2.4 Euler’s Theorem Theorem: If θ be real or complex, then prove that Proof:

e iθ = cos θ + i sin θ.

If z is any complex number, then by definition of e z , we have ez = 1 +

z z2 z2 + + +… 1! 2 ! 3!

Replacing z by i θ, we have ei θ = 1 +

iθ 1!

+

(i θ)2

+

2!

(i θ)3 3!

+

(i θ)4 4!

+…

    θ2 θ4 θ3 θ5 = 1 − + − … + i θ − + − … 2! 4! 3! 5!     = cos θ + i sin θ, by definitions of cos θ and sin θ. Thus

e

iθ

= cos θ + i sin θ.

…(1)

This result is known as Euler’s theorem. Replacing θ by − θ in (1), we get e − i θ = cos (− θ) + i sin (− θ) or

e − i θ = cos θ − i sin θ.

…(2)

From (1) and (2) on addition and subtraction, we get e

iθ

+ e − i θ = 2 cos θ,

and

e

iθ

− e − i θ = 2i sin θ.

∴

cos θ =

e iθ + e − iθ 2

and

sin θ =

e iθ − e − iθ 2i

…(3)

These results are known as Euler’s exponential values of cos θ and sin θ. From these exponential values of cos θ and sin θ, we see that tan θ =

sin θ cos θ

=

e i θ − e −i θ i (e i θ + e − i θ )

,

cot θ = i

e i θ + e −i θ e i θ − e −i θ

, etc.

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Putting θ = 0 in the formulae (3), we get 1 1 1 cos 0 = (e0 + e0 ) = (1 + 1) = ⋅ 2 = 1, 2 2 2 and

sin 0 = 0.

2.5 Periodicity of Functions (Rohilkhand 2005; Kanpur 05)

(a)

z

Prove that e is a periodic function, where z is a complex quantity.

Proof: If z = x + iy, then e z = e x + iy = e x . e iy = e x (cos y + i sin y), by Euler’s theorem = e x [cos (2nπ + y) + i sin (2nπ + y)], x

= e .e Hence e

z

i (2 nπ + y)

=e

x + iy + 2 nπi

=e

z + 2 nπ i

where n is any integer .

is a periodic function of period 2πi.

(b) Prove that sin z, cos z, tan z etc. are periodic functions, where z is a complex quantity. (Kanpur 2005)

Proof: We have cos (z + 2nπ) = cos z cos 2nπ − sin z sin 2nπ = cos z , n being any integer sin (z + 2nπ) = sin z cos 2nπ + cos z sin 2nπ

and

= sin z , n being any integer sin (z + nπ) tan (z + nπ) = cos (z + nπ) =

sin z cos nπ + cos z sin nπ cos z cos nπ − sin z sin nπ

= tan z , n being any integer. From these we conclude that cos z and sin z are periodic functions of period 2π and tan z is a periodic function of period π.

2.6 De Moivre’s Theorem for Complex Argument We have already mentioned De Moivre’s theorem for real values of θ. Now we extend that theorem for complex values of θ. We have (e i θ ) n = e i nθ , whether θ be real or complex ∴

(cos θ + i sin θ) n = cos nθ + i sin nθ, by Euler’s theorem.

This shows that De Moivre’s Theorem is true whether θ be real or complex.

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2.7 Standard Trigonometrical Results for Complex Arguments Since our trigonometric functions of a complex variable are generalizations of trigonometric functions of a real variable, therefore most of our results for trigonometric functions of a real variable also hold good for trigonometric functions of a complex variable. To prove that for all values of x, y, real or complex, the following are true : (i) cos 2 x + sin2 x = 1, (ii) cos (− x) = cos x, (iii) cos 2 x = cos 2 x − sin2 x,

(iv) sin 3 x = 3 sin x − 4 sin3 x,

cos 3 x = 4 cos 3 x − 3 cos x, 1 1 (vi) sin x + sin y = 2 sin ( x + y) cos ( x − y), 2 2 1 1 (vii) cos x − cos y = 2 sin ( x + y) sin ( y − x) 2 2 (viii) sin ( x ± y) = sin x cos y ± cos x sin y, (ix) cos ( x ± y) = cos x cos y + sin x sin y. (v)

Proof: (i) To prove that cos 2 x + sin2 x = 1. L.H.S. = cos 2 x + sin2 x 2

= = = = Aliter:

2

 e ix + e − ix   e ix − e − ix  +     [Refer article 2.4] 2 2i     1 [(e ix + e − ix )2 − (e ix − e − ix )2 ] [ ∵ i 2 = − 1] 4 1 2 ix [(e + e −2 ix + 2e ix . e − ix ) − (e2 ix + e −2 ix − 2e ix . e − ix )] 4 1 [4e ix . e − ix ] = e i x − ix = e0 = 1 = R.H.S. 4

We have cos 2 x + sin2 x = (cos x + i sin x)(cos x − i sin x) = e ix e − ix = e0 = 1.

(ii)

To prove that cos ( − x) = cos x . L.H.S. = cos (− x) = e − ix + e i x

e i (− x) + e − i (− x) e ix

[By article 2.4]

2 + e −i x

= cos x = R.H.S. 2 2 (iii) To prove that cos 2 x = cos 2 x − sin2 x. =

=

2

2

 e ix + e − ix   e ix − e − ix  R.H.S. = cos 2 x − sin2 x =   −  , 2 2i     by Euler’s exponential values of cos x and sin x

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1 [(e ix + e − ix )2 + (e ix − e − ix )2 ] 4 1 = [(e2 ix + e −2 ix + 2e ix . e − ix ) + (e2 ix + e −2 ix − 2e 4 1 1 = [2 (e2 ix + e −2 ix )] = (e2 ix + e −2 ix ) 4 2 1 i (2 x) = [e + e − i (2 x) ] = cos 2 x = R.H.S. 2

[ ∵ i 2 = − 1]

=

ix

. e − ix )]

(iv) To prove that sin 3 x = 3 sin x − 4 sin3 x. L.H.S. = sin 3 x = ix 3

(e )

=

i (3 x)

e

− e − i(3 x)

=

2i

− (e

− ix 3

)

=

2i

(e

ix

e 3 i x − e −3 i x 2i

− e

− ix 3

) + 3e ix . e − ix (e

ix

− e − ix )

2i (Note) [ ∵ a3 − b 3 = (a − b)3 + 3ab (a − b)]  e ix − e − ix   ∵ sin x =  2i  

=

1 [(2i sin x)3 + 3. e0 .(2i sin x)] 2i

=

1 1 [8i3 sin3 x + 6i sin x] = [2i (−4 sin3 x + 3 sin x)] 2i 2i

= − 4 sin3 x + 3 sin x = 3 sin x − 4 sin3 x = R.H.S. (v)

To prove that cos 3 x = 4 cos 3 x − 3 cos x. L.H.S. = cos 3 x = =

1 [e 2

i(3 x)

+ e − i(3 x) ] =

1 [(e ix )3 + (e − ix )3 ] 2

1 [(e ix + e − ix )3 − 3e ix . e − ix (e ix + e − ix )] 2 [ ∵ a3 + b 3 = (a + b)3 − 3ab (a + b)]

1  ∵ cos x = 1 (e [(2 cos x)3 − 3. e0 .(2 cos x)]  2 2 1 = [8 cos 3 x − 6 cos x] = 4 cos 3 x − 3 cos x = R.H. S. 2 1 1 (vi) To prove that sin x + sin y = 2 sin ( x + y) cos ( x − y). 2 2 =

ix

+ e − ix ) 

L.H.S. = sin x + sin y =

e

ix

− e − ix 2i

= (1 / 2i) [{ e

+

e

iy

− e − iy 2i

=

i [( x + y) /2 + ( x − y) /2 ]

1 ix iy [(e − e − ix ) + (e − e − iy )] 2i − e − i [( x + y) /2 + ( x − y) /2 ]}

+ { e i [( x + y) /2 − ( x − y) /2 ] − e − i [( x + y) /2 − ( x − y) /2 ]}] (Note)

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= (1 / 2i) [{e i [( x + y) /2 + ( x − y) /2 ] + e i [( x + y) /2 − ( x − y) /2 ] − { e − i [( x + y) /2 + ( x − y) /2 ] + e − i [( x + y) /2 − ( x − y) /2 ]}] (Note) = (1 / 2i) [e i ( x + y) /2 { e i ( x − y) /2 + e − i ( x − y) /2 } − e − i ( x + y) /2 { e − i ( x − y) /2 + e i ( x − y) /2 }] =

1 i ( x − y) /2 [e + e − i ( x − y) /2 ] [e i ( x + y) /2 − e − i ( x + y) /2 ] 2i

=

1 2i

 x − 2 cos  2  

y   x +   ⋅ 2i sin     2

y   

1 1 ( x + y) cos ( x − y) = R.H. S. 2 2 1 1 (vii) To prove that cos x – cos y = 2 sin ( x + y) . sin ( y – x) . 2 2 1 1 R.H.S. = 2 sin ( x + y) . sin ( y − x) 2 2 = 2 sin

 e i ( x + y) /2 − e − i ( x + y) /2   e i ( y − x) /2 − e − i ( y − x) /2  =2   2i 2i     1 = − [e i ( x + y + y − x) /2 − e i ( x + y − y + x) /2 − e − i( x + y − y + x) /2 2 + e − i ( x + y + y − x) / 2 ] (simply by multiplication) 1 1 1 = − [e iy − e ix − e − ix + e − iy ] = (e ix + e − ix ) − (e iy + e − iy ) 2 2 2 = cos x − cos y = L.H.S. (viii) To prove that sin ( x + y) = sin x cos y + cos x sin y. R.H.S. =

e ix − e − ix 2i

= (1 / 4i) [{ e

×

e iy + e − iy

i ( x + y)

2 +e

+

i ( x − y)

e ix + e − ix 2 −e

×

− i ( x − y)

e iy − e − iy 2i −e

− i ( x + y)

}

+ { e i ( x + y) − e i ( x − y) + e − i ( x − y) − e − i ( x + y)}] =

e i ( x + y) − e − i ( x + y) 1 [2e i ( x + y) − 2e − i ( x + y) ] = 4i 2i

= sin ( x + y) = L.H.S. Similarly, sin ( x − y) = sin x cos y − cos x sin y. ∴

sin ( x ± y) = sin x cos y ± cos x sin y.

(ix) To prove that cos ( x ± y) = cos x cos y ∓ sin x sin y. We have cos x cos y + sin x sin y

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=

e ix + e − ix

⋅

e iy + e − iy

+

e ix − e − ix

⋅

e iy − e − iy

2 2 2i 2i 1 i ( x + y) i ( x − y) − i ( x − y) − i ( x + y) = [e +e +e +e ] 4 1 − [e i ( x + y) − e i ( x − y) − e − i ( x − y) + e − i ( x + y) ] [ ∵ i 2 = − 1] 4 1 1 = [2e i ( x − y) + 2e − i ( x − y) ] = [e i ( x − y) + e − i ( x − y) ] = cos ( x − y). 4 2 Similarly we can prove that cos x cos y − sin x sin y = cos ( x + y). cos ( x ± y) = cos x cos y + sin x sin y.

∴

Remark : If u and v are any complex numbers, we can easily prove that 2 sin u cos v = sin (u + v) + sin (u − v), 2 cos u sin v = sin (u + v) − sin (u − v) 2 cos u cos v = cos (u + v) + cos (u − v), 2 sin u sin v = cos (u − v) − cos (u + v).

Example 1: Show that exp (± i π / 2) = ± i.

(Avadh 2014)

Solution: Since exp (± iθ) = cos θ ± i sin θ, we have exp (± i π / 2) = cos (π / 2) ± i sin (π / 2) = 0 ± i.1 = ± i. Example 2: Prove that { sin (α − θ) + e − α i sin θ} n = sin n −1 α { sin (α − nθ) + e − αi sin nθ}. Solution:

L.H.S. = {sin α cos θ − cos α sin θ + (cos α − i sin α) sin θ} n = {sin α cos θ − i sin α sin θ} n = sin n α (cos θ − i sin θ) n = sin n α (cos nθ − i sin nθ), by De Moivre’s theorem

and

R.H.S. = sin n −1 α {sin (α − nθ) + e − α i sin nθ} = sin n −1 α {sin α cos nθ − cos α sin nθ + (cos α − i sin α) sin nθ} = sin n −1 α {sin α cos nθ − i sin α sin nθ} = sin n −1 α . sin α . (cos nθ − i sin nθ) = sin n α (cos nθ − i sin nθ).

∴

L.H.S. = R.H.S.

2.8 Hyperbolic Functions Definition: For all values of x, real or complex, the quantity

e x + e− x 2

is called the

hyperbolic cosine of x and is written as cosh x. Similarly the quantity

e x − e− x 2

is

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called the hyperbolic sine of x and is written as sinh x (read as shine x). Thus we define e x − e−x

sinh x =

e x + e−x

and cosh x =

2

2

⋅

The hyperbolic tangent, cotangent, secant and cosecant are obtained from hyperbolic cosine and sine just as circular tangent, cotangent, secant and cosecant are obtained from circular cosine and sine. Thus sinh x

tanh x =

coth x = sech x = and

=

cosh x cosh x sinh x

(e x − e − x ) / 2 (e x + e − x ) / 2 (e x + e − x ) / 2

=

(e x − e − x ) / 2

e x − e− x

=

=

e x + e− x e x + e− x e x − e− x

1 1 2 = x = x − x cosh x (e + e ) / 2 e + e − x

cosech x =

1 1 2 = x = x ⋅ − x sinh x (e − e ) / 2 e − e − x

2.9 Relations between Hyperbolic and Circular Functions (Bundelkhand 2006)

The hyperbolic functions can be expressed in terms of circular functions as follows: We have sin ( ix) =

e i (i x) − e − i (i x)

= i.

2i e

−x

−e

2i

=

x

ei

2

x

e −e

= i.

2

x

2i x

2

− e −i

=

e− x − e x 2i

−x

[ ∵ i 2 = − 1]

,

2

= i sinh x. From this relation sin (ix) = i sinh x, 1 i we have sinh x = sin (ix) = 2 sin (ix) = − i sin (ix). i i Similarly

cos ( ix ) = =

Now

tan ( ix ) = cot ( ix ) =

e i (i x) + e − i (i x) 2 e

−x

+e

2 sin (ix) cos (ix)

x

=

ei

x

= =

e +e

2 i sinh x cosh x

2

x

+ e −i

2

x

2 −x

= cosh x.

(Lucknow 2007, 08)

= i tanh x,

1 1 −i = = = − i coth x, tan (ix) i tanh x tanh x

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sec ( ix ) = cosec ( ix ) =

and

1 1 = = sech x, cos (ix) cosh x 1 1 −i = = = − i cosech x. sin (ix) i sinh x sinh x

Similarly we can prove that sinh (ix) = i sin x ;

(Rohilkhand 2005)

cosh (ix) = cos x ; tanh (ix) = i tan x and coth (ix) = − i cot x, etc.

2.10 Properties of Hyperbolic Functions Formulae involving hyperbolic functions can be easily deduced with the help of the above relations. A list of some such formulae is given below. (a)

sinh 0 = 0, cosh 0 = 1, tanh 0 = 0

(b)

cosh2 x − sinh2 x = 1

(c)

sinh 2 x = 2 sinh x cosh x

(Bundelkhand 2008)

(d) cosh 2 x = cosh2 x + sinh2 x = 1 + 2 sinh2 x = 2 cosh2 x − 1 (Lucknow 2007, 08)

(e)

sec h2 x = 1 − tanh2 x

(f)

tanh 2 x = 2 tanh x / (1 + tanh2 x)

(g)

(i) sinh ( x + y) = sin x cosh y + cosh x sinh y (ii) cosh ( x + y) = cosh x cosh y + sinh x sinh y

(h)

x

e = cosh x + sinh x and e

−x

(Rohilkhand 2006)

= cosh x − sinh x

3

(i)

sinh 3 x = 3 sinh x + 4 sinh x

(j)

cosh 3 x = 4 cosh3 x − 3 cosh x

(k)

tanh 3 x =

3 tanh x + tanh3 x 1 + 3 tanh2 x

⋅

Proof: (a) sinh 0 = 0 , cosh 0 =1, tanh 0 = 0 . By definition, 1 x 1 x (e − e − x ) and cosh x = (e + e − x ). 2 2 1 0 1 −0 sinh 0 = (e − e ) = (1 − 1) = 0 2 2 1 1 1 cosh 0 = (e0 + e −0 ) = (1 + 1) = . 2 = 1. 2 2 2 sinh 0 0 tanh 0 = = = 0. cosh 0 1 sinh x = ∴ and Also

[ ∵ e0 = e −0 = 1]

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(b)

cosh2 x − sinh2 x = 1

(Kashi 2014) x

−x

2

x

2

−x

e + e  e − e   −   We have cosh2 x − sinh2 x =  2 2     1 1 1 1 = (e2 x + 2 + e −2 x ) − (e2 x − 2 + e −2 x ) = + = 1. 4 4 2 2 (c)

sinh 2 x = 2 sinh x cosh x.

The L.H.S. = sinh 2 x = (1 / i) sin (2ix)

(Note)

= (1 / i) 2 sin (ix) cos (ix) = (1 / i) 2 . (i sinh x) . (cosh x) = 2 sinh x cosh x = R.H.S. (d) cosh 2 x = cosh2 x + sinh2 x =1 + 2 sinh2 x = 2 cosh2 x − 1. We have cosh 2 x = cos (2ix) = cos 2 (ix) 2

2

(Note) 2

= cos (ix) − sin (ix) = cosh x − i

2

2

sinh x

= cosh2 x + sinh2 x 2

…(1) 2

2

= 1 + sinh x + sinh x = 1 + 2 sinh x. Also

…(2)

cosh 2 x = cosh2 x + sinh2 x = cosh2 x + (cosh2 x − 1) = 2 cosh2 x − 1.

(e)

2

…(3)

2

sech x =1 − tanh x.

We have cosh2 x − sinh2 x = 1, by property (b). Dividing by cosh2 x, we get 1 − tanh2 x = sech2 x. Similarly, coth2 x − 1 = cosech2 x. (f)

tanh 2 x = 2 tanh x/ (1 + tanh2 x) .

We have i tanh 2 x = tan (2ix). ∴

tanh 2 x = (1 / i) tan (2ix) = (1 / i) tan 2(ix) 2i tanh x 1 2 tan ix 1 = ⋅ = ⋅ 2 i (1 − tan ix) i (1 − i 2 tanh2 x)

(Note)

= 2 tanh x / (1 + tanh2 x). Similarly, we find sinh 2 x = (g)

2 tanh x 1 − tanh2 x

and

cosh 2 x =

1 + tanh2 x 1 − tanh2 x

⋅

(i) sinh ( x +y) = sinh x cosh y + cosh x sinh y. 1 x 1 1 1 (e − e − x ). (e y + e − y ) + (e x + e − x ). (e y − e − y ) 2 2 2 2 1 x+y x−y −( x − y) −( x + y) x+y = [e +e −e −e +e − e x − y + e −( x − y) 4

R.H.S. =

− e −( x + y) ]

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=

1 1 [2e( x + y ) − 2e −( x + y ) ] = [e( x + y ) − e −( x + y ) ] 4 2

= sinh ( x + y) = L.H.S. (ii) cosh ( x +y) = cosh x cosh y + sinh x sinh y.

(Rohilkhand 2006)

L.H.S. = cosh ( x + y) = cos i ( x + y) = cos (ix + iy) = cos (ix) cos (iy) − sin (ix) sin (iy) = cosh x cosh y − i sinh x . i sinh y = cosh x cosh y − i 2 sinh x sinh y = cosh x cosh y + sinh x sinh y = R.H. S. e = cosh x + sinh x and e − x = cosh x − sinh x. 1 1 We have cosh x = (e x + e − x ) ; sinh x = (e x − e − x ). 2 2 1 x 1 Adding, we get cosh x + sinh x = (e + e − x + e x − e − x ) = . 2e x = e x 2 2 (h)

x

and subtracting, we get cosh x − sinh x = (i)

1 x 1 (e + e − x − e x + e − x ) = . 2e − x = e − x . 2 2

sinh 3 x = 3 sinh x + 4 sinh3 x. L.H.S. = sinh 3 x = (1 / i) sin (3ix) = (1 / i) sin 3 (ix) = (1 / i) [3 sin (ix) − 4 sin3 (ix)] = (1 / i) [3 (i sinh x) − 4 (i sinh x)3 ]

[ ∵ sin (ix) = i sinh x ]

3

[∵ i 2 = − 1 ; i3 = − i]

= (1 / i) [3i sinh x + 4i sinh x] = 3 sinh x + 4 sinh3 x = R.H. S. (j)

cosh 3 x = 4 cosh3 x − 3 cosh x. L.H.S. = cosh 3 x = cos (3ix) = cos 3 (ix) = 4 cos 3 (ix) − 3 cos (ix) = 4 cosh3 x − 3 cosh x.

(k)

3

[ ∵ cos (ix) = cosh x] 2

tanh 3 x = (3 tanh x + tan x) / (1 + 3 tanh x) . 1 L.H.S. = tanh 3 x = (1 / i) tan (3ix) =    i 3i tanh x − i3 tanh3 x    2 2  1 − 3i tanh x 

(Kanpur 2008)

3 tan (ix) − tan3 (ix)   2  1 − 3 tan (ix) 

=

1 i

=

3 3 1 3i tanh x + i tanh x  3 tanh x + tanh x = = R.H.S.   i  1 + 3 tanh2 x 1 + 3 tanh2 x 

[ ∵ tan (ix) = i tanh x]

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2.11 Expansions in Series for sinh x and cosh x We have sinh x =

Thus Also

Thus

1 x [e − e − x ] 2

=

1 2

    x2 x3 x2 x3 + + … − 1 − x + − + …  1 + x + 2! 3! 2! 3!     

=

1 2

  2 x3 2 x5 x3 x5 + + … = x + + + … ad. inf. 2 x + 3! 5! 3! 5!  

sinh x = x + cosh x =

x3 x5 + + ... ad. inf . 3! 5!

1 x [e + e − x ] 2

=

1 2

    x2 x3 x2 x3 + + … + 1 − x + − + …  1 + x + 2! 3! 2! 3!     

=

1 2

  2 x2 2 x4 x2 x4 + + … = 1 + + + … ad.inf. 2 x + 2! 4! 2! 4!  

cosh x = 1 +

x2 x4 + + ... ad. inf. 2! 4!

Remark: The students should note that to change a formula for the circular functions into a formula for the hyperbolic functions, we must replace cos x by cosh x and sin x by i sinh x i. e.,we must write cosh2 x for cos 2 x and − sinh2 x for sin2 x etc.

2.12 Periods of Hyperbolic Functions By Euler’s theorem, we know that e2 nπ i = cos 2nπ + i sin 2nπ = 1 + i . 0 = 1, n being any integer and

e −2 nπ i = cos 2nπ − i sin 2nπ = 1 − i . 0 = 1.

∴

sinh ( x + 2nπi) =

Also,

1 x +2 n π i [e − e −( x + 2 nπ i) ], by definition 2 1 1 = [e x e2 nπ i − e − x e −2 nπ i ] = [e x − e − x ] 2 2 [ ∵ e ± 2 nπ i = 1, n being any integer]

= sinh x. 1 cosh ( x + 2nπi) = [e x + 2 nπ i + e − ( x + 2 nπ i) ] 2 1 = [e x . e2 nπ i + e − x . e −2 nπ i ] 2

[By definition]

T-31

=

1 x [e + e − x ] 2

[∵ e

± 2 nπ i

= 1]

= cosh x. It follows that sinh x and cosh x are periodic functions and both have a period 2πi. (Lucknow 2009)

Hence hyperbolic functions differ from circular functions in having no real period. Their periods are imaginary. sinh ( x + nπi) Again, tanh ( x + nπi) = cosh ( x + nπi) 1 ( x + nπ i) [e − e −( x + nπ i) ] = 2 1 ( x + nπ i) [e + e −( x + nπ i) ] 2 =

= = Thus

[By def.]

e nπ i [e x − e − x . e −2 nπ i ] e nπ i [e

x

(Note)

+ e − x . e −2 nπ i ]

[e x − e − x ]

[ ∵ e −2 nπ i = 1]

[e x + e − x ] sinh x cosh x

= tanh x.

tanh ( x + nπi) = tanh x,

i. e., tanh x is a periodic function with period πi. Therefore, sinh x and cosh x are periodic functions of period 2πiand tanh x is a periodic function of period π i. Remark: Note that the hyperbolic functions have imaginary periods while the circular functions have real periods.

1 Example 3: Show that sinh ( x + y) cosh ( x − y) = (sinh 2 x + sinh 2 y). 2 (Meerut 2008, 12, 12B)

Solution: The L.H.S. = sinh ( x + y) . cosh ( x − y) 1 1 = [e( x + y) − e −( x + y) ] . [e( x − y) + e −( x − y) ] 2 2 1 2x −2 y 2y = [e − e +e − e −2 x ] 4 1 1 1 = . [ (e2 x − e −2 x ) + (e2 y − e −2 y )] 2 2 2 1 = [sinh 2 x + sinh 2 y] = R.H.S. 2

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Example 4:

Show that cos (α + iβ) + i sin (α + iβ) = e − β (cos α + i sin α).

(Meerut 2012B)

Solution: The L.H.S. = cos α cos iβ − sin α sin iβ + i sin α cos iβ + i cos α sin iβ = cos α (cos iβ + i sin iβ) + i sin α (cos iβ + i sin iβ) = (cos iβ + i sin iβ) (cos α + i sin α) = (cosh β − sinh β) (cos α + i sin α) [ ∵ cos iβ = cosh β, sin iβ = i sinh β] = e − β (cos α + i sin α) = R.H. S. Note:

Similarly we can prove that cos (α − iβ) − i sin (α − iβ) = e −β (cos α − i sin α).

Example 5: If cosh α = sec θ, prove that tanh2

1 1 α = tan2 θ. 2 2 (Avadh 2010; Meerut 13, 13B)

Solution: We have cosh α = sec θ. cosh α 1 ∴ = ⋅ 1 cos θ Applying componendo and dividendo, we get 1 2 sinh2 α 2 sin2 cosh α − 1 1 − cos θ 2 = = or 2 1 cosh α + 1 1 + cos θ 2 cosh α 2 cos 2 2 1 1 or tanh2 α = tan2 θ. 2 2 Note: The componendo and dividendo is that if a−b c −d a c = , then = ⋅ b d a+b c +d

1 θ 2 1 θ 2

Example 6: If tan θ = tanh x cot y and tan φ = tanh x tan y, sin 2θ cosh 2 x + cos 2 y show that = ⋅ sin 2φ cosh 2 x − cos 2 y Solution:

L.H.S. =

sin 2θ sin 2φ

=

2 tan θ / (1 + tan2 θ) 2 tan φ / (1 + tan2 φ)

tan θ 1 + tan2 φ ⋅ tan φ tan φ 1 + tan2 θ 1 + tan2 θ tanh x cot y 1 + tanh2 x tan2 y = × , tanh x tan y 1 + tanh2 x cot 2 y putting the given val ues of tan θ and tan φ cos 2 y cosh2 x cos 2 y + sinh2 x sin2 y = ⋅ sin2 y cosh2 x cos 2 y cosh2 x sin2 y × cosh2 x sin2 y + sinh2 x cos 2 y

=

tan θ

×

1 + tan2 φ

=

T-33

=

= =

cosh2 x cos 2 y + sinh2 x sin2 y cosh2 x sin2 y + sinh2 x cos 2 y (2 cosh2 x) (2 cos 2 y) + (2 sinh2 x) (2 sin2 y)

(Note)

(2 cosh2 x) (2 sin2 y) + (2 sinh2 x) (2 cos 2 y) (1 + cosh 2 x)(cos 2 y + 1) + (cosh 2 x − 1)(1 − cos 2 y) (1 + cosh 2 x)(1 − cos 2 y) + (cosh 2 x − 1)(1 + cos 2 y)

[ ∵ 2 cosh2 x = 1 + cosh 2 x ; 2 sinh2 x = cosh 2 x − 1] =

2 (cosh 2 x + cos 2 y) 2 (cosh 2 x − cos 2 y)

=

cosh 2 x + cos 2 y

= R.H.S.

cosh 2 x − cos 2 y

Comprehensive Exercise 1 1.

Prove that sin (α + nθ) − e iα sin nθ = e − i nθ sin α.

2.

Prove that {sin (α + θ) − e αi sin θ} n = sin n α e − nθi .

3.

Show that

4.

Show that tanh ( x + y) =

5.

Show that

1 + tan h x 1 − tan h x

= cos h 2 x + sin h 2 x. tanh x + tanh y 1 + tanh x tanh y

(Gorakhpur 2005)

⋅

1 1 ( x + y) cosh ( x − y) 2 2 1 1 (ii) cosh x − cosh y = 2 sinh ( x + y) sinh ( x − y). 2 2 (i) sinh x + sinh y = 2 sinh

6.

If α = log tan [π / 4 + θ / 2], prove that (i) tanh α / 2 = tan θ / 2 ;

(ii) cos θ cosh α = 1 ;

(iii) sinh α = tan θ ; (Bundelkhand 2006)

(iv) tanh α = sin θ ;

3

5

(v) If α = θ + a3 θ + a5 θ + …, show that θ = α − a3 α 3 + a5 α 5 − ... (Kanpur 2008)

7.

If α, β be the imaginary cube roots of unity, prove that  √3 √3 αe αx + βeβx = − e − x /2  √ 3 sin x + cos 2 2 

 x ⋅ 

2.13 Separation into Real and Imaginary Parts By separation or resolving into real and imaginary parts, we mean to put a complex quantity in the form x + iy where x and y are real quantities.

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When we are given a fraction to separate into real and imaginary parts whose numerator and denominator are both complex functions, we multiply the numerator and denominator by the conjugate of the denominator and thus make the denominator real. The conjugate complex is obtained by putting −i for i , i. e., x + iy and x − iy are conjugate complex quantities.

Example 7: Resolve into real and imaginary parts : (i)

sin (α ± iβ) (Bundelkhand 2004)

(ii) cos (α ± iβ)

(iii) tan (α + iβ) Solution: (i)

(Rohilkhand 2007)

(Bundelkhand 2006; Meerut 12)

We have

sin (α ± iβ) = sin α cos (iβ) ± cos α sin (iβ) = sin α cosh β ± i cos α sinh β. [∵ sin (iβ) = i sinh β and cos (iβ) = cosh β] (ii)

cos (α ± iβ) = cos α cos (iβ) + sin α sin (iβ) = cos α cosh β + sin α (i sinh β) = cos α cosh β + i sin α sinh β.

Remark: From the expressions for sin (α + iβ) and sin (α − iβ) into real and imaginary parts, we observe that sin (α − iβ) is the conjugate complex of sin (α + iβ).Similarly cos (α + iβ) and cos (α − iβ) are conjugate complex numbers. sin (α + iβ) (iii) We have tan (α + iβ) = ⋅ cos (α + iβ) In this case, the denominator is a complex quantity so we have to multiply the numerator and denominator by the conjugate of the denominator. The conjugate complex of cos (α + iβ) is cos (α − iβ). Therefore multiplying the Nr. and Dr. by cos (α − iβ), we have sin (α + iβ) 2 cos (α − iβ) tan (α + iβ) = × (Note) cos (α + iβ) 2 cos (α − iβ) = =

sin [(α + iβ) + (α − iβ)] + sin [(α + iβ) − (α − iβ)] cos [(α + iβ) + (α − iβ)] + cos [(α + iβ) − (α − iβ)] sin (2α) + sin (2iβ) cos (2α) + cos (2iβ)

=

sin 2α + i sinh 2β cos 2α + cosh 2β

    sin 2α sinh 2β =  + i ⋅ cos 2 α + cosh 2 β cos 2 α + cosh 2 β    

T-35

Remark: As proved in part (iii), we have     sin 2α sinh 2β tan (α + iβ) =   + i ⋅  cos 2α + cosh 2β   cos 2α + cosh 2β  Replacing β by − β on both sides, we have     sin 2α sinh 2β tan (α − iβ) =   − i ⋅  cos 2α + cosh 2β   cos 2α + cosh 2β  [ ∵ cosh (− x) = cosh x, and sinh (− x) = − sinh x] From these expressions for tan (α + iβ) and tan (α − iβ) into real and imaginary parts, we observe that tan (α − iβ) is the conjugate complex of tan (α + iβ). Thus tan (α + iβ) and tan (α − iβ) are conjugate complex numbers. cos ( x + iy) Example 8: Separate into real and imaginary parts. ( x + iy) + 1 Solution:

We have cos ( x + iy) ( x + iy) + 1

=

= =

=

cos ( x + iy) ( x + 1) + iy

=

cos ( x + iy) {( x + 1) − iy } {( x + 1) + iy }{( x + 1) − iy)}

,

multiplying the Nr. and the Dr. by the conjugate complex of the Dr. [cos x cos (iy) − sin x sin (iy)][( x + 1) − iy] ( x + 1)2 − i2 y 2 (cos x cosh y − i sin x sinh y) [( x + 1) − iy)] ( x + 1)2 + y 2 [( x + 1) cos x cosh y − y sin x sinh y] − i [( x + 1) sin x sinh y + y cos x cosh y] ( x + 1)2 + y 2

= a − ib where a = [( x + 1) cos x cosh y − y sin x sinh y] / {( x + 1)2 + y 2 } and b = [( x + 1) sin x sinh y + y cos x cosh y] / {( x + 1)2 + y 2 }. Example 9: Resolve e sin ( x + iy) into real and imaginary parts. Solution:

We have e sin ( x + iy) = e sin = e sin

x cosh y + i cos x sinh y

= e sin

x cosh y

x cos (iy) + cos x sin (iy)

= e sin

x cosh y

. e i cos x sinh

y

[cos (cos x sinh y) + i sin (cos x sinh y)], [ ∵ e i θ = cos θ + i sin θ]

which is of the form a + ib. Example 10:

Resolve sin2 ( x + iy) into real and imaginary parts.

1 Solution: We have sin2 ( x + iy) = [2 sin2 ( x + iy)] 2

(Meerut 2009)

T-36

=

1 [1 − cos 2 ( x + iy)] 2

[ ∵ 2 sin2 θ = 1 − cos 2θ]

1 1 [1 − cos (2 x + 2iy)] = [1 − {cos 2 x cos (2iy) − sin 2 x sin (2iy)}] 2 2 1 = [1 − cos 2 x cosh 2 y + i sin 2 x sinh 2 y] 2 1 1 = (1 − cos 2 x cosh 2 y) + i (sin 2 x sinh 2 y). 2 2 =

Comprehensive Exercise 2 1.

Resolve into real and imaginary parts : (i) cot (α + iβ)

(ii) sec (α + iβ)

(iii) cosec (α − iβ) 2.

3. 4.

Resolve into real and imaginary parts : (i) sinh (α + iβ) (Bundelkhand 2009)

(ii) cosh (α + iβ)

(iii) tanh (α + iβ)

(iv) coth (α + iβ).

Resolve e

cosh ( x ± iy)

Resolve e

sinh ( x + iy)

into real and imaginary parts.

into real and imaginary parts. iθ

5.

Split into real and imaginary parts e

6.

 x − a + iy  If E   = P + iQ, find P and Q.  x + a + iy 

iφ

/ (1 − ke ).

(Kumaun 2008) (Meerut 2010B)

A nswers 2 1.

    sin 2α sinh 2β (i)   − i  cosh 2 β − cos 2 α cosh 2 β − cos 2 α      2 cos α cosh β   2 sin α . sinh β  (ii)   + i   cos 2α + cosh 2β   cos 2α + cosh 2β   2 sin α . cosh β   2 cos α . sinh β  (iii)   + i   cosh 2β − cos 2α   cosh 2β − cos 2α 

2.

(i) (sinh α cos β) + i (cosh α sin β) sinh α cos β − i cosh α sin β (ii) (cosh α cos β) + i (sinh α sin β) cosh α cos β − i sinh α sin β sinh 2α sin 2β (iii) +i cosh 2α + cos 2β cosh 2α + cos 2β

T-37

    sinh 2α sin 2β (iv)   − i ⋅ cosh 2 α − cos 2 β cosh 2 α − cos 2 β     3.

e cosh

x cos y

[cos (sinh x sin y) ± i sin (sinh x sin y),

sinh x cos y

[cos (cosh x sin y) + i sin (cosh x sin y)]

4.

e

5.

 cos θ − k cos (θ − φ)  sin θ − k sin (θ − φ)      1 − 2k cos φ + k 2  + i  1 − 2k cos φ + k 2     

6.

P = e p cos q and Q = e p sin q , where p =

x 2 + y 2 − a2 2

( x + a) + y

and q =

2

2ay ( x + a)2 + y 2

⋅

Problems Involving Trigonometric and Hyperbolic Funtions: The problems involving trigonometric and hyperbolic functions can be solved by using the relations given in article 2.9. Example 11:

If tan (θ + iφ) = tan α + i sec α, then prove that 1 1 e2 φ = ± cot α and 2θ = nπ + π + α. 2 2 (Meerut 2010; Bundelkhand 10; Purvanchal 14; Kashi 14)

Solution:

We have tan (θ + iφ) = tan α + i sec α,

so that tan (θ − iφ) = tan α − i sec α. Now tan 2θ = tan [(θ + iφ) + (θ − iφ)] = = = Thus ∴ or Again,

or

1 − tan (θ + iφ) tan (θ − iφ)

(tan α + i sec α) + (tan α − i sec α) 1 − (tan α + i sec α)(tan α − i sec α) 2 tan α −2 tan2 α

=−

=

2 tan α 1 − (tan2 α + sec2 α)

1 = − cot α. tan α

1 tan 2θ = − cot α = tan ( π + α). 2 1 2θ = π + α 2 1 2θ = nπ + π + α 2

(principal value) (general value)

tan 2iφ = tan [(θ + iφ) − (θ − iφ)] = =

∴

tan (θ + iφ) + tan (θ − iφ)

tan (θ + iφ) − tan (θ − iφ) 1 + tan (θ + iφ) tan (θ − iφ)

(tan α + i sec α) − (tan α − i sec α) 2

2

1 + (tan α + sec α)

i tanh 2φ = i cos α, or tanh 2φ = cos α e 2 φ − e −2 φ e

2φ

+e

−2 φ

=

cos α 1

⋅

=

2i sec α 2 sec2 α

= i cos α.

T-38

Applying componendo and dividendo, we have 1 2 cos 2 α 1 + cos α 2e2 φ 1 4φ 2 or = , or e = = cot 2 α. −2 φ 1 2 1 − cos α 2 2e 2 sin α 2 1 ∴ e2 φ = ± cot α. 2 Example 12:

If sin (θ + iφ) = cos α + i sin α, then prove that cos2 θ = sinh2 φ = ± sin α.

(Meerut 2009B)

Solution: We have sin (θ + iφ) = cos α + i sin α or sin θ cosh φ + i cos θ sinh φ = cos α + i sin α. Equating real and imaginary parts, we have sin θ cosh φ = cos α cos θ sinh φ = sin α. First we shall eliminate φ between (1) and (2). We have cosh2 φ − sinh2 φ = 1. ∴

cos 2 α sin2 θ

−

sin2 α cos 2 θ

or

=1

cos 2 α cos 2 θ − sin2 α sin2 θ sin2 θ cos 2 θ

…(1) …(2)

=1

or

cos 2 α cos 2 θ − sin2 α sin2 θ = sin2 θ cos 2 θ

or

cos 2 θ (1 − sin2 α) − sin2 α (1 − cos 2 θ) = cos 2 θ (1 − cos 2 θ) [changing all the terms to cos θ and sin α] 2

2

2

or

cos θ − cos θ sin α − sin2 α + sin2 α cos 2 θ = cos 2 θ − cos 4 θ

or

cos 4 θ = sin2 α.

∴

cos 2 θ = ± sin α.

Now we shall eliminate θ between (1) and (2) . We have

sin2 θ + cos 2 θ = 1.

Therefore from (1) and (2), we have cos 2 α cosh2 φ

+

sin2 α sinh2 φ

=1

or

cos 2 α sinh2 φ + cosh2 φ sin2 α = cosh2 φ sinh2 φ

or

sinh2 φ (1 − sin2 α) + sin2 α (1 + sinh2 φ) = (1 + sinh2 φ) sinh2 φ [changing all the terms to sinh φ and sin α] 2

2

2

or

sinh φ − sinh φ sin α + sin2 α + sinh2 φ sin2 α = sinh2 φ + sinh4 φ

or

sinh4 φ = sin2 α.

∴

sinh2 φ = ± sin α.

Thus

cos 2 θ = sinh2 φ = ± sin α.

T-39

Example 13:

If sin (θ + iφ) = ρ (cos α + i sin α), prove that ρ2 =

Solution:

1 [cosh 2φ − cos 2θ] and tan α = tanh φ cot θ. 2 (Garhwal 2001)

We have sin (θ + iφ) = ρ cos α + iρ sin α sin θ cosh φ + i cos θ sinh φ = ρ cos α + iρ sin α. sin θ cosh φ = ρ cos α cos θ sinh φ = ρ sin α.

or ∴ and

…(1) …(2)

Squaring (1) and (2) and adding, we get ρ2 = sin2 θ cosh2 φ + cos 2 θ sinh2 φ . 2 ρ2 = 2 sin2 θ cosh2 φ + 2 cos 2 θ sinh2 φ

∴

= (1 − cos 2θ) cosh2 φ + (1 + cos 2θ) sinh2 φ , changing to double angles = (cosh2 φ + sinh2 φ) − cos 2θ (cosh2 φ − sinh2 φ) [ ∵ cosh2 φ + sinh2 φ = cosh 2φ] = cosh 2φ − cos 2θ. ρ2 =

∴

1 (cosh 2φ − cos 2θ). 2

Again, dividing (2) by (1), we get cos θ sinh φ sin α = or tan α = tanh φ cot θ. sin θ cosh φ cos α

Comprehensive Exercise 3 1. If sin (α + iβ) = x + iy, prove that (i)

x 2 cosec 2 α − y 2 sec 2 α = 1,

(ii) x 2 sech2 β + y 2 cosech2 β = 1.

(Bundelkhand 2005, 14; Avadh 11)

(Kanpur 2005, 12; Avadh 11; Bundelkhand 14)

2. If cos −1 ( x + iy) = A + iB, prove that (i)

x 2 sec 2 B + y 2 cosech2 B = 1

(ii) x 2 sec 2 A − y 2 cosec 2 A = 1.

(Garhwal 2000)

3. If tan (θ + iφ) = sin ( x + iy), then prove that coth y sinh 2φ = cot x sin 2θ.

(Purvanchal 2011; Agra 14)

4. If tanh (α + iβ) = sin ( x + iy), prove that sinh 2α cosec 2β = tan x coth y. 5. If tan (α + iβ) = x + iy, prove that (i)

x 2 + y 2 + 2 x cot 2α = 1;

(ii) x 2 + y 2 − 2 y coth 2β + 1 = 0; (iii) x coth 2α + y coth 2β = 1.

(Kanpur 2010) (Kanpur 2010)

T-40

6. If tan (θ + iφ) = cos α + i sin α, prove that θ=

nπ π 1 π α  π α + and φ = log tan  +  or e2 φ = tan  +  ⋅    2 4 2 4 2 4 2 (Garhwal 2002; Agra 07; Gorakhpur 07; Meerut 08; Purvanchal 10)

7. If tan (α + i β) = i, where α and β are real, then prove that α is indeterminate and β is infinite.

(Bundelkhand 2007, 14; Purvanchal 14)

8. If sin (θ + iφ) = tan α + i sec α, then prove that cos 2θ cosh 2φ = 3. (Garhwal 2001; Avadh 07, 09; Purvanchal 08; Agra 09, 10; Kashi 13)

9. If cos ( x + iy) = cos α + i sin α, prove that (ii) sin4 x = sin2 α

(i) cosh 2 y + cos 2 x = 2 (iii) sinh4 y = sin2 α.

10. If cos (θ + iφ) = r (cos α + i sin α), then prove that φ =

sin (θ − α) 1 log ⋅ 2 sin (θ + α)

(Bundelkhand 2003; Avadh 05; Kanpur 06; Kashi 12; Rohilkhand 09)

11. If cosh u = sec θ, show that

and

1 1 u = log tan ( π + θ) 4 2 2 1 2 1 tanh u = tan θ. 2 2

(Meerut 2013) (Meerut 2011; Avadh 10)

12. If A + iB = C tan ( x + iy), then prove that tan 2 x =

2CA 2

C − A2 − B2

sinh 2 y

13. If u + iv = cot ( x + iy), show that v = −

cosh 2 y − cos 2 x

⋅

⋅

14. If cos (u + iv) = x + iy, prove that (1 + x)2 + y 2 = (cosh v + cos u)2

and (1 − x)2 + y 2 = (cosh v − cos u)2 ,

where x, y, u, v are all real. 15. If cosh ( x + iy) = p + iq, where x, y, p and q are real, show that p2 cos 2 y

−

q2 sin2 y

= 1.

16. If sin ( x + iy) sin (θ + iφ) = 1, show that (i) tanh2 y cosh2 φ = cos 2 θ ; 2

(ii) tanh φ cosh

2

2

y = cos x.

(Kanpur 2010) (Kanpur 2010)

17. If cos (θ + iφ) cos (α + iβ) = 1, prove that tanh2 φ cosh2 β = sin2 α and tanh2 β cosh2 φ = sin2 θ.

(Kanpur 2009)

T-41

18. If x = 2 cos α cosh β and y = sin α sinh β, prove that 4x (i) sec (α + iβ) + sec (α − iβ) = 2 ⋅ x + y2 4iy (ii) sec (α + iβ) − sec (α − iβ) = 2 ⋅ x + y2

O bjective T ype Q uestions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1. If sin ( x + iy) = p + iq, where p and q are real, then

2.

(a) q = sin x cos y

(b) q = cos x sin y

(c) q = sin x cosh y

(d) q = cos x sinh y.

If θ is real, then (a) cos (iθ) = i cosh θ

(b) sin (iθ) = i sinh θ

(c) tan (iθ) = tanh θ

(d) cot (iθ) = i coth θ.

Fill in the Blanks Fill in the blanks “……” so that the following statements are complete and correct. 1. The Euler’s theorem states that e iθ = …… . 2. The exponential value of cos θ is …… . 3. The exponential value of sin θ is …… . 4. If e x + iy = a + ib, where a and b are real, then b = …… . 5. If cos ( x + iy) = a + ib, where a and b are real, then a = …… . 6. If cos ( x + iy) = a + ib, where a and b are real, then b = …… . 7. The complex conjugate of cos ( x − iy) is …… . 8. If esin ( x + iy) = a + ib, where a and b are real, then b = …… .

True or False Write ‘T’ for true and ‘F’ for false statement. 1. cot (iθ) = − i coth θ. 2.

If sin ( x − iy) = a + ib, where a and b are real, then b = cos x sinh y.

3.

If sin (α + iβ) = x + iy, then

4.

cos (α + iβ) = cos α cosh β + i sin α sinh β.

x2 sin2 α

+

y2 cos 2 α

= 1.

T-42

5.

    sin 2α sinh 2β tan (α + iβ) =   + i ⋅  cos 2α + cosh 2β   cos 2α + cosh 2β 

A nswers Multiple Choice Questions 1. (d)

2.

(b)

Fill in the Blanks 1. cos θ + i sin θ 4.

2.

e iθ + e − iθ

2 5. cos x cosh y

e x sin y

7. cos ( x + iy)

8.

esin

x cosh y

3.

e iθ − e − iθ

2i 6. − sin x sinh y

sin (cos x sinh y)

True or False 1.

T

2.

F

3.

F

4.

F

5.

T

¨

T-43

3 L ogarithms

of

C omplex N umbers

3.1 Logarithms in the Set of Real Numbers e know that if a and x are two real numbers such that e x = a, then x is called the logarithm of a to the base e and we write it as

W

x = log e a. Since for all real numbers x, we have e x > 0, therefore, in the case of a real variable, log e a exists if and only if a is positive. Also if a is a positive real number, log e a is a unique real number.

3.2 Logarithms of Complex Numbers (Gorakhpur 2006)

Now we shall extend our definition of logarithmic function to the set of complex numbers. Definition: Let z and w be two complex numbers such that e z = w. Then z is called a logarithm of w to the base e and we write it as z = log e w or simply as z = log w, the base e remaining understood.

T-44

Thus by definition, z = log w if and only if w = e z . Since e z ≠ 0 for any complex number z, therefore log w does not exist if w = 0.

Logarithm of a complex number is a many-valued function: Let log e w = z . Then e z = w. If n is any integer, we have e2 nπi = cos 2 nπ + i sin 2nπ = 1 + i0 = 1. ∴

z

z

z

e = e .1 = e e

2 nπi

=e

z + 2 nπi

(Bundelkhand 2005)

,

which means that if log w = z , then we also have log w = z + 2nπi. Thus logarithm of a complex number is a many-valued function.

3.3 Principal and General Values of Logarithm of a Non-Zero Complex Number (Gorakhpur 2006)

Let z = x + iy be a non-zero complex number. Suppose log e z = α + iβ, where α and β are real. Then z = e α + iβ = e α e iβ = e α (cos β + i sin β). Since e α is a positive real number, therefore z = e α (cos β + i sin β) is a representation of z in a modulus-argument form. We have e α = |z | = √ ( x 2 + y 2 ) and β = arg z = tan −1 ( y / x). The equation e α = |z | has a unique solution α = log |z |, the real logarithm of the positive number |z |. ∴

log e z = log |z | + i arg z

i. e.,

log e ( x + iy) = log √ ( x 2 + y 2 ) + i tan −1 ( y / x).

Now arg z is a many-valued function and consequently log z is a many-valued function. If β 0 is the principal value of arg z i. e., the value of arg z lying between −π and π, then log |z | + iβ 0 is called the principal value of log z . Again if β 0 is the principal value of arg z, then 2nπ + β 0 is the general value of arg z and log |z | + iβ 0 + 2nπi is called the general value of log z.Thus every non-zero complex number has infinitely many logarithms which differ from one another by an integral multiple of 2πi. It is usual to denote the general value of log e z by Log e z, using the first letter L as the capital letter, and the principal value by log e z ,using the first letter l as the small letter. Thus we have

T-45

Log e z = log

e

z + 2nπi, where n is any integer.

If in the general value, we put n = 0, we get the principal value. Remember : log e ( x + iy) = log √ ( x 2 + y 2 ) + i tan −1 ( y / x), where tan −1 ( y / x) represents the principal value of arg ( x + iy) Log e ( x + iy) = log √ ( x 2 + y 2 ) + i tan −1 ( y / x) + 2nπi,

and

where n is any integer.

3.4 Properties of the Logarithmic Function If u and v are two non-zero complex numbers, then and Proof:

Log (uv) = Log u + Log v

…(1)

Log (u / v) = Log u − Log v.

…(2)

Let Log u = z1 and Log v = z 2 .

Then e z1 = u and e z2 = v.

∴

e z1 e z2 = uv, or e z1 + z2 = uv.

By the definition of logarithm, we have Log uv = z1 + z 2 = Log u + Log v, which proves the result (1). The result (2) can be proved similarly. The equality (1) does not mean that the principal value of log uv is equal to the sum of the principal values of log u and log v. Note that the sum of the principal arguments of u and v need not be equal to the principal argument of uv. The equality (1) implies that each value of log u + log v is equal to some value of log uv and each value of log uv is equal to some value of log u + log v. A similar remark holds for the equality (2) also. Remark: If z be a non-zero complex number and n be a positive integer, the equality log z n = n log z may not be true even for the general values. As an illustration, we have Log i3 = Log (− i) = (2nπ −

1 1 π) i, while 3 Log i = 3 (2mπ + π) i. 2 2

Now it is not correct to say that 1 1 3 (2mπ + π) = 2nπ − π. 2 2 For, the left hand side may be written as 2 (3m + 1) π −

1 π, 2

which shows that the solution set on the left is only a subset of the solution set on the right. Thus

Log i3 ≠ 3 Log i.

T-46

3.5 Working Rule to Evaluate Log ( x + iy) i. e., to Express Log ( x + iy) in the Form A + iB (Meerut 2010)

First we put x + iy in the modulus-argument form. So let

x + iy = r (cos θ + i sin θ).

Then x = r cos θ, y = r sin θ.

∴ r = √ ( x 2 + y 2 ) and θ = tan −1 ( y / x).

Now Log ( x + iy) = Log (r cos θ + ir sin θ) = Log r (cos θ + i sin θ) = log [r {cos (2nπ + θ) + i sin (2nπ + θ)}] = log r e i (2 nπ + θ) = log r + log e i (2 nπ + θ) = log r + i (2nπ + θ)

where

= log √ ( x 2 + y 2 ) + i {2nπ + tan −1 ( y / x)} 1 = log ( x 2 + y 2 ) + i {2nπ + tan −1 ( y / x)} = A + iB, 2 1 A = log ( x 2 + y 2 ), B = 2nπ + tan −1 ( y / x). 2

If we put n = 0, we obtain the principal value of Log ( x + iy) written as log ( x + iy). 1 Thus log ( x + iy) = log ( x 2 + y 2 ) + i tan −1 ( y / x). 2 (Rohilkhand 2006, 08)

Putting − y for y on both sides, we get 1 log ( x − iy) = log ( x 2 + y 2 ) − i tan −1 ( y / x). 2

3.6 Logarithm of a Positive Real Number in the Set of Complex Numbers Let x be a positive real number. Let

x = x + i0 = r (cos θ + i sin θ).

Then x = r cos θ, 0 = r sin θ.

∴

r = x and θ = 0.

∴

Log x = log x + i0 + 2nπi = log x + 2nπi.

Obviously only n = 0 gives a real value of Log x. Thus in the set of complex numbers a positive real number has an infinite number of logarithms out of which only one is real.

3.7 Logarithm of a Negative Real Number Let x be a positive real number so that − x is a negative real number. We have to find Log (− x).

T-47

Let

− x = − x + i0 = r (cos θ + i sin θ). Then

∴

r = x, and θ = π.

∴

Log (− x) = Log x (cos π + i sin π)

− x = r cos θ, 0 = r sin θ.

= log x {cos (2nπ + π) + i sin (2nπ + π)} = log xe(2 nπ + π ) i = log x + log e(2 nπ + π ) i = log x + (2n + 1) πi, which is never real. The principal value of log (− x) is log x + iπ. Hence the principal value of the logarithm of a negative real number is the logarithm of the corresponding positive number added with πi.

Example 1: Find the principal and general value of log (−1 + i). Solution:

Let −1 + i = r (cos θ + i sin θ), so that r cos θ = − 1 and r sin θ = 1.

Squaring and adding, we have r 2 = 2, i. e., r = √ 2. Now

cos θ = − 1 / √ 2 and sin θ = 1 / √ 2, so that

∴

−1 + i = √ 2 (cos

∴

θ=

3 π. 4

3 3 π + i sin π) = √ 2 e(3 π /4) i . 4 4

the general value is Log (−1 + i) = log { √ 2 e(3 π /4) i e2 nπi} = log √ 2 +

3 1 3 πi + 2nπi = log 2 + (2n + ) πi. 4 2 4

Putting n = 0, the principal value is given by 1 3 log (−1 + i) = log 2 + πi. 2 4 Example 2: Find the general value of log (−3). (Bundelkhand 2006; Rohilkhand 06, 08)

Solution: Let −3 = − 3 + i0 = r (cos θ + i sin θ), so that These give r

−3 = r cos θ and 0 = r sin θ. 2

= 9 i. e., r = 3. Putting r = 3, we get cos θ = − 1 and sin θ = 0, giving θ = π.

∴

−3 = 3 (cos π + i sin π) = 3 . e iπ .

∴

Log (−3) = log {3 . e iπ . e2 nπi} = log 3 + log e(2 n π + π ) i = log 3 + (2n + 1) πi.

[ ∵ e2 nπi = 1]

T-48

The principal value of Log (−3) i. e., log (−3) is obtained by putting n = 0 in the above result. Thus

log (− 3) = log 3 + iπ.

Example 3:

Find the general value of log (− i).

Solution: We have − i = (cos

(Agra 2007)

1 1 π − i sin π) = e − iπ /2 2 2

so that

log (− i) = log e − iπ /2 = − iπ / 2, giving the principal value.

∴

Log (− i) = log (− i) + 2 nπi = − (iπ / 2) + 2nπi =

1 (4n − 1) πi. 2

Example 4: Express log (1 + i)(1− i) in the form A + iB. Solution:

We have log (1 + i)(1− i) = (1 − i) log (1 + i) 1 = (1 − i) [ log (12 + 12 ) + i tan −1 1] 2 1 1 = (1 − i) [ log 2 + i π] 2 4 1 1 1 1 = ( log 2 + π) + i ( π − log 2), 2 4 4 2 which is of the form A + iB.

Example 5: Prove that

 1 log  iα  1− e

 1 α π α   = log  cosec  + i  − ⋅     2 2 2 2   (Meerut 2009B; Kashi 14)

Solution: We have 1 1 log = log iα 1 − cos α − i sin α 1− e = log

= log

= log

= log

= log

[ ∵ e i α = cos α + i sin α ]

1 1 1 1 α − 2i sin α cos α 2 2 2 1 1 1 1 2 sin α (sin α − i cos α) 2 2 2 1 1 1 1 1 1 2 sin α {cos ( π − α) − i sin ( π − α)} 2 2 2 2 2 1 [ ∵ e − i θ = cos θ − i sin θ ] 1 − i (π / 2 − α / 2 ) 2 sin α e 2 1 1 [( cosec α) . e i (π /2 − α /2) ] 2 2 2 sin2

T-49

1 = log ( cosec 2 1 = log ( cosec 2 Example 6:

Prove that

1 α) + log e i (π /2 − α /2) 2 1 1 1 α) + i ( π − α). 2 2 2 1 1 log tan ( π + i α) = i tan −1 (sinh α). 4 2

(Agra 2005, 06; Meerut 07, 08, 11, 13B; Bundelkhand 10; Avadh 09, 11; Kanpur 11; Kashi 13)

 sin (1 π + i 1 α)   1 1 4 2  Solution: L.H.S. = log tan ( π + i α) = log  1 1  4 2  cos ( π + i α)  4 2   2 sin (1 π + 1 αi) cos (1 π − 1 αi)    4 2 4 2 multiplying the Nr. = log  , 1 1 1 1  2 cos ( π + αi) cos ( π − αi)   4 2 4 2   = log      = log   

and the Dr. by the conjugate complex of the Dr. 1 1 1 1 sin ( π + π) + sin ( α i + α i)   4 4 2 2  1 1 1 1 cos ( π + π) + cos ( α i + α i)   4 4 2 2 1 sin π + sin (αi)   1 + i sinh α   2 = log    1  cosh α  cos π + cos (αi)   2 [ ∵ sin (iα) = i sinh α, cos (iα) = cosh α]

 1 sinh α  = log  +i  cosh α cosh α   =

  sinh α / cosh α  sinh2 α  1 1  + i tan −1  log  +   2 2 2 1 / cosh α    cosh α cosh α  [ By article 3.5, here x = 1 / cosh α, y = sinh α / cosh α ]

=

 1 + sinh2 α 1 log  2 2  cosh α

=

 cosh2 α  1  + i tan −1 (sinh α) log   2 2  cosh α 

=

1 log 1 + i tan −1 (sinh α) = i tan −1 (sinh α) = R.H. S. 2

  + i tan −1 (sinh α)  

Example 7: Separate Log sin ( x + iy) into real and imaginary parts. Solution:

We have Log sin ( x + iy) = Log (sin x cos iy + cos x sin iy)

T-50

= Log (sin x cosh y + i cos x sinh y) 1 = log (sin2 x cosh2 y + cos 2 x sinh2 y) 2  cos x sinh y  + i tan −1   + 2 nπi [By article 3.5]  sin x cosh y  =

1 1 1 log { (1 − cos 2 x) cosh2 y + (1 + cos 2 x) sinh2 y} 2 2 2 + i tan −1 (cot x tanh y) + 2nπi

=

1 1 1 log { (cosh2 y + sinh2 y) − cos 2 x (cosh2 y − sinh2 y)} 2 2 2 + i tan −1 (cot x tanh y) + 2nπi

1 1 1 log ( cosh 2 y − cos 2 x) + i [2 nπ + tan −1 (cot x tanh y)] 2 2 2 1 1 = log [ (cosh 2 y − cos 2 x)] + i [2nπ + tan −1 (cot x tanh y)], 2 2 =

which is of the form P + iQ. To find out the principal value, we put n = 0 in the above result. So we have 1 1 log sin ( x + iy) = log [ (cosh 2 y − cos 2 x)] + i tan −1 (cot x tanh y). 2 2 Example 8: If log sin ( x + iy) = α + iβ, show that (i) (ii)

α=

1 1 log (cosh 2 y − cos 2 x), 2 2

2 cos 2 x = e

2y

+e

−2 y

− 4e

2α

,

(Meerut 2013B)

(iii)

β = tan

−1

(cot x tanh y),

(iv) cos ( x − β) = e2 y cos ( x + β). ⇒ sin ( x + iy) = e α + iβ = e α e

Solution:

log sin ( x + iy) = α + iβ,

or

sin x cos iy + cos x sin iy = e

or

sin x cosh y + i cos x sinh y = e

α

iβ

(cos β + i sin β) α

cos β + i e

α

sin β.

Equating real and imaginary parts, we have sin x cosh y = e α cos β and

cos x sinh y = e

α

sin β.

…(1) …(2)

Squaring and adding (1) and (2), we get sin2 x cosh2 y + cos 2 x sinh2 y = e2 α (cos 2 β + sin2 β) or or or

1 1 (1 − cos 2 x) cosh2 y + (1 + cos 2 x) sinh2 y = e2 α 2 2 1 1 2 2 (cosh y + sinh y) − cos 2 x (cosh2 y − sinh2 y) = e2 α 2 2 1 1 1 cosh 2 y − cos 2 x = e2 α or (cosh 2 y − cos 2 x) = e2 α . 2 2 2

…(A)

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∴ or

1 (cosh 2 y − cos 2 x), 2 1 1 α = log (cosh 2 y − cos 2 x), which proves the result (i). 2 2 2α = log

From the result (A), we have cosh 2 y − cos 2 x = 2e2 α

or cosh 2 y = cos 2 x + 2e2 α

or

2 cosh 2 y = 2 cos 2 x + 4e2 α

or

2 cos 2 x = 2 cosh 2 y − 4e2 α = e2 y + e −2 y − 4e2 α , which proves the result (ii).

Dividing (2) by (1), we get cos x sinh y tan β = = cot x tanh y. sin x cosh y ∴

…(B)

β = tan −1 (cot x tanh y), which proves the result (iii).

From the result (B), we have tan β e tanh y = or cot x e

y

− e− y

y

−y

+e

=

sin β sin x cos β cos x

⋅

Applying componendo and dividendo, we have (e y + e − y ) + (e y − e − y ) cos β cos x + sin β sin x = (e y + e − y ) − (e y − e − y ) cos β cos x − sin β sin x or or

2e 2e

y

−y

=

cos ( x − β) cos ( x + β)

or e2 y =

cos ( x − β) cos ( x + β)

cos ( x − β) = e2 y cos ( x + β), which proves the result (iv).

Comprehensive Exercise 1 1. (i) Prove that Log (1 + i) =

1 1 log 2 + i (2nπ + π). 2 4 (Meerut 2012; Avadh 14)

(ii) Prove that Log (−5) = log 5 + (2nπ + π) i. (iii) Find the general value of log i.

(Kanpur 2005)

2. (i) Find the general value of log √ i.

(Rohilkhand 2007)

1 1 (ii) Show that log (1 + e iθ ) = log (2 cos θ) + iθ, if − π < θ < π. 2 2  a + ib  −1  b  3. (i) Prove that log   = 2i tan   ⋅  a  a − ib  (Avadh 2009; Purvanchal 11) x−i (ii) Show that i log = π − 2 tan −1 x. (Kanpur 2008; Avadh 08, 10; x+i

Purvanchal 11; Kashi 13)

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 a − ib  2ab 4. (i) Show that tan  i log ⋅  = 2 a + ib  a − b 2 

(Meerut 2004, 10, 10B; Avadh 05; Purvanchal 07, 09)

(ii) Prove that log (1 + i tan α) = log sec α + αi. 5. Prove that sin (log i i ) = − 1. π θ  6. If u = log tan  +  , prove that  4 2 u θ (i) tanh = tan 2 2 (Kanpur 2007; Meerut 10, 12B; Kashi 11; Agra 14) π iu  (ii) θ = − i log tan  + ⋅  4 2  (Gorakhpur 2005; Purvanchal 08; Avadh 13)

7. Prove that log (1 + cos 2θ + i sin 2θ) = log (2 cos θ) + iθ, if − π < θ ≤ π. sin ( x + iy) −1 8. Prove that log   = 2i tan (cot x tanh y). sin ( x − iy)

(Meerut 2012, 13)

9. Prove that log cos ( x + iy) =

1 1 log (cosh 2 y + cos 2 x) − i tan −1 (tan x tanh y). 2 2 (Kumaun 2008)

 cos ( x − iy)  −1 10. Prove that log   = 2 i tan (tan x tanh y). cos ( x + iy )   (1 + i) (1 + i √ 3)

11. Show that one of the values of log

√3 + i

is

1 5 log 2 + i π. 2 12

12. If tan log ( x + iy) = a + ib, where a2 + b 2 ≠ 1, prove that tan { log ( x 2 + y 2 )} =

2a 1 − a2 − b 2

⋅

(Meerut 2005B; Avadh 13)

13. If log log ( x + iy) = p + iq, prove that y = x tan [tan q log ( x 2 + y 2 )1 /2 ]. (Purvanchal 2006; Avadh 10) 14. If log log log (α + iβ) = p + iq, prove that p 1 (i) e e cos q . cos (e p sin q) = log (α 2 + β 2 ), 2 (ii) e

e p sin q

. sin (e

p

sin q) = tan

−1

(Kanpur 2011)

( β / α).

15. If (a1 + ib1 )(a2 + ib2 )……(a n + ib n ) = A + iB, prove that  b   b   b  B tan −1  1  + tan −1  2  + … + tan −1  n  = tan −1 , A  an   a1   a2  (Avadh 2009)

and

(a12 + b12 ) (a2 2 + b2 2 )…(a n2 + b n2 ) = A2 + B2 .

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A nswers 1 1. (iii)

1 (4n + 1) πi 2

2. (i)

1 (8n + 1) πi 4

3.8 The General Exponential Function a z If a and z are any two complex numbers, the general exponential function a z is defined as a z = e z Log a = exp (z Log a). Since Log a is a many-valued function, a z

a =e

z Log a

=e

z (log a + 2 nπi)

z

is also a many-valued function. Thus

= exp [z (log a + 2 nπi)].

This gives the general value of a z . The principal value of a z is obtained by putting n = 0. Hence the principal value of a z = e z log

a

= exp (z log a).

3.9 To Separate (α + iβ ) p + iq into Real and Imaginary Parts By the definition of the general exponential function, we have (α + iβ) p + iq = e( p + iq) Log (α + iβ) = exp [( p + iq) Log (α + iβ)] 1 = exp [( p + iq) { log (α 2 + β 2 ) + i tan −1 ( β / α) + 2 nπi }] 2 [By article 3.5] 1 2 2 −1 = exp [{ p log (α + β ) − q tan ( β / α) − 2 nqπ } 2 1 + i { q log (α 2 + β 2 ) + p tan −1 ( β / α) + 2 npπ }] 2 = exp ( A + iB), 1 where A = p log (α 2 + β 2 ) − q { tan −1 ( β / α) + 2 nπ } 2 1 and B = q log (α 2 + β 2 ) + p { tan −1 ( β / α) + 2 nπ } 2 = e A + iB = e A e iB = e A (cos B + i sin B). Therefore the real part is e A cos B and the imaginary part is e A sin B. The principal value is obtained by putting n = 0. Note: In the following examples it is sometimes found convenient to write e x as exp ( x).

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Example 9:

Prove that ii = e −(4 n +1) π /2 .

(Meerut 2004B; Gorakhpur 06; Agra 08; Purvanchal 10; Rohilkhand 09; Bundelkhand 14)

Solution: We have i i = e

i Log i

, by def. of a z

= exp (i Log i) = exp [i (log i + 2 nπi)]. …(1) 1 1 1 1 log i = log (cos π + i sin π) [ ∵ i = cos π + i sin π] 2 2 2 2 = log e iπ /2 = i π / 2.

Now

Substituting for log i in (1), we have 1 1 i i = exp [i (i π + 2 nπi)] = exp [i2 (2n + ) π] 2 2 = exp [− (4n + 1) π / 2] = e −(4 n +1) π /2 .

…(2) i

Note 1:

Putting n = 0 in (2), we get the principal value of i = e

Note 2:

Putting n = 0, 1, 2,… in (2), the various values of ii are

−π /2

.

e − π /2 , e −5 π /2 , e −9 π /2 , e −13 π /2 ,… which is a geometrical progression with common ratio e −2 π . Example 10: (i)

tan

If i i

i … ad inf .

1 π A = B / A, 2

Solution: (i) We have i A + iB

= A + iB,principal values only being considered, prove that A2 + B2 = e −πB .

and (ii)

(Bundelkhand 2007; Kanpur 09; Purvanchal 14) i i … ad inf .

= A + iB.

∴

i

or

exp [( A + iB) log i ] = A + iB, taking the principal value of i A + iB 1 1 exp [( A + iB) log (cos π + i sin π)] = A + iB 2 2 exp [( A + iB) log e iπ /2 ] = A + iB 1 exp [( A + iB) i π] = A + iB 2 1 1 exp [− πB + i πA] = A + iB or e − Bπ /2 e πAi /2 = A + iB 2 2 1 1 − Bπ / 2 e (cos πA + i sin πA) = A + iB. 2 2

or or or or or

= A + iB

Equating real and imaginary parts on both sides, we have 1 e − Bπ /2 cos πA = A , 2 and

e − Bπ /2 sin

1 πA = B. 2

…(1) …(2)

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Dividing (2) by (1), we have tan

1 πA = B / A . 2

Squaring and adding (1) and (2), we get 1 1 (ii) (e − Bπ /2 )2 (cos 2 πA + sin2 πA) = A2 + B2 2 2 or

e − Bπ = A2 + B2 .

Example 11:

If pα + iβ = ( x + iy) m + in and the principal values are considered, prove that

(i)

α=

1 m log p ( x 2 + y 2 ) − n tan −1 ( y / x) log p e, 2

(Meerut 2006B; Kanpur 08)

(ii)

log p ( x 2 + y 2 ) =

2 (αm + βn) m2 + n2

⋅

(Kanpur 2008)

Solution: We have pα + iβ = ( x + iy) m + in . Taking logarithm of both sides, we have (α + iβ) log e p = (m + in) log e ( x + iy) 1 = (m + in) [ log e ( x 2 + y 2 ) + i tan −1 ( y / x)], refer article 3.5 2 1 = [ m log e ( x 2 + y 2 ) − n tan −1 ( y / x)] 2 1 + i [ n log e ( x 2 + y 2 ) + m tan −1 ( y / x)]. 2 Equating real and imaginary parts on both sides, we get 1 α log e p = m log e ( x 2 + y 2 ) − n tan −1 ( y / x) 2 and (i)

β log e p =

…(1)

1 m log e ( x 2 + y 2 ) + m tan −1 ( y / x). 2

…(2)

From (1), we have 1 m log e ( x 2 + y 2 ) − n tan −1 ( y / x) 2 1 e) = [ m log e ( x 2 + y 2 ) − n tan −1 ( y / x)]. log 2

α log e p = or

or

(α log e p) . (log

p

e

e,

multiplying both sides by log p e 1 α = m log p ( x 2 + y 2 ) − n tan −1 ( y / x) . log p e. 2 [ ∵ log b a × log a b = 1 ; log a m × log b a = log b m]

∴

α=

1 m log 2

p

( x 2 + y 2 ) − n tan −1 ( y / x) log p e.

Proved.

T-56

(ii)

or

Multiplying (1) by m and (2) by n and adding, we get 1 (αm + βn) log e p = (m2 + n2 ) log e ( x 2 + y 2 ) 2 1 (αm + βn) log e p × log p e = (m2 + n2 ) log e ( x 2 + y 2 ) × log p e, 2 multiplying both sides by log p e 1 2 (m + n2 ) log p ( x 2 + y 2 ) 2 2 (αm + βn) + y2 ) = ⋅ (m2 + n2 )

or

(αm + βn) . 1 =

or

log p ( x 2

Example 12:

If (a + ib) p = m x + iy , then prove that

y x

=

2 tan −1 (b / a) log (a2 + b 2 )

where only principal values are considered .

,

(Kanpur 2007)

Solution: Given that (a + ib) p = m( x + iy) . Taking logarithm of both sides, we have p log (a + ib) = ( x + iy) log m 1 p [ log (a2 + b 2 ) + i tan −1 (b / a)] = x log m + iy log m. 2

or

Equating real and imaginary parts, we get 1 p log (a2 + b 2 ) = x log m 2 and p tan −1 (b / a) = y log m.

…(1 ) …(2)

Dividing (2) by (1), we get y p tan −1 (b / a) 2 tan −1 (b / a) = = ⋅ x ( p / 2) log (a2 + b 2 ) log (a2 + b 2 ) Example 13: (sec α) sec

2

β

Prove that if (1 + i tan α)1 + i tan β can have real values, one of them is

.

(Kanpur 2008, 10; Rohilkhand 05)

Solution: We have (1 + i tan α)1 + i tan β = exp [(1 + i tan β) log (1 + i tan α)], taking the principal value = exp [(1 + i tan β) { log √ (1 + tan2 α) + i tan −1 (tan α)}] = exp [(1 + i tan β)(log sec α + iα)] = exp [log sec α − α tan β + i (α + tan β log sec α)] = exp [log sec α − α tan β] . exp [i (α + tan β log sec α)] = exp [log sec α − α tan β] . [cos {α + tan β log sec α } + i sin {α + tan β log sec α}].

T-57

Now if this value is real, the imaginary part is equal to zero. ∴

sin {α + tan β log sec α } = 0

or

α = − tan β log sec α.

or α + tan β log sec α = 0 …(1)

Also then (1 + i tan α)1 + i tan β = exp [log sec α − α tan β] . (cos 0 + i sin 0) = exp [log sec α + tan β . tan β log sec α], substituting for α from (1) = exp [(1 + tan2 β) log sec α] = exp [sec 2 β log sec α] = exp [log (sec α) sec

2

β

] = (sec α) sec

2

β

.

This is one of the values since (1 + i tan α)1 + i tan β can have infinite number of values. Example 14: Solution:

Find the general value of log4 (−2).

We have Log 4 (−2) = =

= = =

=

=

log e (−2) + 2n πi log e 4 + 2m πi

(Meerut 2009)

Log e (−2) Log e 4 , where m and n are any integers

log e {2 (cos π + i sin π)} + 2nπi log e 4 + 2mπi log e 2 + iπ + 2nπi log e 4 + 2mπi

=

=

log e (2e iπ ) + 2nπi log e 4 + 2 mπi

log e 2 + i (2nπ + π) 2 log e 2 + 2mπi

[log e 2 + i (2nπ + π)] [2 log e 2 − 2mπi] (2 log e 2 + 2mπi) (2 log

e

2 − 2mπi)

2 (log e 2)2 + 2mπ (2nπ + π) + 2i {(2n + 1) π log e 2 − mπ log e 2} 4 (log e 2)2 + 4m2 π 2 (log e 2)2 + (2n + 1) mπ 2 2

2

2 (log e 2) + 2m π

2

+i

(2n + 1 − m) π log e 2 2 (log e 2)2 + 2m2 π 2

,

which is required general value of log 4 (−2). Example 15: Solution:

i

If sin log (i ) = a + ib, find a and b. Hence find cos (log i i ).

First, we have i i = exp (i log i) = exp [i log (cos

1 1 π + i sin π)] 2 2

= exp [i log e iπ /2 ] = exp [i (iπ / 2)] = exp (− π / 2) = e − π /2 .

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∴

log i i = log e −π /2 = − π / 2.

∴

sin (log i i ) = sin (− π / 2) = − 1.

Hence

sin (log ii ) = a + ib ⇒ −1 = a + ib.

Equating real and imaginary parts, we get a = − 1, b = 0. ∴

sin (log i i ) = − 1.

Now

cos (log i i ) = √ {1 − sin2 (log ii )} = √ {1 − 1} = 0.

Comprehensive Exercise 2 1. If iα + iβ = α + iβ, show that α 2 + β 2 = e −(4 n +1) πβ . (Agra 2005; Kanpur 09; Bundelkhand 11)

1 1 2. Prove that i a = cos [(2m + ) πa] + i sin [(2m + ) πa]. 2 2 3.

i

If ii = cos θ + i sin θ, prove that 1 1 θ = (2m + ) π exp [− (2n + ) π]. 2 2

4.

If (ii ) i = cos θ − i sin θ, show that θ =

(Kanpur 2008)

1 π (4n + 1). 2

5. Show that the sum of the moduli of the values of (1 + i)1+ i , which are less than unity is (1 / 2) . e3 π /4 . cosech π. 6. Show that the principal value of (a + ib) p + iq / (a − ib) p − iq is cos 2( pα + q log r) + i sin 2( pα + q log r), where

r = √ (a2 + b 2 ) and α = tan −1 (b / a).

7. Prove that the real part of the principal value of 1 ( i) log (1 + i) is exp (− π 2 / 8) × cos ( π log 2). 4 8. Prove that ( x + ix tan y) log ( x sec

y) − iy

(Purvanchal 2008)

is real, when only principal values are

considered. 9. Prove that the principal value of (a + ib) c + id is wholly real or wholly imaginary according as

1 d log (a2 + b 2 ) + c tan −1 (b / a) is an even or an 2

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odd multiple of

1 π. 2

(Gorakhpur 2005; Avadh 07; Rohilkhand 10)

In case it is wholly real, prove that (a + ib) c + id = (a2 + b 2 )(c 10. Prove that the general value of (1 + i tan α) − i is

2

+ d2 ) / 2 c

.

(Purvanchal 2006)

exp (α + 2mπ). [cos (log cos α) + i sin (log cos α)]. 4m + 1

11. Prove that Log i i =

4n + 1

, where m and n are integers. (Rohilkhand 2005; Kanpur 06)

12. Prove that  (a − x)2 + b 2   a + ib − x  1 log  + i tan −1  = log  2 2 a + ib + x 2    (a + x) + b  13. If

(1 + i) p + iq

= α + iβ ,

(1 − i) p − iq

prove that

one

 b b  − tan −1  ⋅ a − x a + x  of tan −1 ( β / α) is

value

(πp / 2) + q log 2. 14. If [cos (θ − iφ)] x + iy = A + iB and principal values are taken into consideration, then tan −1 ( B / A) =  a + x + iy  15. If    a − x − iy 

1 y log (cosh2 φ − sin2 θ) + x tan −1 (tan θ tanh φ). 2

λ + µi

= X + iY , prove that one of the values of

2 2 2ay   µ  (a + x) + y  tan −1 (Y / X ) is λ tan −1  2 + log   ⋅  a − x 2 − y 2  2  (a − x)2 + y 2 

16. If iα + iβ = e x (cos y + i sin y), then prove that x=−

1 1 (4n + 1) πβ and y = (4n + 1) πα. 2 2

O bjective T ype Q uestions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1. The general value of log (−i) is  2n + 1 (a)   πi  2 

 4n − 1 (b)   πi  2 

 2n − 1 (c)   πi  2 

 3n − 1 (d)   πi  2 

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2. If log ( x − iy) = A + iB, where A and B are real, then 1 log ( x 2 − y 2 ) 2 1 (c) A = log ( x 2 + y 2 ) 2

(b) A = log ( x 2 + y 2 )

(a) A =

(d) A = log ( x 2 − y 2 )

Fill in the Blanks Fill in the blanks “……” so that the following statements are complete and correct. 1. If only the principal value is considered, then 1 log e ( x + iy) = log e ( x 2 + y 2) + …… . 2 2. If the general value is taken into account, then y 1 Log e ( x + iy) = log ( x 2 + y 2 ) + i tan −1 + …… . 2 x 3. The principal value of log (−3) is …… . 4. The general value of Log (−5) is …… . 5. log (1 + i tan α) = log sec α + …… .

True or False Write ‘T’ for true and ‘F’ for false statement. 1. Logarithm of a complex number is a many-valued function. x−i 2. i log = π − 2 tan −1 x. x+i 1 3. The general value of log √ i is (8n + 1) πi. 2

A nswers Multiple Choice Questions 1. (b)

2. (c)

Fill in the Blanks 1. i tan −1

y

2. 2nπi, where n is any integer

x 3. log 3 + iπ

4. log 5 + (2n + 1) πi, where n is any integer

5. iα

True or False 1.

T

2. T

3.

F

¨

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4 I nverse C ircular and H yperbolic F unctions of C omplex N umbers

4.1 Inverse Circular Functions of Complex Numbers

I if

f cos ( x + iy) = u + iv, then x + iy is called an inverse cosine of u + iv and is denoted by cos −1 (u + iv). Thus cos ( x + iy) = u + iv, then cos −1 (u + iv) = x + iy.

Also, if cos ( x + iy) = u + iv, then u + iv = cos {2nπ ± ( x + iy)}, where n is any integer. Therefore by the above definition the general value of inverse cosine of u + iv is 2nπ ± ( x + iy), and is denoted by Cos −1 (u + iv) i. e., by writing the first letter ‘ C ’ as capital. It follows that the inverse cosine of u + iv is a many-valued function. Its principal value is that value of 2nπ ± ( x + iy) in which the real part lies between 0 and π. It is denoted by cos −1 (u + iv) i. e., by writing the first letter ‘ c ’ as small. Thus Cos −1 (u + iv) = 2nπ ± ( x + iy) = 2nπ ± cos −1 (u + iv).

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In a like manner if sin ( x + iy) = u + iv, then x + iy is called an inverse sine of u + iv and is denoted by sin −1 (u + iv). Thus if

sin ( x + iy) = u + iv, then sin −1 (u + iv) = x + iy.

Also, if sin ( x + iy) = u + iv, then u + iv = sin { nπ + (−1) n ( x + iy)}, where n is any integer. Therefore the general value of inverse sine of u + iv is nπ + (−1) n ( x + iy), and is denoted by Sin −1 (u + iv), the first letter ‘ S’ being written capital. Therefore the inverse sine of u + iv is a many-valued function. Its principal value is 1 1 that value of nπ + (−1) n ( x + iy) for which the real part lies between − π and π. It 2 2 −1 is denoted by sin (u + iv), the first letter ‘s’ being written small. Thus Sin −1 (u + iv) = nπ + (−1) n sin −1 (u + iv). The other inverse circular functions may be defined similarly. For example, if tan ( x + iy) = u + iv, then

Tan −1 (u + iv) = nπ + ( x + iy) = nπ + tan −1 ( x + iy),

the principal value being that for which the real part lies between −

1 1 π and π. 2 2

The relations connecting the general and principal values of the remaining inverse circular functions may be written as follows : Sec −1 (u + iv) = 2nπ ± sec −1 (u + iv), Cosec −1 (u + iv) = nπ + (−1) n cosec −1 (u + iv), and

Cot −1 (u + iv) = nπ + cot −1 (u + iv).

It should be noted that the principal value for the case of sin, cosec, tan and cot is 1 1 that value for which the real part lies between − π and π, while for the case of 2 2 cos and sec it is that value for which its real part lies between 0 and π.

4.2 Inverse Hyperbolic Functions Let z and w be two complex numbers. If sinh w = z , then w is called the inverse hyperbolic sine of z and is written as w = sinh −1 z . The other inverse hyperbolic functions cosh −1 z , tanh −1 z , cosech −1 z , sec h −1 z and coth −1 z are defined similarly.

T-63

The values of inverse hyperbolic functions can also be expressed in terms of the logarithmic functions as shown below. (i)

To prove that sinh −1 z = log [z + √ (z 2 + 1)]. (Meerut 2004; Agra 05; Kumaun 08; Kashi 12)

Proof:

Let sinh −1 z = w. Then sinh w = z .

…(1) 2

2

cosh w = √ (1 + sinh w) = √ (1 + z ).

∴

…(2)

Adding (1) and (2), we get sinh w + cosh w = z + √ (1 + z 2 ) i. e.,

1 w 1 (e − e − w ) + (e w + e − w ) = z + √ (1 + z 2 ) 2 2

∴

w = log e [z + √ (1 + z 2 )].

Hence

sinh −1 z = log [ z + √ ( z2 + 1)].

(ii)

i. e.,

e w = z + √ (1 + z 2 ).

To prove that cosh −1 z = log [z + √ (z 2 − 1)].

Proof:

Let cosh −1 z = w. Then cosh w = z .

…(1) 2

sinh w = √ (cosh w − 1) = √ (z

∴

2

− 1).

…(2)

Adding (1) and (2), we get cosh w + sinh w = z + √ (z 2 − 1) ∴

w = log [z + √ (z 2 − 1)].

Hence

cosh −1 z = log [ z + √ ( z2 − 1)]. 1+ z 1 log ⋅ 2 1− z

(iii) To prove that tanh −1 z = Proof: or

e w = z + √ (z 2 − 1).

or

(Bundelkhand 2005)

Let tanh −1 z = w. Then tanh w = z e w − e −w e

w

+e

=

−w

z ⋅ 1

Applying componendo and dividendo, we have (e w + e − w ) + (e w − e − w ) (e or ∴

w

+e

2e w 2e

−w

=

−w

) − (e

1+ z 1− z

2w = log

1+ z 1− z

w

−e

or or

−w

)

=

1+ z 1− z

e2 w = w=

1+ z 1− z

⋅

1+ z 1 log ⋅ 2 1− z

T-64

Hence (iv)

tanh −1 z =

z +1 1 log . 2 z −1

Similarly, we can prove that z +1 1 coth −1 z = log . 2 z −1

(Meerut 2007B)

4.3 Relations between Inverse Hyperbolic Functions and Inverse Circular Functions (i) To prove that sinh −1 x = − i sin −1 (ix). Proof:

Let sinh −1 x = y. Then x = sinh y.

∴

ix = i sinh y = sin (iy).

∴

iy = sin −1 (ix)

Hence

sinh −1 x = − i sin −1 ( ix).

(ii)

y = (1 / i) sin −1 (ix) = − i sin −1 (ix).

or

To prove that cosh −1 x = − i cos −1 ( x).

Proof:

Let cosh −1 x = y. Then x = cosh y = cos (iy).

∴

iy = cos −1 x

Hence

cosh −1 x = − i cos −1 x .

(iii)

or

y = (1 / i) cos −1 x = − i cos −1 x.

To prove that tanh −1 x = − i tan −1 (ix).

Proof:

Let tanh −1 x = y. Then x = tanh y.

∴

ix = i tanh y = tan (iy).

∴

iy = tan −1 (ix)

Hence

tanh −1 x = − i tan −1 ( ix).

or

y = (1 / i) tan −1 (ix) = − i tan −1 (ix).

Example 1: Express cos −1 ( x + iy) in the form A + iB. (Kanpur 2009; Avadh 08) Solution: Let cos −1 ( x + iy) = A + iB. Then

cos ( A + iB) = x + iy

or

cos A cos (iB) − sin A sin (iB) = x + iy

or

cos A cosh B − sin A sinh B = x + iy.

T-65

Equating real and imaginary parts on both sides, we get

and

cos A cosh B = x,

…(1)

sin A sinh B = − y.

…(2)

Let us first eliminate B between (1) and (2). From (1) and (2), we have cosh B = x / cos A and sinh B = − y / sin A. ∴

x2 cos 2 A

y2

−

sin2 A

= cosh2 B − sinh2 B = 1

or

x 2 sin2 A − y 2 cos 2 A = cos 2 A sin2 A

or

x 2 sin2 A − y 2 (1 − sin2 A) = (1 − sin2 A) sin2 A

or

x 2 sin2 A − y 2 + y 2 sin2 A = sin2 A − sin4 A

or

sin4 A + ( x 2 + y 2 − 1) sin2 A − y 2 = 0,

which is a quadratic in sin2 A. ∴

sin2 A =

− ( x 2 + y 2 − 1) ± √ [( x 2 + y 2 − 1)2 + 4 y 2 ]

⋅

2

Since sin2 A must be positive, therefore neglecting the –ive sign, we have 2

sin A =

∴

or

√ {( x 2 + y 2 − 1)2 + 4 y 2 } − ( x 2 + y 2 − 1) 2

⋅

 √ {( x 2 + y 2 − 1)2 + 4 y 2 } − ( x 2 + y 2 − 1) sin A = ±  2  A = ± sin

−1

1 /2

  

 √ {( x 2 + y 2 − 1)2 + 4 y 2 } − ( x 2 + y 2 − 1)  2 

1 /2

  

Now let us eliminate A between (1) and (2). From (1) and (2), we have cos A = x / cosh B ∴

x

2

cosh

2

B

+

y

2

sinh2 B

and

sin A = − y / sinh B.

= cos 2 A + sin2 A = 1

or

x 2 sinh2 B + y 2 cosh2 B = cosh2 B sinh2 B

or

x 2 sinh2 B + y 2 (1 + sinh2 B) = (1 + sinh2 B) sinh2 B

or

x 2 sinh2 B + y 2 + y 2 sinh2 B = sinh2 B + sinh4 B

or

sinh4 B + (1 − x 2 − y 2 ) sinh2 B − y 2 = 0,

which is a quadratic in sinh2 B.

…(3)

T-66

sinh2 B =

∴

−(1 − x 2 − y 2 ) ± √ {(1 − x 2 − y 2 )2 + 4 y 2 } 2

⋅

But sinh2 B is positive, so neglecting the –ive sign, we have sinh2 B =

√ {(1 − x 2 − y 2 )2 + 4 y 2 } − (1 − x 2 − y 2 ) 2

⋅ 1 /2

 √ {(1 − x 2 − y 2 )2 + 4 y 2 } − (1 − x 2 − y 2 ) sinh B = ±   2  

∴

or

B = ± sinh

−1

 √ {(1 − x 2 − y 2 )2 + 4 y 2 } − (1 − x 2 − y 2 )  2 

1 /2

  

⋅ …(4)

Remark: The general value of cos

−1

( x + iy) i. e., Cos

−1

( x + iy) is given by

Cos −1 ( x + iy) = 2nπ ± cos −1 ( x + iy) = 2nπ ± ( A + iB), where A and B are as found in (3) and (4). Explain the meaning of sin −1 x, when x is real and greater than1, and find the

Example 2:

value of Sin −1 2. Solution:

We know that if α is a real number, then sin α lies between −1 and 1.

Therefore if x is a real number greater than 1, then sin −1 x is not a real number, but is a complex number. We have Sin −1 2 = nπ + (−1) n sin −1 2. Let sin −1 2 = u + iv, where u and v are real. Then

sin (u + iv) = 2

or

sin u cos iv + cos u sin iv = 2

or

sin u cosh v + i cos u sinh v = 2.

Equating real and imaginary parts, we get sin u cosh v = 2, and

…(1)

cos u sinh v = 0.

…(2)

From (2), cos u sinh v = 0. But sinh v = 0 gives v = 0 which makes sin we have sinh v ≠ 0. ∴

cos u = 0, or u = ±

Putting u = ±

1 π in (1), we get 2 cosh v = ± 2.

1 π. 2

−1

2 real. So

T-67

v = cosh −1 (± 2) = log (√ 3 ± 2).

∴

[ ∵ cosh −1 x = log { x + √ ( x 2 − 1)}] 1 π + i log (√ 3 ± 2). 2 1 2 = nπ + (−1) n [± π + i log (√ 3 ± 2)]. 2

∴

sin −1 2 = u + iv = ±

∴

Sin −1

Example 3:

Express tan −1 ( x + iy) as the sum of real and imaginary parts. (Rohilkhand 2007; Agra 08; Bundelkhand 08; Meerut 10B; Purvanchal 10; Avadh 12)

Solution: Let tan −1 ( x + iy) = A + iB. Then

tan ( A + iB) = x + iy,

and

tan ( A − iB) = x − iy, by equating complex conjugates.

Now

tan 2 A = tan [( A + iB) + ( A − iB)] tan ( A + iB) + tan ( A − iB) ( x + iy) + ( x − iy) = = 1 − tan ( A + iB) . tan ( A − iB) 1 − ( x + iy) ( x − iy) =

2x 1 − (x

2

2

+ y )

2x

=

2x 1− x

2

− y2

⋅

1 2x tan −1 ⋅ 2 2 1 − x − y2

∴

2 A = tan −1

Again

tan (2iB) = tan [( A + iB) − ( A − iB)] tan ( A + iB) − tan ( A − iB) ( x + iy) − ( x − iy) = = 1 + tan ( A + iB) tan ( A − iB) 1 + ( x + iy) ( x − iy)

1− x

= ∴

i tanh 2 B

=

or

tanh 2 B

=

or Hence

B=

− y

or

2

2iy 1 + x2 + y2 2iy 1 + x2 + y2 2y 1+ x

2

+ y

2

A=

⋅ [∵ tan (iθ) = i tanh θ] or

2 B = tanh −1

2y 1 + x2 + y2

2y 1 tanh −1 ⋅ 2 1 + x2 + y2

tan −1 ( x + iy) = A + iB =

Note:

2

2y 1 2x 1 −1 tan −1 + i tanh ⋅ 2 2 1 − x2 − y2 1 + x2 + y2

If we are to find the general value Tan −1 ( x + iy), then Tan −1 ( x + iy) = nπ + tan −1 ( x + iy) = nπ +

2y 1 2x 1 tan −1 + i tanh −1 ⋅ 2 2 2 2 1− x − y 1 + x2 + y2

T-68

Example 4:

Prove that sinh −1 (cot x) = log (cot x + cosec x).

Solution: Let sinh Then

−1

(Kanpur 2014)

(cot x) = y.

sinh y = cot x.

…(1)

cosh y = √ (1 + sinh

∴

= √ (1 + cot

2

2

y)

x) = cosec x.

…(2)

Adding (1) and (2), we have sinh y + cosh y = cot x + cosec x y

or

e

or

y = log (cot x + cosec x).

= cot x + cosec x

sinh −1 (cot x) = log (cot x + cosec x).

∴

Example 5: Express tanh −1 ( x + iy) into the form α + iβ and hence deduce the value of tanh −1 (iy).

(Avadh 2014)

Solution: Let tanh

−1

( x + iy) = α + iβ

so that

x + iy = tanh (α + iβ) = (1 / i) tan [i (α + iβ)]. [ ∵ tan iθ = i tanh θ]

∴

i ( x + iy) = tan (iα − β) or

or

− ( y − ix) = − tan (β − iα)

or

tan (β − iα) = y − ix.

ix − y = tan {− (β − iα)} [ ∵ tan (− z ) = − tan z ] …(1)

Equating the complex conjugates of both sides of (1), we have tan (β + iα) = y + ix. Now

…(2)

tan 2β = tan [(β − iα) + (β + iα)] tan (β − iα) + tan (β + iα) ( y − ix) + ( y + ix) = = , 1 − tan (β − iα) tan (β + iα) 1 − ( y − ix) ( y + ix) from (1) and (2) =

∴ Again,

1 − ( y2 + x2 )

2β = tan −1

=

2y 1 − x2 − y2

2y 1 − x2 − y2

or

⋅

β=

1 tan −1 2

2y     1 − x 2 − y 2   

tan (2iα) = tan [(β + iα) − (β − iα)] tan (β + iα) − tan ( β − iα) ( y + ix) − ( y − ix) = = 1 + tan (β + iα) tan (β − iα) 1 + ( y + ix) ( y − ix) =

∴

2y

2ix 1 + y2 + x2

i tanh 2α =

⋅ 2ix

1+ x

2

+ y

2

or

tanh 2α =

2x 1 + x2 + y2

…(3)

T-69

1 tanh −1 2

  2x   1 + x 2 + y 2  ⋅  

or

α=

∴

tanh −1 ( x + iy) = α + iβ =

1 tanh −1 2

…(4)

2y   2x 1   + i tan −1 ⋅ 1 + x 2 + y 2  2 (1 − x 2 − y 2 )  

…(5)

[From (3) and (4)] To deduce the value of tanh tanh −1 (iy) =

Then

=0 +i

–1

( iy ), put x = 0 on both sides of (5).

2y 1 1 tanh −1 0 + i tan −1 2 2 1 − y2 2y   −1 y = tan −1  ∵ 2 tan  1 − y 2  

1 . 2 tan −1 y 2

= i tan −1 y. tanh −1 (iy) = i tan −1 y.

∴

Example 6: If sin −1 (θ + iφ) = α + iβ, then prove that sin2 α and cosh2 β are the roots of the equation x 2 − x (1 + θ2 + φ2 ) + θ2 = 0. Solution:

We have sin −1 (θ + iφ) = α + iβ so that θ + iφ = sin (α + iβ)

or

θ + iφ = sin α cosh β + i cos α sinh β.

Equating real and imaginary parts, we have sin α cosh β = θ, and

…(1)

cos α sinh β = φ. 2

2

2

…(2) 2

2

2

Now 1 + θ + φ = 1 + sin α cosh β + cos α sinh β, from (1) and (2) = 1 + sin2 α cosh2 β + (1 − sin2 α) (cosh2 β − 1) = 1 + sin2 α cosh2 β + cosh2 β − 1 − sin2 α cosh2 β + sin2 α = sin2 α + cosh2 β. Thus

sin2 α + cosh2 β = 1 + θ2 + φ2 .

Also from (1), sin2 α cosh2 β = θ2 . ∴

sin2 α and cosh2 β are the roots of the equation x 2 − (sin2 α + cosh2 β) x + sin2 α cosh2 β = 0

or

x 2 − (1 + θ2 + φ2 ) x + θ2 = 0.

T-70

Comprehensive Exercise 1 1. Show that sin −1 (cos θ + i sin θ) = cos −1 √ (sin θ) + i log [√ (sin θ) + √ (1 + sin θ)], where θ is a positive acute angle. (Purvanchal 2006; Rohilkhand 07; Agra 09; Avadh 10)

2. (i)

Express sin

−1

(ii) Express cosh 3. (i)

( x + iy) in the form A + iB.

−1

(Purvanchal 2007)

( x + iy) in the form α + iβ.

Show that cos −1 (cos θ + i sin θ) = sin −1 √ (sin θ) + i log { √ (1 + sin θ) − √ (sin θ)}, where θ is a positive acute angle.

(ii) Separate cos −1 (cos θ + i sin θ) into real and imaginary parts. 4. If cos −1 (α + iβ) = θ + iφ, then prove that (i)

α 2 sech2 φ + β 2 cosech2 φ = 1,

(ii) α 2 sec 2 θ − β 2 cosec 2 θ = 1.

5. If sin −1 ( x + iy) = tan −1 (u + iv), show that [( x − 1)2 + y 2 ] [( x + 1)2 + y 2 ] = 6.

Prove that Sin −1 (cosec θ) = {2n + (−1) n}

( x 2 + y 2 )2 (u2 + v2 )2

⋅

1 1 π + i (−1) n log cot θ. 2 2 (Gorakhpur 2007)

7. Show that Sin

−1

n

2

(ix) = nπ + i (−1) log {x + √ (1 + x )}.

8. (i) Prove that Tan −1 (cos θ + i sin θ) = nπ +

1 1 1 1 π − i log tan ( π − θ). 4 2 4 2

(Gorakhpur 2005; Avadh 09; Rohilkhand 14)

(ii) Prove that Tan −1 (cos θ + i sin θ) = nπ +

1 1 1 1 π + i log tan ( π + θ). 4 2 4 2 (Bundelkhand 2009)

9. If x > y, then show that tan

−1

 x + iy  π x+ y i ⋅   = + log x− y  x − iy  4 2 (Meerut 2011, 12, 12B)

 x − a 1  a 10. Prove that tan −1 i  = − i log   ⋅ 2 x (Meerut 2004B; Purvanchal 09)  x + a 11. If cosh x = sec θ, then prove that x = log (sec θ ± tan θ).

T-71

12. (i) Show that sinh −1 x = tanh −1

x √ (1 + x 2 )

⋅

(ii) Prove that tanh −1 x = sinh −1 { x / √ (1 − x 2 )}. (iii) Prove that coth

−1

(2 / x) = sinh

−1

(Kashi 2012) 2

{ x / √ (4 − x )}.

13. Prove that  tan 2θ + tanh 2φ −1 −1  tan θ − tanh φ Tan −1   + Tan   = Tan (cot θ coth φ).  tan 2θ − tanh 2φ  tan θ + tanh φ 14. If cos −1 (u + iv) = α + iβ, prove that cos2 α and cosh2 β are the roots of the equation x 2 − (1 + u2 + v2 ) x + u2 = 0. 15. If cosh

−1

( x + iy) + cosh

show that 2 (a − 1) x

2

−1

(Agra 2007; Kanpur 09)

( x − iy) = cosh

+ 2 (a + 1) y

2

−1

a,

2

= a − 1. (Meerut 2006; Kanpur 10)

A nswers 1

2. (i)

1 /2

 √ {( x 2 + y 2 − 1)2 + 4 y 2 } − ( x 2 + y 2 − 1) 1 π ± sin −1  2 2  ± i sinh

−1

 √ {(1 − x 2 − y 2 )2 + 4 y 2 } − (1 − x 2 − y 2 )  2 

  

1 /2

  

1 /2

(ii) α = ± sinh

−1

 √ {(1 − x 2 − y 2 )2 + 4 y 2 } − (1 − x 2 − y 2 )   2  

1 /2

β = ± sin

−1

 √ {( x 2 + y 2 − 1)2 + 4 y 2 } − ( x 2 + y 2 − 1)   2  

O bjective T ype Q uestions

Fill In The Blanks Fill in the blanks “……” so that the following statements are complete and correct. 1. If w and z are any complex numbers then we define w = sinh −1 z if …… . 2. sinh −1 z = log [z + ……]. 3. cosh −1 z = log [z + ……].

T-72

4. tanh −1 z = …… log

1+ z 1− z

⋅

5. If w and z are any complex numbers then we define w = tanh −1 z if …… .

True or False Write ‘T’ for true and ‘F’ for false statement. 1. sinh −1 x = i sin −1 (ix). 2. coth −1 z =

z +1 1 log ⋅ 2 z −1

3. cosh −1 z = log [z + √ (1 − z 2 )]. 4. tanh −1 x = − i tan −1 (ix). 5.

cosh −1 x = i cos −1 x.

A nswers Fill in the Blanks 2. √ (z 2 + 1)

1. sinh w = z 1 4. 2

3. √ (z 2 − 1)

5. tanh w = z

True or False 1.

F

2. T

3.

F

4. T

5.

F

¨

T-73

5 G regory's S eries

5.1 Gregory’s Series o prove that, if θ lies within the closed interval [−π / 4, π / 4] , i.e., if − π / 4 ≤ θ ≤ π / 4, then 1 1 1 θ = tan θ − tan3 θ + tan5 θ − tan7 θ + …… ad. inf . 3 5 7

T

(Kanpur 2005, 08; Meerut 10B; Avadh 06; Bundelkhand 14; Agra 14)

Proof:

We have  sin θ  1 (1 + i tan θ) = 1 + i (cos θ + i sin θ) = sec θ . e i θ .  = cos θ cos θ 

Taking logarithm of both sides, we have log (1 + i tan θ) = log sec θ + log e i θ , (considering only principal values) or

log (1 + i tan θ) = log sec θ + iθ.

Now since θ lies between −π / 4 and π / 4, tan θ lies between −1 and 1, i. e., tan θ is numerically not greater than unity.

T-74

Therefore, log sec θ + iθ = log (1 + i tan θ) = i tan θ −

i2 tan2 θ 2

+

i3 tan3 θ 3

−

i4 tan4 θ 4

+ ……,

…(1)

(expanding the R.H.S. by logarithmic series which is justified because |i tan θ | = | tan θ | ≤ 1) 1 1 1 1 2 = i tan θ + tan θ − i tan3 θ − tan4 θ + i tan5 θ + … 2 3 4 5 Equating the imaginary parts on both sides, we have 1 1 θ = tan θ − tan 3 θ + tan 5 θ − ...... 3 5 The general term on the R.H.S. is

(−1)

n −1

2n − 1

…(2)

tan2 n −1 θ.

This expansion (2) is known as Gregory’s series after the name of James Gregory (1638 – 1675). Another Form of Gregory’s Series. In the series (2) if we put tan θ = x so that θ = tan −1 x, then we have another form of the Gregory’s series as x3 x5 x7 + − + ...... ad. inf. 3 5 7 where – 1 ≤ x ≤ 1 i. e., | x | ≤ 1. tan −1 x = x −

Corollary:

Equating real parts from both sides of (1), we have 1 1 1 log sec θ = tan2 θ − tan4 θ + tan6 θ − …… 2 4 6

(Bundelkhand 2008)

5.2 General Theorem on Gregory’s Series 1 1 π and nπ + π, both limits being inclusive, then 4 4 1 1 θ − nπ = tan θ − tan3 θ + tan5 θ − …… ad. inf . 3 5

If θ lies between nπ −

Proof:

Let

θ − nπ = φ

Then φ lies between − Now

i. e.,

θ = nπ + φ.

π π and ⋅ 4 4

1 + i tan θ = 1 + i tan (nπ + φ) = 1 + i tan φ = sec φ (cos φ + i sin φ) = sec φ . e iφ .

Taking logarithm of both sides, we get log (1 + i tan θ) = log sec φ + log e iφ

T-75

or

log sec φ + iφ = log (1 + i tan θ),

the expansion of which is valid because θ lies between nπ −

1 1 π and nπ + π 4 4

implies that tan θ is not numerically greater than 1. 1 1 ∴ log sec φ + iφ = i tan θ − i 2 tan2 θ + i 3 tan3 θ 2 3 1 4 − i tan4 θ + …… ∞, 4 1 1 [∵ log (1 + z ) = z − z 2 + z 3 − … ∞, if |z | ≤ 1] 2 3 1 1 1 = i tan θ + tan2 θ − i tan3 θ − tan4 θ + …… ∞. 2 3 4 Equating imaginary parts from both sides, we have 1 1 φ = tan θ − tan3 θ + tan5 θ − …… ad. inf. 3 5 1 1 ∴ θ − nπ = tan θ − tan 3 θ + tan 5 θ − ...... ad. inf. 3 5

5.3 Value of π The main use of Gregory’s series is to find the value of π to various decimal places. Some mathematicians have designed different expressions based on Gregory’s series for finding the value of π. Some important cases are given below : (a)

Gregory’s series:

In the Gregory’s series tan −1 x = x −

1 3 1 5 1 7 x + x − x + …… ∞, 3 5 7

if we put x = 1, then we have 1 1 1 1 tan −1 1 = 1 − + − + − …… ∞ or 3 5 7 9

1 1 1 1 1 π = 1 − + − + − …… ∞. 4 3 5 7 9

From this the value of π can be calculated. But this series does not converge rapidly and so a large number of terms will have to be taken in order to evaluate π correct to a certain decimal place. So several other series have been designed for this purpose. (b)

Euler’s series: tan −1

We have 1 1 + 1 1 5/6 π + tan −1 = tan −1 2 3 = tan −1 = tan −1 1 = ⋅ 1 2 3 5/6 4 1− 6

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Thus

π 1 1 = tan −1 + tan −1 4 2 3 1 1 1 1 1   1 1 1 1 1  =  − ⋅ 3 + ⋅ 5 − ...... +  − ⋅ 3 + ⋅ 5 − ...... , 2 3 2  3 3 3  5 2 5 3 1 1 expanding both tan −1 and tan −1 2 3 1 1 by Gregory’s series because < 1 and < 1 2 3 1 1 1 1 1 1 1 1 =  +  −  3 + 3  +  5 + 5  − ……  2 3 3  2 3  5 2 3 

From this the value of π can be calculated. This series is more rapidly convergent than the preceding one. (c) Machin’s series: 4 tan −1

We have 2 /5 1 1 5 = 2 . 2 tan −1 = 2 tan −1 = 2 tan −1 5 5 1 − (1 / 25) 12

120  2 . (5 / 12)  = tan −1  = tan −1 ⋅ 2 119 1 − (5 / 12)  1 1 120 1 Now 4 tan −1 − π = tan −1 − tan −1 1 [ ∵ π = tan −1 1] 5 4 119 4 120 −1 1 −1 119 = tan = tan −1 ⋅ 120 239 1+ ×1 119 π −1 1 −1 1 Therefore, = 4 tan − tan 4 5 239  π 1 1 1 1 1 1 1 1 1   1 or = 4  − ⋅ 3 + ⋅ 5 − …… −  − ⋅ + ⋅ − ……, 3 5 4 5 5 5 (239) 5 3 5  239 3 (239)  1 expanding both the terms on the R.H.S. by Gregory’s series because < 1 and 5 1 < 1. 239 The above is Machin’s series and it is more rapidly convergent than Euler’s series. (d) Rutherford’s Series: tan −1

We have

1 1 − 99 − 70 1 1 − tan −1 = tan −1 70 99 = tan −1 1 1 70 99 70 . 99 + 1 1+ ⋅ 70 99 29 1 1 π = tan −1 = tan −1 = 4 tan −1 − , 6931 239 5 4 as shown in Machin’s series.

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∴

π 1 1 1 = 4 tan −1 − tan −1 + tan −1 4 5 70 99

i. e.,

π 1 1 1 1 1 1 1 1 1   1  = 4  − ⋅ 3 + ⋅ 5 −… −  − ⋅ + ⋅ −… 3 5 4 5 5 5 (70) 5 3 5   70 3 (70)   1 1  1 1 1 + − ⋅ + ⋅ − ...... . 3 5 5 (99)  99 3 (99) 

It is Rutherford’s series. This is more convenient for expansion than Machin’s series and converges equally rapidly.

Example 1:

Assuming that θ − nπ = tan θ −

1 1 tan3 θ + tan5 θ − … , when θ lies 3 5

1 1 π) and (nπ + π), write down the value of n when θ lies between 4 4 7π 9π − 3π − 5π 11π 13π (i) (ii) (iii) and and and 4 4 4 4 4 4 −15π −17π 19π 21π (iv) and (v) and ⋅ 4 4 4 4 between (nπ −

Solution: (i) θ lies between (7π / 4) and (9π / 4) 1 1 π) and (2π + π). ∴ n = 2. 4 4 (ii) θ lies between −(3π / 4) and −(5π / 4) 1 1 i. e., between (− π + π) and (− π − π). ∴ n = − 1. 4 4 (iii) θ lies between (11π / 4) and (13π / 4) 1 1 i. e., between (3π − π) and (3π + π). ∴ n = 3. 4 4 (iv) θ lies between −(15 π / 4) and −(17π / 4) 1 1 i. e., between (− 4π + π) and (− 4π − π). ∴ n = 4. 4 4 (v) θ lies between (19π / 4) and (21π / 4) 1 1 i. e., between (5π − π) and (5π + π). ∴ n = 5. 4 4 1 1 1 Example 2: Sum the series − + − …… ad. inf . 3 7 2 3 .2 5 . 211 (Meerut 2013B; Kanpur 14) 1 1 1 Solution: We have 3 − + + …… 7 2 3 .2 5 . 211 i. e., between (2π −

=

1 2

1  1 1 + − ……,  2 − 6 10 3 .2 5 .2 2 

(taking

1 common) 2

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=

1 2

 1  1 1  1  + …… = tan −1  2  ,  2 − 2 3  2  (2 ) 3 (2 )  2 by Gregory’s series because 1 / 22 < 1. tan

−1

[Note that by Gregory’s series 1 1 x = x − x 3 + x 5 − … ∞, if | x| ≤ 1] 3 5

1 1 tan −1 ⋅ 2 4 Example 3: Sum to infinity the series 1 1 (i) 1 − + − … ad. inf . ; 2 3 .4 5 . 44 (Meerut 2013) 1 1 1 (ii) 1 − 2 + − + … ad. inf . 3 5 . 32 7 . 33 1 1 Solution: (i) The given series 1 − + − … ad. inf. 2 3 .4 5 . 44 =

1  1 1 =4 − + − … ad. inf. , (taking 4 common) 3 5 . 45 4 3 . 4  1 = 4 tan −1 , by Gregory’s series because 1 / 4 < 1. 4 (ii) The given series 1 1 1 1 1 1 1− 2 + − + …= 1 − + − +… 2 3 2 3 .3 5 .3 3 5 .3 7 .3 7 . 33  1  1 1 1 = √3  − + − + …  3 5 . (√ 3)5 7 . (√ 3)7  √ 3 3 . (√ 3)  −1 by Greg ory’s series because 1 / √ 3 < 1 = √ 3 { tan (1 / √ 3)}, = √ 3 . (π / 6) = (π √ 3) / 6. Note: This question can also be written as   1 1 1 Prove that π = 2 √ 3 1 − 2 + − + … ⋅ 2 3 3 5 .3 7 .3   (Agra 2005; Rohilkhand 08; Kashi 12)

Example 4: Prove that (−1) n +1  2 1− n π 17 713 = − + …… + + 71−2 n  + …  9  4 21 81 × 343 2n − 1  3 (Kanpur 2008, 10; Agra 08)

Solution: The nth term of the given series on the R.H.S. is Tn =

(−1) n +1  2 (−1) n +1  2 2 −2 n 2 1− n + 71−2 n  = + 71−2 n   ⋅ (3 )  ⋅3   2n − 1  3 2n − 1  3

T-79

=

(−1) n +1  2 1  + 2 n −1  ⋅  2 n − 1 (2n − 1) 3  7

Putting n = 1, 2, 3, …… etc., we have 2 1 first term = T1 =  +  ; 3 7  1 2 1 2 1 second term = T2 = −  3 + 3  = − − ; 3 3 7  33 . 3 37 . 3 1 2 1  2 1 1 third term = T3 =  5 + 5  = ⋅ 5 + ; and so on.  5 3 5 7 3 57 . 5 ∴ the sum of the series on the R.H.S. = T1 + T2 + T3 + … ad. inf. 2 1 1   2 1   2 =  +  −  + + +  −… 3 3 5  3 7   33   . 37 . 53 . 57 . 5 1 1 1 1 1  1  =2 − + − … +  − + − … 3 5 3 5  3 33    7 37 . 53 . . 57 . −1 −1 (by Gregory’s series) = 2 tan (1 / 3) + tan (1 / 7), 1 1   +  1 3 1 −1  3 3 = tan  + tan −1 = tan −1 + tan −1 1 1 7 4 7 1 − ⋅   3 3  3+1  −1  4 7  = tan −1 1 = π ⋅ = tan  3 1 4 1 − ⋅   4 7 Express tan −1 (cos θ + i sin θ) in the form A + iB and deduce that 1 1 1 (i) cos θ − cos 3θ + cos 5θ − … = ± π ; and (Kanpur 2014) 3 5 4 1 1 1 1 1 (ii) sin θ − sin 3θ + sin 5θ − … = log { ± tan ( π + θ)}. 3 5 2 4 2 Solution: Let tan −1 (cos θ + i sin θ) = A + iB Example 5:

so that

tan −1 (cos θ − i sin θ) = A − iB,

∴

2 A = ( A + iB) + ( A − iB) = tan −1 (cos θ + i sin θ) + tan −1 (cos θ − i sin θ)

(complex conjugates).

 (cos θ + i sin θ) + (cos θ − i sin θ)  = tan −1   1 − (cos θ + i sin θ)(cos θ − i sin θ) 2 cos θ   −1  2 cos θ = tan −1    = tan  2 2  1−1  1 − (cos θ + sin θ) 1  2 cos θ −1 = tan −1   = tan (± ∞) = ± π.  0  2 [ ∵ θ lies between – π and π]

T-80

1 π. 4 2iB = ( A + iB) − ( A − iB) = tan −1 (cos θ + i sin θ) − tan −1 (cos θ − i sin θ) Therefore,

Again,

or or

A= ±

…(1)

 (cos θ + i sin θ) − (cos θ − i sin θ)  = tan −1   1 + (cos θ + i sin θ)(cos θ − i sin θ)   2i sin θ −1 −1  2i sin θ = tan −1   = tan   = tan (i sin θ) 2 2 1 + (cos θ + sin θ)  1+1  tan (2iB) = i sin θ or tanh 2B = sin θ e 2 B − e −2 B sin θ 1 + sin θ 2e2 B or = = 2B −2 B − 2 B 1 1 − sin θ e +e 2e

i. e.,

e4 B =

or

e2 B

∴ Now tan −1

1 + sin θ

[By componendo and dividendo] {cos (θ / 2) + sin (θ / 2)}2

= 1 − sin θ {cos (θ / 2) − sin (θ / 2)}2 {cos (θ / 2) + sin (θ / 2)} {1 + tan (θ / 2)} π θ =± =± = ± tan  +  ⋅  {cos (θ / 2) − sin (θ / 2)} {1 − tan (θ / 2)} 4 2

1 π θ   log  ± tan  +    4 2  2  (cos θ + i sin θ) = A + iB B=

…(2)

1 3i θ 1 5i θ e + e − … = A + iB, 3 5 using Gregory’s series 1 1 or (cos θ + i sin θ) − (cos 3θ + i sin 3θ) + (cos 5θ + i sin 5θ) − … = A + iB 3 5 1 1 1 1 or (cos θ − cos 3θ + cos 5θ − …) + i (sin θ − sin 3θ + sin 5θ − …) 3 5 3 5 1 1 1 1 by (1) and (2). = A + iB = ± π + i log {± tan ( π + θ)}, 4 2 4 2 Equating real and imaginary parts, we have 1 1 1 cos θ − cos 3θ + cos 5θ − … = ± π 3 5 4 1 1 1 1 1 and sin θ − sin 3θ + sin 5θ − … = log { ± tan ( π + θ)}. 3 5 2 4 2 or

tan −1 (e i θ ) = A + iB or

ei θ −

Comprehensive Exercise 1

1. (i) Prove that

π 2 1 1  2 1  1 2 1 = + − + + + − ... 4 3 7 3 33 73  5 35 75  (Kanpur 2005, 08, 10; Agra 08; Bundelkhand 09; Purvanchal 10; Meerut 12; Kashi 13)

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(ii) Prove that

π 1 1 1 = + + + …ad. inf. 8 1.3 5 .7 9 .11 (Meerut 2004B, 07, 09, 13B; Agra 07, 08; Kanpur 12)

(iii) Show that π 1  1  = 1 − 1 /2  − 12  3  2

1  1  1 − 3 /2  +   5 3

1   1 − 5 /2  − … ∞.   3

(Meerut 2012B, 13)

(iv) Show that π  1 1 1 1  1 1 1 11 1 1 = + + −  3 + 3 + 3  +  5 + 5 + 5  − ……   4 2 5 8 3 2 5 8  5 2 5 8  1 3 1 5 2. If x < (√ 2 − 1), prove that 2 ( x − x + x − … ad. inf.) 3 5  2x  1  2x  = −   1 − x 2  3 1 − x 2  3. If x > 0, prove that tan −1 x =

3

+

1  2x    5 1 − x 2 

5

− ... (Kashi 2014)

π x − 1 1  x − 1 + −   4 x + 1 3  x + 1

3

+… (Garhwal 2000)

1 4. When θ lies between 0 and π, prove that 2 1 θ 1 θ −1 1 − cos θ 2 θ tan  − tan6 + tan10 − …… ∞.  = tan 1 + cos θ 2 3 2 5 2 (Bundelkhand 2010)

1 5. When both θ and tan (sec θ) lie between 0 and π , prove that 2 1 1 1 1 1 1 tan −1 (sec θ) = π + tan2 θ − tan6 θ + tan10 θ − … 4 2 3 2 5 2 −1

(Kanpur 2006)

6. Prove that  cos θ + sin θ 1 3 5 tan −1   = nπ + π + tan θ − (1 / 3) tan θ + (1 / 5) tan θ − … 4  cos θ − sin θ (Meerut 2006; Kanpur 14)

7. (i) If x lies between −π / 4 and π / 4, show that 1 1 1 tan x − tan3 x + tan5 x − … ad. inf. = tanh x + tanh3 x 3 5 3 1 5 + tanh x + … ad. inf. 5 (Kumaun 2000) (ii) If φ lies between (π / 4) and (3π / 4), show that 1 1 1 φ = π − cot φ + cot 3 φ − cot 5 φ + …… ∞. 2 3 5 (Kanpur 2005) 8. If tan x < 1, show that tan2 x −

1 1 1 1 tan4 x + tan6 x − … = sin2 x + sin4 x + sin6 x + … 2 3 2 3 (Meerut 2011)

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9. Prove that if x, y, z are cube roots of unity, tan −1 y tan −1 z tan −1 x + + =3 x y z

1 − 1 + 1 − 1 + 1 − … ⋅  7 13 19 25  (Kanpur 2009)

O bjective T ype Q uestions

Fill in the Blanks Fill in the blanks “……” so that the following statements are complete and correct. π π 1. If − ≤ θ ≤ , then by Gregory’s series, we have θ = …… . 4 4 2. If | x | ≤ 1, then by Gregory’s series, we have tan −1 x = …… . 3. The sum of the infinite series 1 1 1 − + − …… ∞ is …… . 3 7 2 3 .2 5 . 211

True or False Write ‘T’ for true and ‘F’ for false statement. 1 1 1 1. 1 − + − ...... ∞ = 4 tan −1 2 4 4 3 .4 5 .4 2. If | x | ≤ 1, then tan −1 x = x +

x3 x5 x7 + + + ...... ∞. 3 5 7

A nswers Fill in the Blanks 1. tan θ −

tan3 θ

+

tan5 θ

−

3 5 3 5 7 x x x 2. x − + − + …… ∞ 3 5 7 1 1 3. tan −1 2 4

tan7 θ 7

+ …… ∞

True or False 1. T

2.

F

¨

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6 S ummation of T rigonometrical S eries

6.1 Introduction n this chapter we shall give some important methods for summing up trigonometric series which may be finite or infinite. There are two main methods for summation, known as the C + iS method and the difference method. First we shall discuss the C + iS method.

I

6.2 C + iS Method for Summing Up Trigonometric Series Consider the series and

C = a0 cos α + a1 cos (α + β) + a2 cos (α + 2β) + …

…(1)

S = a0 sin α + a1 sin (α + β) + a2 sin (α + 2β) + …

…(2)

These series may be finite or infinite. The coefficients a0 , a1 , a2 etc. and α, β may be any numbers real or complex. The series (1) and (2) are companion series because each is summed up with the help of the other.

T-84

In the series (1) we have terms which contain cosines of numbers which increase in arithmetic progression. It is called cosine series and its sum is denoted by C.The series (2) is obtained from the series (1) by simply replacing cos by sin in various terms where we have cosines of numbers which increase in arithmetic progression, the coefficients a0 , a1 , a2 etc. are kept as they are. The series (2) is called sine series and its sum is denoted by S. In case we are given one of these two series and are required to find its sum, we first write the other series called its companion series or its auxiliary series. After writing both C and S, we find C + iS and C − iS by making use of the Euler’s theorem e iz = cos z + i sin z , and e − iz = cos z − i sin z . Thus we get

and

C + iS = a0 e iα + a1 e i (α + β) + a2 e i (α + 2β) + …

…(3)

C − iS = a0 e − iα + a1 e − i (α + β) + a2 e − i (α + 2β) + …

…(4)

The series in (3) and (4) can generally be summed up easily. Depending upon the coefficients a0 , a1 , a2 etc., these series are generally in one of the following forms : (i)

Series in geometric progression, or arithmetico-geometric series.

(ii)

Binomial series.

(iii) Exponential series or its sub-case sine or cosine series. (iv) Logarithmic series or its sub-case Gregory’s series. Having found the sums of the series (3) and (4), we use 1 C = [(C + iS) + (C − iS)] and S = (1 / 2i) [(C + iS) − (C − iS )] 2 to find the values of C and S respectively. In the above discussion the coefficients a0 , a1 , a2 etc. and α, β are real or complex. However, if these are real, we need not find C − iS. To get C and S we simply find the sum of the series (3) i. e., we find C + iS. After separating C + iS into real and imaginary parts, we get the values of C and S by equating the real and imaginary parts. Remark: We are always to find C + iS and never S + iC whether we first write C or S. Now we shall illustrate C + iS method by taking examples based upon series of different types.

6.3 Series Based on Geometric Progression or Arithmetico-Geometric Series We know that a geometric progression (G.P.) is of the form a, ar, ar 2 , ar 3 , …, ar n , … ∞.

T-85

Its common ratio is r and nth term is ar n −1 . Sum of n terms in G.P. i. e.,

1 − r n   a + ar + ar 2 + ar 3 + … + ar n −1 = a    1− r 

or

 r n − 1 ⋅ = a    r −1

Sum of an infinite series in geometrical progression i. e., a + ar + ar 2 + ar 3 + … + ar n + … ad. inf. = a / (1 − r), provided |r | < 1. An arithmetico-geometric progression is of the form a, (a + d ) r, (a + 2d ) r 2 , (a + 3d ) r 3 , … .

Example 1: Sum the series sin α + c sin (α + β) + c 2 sin (α + 2β) + … upto n terms. Deduce the sum to infinity if |c | < 1 i. e., − 1 < c < 1. Solution: Let the given series be denoted by S. Then S = sin α + c sin (α + β) + c 2 sin (α + 2β) + … to n terms. So the auxiliary series is given by C = cos α + c cos (α + β) + c 2 cos (α + 2β) + … to n terms. Now multiplying the first series by i and adding to the second, we get C + iS = (cos α + i sin α) + c {cos (α + β) + i sin (α + β)} + c 2 {cos (α + 2β) + i sin (α + 2β)} + … to n terms = e iα + ce i (α + β) + c 2 e i (α + 2β) + …

to n terms.

…(1)

The series on the R.H.S. is a geometrical progression having n terms. The first term is e iα and the common ratio is ce iβ . Therefore summing up the G.P., we have C + iS =

=

e iα {1 − (ce iβ ) n} 1 − ce iβ [e iα − c

n

=

e i α − c n e i (α + nβ)

e i (α + nβ) ]

(1 − ce iβ )

(1 − ce iβ ) ×

(1 − ce − iβ ) (1 − ce − iβ )

,

multiplying the Nr.

and the Dr. by the conjugate complex of the Dr. =

e iα − c n e i (α + nβ) − ce i (α − β) + c n +1 e i [α + (n −1) β] 1 − c (e iβ + e − iβ ) + c 2 . e iβ . e − iβ

⋅

Now the numerator of the R.H.S. = [(cos α + i sin α) − c n {cos (α + nβ) + i sin (α + nβ)} − c {cos (α − β) + i sin (α − β)} +c

n +1

{cos (α + n − 1 β) + i sin (α + n − 1 β)}]

T-86

= [cos α − c cos (α − β) − c n cos (α + nβ) + c n + 1 cos {α + (n − 1 ) β }] + i [sin α − c sin (α − β) − c n sin (α + nβ) + c n + 1 sin {α + (n − 1) β }], (putting into the form A + iB) and the denominator of the R.H.S. 1 = 1 − 2c . (e i β + e − iβ ) + c 2 e 2

iβ

. e − iβ

= 1 − 2c cos β + c 2 , which is real. Now we are required to find the sum of the sine series (S) which is the imaginary part of C + iS. Hence equating imaginary parts from both sides, we have the required sum S=

sin α − c sin (α − β) − c n sin (α + nβ) + c n +1 sin {α + (n − 1) β } 1 − 2c cos β + c 2

⋅

…(2)

To find sum upto infinity, c must be numerically less than 1, and then c n → 0 as n → ∞. ∴

from (2), as c n → 0, the required sum sin α − c sin (α − β) − 0 . sin (α + nβ) + 0. sin {α + (n − 1) β } S∞ = 1 − 2c cos β + c 2 =

sin α − c sin (α − β) 1 − 2c cos β + c 2

⋅

Example 2: Sum the series 1 + c cosh θ + c 2 cosh 2θ + c 3 cosh 3θ + … to n terms. Solution:

The given series may be written as  e θ + e −θ   e 2 θ + e −2 θ   + c2   + c3 1 + c     2 2    

 e 3 θ + e −3 θ    + … to n terms   2  

(Note) 1 = [2 + c (e θ + e − θ ) + c 2 (e2 θ + e −2 θ) + c 3 (e3 θ + e −3 θ ) + … to n terms] 2 1 = [(1 + ce θ + c 2 e2 θ + c 3 e3 θ + … to n terms) 2 + (1 + ce − θ + c 2 e −2 θ + c 3 e −3 θ + … to n terms)] =

1 2

1 − c n e nθ  1 − c n e − nθ    +  ,   1 − ce θ   1 − ce − θ       summing up both the geometric series upto n terms n nθ

=

1 (1 − c e  2 

) (1 − ce − θ ) + (1 − c n e − nθ ) (1 − ce θ )    (1 − ce θ ) (1 − ce − θ ) 

T-87 n nθ

− ce − θ + c n +1 e(n − 1) θ + 1 − c n e − nθ − ce θ + c n + 1 e −(n −1) θ   1 − ce θ − ce − θ + c 2 e θ . e − θ 

=

1 2

1 − c  

=

1 2

2 − c (e θ + e − θ ) − c n (e nθ + e − nθ ) + c n + 1 { e(n − 1) θ + e − (n − 1) θ }   1 − c (e θ + e − θ ) + c 2  

=

n n +1 {2 cosh (n − 1) θ} 1 2 − c (2 cosh θ) − c (2 cosh nθ) + c   2 2  1 − c (2 cosh θ) + c 

=

e

1 − c cosh θ − c n cosh nθ + c n + 1 cosh (n − 1) θ 1 − 2c cosh θ + c 2

⋅

Example 3: Sum the series 1 − 2 cos α + 3 cos 2α − 4 cos 3α + … to n terms. Solution:

Let C = 1 − 2 cos α + 3 cos 2α − 4 cos 3α + … + (−1) n − 1 n cos {(n − 1) α },

and ∴

S = − 2 sin α + 3 sin 2α − 4 sin 3α + … + (−1) n − 1 n sin {(n − 1) α }. C + iS = 1 − 2 (cos α + i sin α) + 3 (cos 2α + i sin 2α) − 4 (cos 3α + i sin 3α) + …… + (−1) n − 1 n [cos {(n − 1) α } + i sin {(n − 1) α }] = 1 − 2e

or

iα

+ 3e

2 iα

− 4e

3 iα

+ … + (−1)

n −1

ne

i (n − 1) α

C + iS = 1 − 2 x + 3 x 2 − 4 x 3 + … + (−1) n − 1 nx n − 1 ,

…(1) where x = e iα

The right hand side of (1) is an arithmetico-geometrical progression. To sum it up multiply both sides of (1) by − x which is the common ratio of the geometric factors of the terms. So multiplying both sides of (1) by − x, we have (C + iS) (− x) = − x + 2 x 2 − 3 x 3 + 4 x 4 − … + (−1) n nx n .

…(2)

Subtracting (2) from (1), we have (C + iS) (1 + x) = (1 − x + x 2 − x 3 + x 4 − … to n terms) − (−1) n nx n . There are (n + 1) terms in this right hand side series and the first n terms form a geometric progression of common ratio − x. (C + iS) (1 + x) =

∴

1 − (− x) n 1 − (− x)

− (−1) n nx n .

Dividing both sides by (1 + x) and putting x = e (C + iS) =

1 − (−1) n e niα (1 + e iα )2

−

(−1) n ne niα (1 + e iα )

iα

, we have

T-88

=

=

=

=

1 + (−1) n − 1 e ni α (1 + e iα )2

+

(−1) n − 1 ne niα (1 + e iα )

,

[Putting (−1) n = (−1) . (−1) n − 1 ]

[1 + (−1) n − 1 e ni α ] + (1 + e iα ) [(−1) n −1 ne niα ] (1 + e iα )2 1 + (−1) n − 1 [(n + 1) e inα + ne i (n +1) α ]

(Note)

e iα [e − iα /2 + e iα /2 ]2 e − iα + (−1) n − 1 [(n + 1) e i (n − 1) α + ne inα ] (2 cos

1 2 α) 2

,

dividing Nr. and Dr. by e iα

(cos α − i sin α) + (−1) n − 1 {(n + 1) cos (n − 1) α + i (n + 1) sin (n − 1) α + n cos nα + in sin nα } = 2 1 4 cos α 2 [cos α + (−1) n −1 {(n + 1) cos (n − 1) α + n cos nα }] + i [(−1) n − 1 {(n + 1) sin (n − 1) α + n sin nα } − sin α ]

=

2 (1 + cos α)

⋅

Now equating real parts on both sides, we have C=

cos α + (−1) n − 1 {(n + 1) cos (n − 1) α + n cos nα } 2 (1 + cos α)

⋅

Comprehensive Exercise 1 Sum the following series. In the case of infinite series it may be assumed that −1 < c < 1. 1.

cos α + c cos (α + β) + c 2 cos (α + 2β) + … to n terms. Deduce the sum to infinity if |c | < 1 i. e., − 1 < c < 1.

2. (i) cos α + cos (α + β) + cos (α + 2β) + … to n terms. (Purvanchal 2006; Bundelkhand 09; Agra 14)

(ii) sin α + sin (α + β) + sin (α + 2β) + … to n terms. 2

3

3. 1 + c cos α + c cos 2α + c cos 3α + … to n terms.

(Agra 2014) (Purvanchal 2008)

2

4. (i) 1 + c cos α + c cos 2α + … ad. inf. (ii) c sin α + c 2 sin 2α + c 3 sin 3α + … ad. inf. 1 5. sin α + sin 2α + (1 / 22 ) sin 3α + …ad. inf. 2

(Purvanchal 2008)

(Kumaun 2000; Kanpur 07)

T-89

6. (i) cos α sin α + cos 2α sin2 α + cos 3α sin3 α + … ad. inf. (Meerut 2012) 1 (ii) sin α sin α + sin 2α sin2 α + sin 3α sin3 α + … ad. inf ., where α ≠ ± π. 2 (Meerut 2012B) 2

3

7. cos α cos α + cos α cos 2α + cos α cos 3α + … ad. inf . (Bundelkhand 2005, 08)

8. 1 +

cos θ cos θ

+

cos 2θ cos 2 θ

+

cos 3θ cos 3 θ

+ … to n terms.

Or 1 + sec θ cos θ + sec 2 θ cos 2θ + sec 3 θ cos 3θ + … to n terms. 9. 1 + cos θ cos θ + cos 2 θ cos 2θ + cos 3 θ cos 3θ + … ad. inf. sin θ

10. (i)

sin 2θ

sin 3θ

− … ad. inf. π3 sin 3θ (ii) − + − … ad. inf., (tan φ > 1). 2 tan φ tan φ tan3 φ π sin θ

−

2

+

π sin 2θ

11. c sinh θ + c 2 sinh 2θ + c 3 sinh 3θ + … ad. inf.

(Kumaun 2008)

12. 2 sin α − 3 sin 2α + 4 sin 3α − … to n terms. 13. (i) 3 cos θ + 5 cos 2θ + 7 cos 3θ + … to n terms (Meerut 2009; Kanpur 11) (ii) 3 sin θ + 5 sin 2θ + 7 sin 3θ + … to n terms. n (n − 1) 14. (i) sin α + n sin (α + β) + sin (α + 2β) + … to (n + 1) terms. 1.2 (Kumaun 2008) n (n − 1) n (n − 1)(n − 2) (ii) n sin α + sin 2α + sin 3α + … to n terms. 1.2 1.2 .3

A nswers 1 1.

C=

cos α − c cos (α − β) − c n cos (α + nβ) + c n + 1 cos {α + (n − 1) β } 1 − 2c cos β + c 2

Sum to infinity = 2.

cos α − c cos (α − β) 1 − 2c cos β + c 2

(i) C = cos {α + (n − 1) β / 2} sin (nβ / 2) cosec (β / 2) (ii) S = sin {α + (n − 1) β / 2) sin (nβ / 2) cosec (β / 2)

3.

C=

4.

(i)

1 − c cos α − c n cos nα + c n + 1 cos (n − 1) α 1 − 2c cos α + c 2 1 − c cos α

1 − 2c cos α + c

2

(ii)

c sin α 1 − 2c cos α + c 2

T-90

5.

(4 sin α) / (5 − 4 cos α)

6.

(i)

7.

0

cos α sin α − sin2 α 1 − sin 2α + sin2 α

8. π sin θ

10. (i) 11.

12.

(ii)

sin2 α 1 − sin 2α + sin2 α sec n θ sin nθ tan θ (ii)

2

π + 2π cos θ + 1

9. 1

tan φ . sin θ 2

tan φ + 2 tan φ cos θ + 1

c sinh θ 1 − 2c cosh θ + c 2 sin α + (−1) n + 1 [(n + 2) sin nα + (n + 1) sin (n + 1) α] 2 (1 + cos α)

13. (i)

cos θ + (2n + 3) cos nθ − (2n + 1) cos (n + 1) θ − 3

(ii)

2 (1 − cos θ) sin θ + (2n + 3) sin nθ − (2n + 1) sin (n + 1) θ 2 (1 − cos θ)

14. (i) (2 cos

1 n 1 β) sin (α + nβ) 2 2

1 n 1 (ii) 2 cos α sin nα   2 2

6.4 Series Based on Binomial Expansions Remember the following formulae : (i)

When n is a positive integer and x, a are any complex numbers, we have n (n − 1) n − 2 2 ( x + a) n = x n + nx n − 1 . a + x a +… to (n + 1) terms, 2! n (n − 1) n − 2 2 ( x − a) n = x n − nx n − 1 . a + x a − … to (n + 1) terms, 2! n (n − 1) 2 n (n − 1) (n − 2) 3 (1 + x) n = 1 + nx + x + x + … to (n + 1) terms, 1.2 1.2 .3 n (n − 1) 2 n (n − 1) (n − 2) 3 and (1 − x) n = 1 − nx + x − x + … to (n + 1) terms. 1.2 1.2.3 (ii) When n is any rational index and x is a complex number such that | x | < 1, we have n (n − 1) 2 n (n − 1) (n − 2) 3 (1 + x) n = 1 + nx + x + x + … ∞. 1.2 1.2 .3 When | x | = 1, this result is still true if either n > 0 or if −1 < n < 0 and x ≠ − 1. Also remember that with suitable restrictions on the values of x as mentioned above, we have

T-91

(1 + x) − n = 1 − nx +

1.2

(1 − x) − n = 1 + nx +

x2 +

x 3 + … ∞,

n (n + 1) (n + 2) 1.2 .3

x 3 + … ∞,

1.4 2 1.4 .7 3 1 x+ x + x + … ∞. 3 3 .6 3 .6 .9

1 1 π ≤ θ ≤ π. 2 2

Solution: Let C = 1 +

1.2 .3

x3 + … ∞

1.3 2 1.3 .5 3 1 x+ x + x + … ∞, 2 2 .4 2 .4 .6

Example 4: Sum the series 1 + where −

1.2 .3

1.2 2 1.2 .5 3 1 x− x + x − … ∞, 3 36 . 3 .6 .9

(1 − x) −1 /3 = 1 +

and

n (n + 1) (n + 2)

n (n − 1) (n − 2)

x2 −

n (n + 1) 1.2

x2 −

1 1 2 1.3 3 x− x + x − … ∞, 2 2 .4 2 .4 .6

(1 − x) −1 /2 = 1 + (1 + x)1 /3 = 1 +

1.2

n (n − 1)

(1 − x) n = 1 − nx +

(1 + x)1 /2 = 1 +

n (n + 1)

1 1 1.3 cos 2θ − cos 4θ + cos 6θ − … ad. inf ., 2 2 .4 2 .4 .6 (Garhwal 2001; Rohilkhand 06; Meerut 09)

1 1 1 cos 2θ − cos 4θ + cos 6θ − … ad. inf. 2 2 .4 2 .4 .6

The companion sine series is 1.3 1 1 S = sin 2θ − sin 4θ + sin 6θ − … ad. inf. 2 2 .4 2 .4 .6 ∴

C + iS = 1 +

1 1 (cos 2θ + i sin 2θ) − (cos 4θ + i sin 4θ) 2 2 .4 +

=1+

1.3 2 .4 .6

(cos 6θ + i sin 6θ) − … ad. inf.

1 2i θ 1 4 iθ 1.3 6 iθ e − e + e − … ad. inf. 2 2 .4 2 .4 .6

= [1 + e2 iθ ]1 /2 ,

summing the binomial series

= (1 + cos 2θ + i sin 2θ)1 /2 = (2 cos 2 θ + i . 2 sin θ cos θ)1 /2 = (2 cos θ)1 /2 (cos θ + i sin θ)1 /2 = (2 cos θ)1 /2 (cos

1 1 θ + i sin θ). 2 2

T-92

Hence C, the sum of the given series = real part of C + iS = (2 cos θ)1 /2 . cos = [cos θ . 2 cos 2

1 1 θ = [2 cos θ . cos 2 θ]1 /2 2 2

1 1 /2 θ] = √ {(cos θ (1 + cos θ)}. 2

Example 5: Find the sum of the following series : n (n + 1) n (n + 1)(n + 2) n sin α + sin 2α + sin 3α + … ad. inf . 1.2 1.2.3 (Bundelkhand 2010)

Solution: Let S = n sin α +

n (n + 1) 1.2

sin 2α +

n (n + 1)(n + 2) 1.2.3

sin 3α + … ad. inf.

The companion cosine series is n (n + 1) n (n + 1)(n + 2) C = 1 + n cos α + cos 2α + cos 3α + 1.2 1.2 .3 … ad. inf. ∴

C + iS = 1 + n (cos α + i sin α) + + = 1 + ne iα +

n (n + 1) 1.2

n (n + 1)

1.2 n (n + 1)(n + 2)

e2 iα

(Note)

(cos 2α + i sin 2α)

(cos 3α + i sin 3α) + ... ad. inf. 1.2 .3 n (n + 1) (n + 2) 3 iα + e + … ad. inf. 1.2 .3

= (1 − e iα ) − n , by binomial theorem = [1 − (cos α + i sin α)] − n = [(1 − cos α) − i sin α] − n 1 1 1 = [2 sin2 α − i . 2 sin α cos α] − n 2 2 2 1 1 1 = (2 sin α) − n [sin α − i cos α] − n 2 2 2 1 −n 1 1 1 1 = (2 sin α) [cos ( π − α) − i sin ( π − α)] − n 2 2 2 2 2 1 −n 1 1 1 1 = (2 sin α) [cos {− n ( π − α)} − i sin { − n ( π − α)}], 2 2 2 2 2

= (2 sin

1 −n α) 2

by De Moivre’s theorem 1 1 1 1 [cos { n ( π − α)} + i sin { n ( π − α)}]. 2 2 2 2

Hence equating imaginary parts on both sides, we have the required sum 1 1 1 S = (2 sin α) − n sin { n ( π − α)}. 2 2 2

T-93

Comprehensive Exercise 2 Sum the following series: 1. cos n α − n cos n − 1 α cos α +

n (n − 1) 1. 2

cos

n−2

α cos 2α − … to (n + 1) terms . (Meerut 2006B; Agra 07)

1 1.3 2. sin α + sin 3α + sin 5α + … to ∞. 2 2.4 1.3 1.3 .5 1 3. sin θ + sin 2θ + sin 3θ + … ad. inf. 2 2 .4 2 .4 .6

(Rohilkhand 2006)

1.3 1.3 .5 1 cos θ + cos 2θ − cos 3θ + … ad. inf . (− π < θ < π). 2 2 .4 2 .4 .6 (Meerut 2005; Purvanchal 09) 1.3 1.3 .5 1 (ii) 1 + cos α + cos 2α + cos 3α + … ad. inf . 2 2 .4 2 .4 .6 (Rohilkhand 2006)

4. (i) 1 −

5. 1 +

1.4 2 1.4 .7 3 1 y cos α + y cos 2α + y cos 3α + … to ∞, if y < 1. 3 3 .6 3 .6 .9

6. 1 + n cos α +

n (n + 1) 1.2

7. sinh α + n sinh 2α +

cos 2α +

n (n − 1) 2!

n (n + 1)(n + 2) 1.2 .3

cos 3α + … ad. inf.

sinh 3α + … to (n + 1) terms,

where n is a positive integer.  cos nA n (n + 1) a2 1  a 8. n 1 + n cos B + , ⋅ 2 cos 2 B + … ad. inf .  = c 1.2 c  bn c  where a, b, c are the sides of the triangle ABC and a < c . (Meerut 2004; Kanpur 10)

Also show that  sin nA n (n + 1) a2 1  na sin B + ⋅ 2 sin 2 B + … = ⋅ n  c 1.2 c  b n (Kanpur 2010) c 

A nswers 2 1.

(−1) n /2 sin n α when n is even; 0, when n is odd

2.

1 1 (2 sin α) −1 /2 sin ( π + α) 4 2

4.

(i) (cos

1 1 θ) / (2 cos θ)1 /2 4 2

1 −1 / 2 1 1 θ) sin ( π − θ) 2 4 4 1 cos α 4 (ii) 1 (− 2 cos α)1 /2 2 3. (2 sin

T-94

5.

6.

r −1 /3 cos

1 φ, 3

 y sin α  where r = (1 − 2 y cos α + y 2 )1 /2 and φ = tan −1    (1 − y cos α) 1 2 n cosh n (α / 2) . sinh { (n + 2) α } 2

6.5 Series Based on Exponential Series We know that if x is any complex number, then x2 x3 + + … ad. inf. 2! 3!

(i)

ex =1+ x +

(ii)

e− x = 1 − x +

x2 x3 − + … ad. inf. 2! 3!

(iii) sin x = x −

x3 x5 + − … ad. inf. 3! 5!

(iv) cos x = 1 −

x2 x4 + − … ad. inf. 2! 4!

(v)

sinh x = x +

(vi) cosh x = 1 +

x3 x5 + + … ad. inf. 3! 5!

x2 x4 + + … ad. inf. 2! 4!

Example 6: Sum the series (i) (ii)

1 2 c cos (α + 2β) + … ad. inf . 2! 1 2 sin α + c sin (α + β) + c sin (α + 2β) + … ad. inf . 2! cos α + c cos (α + β) +

(Meerut 2012) (Meerut 2012B)

Solution: Let 1 2 c cos (α + 2β) + … ad. inf. 2! 1 2 and S = sin α + c sin (α + β) + c sin (α + 2β) + … ad. inf. 2! Multiplying the second series by i, and adding to the first, we get C + iS = (cos α + i sin α) + c {cos (α + β) + i sin (α + β)} C = cos α + c cos (α + β) +

+ (1 / 2 !) c 2 {cos (α + 2β) + i sin (α + 2β)} + … ad. inf. = e iα + ce i (α + β) + (1 / 2 !) c 2 e

i (α +2β)

+ … ad. inf.

T-95

= e iα [1 + ce iβ + (1 / 2 !) c 2 e2 iβ + … ad. inf.] iβ = e iα e ce ,  

summing up the exponential series

= e iα [e c (cos β + i sin β) ] = e iα . e c cos β . e ic sin β = e c cos β . e i (α + c sin β) = e c cos β [cos (α + c sin β) + i sin (α + c sin β)]. Equating real and imaginary parts on both sides, we have C = the sumof the series (i) = e c cos β . cos (α + c sin β) S = the sum of the series (ii) = e c cos β . sin (α + c sin β).

and

Example 7: Sum the series 1 2 cos α e cos (2 sin α) + … ad. inf . 2! 1 2 cos α If C = 1 + e cos α cos (sin α) + e cos (2 sin α) + … ad. inf. 2! 1 2 cos α S = e cos α sin (sin α) + e sin (2 sin α) + … ad. inf. 2! 1 + e cos α cos (sin α) +

Solution: then

C + iS = 1 + e cos α [cos (sin α) + i sin (sin α)] 1 2 cos α + e [cos (2 sin α) + i sin (2 sin α)] + … ad. inf. 2! 1 1 2 cos α 2 i sin α = 1 + e cos α . e i sin α + e .e + … ad. inf. 1! 2! 1 1 2 2 i sin α = 1 + ⋅ y e i sin α + y e + … ∞, where y = e cos α 1! 2!

∴

=e

y . e i sin α

=e

=e

y cos (sin α )

=e

y cos (sin α )

y { cos (sin α ) + i sin (sin α )}

. e iy sin (sin α ) . [cos { y sin (sin α)} + i sin { y sin (sin α)}].

Equating real parts on both sides, we have C=e

y cos (sin α )

. cos { y sin (sin α)}, where y = e cos α .

Example 8: (i) (ii)

Sum the series cos (α + 2β) cos (α + 4β) cos α − + − … ad. inf . 3! 5! sin (α + 2β) sin (α + 4β) sin α − + − … ad. inf . 3! 5!

(Meerut 2010B)

Solution: (i) Let C = cos α − so that

S = sin α −

cos (α + 2β) 3! sin (α + 2β) 3!

+

+

cos (α + 4β)

5! sin (α + 4β) 5!

− … ad. inf.

− … ad. inf.

T-96

C + iS = (cos α + i sin α) − (1 / 3 !) {cos (α + 2β) + i sin (α + 2β)}

∴

+ (1 / 5 !) {cos (α + 4β) + i sin (α + 4β)} − … ad. inf. =e

iα

− (1 / 3 !) e i (α + 2β) + (1 / 5 !) e i (α + 4β) − … ad. inf.

= e iα [1 − (1 / 3 !) e2 iβ + (1 / 5 !) e4 iβ + … ad. inf.] = e iα . e − i β [e iβ − (1 / 3 !) e3 iβ + (1 / 5 !) e5 iβ − … ad. inf.] =e

i (α − β)

iβ

. sin (e ) = e

i (α − β)

(Note)

. sin (cos β + i sin β)

= [cos (α − β) + i sin (α − β)] [sin (cos β) . cos (i sin β)] + cos (cos β) sin (i sin β)] = [cos (α − β) + i sin (α − β)] [sin (cos β) cosh (sin β) + i cos (cos β) sinh (sin β)]. Hence the sum of the given series = the real part of C + iS = cos (α − β) sin (cos β) cosh (sin β) − sin (α − β) cos (cos β) sinh (sin β). (ii)

Do yourself.

Comprehensive Exercise 3 Sum the following series : sin θ sin 2θ sin 3θ 1. (i) − + − … ad. inf. 1! 2! 3! sin θ sin 2θ sin 3θ (ii) + + + … ad. inf. 1! 2! 3! cos φ cos (θ + φ) cos 2 φ cos (θ + 2φ) 2. (i) cos θ + + + … to ∞. 1! 2! cos θ cos 2 θ (ii) cos θ + cos 2θ + cos 3θ + … to ∞. 1! 2! sin θ sin2 θ 3. cos θ + cos 2θ + cos 3θ + … ad. inf. 1! 2! (Rohilkhand 2007; Kashi 14; Rohilkhand 14)

4. (i) 1 + cos θ cos θ + (ii) sin θ cos θ + 5. (i) 1 + (ii)

cos α cos α

cos α cos α

+

+

cos 2θ cos 2 θ 2!

sin 2θ cos 2 θ 2! cos 2α

2 ! cos 2 α

cos 2α 2

2 ! cos α

+

+

+

+

cos 3θ cos 3 θ 3!

sin 3θ cos 3 θ 3!

cos 3α 3 ! cos 3 α

cos 3α 3 ! cos 3 α

+ … to ∞,

+ … to ∞.

(Rohilkhand 2005; Kanpur 05)

+ … ad. inf.

+ … ad. inf.

(Rohilkhand 2006)

T-97

6. 1 − cos α cos β +

cos 2 α cos 3 α cos 2β − cos 3β + … ad. inf. 2! 3! (Bundelkhand 2007; Kanpur 09; Purvanchal 10)

7. 1 + esin α cos (cos α) + sin (α + 2β)

8. sin α −

2!

+

e

2 sin α

cos (2 cos α) +

2! sin (α + 4β)

e3 sin α cos (3 cos α) + … ad. inf. 3!

− … ad. inf.

4!

(Agra 2005, 06)

9. sin θ −

sin 2θ 2!

c 2 cos 2θ

10. (i) 1 +

2! c

(ii) 1 +

2

sin 2θ 2!

5 cos θ

− … ad. inf.

4!

c 4 cos 4θ

+ +

4! c

4

sin 4θ 4!

sin3 θ cos 3θ

11. sin θ cos θ + 12.

+

sin 3θ

3!

7 cos 3θ

+

(Kanpur 2014)

+… ad. inf.

(Avadh 2006; Gorakhpur 07)

+ … ∞.

sin5 θ cos 5θ 5!

+ … ad. inf.

9 cos 5θ

+ … ad. inf. 5! cosh 3α 13. 1 + cosh α + + + … ad. inf. 2! 3! sinh 2α sinh 3α 14. sinh α + + + … ad. inf. 2! 3! +

1!

15. cosh θ +

+

3! cosh 2α

sin θ 1!

cosh 2θ +

(Kanpur 2010)

sin2 θ cosh 3θ + … ad. inf. 2!

A nswers 3 1.

(i) e − cos θ sin (sin θ)

2.

(i) e cos

3.

2

e

φ

sin θ cos θ 2

(ii) e cos θ . sin (sin θ)

. cos (θ + cos φ sin φ)

2

θ

2

θ

(ii) e cos

. cos (θ + cos θ sin θ)

2

. cos (θ + sin θ)

4.

(i) e cos

5.

(i) e. cos (tan α)

6.

e − cos α cos β . cos (cos α sin β)

7.

e

8.

sin α cos (cos β) cosh (sin β) − cos α sin (cos β) sinh (sin β) 1 1 1 1 S = sin θ cos (cos θ) cosh (sin θ) − cos θ sin (cos θ) sinh (sin θ) 2 2 2 2

9.

θ

. cos (sin θ cos θ)

y cos (cos α )

(ii) e cos

. sin (sin θ cos θ)

(ii) [e cos (tan α)] − 1

. cos { y sin (cos α)}, where y = esin α

T-98

10. (i) cosh (c cos θ) cos (c sin θ)

(ii) 1 + sinh (c cos θ) sin (c sin θ).

2

11. sinh (sin θ cos θ) cos (sin θ). 12.

4 sinh (cos θ) cos (sin θ) + cos θ cosh (cos θ) cos (sin θ)

13. e

cosh α

14. e

sin θ cosh θ

cosh (sinh α).14. e

cosh α

− sin θ sinh (cos θ) sin (sin θ). sinh (sinh α).

. cosh (θ + sin θ sinh θ).

6.6 Series Based on Logarithmic Series and its Sub-Case Gregory’s Series Remember the following results : If z is any complex number, then z2 z3 z4 (i) log (1 + z ) = z − + − + … ad. inf. provided |z | ≤ 1 and z ≠ − 1. 2 3 4   z2 z3 z4 (ii) log (1 − z ) = −  z + + + + … ad. inf. provided |z | ≤ 1 and z ≠ 1. 2 3 4     z3 z5 (iii) log (1 + z ) − log (1 − z ) = 2  z + + + … ad. inf. , 3 5   provided |z | ≤ 1 and z ≠ ± 1. This result may also be put in the form 1+ z 1 z3 z5 log =z + + + … ∞. 2 1− z 3 5 (iv) tan −1

z3 z5 z7 z =z − + − + … ad. inf., 3 5 7

(Kumaun 2008)

provided |z | ≤ 1.

This is Gregory’s series. Also remember that if α and β are any real numbers, then 1 (v) log (α + iβ) = log (α 2 + β 2 ) + i tan −1 ( β / α) 2 1 (vi) log (α − iβ) = log (α 2 + β 2 ) − i tan −1 (β / α) 2 (vii) log {(α + iβ) / (α − iβ)} = 2i tan −1 (β / α).

Example 9: Sum the series when c is positive and < 1. 1 1 (i) c cos α − c 2 cos 2α + c 3 cos 3α − … ad. inf . 2 3 1 1 (ii) c sin α − c 2 sin 2α + c 3 sin 3α − … ad. inf . 2 3

(Garhwal 2002)

(Garhwal 2002; Purvanchal 09)

T-99

Solution: and ∴

1 2 1 c cos 2α + c 3 cos 3α − … ad. inf. 2 3 1 2 1 3 S = c sin α − c sin 2α + c sin 3α − … ad. inf. 2 3 1 2 C + iS = c (cos α + i sin α) − c (cos 2α + i sin 2α) 2 1 + c 3 (cos 3α + i sin 3α) − … ad. inf. 3 1 1 = c e iα − c 2 e2 iα + c 3 e3 iα − … ad. inf. 2 3 Let C = c cos α −

= log (1 + c e iα ),

[ ∵ log (1 + x) = x −

1 2 1 3 x + x − … ∞, if | x | < 1. 2 3

Here |c e iα | = |c | which is given to be < 1.] = log [1 + c (cos α + i sin α)] = log [(1 + c cos α) + ic sin α] =

1 log [(1 + c cos α)2 + c 2 sin2 α] + i tan −1 2 [ ∵ log (α + iβ) =

=

 c sin α   , 1 + c cos α 

1 log (α 2 + β 2 ) + i tan −1 (β / α)] 2

1 log [1 + 2c cos α + c 2 cos 2 α + c 2 sin2 α] 2  c sin α  + i tan −1   1 + c cos α 

=

 c sin α  1 log (1 + 2c cos α + c 2 ) + i tan −1  ⋅ 2 1 + c cos α 

Equating real and imaginary parts, we have 1 C = log (1 + 2c cos α + c 2 ), which is the sum of the series (i) 2 and S = tan −1 {(c sin α) / (1 + c cos α)}, which is the sum of the series (ii). Example 10:

Sum the series 1 1 sin α sin β + sin 2α sin 2β + sin 3α sin 3β + … ad. inf . 2 3 Solution: We know that 1 1 sin α sin β = (2 sin α sin β) = [cos (α − β) − cos (α + β)], 2 2 1 sin 2α sin 2β = [cos 2(α − β) − cos 2(α + β)], and so on. 2 the given series may be written as ∴ 1 1 1 [cos (α − β) − cos (α + β)] + ⋅ [cos 2(α − β) − cos 2(α + β)] 2 2 2 + … ad. inf. 1 1 1 = [{cos (α − β) + cos 2(α − β) + cos 3(α − β) + … ad. inf.} 2 2 3

T-100

− {cos (α + β) +

1 1 cos 2(α + β) + cos 3(α + β) + … ad. inf.}]. 2 3

1 of the difference of the two series 2 1 1 cos (α − β) + cos 2(α − β) + cos 3(α − β) + … ad. inf. 2 3 1 1 cos (α + β) + cos 2(α + β) + cos 3(α + β) + … ad. inf. 2 3

Thus the given series is equal to

and

…(1) …(2)

Now to find the sum of the series (1), put 1 1 C = cos (α − β) + cos 2(α − β) + cos 3(α − β) + … ad. inf. 2 3 1 1 so that S = sin (α − β) + sin 2(α − β) + sin 3(α − β) + … ad. inf. 2 3 1 ∴ C + iS = {cos (α − β) + i sin (α − β)} + {cos 2(α − β) + i sin 2(α − β)} 2 1 + {cos 3(α − β) + i sin 3(α − β)} + … ad. inf. 3 1 i 2 (α − β) 1 i 3 (α − β) i (α − β) =e + e + e + … ad. inf. 2 3 = − log [1 − e i (α − β) ] = − log [1 − cos (α − β) − i sin (α − β)] =−

1 log [{1 − cos (α − β)}2 + sin2 (α − β)] 2  sin (α − β)  + i tan −1   1 − cos (α − β)

=−

1 log [2 − 2 cos (α − β)] + i tan −1 2

 sin (α − β)    1 − cos (α − β)

 sin (α − β)    1 − cos (α − β) −1 / 2  sin (α − β)  1 = log 4 sin2 (α − β) + i tan −1   2   1 − cos (α − β)  sin (α − β)  1 1 = log  cosec (α − β) + i tan −1  ⋅ 2 2  1 − cos (α − β) =−

1 1 log 2 . 2 sin2 (α − β) + i tan −1   2 2 

Equating real parts on both sides, we get 1 1 C = log [ cosec (α − β)]. 2 2 1 1 sum of the series (1) = log [ cosec (α − β)]. ∴ 2 2 Now the series (2) differs from the series (1) in having − β in place of β. ∴ sum of the series (2) can be written simply by replacing β by − β in the sum of the series (1).

T-101

1 1 Hence the sum of the series (2) = log [ cosec (α + β)]. 2 2 1 [sum of the series (1) – sum of the series (2) ] ∴ the required sum = 2 1 1 1 1 1 = [log { cosec (α − β)} − log { cosec (α + β)}] 2 2 2 2 2 cosec 1 (α − β) sin 1 (α + β)   1   1 2 2 = log  = log   ⋅ 1 1 2 cosec (α + β) 2 sin (α − β) 2 2     Example 11: (i) (ii)

Sum the series 1 3 1 c cos α + c cos 3α + c 5 cos 5α + … ad. inf . 3 5 1 3 1 c sin α + c sin 3α + c 5 sin 5α + … ad. inf . 3 5

Solution: Let C and S denote the sums of the two given series respectively. Then multiplying the second series by i, and adding it to the first, we get 1 C + iS = c (cos α + i sin α) + c 3 (cos 3α + i sin 3α) 3 1 + c 5 (cos 5α + i sin 5α) + … ad. inf. 5 1 1 = ce iα + c 3 e3 iα + c 5 e5 iα + … ad. inf. 3 5 iα     1 + c e 1+ x 1 1 x3 x5   = log  ∵ log = x + + +…∞    i α 2 2 1− x 3 5 1 − c e    1 1 iα iα = log (1 + c e ) − log (1 − c e ) 2 2 1 1 = log {(1 + c cos α) + i c sin α } − log {(1 − c cos α) − ic sin α } 2 2 =

=

1 2

1 2 2 2 −1  log {(1 + c cos α) + c sin α } + i tan 2 

−

1 2

 c sin α     1 + c cos α  

1 2 2 2 −1  log {(1 − c cos α) + c sin α } − i tan 2 

 c sin α     1 − c cos α  

1 [log (1 + 2c cos α + c 2 ) − log (1 − 2c cos α + c 2 )] 4  c sin α  1 −1  c sin α   + i ⋅ tan −1   + tan   ⋅ 2  1 + c cos α  1 − c cos α  

Equating real and imaginary parts on both sides, we get 1 C = [log (1 + 2c cos α + c 2 ) − log (1 − 2c cos α + c 2 )] 4

T-102

=

and

1 log {(1 + 2c cos α + c 2 ) / (1 − 2c cos α + c 2 )}, 4

which is the sum of the series (i)  c sin α  1 −1  c sin α   S = tan −1   + tan   2  1 + c cos α  1 − c cos α    {(c sin α) / (1 + c cos α)} + {(c sin α) / (1 − c cos α)}  1 = tan −1   2 1 − {(c sin α) / (1 + c cos α)}.{(c sin α) / (1 − c cos α)} (c sin α) {(1 − c cos α) + (1 + c cos α)} 1 = tan −1   2 (1 − c 2 cos 2 α) − c 2 sin2 α    2c sin α  1 = tan −1  , which is sum of the series (ii). 2  2  1 − c 

Comprehensive Exercise 4 Sum the following series. It may be assumed that −1 < c < 1: 1 1 cos 2θ + cos 3θ − … ad. inf. 2 3 1 1 (ii) sin θ − sin 2θ + sin 3θ − … ad. inf. 2 3 1 1 2. (i) cos α cos α − cos 2 α cos 2α + cos 3 α cos 3α − … ad. inf. 2 3 1 1 2 (ii) cos α sin α − cos α sin 2α + cos 3 α sin 3α −… ad. inf. 2 3 1 1 (iii) sin α sin α − sin2 α sin 2α + sin3 α sin 3α − … ad. inf. 2 3 1 1 3. (i) c cos α + c 2 cos 2α + c 3 cos 3α + … ad. inf. 2 3 (Gorakhpur 2005) 1 2 1 3 (ii) c sin α + c sin 2α + c sin 3α + … ad. inf. 2 3 1. (i) cos θ −

(Avadh 2008)

cos α

1 cos 2α 1 cos 3α (iii) + + + … ad. inf. 2 2 22 3 23 1 1 (iv) cos α sin α + cos 2 α sin 2α + cos 3 α sin 3α + … ad. inf. 2 3 π 1 2π 1 3π 4. cos + cos + cos + … ad. inf. 3 2 3 3 3 (Kanpur 2008, 09) 1 2 1 3 5. (i) c cos α + c cos (α + β) + c cos (α + 2β) + … ad. inf. 2 3 1 2 1 (ii) c sin α + c sin (α + β) + c 3 sin (α + 2β) + … ad. inf., 2 3 (Gorakhpur 2005; Bundelkhand 11)

T-103

1 2 1 c cos (α + β) + c 3 cos (α + 2β) − … ad. inf. 2 3 1 2 1 3 (ii) c sin α − c sin (α + β) + c sin (α + 2β) − … ad. inf. 2 3 1 1 c sin2 θ − c 2 sin2 2θ + c 3 sin2 3θ − … ad. inf. 2 3 1 1 2 1 3 1 4 cos π + cos π + cos π + cos π + …ad. inf. 3 3 3 5 3 7 3 1 3 1 5 (i) c cos θ − c cos 3θ + c cos 5θ − … ad. inf. 3 5 1 3 1 5 (ii) c sin θ − c sin 3θ + c sin 5θ − … ad. inf. 3 5 1 1 2 3 cos θ − cos θ cos 3θ + cos 5 θ cos 5θ − … ad. inf. 3 5 1 3α 1 5α α e cos β − e cos 3β + e cos 5β − … ad. inf. 3 5 1 1 (i) cosh θ − cosh 2θ + cosh 3θ − … ad. inf., 2 3 1 1 (ii) sinh θ − sinh 2θ + sinh 3θ − … ad. inf. 2 3 1 1 2 1 If θ − α = tan ( φ) sin 2θ − tan4 ( φ) sin 4θ 2 2 2 1 1 + tan6 ( φ) sin 6θ − … ad. inf., 3 2 show that tan α = tan θ cos φ. (Agra 2010)

6. (i)

7. 8. 9.

10. 11. 12.

13.

c cos α −

14. Prove that 1 1 1 1 tanh x + tanh3 x + tanh5 x + … = tan x − tan3 x + tan5 x − … , 3 5 3 5 where x lies between − (π / 4) and + (π / 4). (Meerut 2010; Kanpur 14) 15. Prove that 1 1 sin θ + sin3 θ + sin5 θ + … ad. inf. 3 5 1 1 1 = 2 [sin θ − sin 3θ + sin 5θ − … ad. inf. ], where θ ≠ (2n + 1) π. 3 5 2 (Kanpur 2007)

A nswers 4 1.

(i) log 2 + log cos

2.

(i)

1 α 2

1 log (1 + 3 cos 2 α) 2

(ii)

1 α 2

(ii) tan −1

sin α cos α 1 + cos 2 α

T-104

(iii) tan −1

sin2 α 1 + sin α cos α

1 log (1 − 2c cos α + c 2 ) (ii) tan −1 {(c sin α) / (1 − c cos α)} 2 1 5 1 (iii) − log ( − cos α) (iv) π − α 4.0 2 4 2

3.

(i) −

5.

(i) − sin (α − β) . tan −1 {(c sin β) / (1 − c cos β)} 1 − cos (α − β) log (1 − 2c cos β + c 2 ) 2 (ii) cos (α − β) . tan −1 {(c sin β) / (1 − c cos β)} −

6.

1 sin (α − β) log (1 − 2c cos β + c 2 ) 2

1 cos (α − β) log (1 + 2c cos β + c 2 ) 2 − sin (α − β) . tan −1 {(c sin β) / (1 + c cos β)}

(i)

1 sin (α − β) log (1 + 2c cos β + c 2 ) 2 + cos (α − β) . tan −1 {(c sin β) / (1 + c cos β)}. (ii)

7.

 (1 + c ) 1 log  2 2  √ (1 + 2c cos 2θ + c )

9.

(i)

1 2c cos θ tan −1 ( ) 2 1− c 2

1 tan −1 (2 cot 2 θ) 2 1 12. (i) log (2 cosh θ) 2 10.

8.

1 [2 √ 3 log (2 + √ 3) − π] 8

1 + 2c sin θ + c 2 1 log 4 1 − 2c sin θ + c 2 1 11. − tan −1 {cos β / sinh α } 2 1 (ii) θ 2 (ii)

6.7 The Difference Method Sometimes in order to sum a series it is convenient to split each term of the series as difference of two terms. The splitting is done in such a way that when all the terms of the series are added together, the component terms cancel in pairs. Ultimately we are left usually with two components, one from the first term and one from the last term. This method is not always easy to apply, since sometimes it requires a considerable amount of mathematical jugglery to split the nth term of the series as the difference of two terms. It is however learnt by practice. Use of certain trigonometric identities is sometimes helpful. If the answer is given, then the mode of splitting can often be found by putting n = 1in the answer, since then it reduces to first term of the series. Suppose the rth term ur of the series is expressed as ur = f (r + 1) − f (r).

T-105

Then

u1 = f (2) − f (1) u2 = f (3) − f (2) … … … … un − 1 = f (n) − f (n − 1) un = f (n + 1) − f (n).

Adding these n relations, we have S n = f (n + 1) − f (1), where

S n = u1 + u2 + … + un

= the sum of the first n terms of the series. If the given series is convergent and sum to infinity is required, we deduce it as follows : lim lim S ∞ = n → ∞ S n = n → ∞ [ f (n + 1) − f (1)]. Thus to find the sum upto infinity we have to first find the sum upto n terms and then proceed to the limits as n → ∞. The following trigonometric identities are often useful : 1 (i) cosec α = cot α − cot α , 2 (ii) tan α = cot α − 2 cot 2α , (iii) tan α sec 2α = tan 2α − tan α, 1 (iv) sin3 α = (3 sin α − sin 3α). 4 If the n th term is of the form tan −1 (a / b) , we put it in the form tan −1 x − tan −1 y.

Example 12: Sum the series Or

sec θ sec 2θ + sec 2θ sec 3θ + sec 3θ sec 4θ + … to n terms. 1 1 1 + + + … to n terms. cos θ cos 2θ cos 2θ cos 3θ cos 3θ cos 4θ (Kanpur 2011; Avadh 12)

Solution: The given series is sec θ sec 2θ + sec 2θ sec 3θ + ... to n terms 1 1 1 = + + + … to n terms. cos θ cos 2θ cos 2θ cos 3θ cos 3θ cos 4θ The first term of the above series is [1 / (cos θ cos 2θ)]. We should try to split it up as a difference of two terms. If T1 , T2 , etc. represent the successive terms of the given series, then

T-106

T1 = sec θ sec 2θ =

 sin θ 1 1  =  , cos θ cos 2θ sin θ cos θ cos 2θ multiplying the Nr. and Dr. by sin θ

=

1 sin θ

 sin (2θ − θ)  1 sin 2θ cos θ − cos 2θ sin θ  =   cos θ cos 2θ cos θ cos 2θ sin θ  

= (1 / sin θ) [tan 2θ − tan θ]. Similarly the second term 1 1 T2 = sec 2θ sec 3θ = = [tan 3θ − tan 2θ]. cos 2θ cos 3θ sin θ The third term T3 = sec 3θ sec 4θ = …

…

1 1 = [tan 4θ − tan 3θ]. cos 3θ cos 4θ sin θ

…

…

…

…

…

…

Lastly, the nth term Tn = sec nθ sec (n + 1) θ =

1 cos nθ cos (n + 1) θ

= (1 / sin θ) [tan (n + 1) θ − tan nθ]. Adding up the above n relations, we have the required sum = T1 + T2 + T3 + … + Tn = (1 / sin θ) [tan (n + 1) θ − tan θ] , the other terms on the R.H.S. cancelling each other. Aliter:

By inspection the nth term of the given series is 1 Tn = sec nθ sec (n + 1) θ = cos nθ cos (n + 1) θ   sin θ  , multiplying the Nr. and Dr. by sin θ cos nθ cos (n + 1) θ

=

1 sin θ

=

1  sin {(n + 1) θ − nθ}   sin θ cos nθ cos (n + 1) θ

=

1 sin (n + 1) θ cos nθ − cos (n + 1) θ sin nθ   sin θ  cos nθ cos (n + 1) θ 

(Note)

= (1 / sin θ) [tan (n + 1) θ − tan nθ]. Putting n = 1, 2, 3,…, n, we get T1 = (1 / sin θ) [tan 2θ − tan θ], T2 = (1 / sin θ) [tan 3θ − tan 2θ], …

…

…

…

…

Tn − 1 = (1 / sin θ) [tan nθ − tan (n − 1) θ], Tn = (1 / sin θ) [tan (n + 1) θ − tan nθ].

T-107

Adding these we get the required sum T1 + T2 + T3 + … + Tn − 1 + Tn = (1 / sin θ) [tan (n + 1) θ − tan θ], other terms cancelling each other. cosec θ + cosec 2θ + cosec 22 θ + … to n terms.

Example 13: Sum the series

(Purvanchal 2011)

Solution: The first term 1 sin θ 1 2 T1 = cosec θ = = sin θ sin 1 θ . sin θ (Note) 2 1 1 1 sin (θ − θ) sin θ cos θ − cos θ sin θ 1 2 2 2 = = = cot θ − cot θ. 1 1 2 sin θ . sin θ sin θ sin θ 2 2 Similarly, T2 = cot θ − cot 2θ, T3 = cot 2θ − cot 22 θ, T4 = cot 22 θ − cot 23 θ, …

…

Tn = cot 2

… n −2

…

θ − cot 2

n −1

θ.

Adding these, we have the required sum 1 S n = T1 + T2 + … + Tn = cot θ − cot 2 n − 1 θ, 2 other terms on the R.H.S. cancelling each other. Example 14 :

Sum the series tan θ + 2 tan 2θ + 22 tan 22 θ + 23 tan 23 θ + … to n terms.

Solution:

Here T1 = tan θ =

sin θ cos θ

=

sin2 θ sin θ cos θ

=

cos 2 θ − cos 2θ sin θ cos θ

(Note)

[ ∵ cos 2θ = cos 2 θ − sin2 θ] =

cos 2 θ sin θ cos θ

−

cos 2θ sin θ cos θ

=

cos θ sin θ

−

Similarly, T2 = 2 tan 2θ = 2 [cot 2θ − 2 cot 2(2θ)] = 2 [cot 2θ − 2 cot 22 θ] = 2 cot 2θ − 22 cot 22 θ, T3 = 22 cot 22 θ − 23 cot 23 θ, … Tn = 2

… n −1

cot 2

… n −1

… n

…

θ − 2 cot 2 n θ.

cos 2θ = cot θ − 2 cot 2θ. 1 sin 2θ 2

T-108

Adding these, we have the required sum = cot θ − 2 n cot 2 n θ , the other terms on the R.H.S. cancelling one another. Example 15:

Sum the series θ θ θ sin3 + 3 sin3 2 + 32 sin3 3 + … to n terms 3 3 3

and also sum to infinity. Solution: We know that sin 3θ = 3 sin θ − 4 sin3 θ giving sin3 θ =

1 (3 sin θ − sin 3θ). 4

…(1)

Now the first term of the given series 1 1 1 1 T1 = sin3 θ = [3 sin θ − sin 3 . θ] 3 4 3 3 1 [Putting θ for θ in (1)] 3 1 1 = [3 sin θ − sin θ]. 4 3 1 3 Similarly, T2 = 3 sin (θ / 32 ) = 3 ⋅ [3 sin (θ / 32 ) − sin 3.(θ / 32 )] 4 [Putting (θ / 32 ) for θ in (1)] 1 2 [3 sin (θ / 32 ) − 3 sin (θ / 3)], 4 1 T3 = 32 sin3 (θ / 33 ) = [33 sin (θ / 33 ) − 32 sin (θ / 32 )], 4 =

…

… … … … 1 n n n −1 Tn = [3 sin (θ / 3 ) − 3 sin (θ / 3 n − 1 )]. 4 Adding up these n relations, we have 1 T1 + T2 + T3 + … + Tn = [3 n sin (θ / 3 n ) − sin θ]. 4 Thus the required sum upto n terms 1 S n = [3 n sin (θ / 3 n ) − sin θ]. 4 The sum to infinity S ∞ is given by 1 θ  S ∞ = lim S n = lim 3 n sin n − sin θ n→ ∞ n→ ∞ 4  3  =

1 lim ⋅ 4 n→ ∞

 sin (θ / 3 n )  − sin θ θ ⋅  (θ / 3 n )  

Let (θ / 3 n ) = h, then as n → ∞, h → 0.

(Note)

T-109

S∞ =

∴

1 lim  sin h  1 θ⋅ − sin θ = [θ − sin θ]. 4 h → 0  h  4 lim   ∵ h → 0 (sin h / h) = 1  

Example 16:

Sum the series

tan α tan (α + β) + tan (α + β) tan (α + 2β) + tan (α + 2β) tan (α + 3β) + … to n terms. Solution:

We have tan β = tan [(α + β) − α] =

tan (α + β) − tan α 1 + tan (α + β) . tan α

or

tan β [1 + tan (α + β) tan α] = tan (α + β) − tan α

or

1 + tan (α + β) tan α = cot β [tan (α + β) − tan α]

or

tan (α + β) tan α = cot β [tan (α + β) − tan α] − 1.

∴

…(1 )

first term of the given series i. e., T1 = cot β [tan (α + β) − tan α] − 1.

Similarly,

T2 = cot β [tan (α + 2β) − tan (α + β)] − 1, putting (α + β) for α in (1) T3 = cot β [tan (α + 3β) − tan (α + 2β)] − 1, …

…

…

…

Tn = cot β [tan (α + nβ) − tan {α + (n − 1) β }] − 1. Adding the above n relations, we have the required sum = cot β [tan (α + nβ) − tan α] − n, the other terms on the R.H.S. cancelling one another. Example 17:

Sum the series 1 1 1 1 tan −1 + tan −1 + tan −1 + tan −1 + … to n terms. 3 7 13 21

(Meerut 2005B, 09B; Kanpur 10; Kashi 11; Avadh 13; Purvanchal 14)

Solution: The given series may be written as 1 1 1 1 tan −1 + tan −1 + tan −1 + … + tan −1 ⋅ 1 + 1. 2 1+ 2 .3 1+ 3 .4 1 + n (n + 1) We have Tn = the nth term of the series (n + 1) − n 1 = tan −1 = tan −1 1 + n (n + 1) 1 + (n + 1) n = tan −1 (n + 1) − tan −1 n. [∵ tan −1 x − tan −1 y = tan −1 {( x − y) / (1 + xy)}]

T-110

Putting n = 1, 2, 3, …, n, we have T1 = tan −1 2 − tan −1 1, T2 = tan −1 3 − tan −1 2, T3 = tan −1 4 − tan −1 3, … … … … …… … … … … …… Tn = tan −1 (n + 1) − tan −1 n. Adding the above n relations , we have the required sum S n = T1 + T2 + … + Tn = tan −1 (n + 1) − tan −1 1, the other terms on the R.H.S. cancelling one another.

Comprehensive Exercise 5 Sum the following series: 1.

(i) sec θ sec (θ + φ) + sec (θ + φ) sec (θ + 2φ) + sec (θ + 2φ) sec (θ + 3φ) + … to n terms. 1 1 1 (ii) + + + …to n terms. cos θ + cos 3θ cos θ + cos 5θ cos θ + cos 7θ (Gorakhpur 2007) (iii) cosec θ cosec 2θ + cosec 2θ cosec 3θ + cosec 3θ cosec 4θ + … to n terms. (Purvanchal 2007)

1 1 1 (iv) + + + … to n terms. cos θ − cos 3θ cos θ − cos 5θ cos θ − cos 7θ 2.

(i) (ii)

sin θ sin 2θ sin 3θ

+

sin θ cos θ + cos 2θ

sin θ sin 3θ sin 4θ +

+

sin θ sin 4θ sin 5θ

sin 2θ cos θ + cos 4θ

+

+ … to n terms.

sin 3θ cos θ + cos 6θ

+ … to n terms.

θ θ θ θ θ θ + 2 cos cos 2 + 22 cos cos 2 cos 3 + … to n terms. 2 2 2 2 2 2 θ θ θ θ θ (ii) tan sec θ + tan 2 sec + tan 3 sec 2 + … to n terms 2 2 2 2 2 (Garhwal 2000)

3.

(i) cos

4.

sin θ sec 3θ + sin 3θ sec 32 θ + sin 32 θ . sec 33 θ + … to n terms.

5. (i) 2 cosec 2θ cot 2θ + 4 cosec 4θ cot 4θ + 8 cosec 8θ cot 8θ + … to n terms. 1 1 sin 4θ cos 2 2θ + sin 8θ cos 2 4θ − … to n terms. 2 4 1 θ 1 θ 1 θ 6. tan θ + tan + 2 tan 2 + 3 tan 3 + … ad. inf. 2 2 2 2 2 2 (ii) sin 2θ cos 2 θ −

(Meerut 2004B; Avadh 07, 10)

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7. 8.

1 θ 1 θ tanh + 2 tanh 2 + … to n terms. 2 2 2 2 1 tan2 θ tan 2θ + tan2 2θ tan 4θ + (1 / 22 ) tan2 4θ tan 8θ + … to n terms. 2 tanh θ +

(Avadh 2008)

9.

cot 2α cot 3α + cot 3α cot 4α + … + cot (n + 1) α cot (n + 2) α.

10. tan −1

1 2 4 2 n −1 + tan −1 + tan −1 + … + tan −1 + … ad. inf. 3 9 33 1 + 22 n − 1 (Purvanchal 2007)

11. tan −1 x + tan −1 12. tan −1

x 1 + 1.2 x 2

+ tan −1

x 1 + 2 .3 x 2

+ … to n terms. (Avadh 2009)

4 6 8 + tan −1 + tan −1 + … + to n terms. 1 + 3.4 1 + 8.9 1 + 15.16 (Meerut 2010; 13)

13. (i) tan −1

1 2

1+1+1

1

+ tan −1

1+ 2 + 2

2

+ tan −1

1 1 + 3 + 32

+ … + to n terms.

(Meerut 2008; 13B) ∞

(ii) Evaluate Σ

n =1

14. tan −1

cot −1 (1 + n + n2 ).

4 4 4 4 + tan −1 + tan −1 + … + tan −1 ⋅ 2 7 19 39 4n + 3

15. tan α tan 2α + tan 2α tan 3α + tan 3α tan 4α + … to (n − 1) terms. (Meerut 2006)

16. tan −1

1 2 . 12

+ tan −1

1

+ tan −1

2 . 22

1 2 . 32

+ … + tan −1

Deduce its sum to infinity. 1 17. cot −1 22 +  + cot −1  2

1 2n2

⋅

(Rohilkhand 2008)

1  3 −1 2 + 2  + cot  2 

1  4 2 + 3  + … ad. inf.  2  (Gorakhpur 2005)

A nswers 5 1. (i) (1 / sin φ) [tan (θ + nφ) − tan θ] (iii) (1 / sin θ) [cot θ − cot (n + 1) θ]

1 cosec θ { tan (n + 1) θ − tan θ} 2 1 (iv) cosec θ [cot θ – cot ( n + 1) θ] 2

(ii)

2. (i) cot 2θ − cot (n + 2) θ (ii) (1 / 4 sin

1 1 1 θ) [sec (2n + 1) θ − sec θ] 2 2 2

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3. (i)

sin nθ  θ θ cot n + 1 − cot  2  2 2

5. (i) cosec 2 θ − 2 n cosec 2 2 n θ 6.

(ii) tan θ (ii)

1 − 2 cot 2θ θ

7. [2 coth 2θ − (1 / 2 n − 1 ) coth (θ / 2 n − 1 )]

8. (1 / 2 n − 1 ) tan 2 n θ − 2 tan θ 10.

1 π 4

1 sin 2θ + (−1) n − 1 (1 / 2 n + 1 ) sin 2 n + 1 θ 2

9. cot α [cot 2α − cot (n + 2) α] − n 11. tan −1 nx

12. tan −1 (n + 1) (n + 2) − tan −1 2 13. (i) tan −1 { n / (n + 2)} 14. tan −1

4n 2n + 5

(ii)

1 π 4

15. cot α tan nα − n.

16. tan −1 (2n + 1) − tan −1 (2n − 1);

1 π 4

17.

cot −1 2

6.8 Angles in Arithmetical Progression (a) To find the sum of the sines of a series of angles, the angles being in arithmetical progression. Let α, α + β, α + 2β, …, {α + (n − 1) β } be n angles in arithmetical progression, the common difference of the angles being β. Then the series of sines of above angles will be sin α + sin (α + β) + sin (α + 2β) + … + sin {α + (n − 1) β } . Suppose S n denotes the sum of the above series, so that S n = sin α + sin (α + β) + sin (α + 2β) + … + sin {α + (n − 1) β }. The common difference of angles is β. 1 Multiplying both sides of (1) by 2 sin β, we have 2 1 1 1 2 sin β . S n = 2 sin α sin β + 2 sin (α + β) . sin β + … 2 2 2

…(1)

… + 2 sin {α + (n − 1) β } sin Now applying the trigonometric formula 2 sin A sin B = cos ( A − B) − cos ( A + B) to each term on the right hand side , we have 1 1 1 2 sin β . S n = cos (α − β) − cos (α + β) 2 2 2 1 3 + cos (α + β) − cos (α + β) 2 2

1 β. 2

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+ cos (α +

3 5 β) − cos (α + β) 2 2

+…

…

+…

…

…

… 3 1 + cos {α + (n − ) β } − cos {α + (n − ) β } 2 2 1 1 = cos (α − β) − cos {α + (n − ) β }, 2 2 the other terms on the right hand side cancelling one another. = 2 sin {α +

1 1 (n − 1) β } sin nβ. 2 2

Therefore 1 1 sin α + ( n + 1) β sin nβ 2 2   Sn = 1 sin β 2

…(2) (Remember)

The sum of the sine series obtained above may be remembered in the following form : sin α + sin (α + β) + sin (α + 2β) + … + sin {α + ( n − 1) β}  first angle + last angle  = sin   2   Particular Cases: I.

diff. sin  n ×   2  diff. sin    2 

If β = α, then the expression denoting the sum of the sine

series (1) becomes S n = sin α + sin 2α + sin 3α + … + sin nα 1 sin nα n + 1 2 = sin  α , from (2),  2  1 sin α 2 II.

If we put β = 2π / n, then since

1 1 nβ = sin π = 0 and sin β ≠ 0, 2 2 2π  4π   we have sin α + sin α +  + sin α +  + … to n terms = 0.   n n sin

In general if β = p. (2π / n), where p is any integer, even then nβ  n2 pπ  sin = sin   = sin pπ = 0.  2n  2 1 Also if p is not divisible by n, then sin β ≠ 0. 2

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Therefore the sum of sines of n angles, which are in arithmetic progression, vanishes if the common difference of the angles is p(2π / n), where p is any integer not divisible by n. (b) To find the sum of the cosines of a series of angles, the angles being in arithmetical progression. Let n angles in arithmetical progression be α, (α + β), (α + 2β), …, {α + (n − 1) β }. Let S n denote the sum of the series of cosines of above angles, so that S n = cos α + cos (α + β) + cos (α + 2β) + … + cos {α + (n − 1) β }. …(3) 1 Multiplying both sides by 2 sin β, we get 2 1 1 1 2 sin β . S n = 2 cos α sin β + 2 cos (α + β) sin β 2 2 2 1 1 + 2 cos (α + 2β) sin β + …… + 2 cos {α + (n − 1) β } sin β. 2 2 Now applying the trigonometric relation 2 cos A sin B = sin ( A + B) − sin ( A − B) to each term on the right hand side, we have 1 1 1 2 sin β . S n = sin (α + β) − sin (α − β) 2 2 2 3 1 + sin (α + β) − sin (α + β) 2 2 5 3 + sin (α + β) − sin (α + β) 2 2 + … … … + … … … 1 3 + sin {α + (n − ) β } − sin {α + (n − ) β } 2 2 1 1 = sin {α + (n − ) β } − sin (α − β), the other terms on the 2 2 right hand side cancelling one another 1 1 = 2 cos {α + (n − 1) β } sin nβ. 2 2 1 1 cos α + (n − 1) β sin nβ 2 2   Therefore Sn = 1 (Remember) …(4 ) sin β 2 Particular Cases: I. If β = α, then the expression denoting the sum of the cosine series (3) becomes S n = cos α + cos 2α + cos 3α + … + cos nα

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1 nα 2 , from (4). 1 sin α 2 1 1 II. If β = 2π / n, then since sin nβ = sin π = 0 and sin β ≠ 0, 2 2 2π  4π    we have cos α + cos α +  + cos α +  + … to n terms = 0.   n n n + 1  = cos  α  2 

sin

In general if β = p (2π / n), where p is any integer not divisible by n, even then the sum of the above cosine series is zero.

Example 18: Sum the series (i) sin α + sin 3α + sin 5α + … + to n terms ; (ii) cos α + cos 3α + cos 5α + … to n terms. (iii) Prove that sin α + sin 3α + sin 5α + … to n terms tan nα = ⋅ cos α + cos 3α + cos 5α + … to n terms Solution:

(i)

Let S n denote the sum of the given series (i) ; then S n = sin α + sin 3α + sin 5α + … to n terms.

…(1)

In the above series (1) the angles α, 3α , 5α, … are in arithmetical progression. The common difference of angles is (3α − α) = 2α. Here we shall multiply both sides of (1) by 1 2 sin (2α) i. e., by 2 sin α. 2 multiplying both the sides of (1) by 2 sin α, we have ∴ S n . 2 sin α = 2 sin α sin α + 2 sin 3α sin α + 2 sin 5α sin α + … + 2 sin {(2n − 1) α } sin α.

…(2)

Now applying the formula 2 sin A sin B = cos ( A − B) − cos ( A + B) to each term on the R.H.S. of (2), we have the first term = cos (α − α) − cos (α + α) = cos 0 − cos 2α, the second term = cos (3α − α) − cos (3α + α) = cos 2α − cos 4α, the third term = cos (5α − α) − cos (5α + α) = cos 4α − cos 6α, … … … … … the nth term = cos [{(2n − 1) α } − α] − cos [{(2n − 1) α + α] = cos (2n − 2) α − cos 2nα. Adding up the above n rows , we have the sum of the n terms on the R.H.S. of (2) = cos 0 − cos 2nα = 1 − cos 2nα, the other terms cancelling one another.

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Hence from (2), we have S n . 2 sin α = 1 − cos 2nα = 2 sin2 nα. S n = sin2 nα / sin α.

∴

Thus the required sum of the series (i) is equal to sin2 nα / sin α. (ii)

Let S n denote the sum of the given series (ii) ; then …(3)

S n = cos α + cos 3α + cos 5α + … + cos (2n − 1) α. The angles in the above series are in A.P. with common difference (3α − α) = 2α. ∴ multiplying both sides of (3) by 2 sin

1 (2α) i. e., by 2 sin α, we have 2

2 sin α . S n = 2 sin α cos α + 2 sin α cos 3α + 2 sin α cos 5α + … + 2 sin α cos (2n − 1) α. Now using the formula 2 cos A sin B = sin ( A + B) − sin ( A − B) term by term on the right hand side of (3), we have the first term = sin (α + α) − sin (α − α) = sin 2α − sin 0, the second term = sin (3α + α) − sin (3α − α) = sin 4α − sin 2α, the third term = sin (5α + α) − sin (5α − α) = sin 6α − sin 4α, … … … … … the nth term = sin {(2n − 1) α + α } − sin {(2n − 1) α − α } = sin 2nα − sin (2n − 2) α. Adding up the above n rows , the sum of the n terms on the R.H.S. of (3) = sin 2nα − sin 0 = sin 2nα, the other terms cancelling one another. Hence from (3), we have 2 sin α . S n = sin 2nα or S n = (sin 2nα) / 2 sin α. Thus the required sum of the series (ii) is equal to (sin 2nα) / (2 sin α). (iii) We have sin α + sin 3α + sin 5α + … to n terms cos α + cos 3α + cos 5α + … to n terms =

(sin2 nα) / (sin α) (sin 2nα) / (2 sin α)

, from the solutions of parts (i) and (ii) (prove here)

2

=

2 sin nα sin 2nα

2

=

2 sin nα 2 sin nα cos nα

=

sin nα cos nα

= tan nα.

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Remark: The above question can also be done by direct application of the formulae derived in article 6.8 as shown below. We know that sin α + sin (α + β) + sin (α + 2β) + … upto n terms 1 sin nβ 1 2 = sin {α + (n − 1) β } ⋅ 1 2 sin β 2 Here β = 3α − α = 2α. sin α + sin 3α + sin 5α + … upto n terms 1 sin { n (2α)} 1 2 = sin {α + (n − 1) 2α } 1 2 sin { (2α)} 2 sin nα sin2 nα = sin nα = ⋅ sin α sin α

∴

Similarly we may find the sum of the cosine series. Example 19:

If β be the exterior angle of a regular polygon of n sides and α is any constant,

prove that (i)

n −1

Σ

r =0

sin (α + βr) = 0,

(ii)

n −1

Σ

r =0

cos (α + βr) = 0.

Solution: We know that the sum of all the exterior angles of a polygon is equal to 2π. Here the polygon is a regular one of n sides. Therefore the sum of the n equal exterior angles = 2π i. e., each exterior angle is equal to (2π / n) i. e., β = (2π / n). (i)

By putting r = 0, 1, 2, …, (n − 1), we find that the series in (i) is sin α + sin (α + β) + sin (α + 2β) + … to n terms.

By article 6.8 (a), we have sin α + sin (α + β) + sin (α + 2β) + … to n terms 1 sin nβ 1 2 = sin α + (n − 1) β 1 2   sin β 2 1 1 sin { n . (2π / n)} sin {α + (n − 1) (2π / n)} ∵ β = 2π  2 2 = 1 n   sin (2π / n) 2 sin π . sin {α + (n − 1) π / n} = = 0. sin (π / n) [ ∵ sin π = 0 and sin (π / n) ≠ 0] (ii)

Putting r = 0, 1, 2, …, (n − 1), we find that the series given in (ii) is cos α + cos (α + β) + cos (α + 2β) + … to n terms.

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By article 6.8 (b), we have cos α + cos (α + β) + cos (α + 2β) + … to n terms 1 sin nβ 1 2 = cos α + (n − 1) β 1 2   sin β 2 1 sin π . cos {α + (n − 1) (2π / n)} 2 = sin (π / n)

[ ∵ sin π = 0 and sin (π / n) ≠ 0]

= 0. Example 20: (i)

 ∵ β = 2π   n 

Find the sum to n terms of the series

2π  4π  4  sin4 α + sin4 α +  + sin α +  +…    n n

2π  4π  4  cos4 α + cos4 α +  + cos α +  +…   n n 1 Solution: (i) We know that sin2 α = [1 − cos 2α]. 2 1 1 ∴ sin4 α = [sin2 α]2 = [1 − cos 2α]2 = [1 + cos 2 2α − 2 cos 2α] 4 4 1 1 = ⋅ [2 + 2 cos 2 2α − 4 cos 2α] 4 2 1 = [2 + (1 + cos 4α) − 4 cos 2α] 8 1 = [cos 4α − 4 cos 2α + 3]. 8 (ii)

2π  1 Similarly sin4 α +  =  n 8

…(1)

8π  4π      cos 4α + n  − 4 cos 2α + n  + 3 , replacing α by α + (2π / n) in (1)

4π  1 sin4 α +  =  n 8

16π  8π      cos 4α + n  − 4 cos 2α + n  + 3, and so on upto n terms.

Adding the above n relations, we have 2π  4π  4  sin4 α + sin4 α +  + sin α +  + … to n terms   n n =

1 8

 8π    cos (4α) + cos 4α + n  + … to n terms  

 4π    − 4 cos 2α + cos 2α +  + … to n terms + 3 (1 + 1 + … to n terms) ⋅   n   

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In the first series on the R.H.S. the common difference of the angles is 8π / n and in the second series it is 4π / n and each of them is a multiple of 2π / n. Therefore the sum of each of these two series vanishes. 1 3 Hence the required sum of the given series = [0 − 0 + 3n] = n. 8 8 (ii)

The given series is

2π  4π  4  cos 4 α + cos 4 α +  + cos α +  + … to n terms.   n n 1 We know that cos 2 α = [1 + cos 2α]. 2 1 1 4 ∴ cos α = [1 + cos 2α]2 = [1 + cos 2 2α + 2 cos 2α] 4 4 1 1 2 = [2 + 2 cos 2α + 4 cos 2α] = [2 + (1 + cos 4α) + 4 cos 2α] 8 8 1 1 = [3 + cos 4α + 4 cos 2α] = [cos 4α + 4 cos 2α + 3]. 8 8 Now proceeding exactly as in part (i) we find that the required sum of the given series 1 3n = [0 + 4 . 0 + 3n] = . 8 8

Comprehensive Exercise 6 1.

Find the sum of n terms of the series (i) sin α − sin (α + β) + sin (α + 2β) − … (ii) cos α − cos (α + β) + cos (α + 2β) − …

2.

Sum the series sin α − sin 2α + sin 3α − sin 4α + … to n terms.

3.

Sum the series cos α − sin 2α − cos 3α + sin 4α + … to n terms.

4.

Sum the series cos

5.

Find the sum of the series sin α + sin 2α + sin 3α + … + sin nα

π 3π 5π + cos + cos + … to n terms. 2n + 1 2n + 1 2n + 1

and hence deduce the sum of the series 1 + 2 + 3 + … + n. 6.

Find the sum of the series (i) sin2 α + sin2 (α + β) + sin2 (α + 2β) + … to n terms (ii) cos 2 α + cos 2 (α + β) + cos 2 (α + 2β) + … to n terms.

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A nswers 6 1.

2.

3.

4. 6.

1 1 1 (n − 1) (π + β)} sin n (π + β) sec β 2 2 2 1 1 1 (ii) cos {α + (n − 1) (π + β)} sin n (π + β) sec β 2 2 2 1 1 1 sin {α + (n − 1) (π + α)} sin n (π + α) sec α 2 2 2 1 1 1 1 sin { (n + 1) ( π + α)} sin { n ( π + α)} 2 2 2 2 ⋅ 1 1 sin { . ( π + α)} 2 2 1 1 5. n (n + 1) 2 2 1 1 cos {2α + (n − 1) β } sin nβ  (i) n −   2 2  sin β  (i) sin {α +

(ii)

n 1 cos {2α + (n − 1) β } sin nβ +  2 2  sin β

  

O bjective T ype Q uestions

Multiple Choice Questions. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

2.

If |c | < 1, then the sum of the infinite series 1 1 ce iβ + c 2 e i2 β + c 3 e i3 β + …… ∞ is 2 3 (a) log (1 + ce iβ )

(b) log (1 − ce iβ )

(c) − log (1 − ce iβ )

(d) − log (1 + ce iβ ).

If tan −1

1

2n2 (a) 2n − 1 (c) n + 1

= tan −1 (2n + 1) − tan −1 y, then the value of y is (b) n − 1 (d) 2n.

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Fill In The Blanks Fill in the blanks “……” so that the following statements are complete and correct. 1 2 1. In C + iS method, if C = cos α + c cos (α + β) + c cos (α + 2β) + …… ∞, 2! then S = …… . iθ

2. If C + iS = 1 − e − e , then S = …… . 3. The sum of the infinite series e

iθ

+

cos φ 1!

e i (θ +φ) +

cos 2 φ 2!

e i (θ +2 φ) + … ∞

is …… . 4. If C + iS = e cos θ e

iθ

− 1, then C = …… .

5. If |c | < 1, then the sum of the infinite series 1 2 i2 α 1 c e + c 3 e i3 α − … ∞ 2 3 is …… . ce iα −

z3 z5 + + …… ∞ is … . 3 5 (c cos θ + ic sin θ), then C = …… .

6. If|z | < 1, then the sum of the infinite series z + 7. If C + iS = tan −1 1 1 8. = [tan 2θ − ……]. cos θ cos 2θ sin θ θ 9. cosec θ = cot − …… . 2 1 10. tan −1 = tan −1 (n + 1) − …… . 1 + n (n + 1)

True or False Write ‘T’ for true and ‘F’ for false statement. 1. 2. 3. 4. 5. 6.

If C + iS = e iθ [e cos φ e

iφ

2

], then C = e cos

φ

. sin (θ + cos φ sin φ).

iα

If C + iS = e sec α . e , then C = e . cos (tan α). 1 sec θ sec (θ + φ) = [tan (θ + φ) − tan θ]. sin φ tan θ = cot θ − 2 cot 2θ. 2 tan −1 = tan −1 (n + 2) − tan −1 (n + 1). (n + 1)2 The sum of the series sec θ sec 2θ + sec 2θ sec 3θ + sec 3θ sec 4θ + …… to n terms is 1 [tan (n + 1) − tan θ]. sin θ

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A nswers Multiple Choice Questions 1.

(c)

2.

(a)

Fill in the Blanks 1. sin α + c sin (α + β) +

1 2 c sin (α + 2β) + …… ∞ 2! iφ

2. e − cos θ sin (sin θ) 2

4.

e cos

6.

1+ z 1 log 2 1− z

θ

3. e iθ [e cos φ e ] 5. log (1 + ce iα )

cos (cos θ sin θ) − 1

7.

8. tan θ 10. tan

−1

1 tan −1 2

 2c cos θ    1 + c2   

9. cot θ n

True or False 1. 6.

F T

2.

T

3.

T

4.

T

5.

F

¨