The Elements of Complex Analysis, 9780471225652, 0471225657

Table of contents :
Preface
Contents
1 METRIC SPACE PRELIMINARIES
1.1 Set theoretic notation and terminology
1.2 Elementary properties of metric spaces
1.3 Continuous functions on metric spaces
1.4 Compactness
1.5 Completeness
1.6 Connectedness
2 THE COMPLEX NUMBERS
2.1 Definitions and notation
2.2 Domains in the complex plane
2.3 The extended complex plane
3 CONTINUOUS AND DIFFERENTIABLE COMPLEX FUNCTIONS
3.1 Continuous complex functions
3.2 Differentiable complex functions
3.3 The Cauchy-Riemann equations
3.4 Harmonic functions of two real variables
4 POWER SERIES FUNCTIONS
4.1 Infinite series of complex numbers
4.2 Double sequences of complex numbers
4.3 Power series functions
4.4 The exponential function
4.5 Branches-of-log
5 ARCS, CONTOURS, AND INTEGRATION
5.1 Arcs
5.2 Oriented arcs
5.3 Simple closed curves
5.4 Oriented simple closed curves
5.5 The Jordan curve theorem
5.6 Contour integration
6 CAUCHY'S THEOREM FOR STARLIKE DOMAINS
6.1 Cauchy's theorem for triangular contours
6.2 Cauchy's theorem for starlike domains
6.3 Applications
7 LOCAL ANALYSIS
7.1 Cauchy's integral formulae
7.2 Taylor expansions
7.3 The Laurent expansion
7.4 Isolated singularities
8 GLOBAL ANALYSIS
8.1 Taylor expansions revisited
8.2 Properties of zeros
8.3 Entire functions
8.4 Meromorphic functions
8.5 Convergence in d(D)
8.6 Weierstrass expansions
8.7 Topological index
8.8 Cauchy's residue theorem
8.9 Mittag-Leffler expansions
8.10 Zeros and poles revisited
8.11 The open mapping theorem
8.12 The maximum modulus principle
9 CONFORMAL MAPPING
9.1 Discussion of the Riemann mapping theorem
9.2 The automorphisms of a domain
9.3 Mappings of the boundary
9.4 Some illustrative mappings
10 ANALYTIC CONTINUATION
10.1 Direct analytic continuations
10.2 General analytic functions
10.3 Complex analytic manifolds
10.4 The gamma and zeta functions
APPENDIX: RIEMANN-STIELTJES INTEGRATION
SUGGESTIONS FOR FURTHER STUDY
BIBLIOGRAPHY
INDEX OF SPECIAL SYMBOLS
SUBJECT INDEX

Citation preview

JOHN WILEY & SONS

.London . New York . Sydn y

The elements of

COMPLEX ANALYSIS

The elements of

COMPLEX ANALYSIS

J. Duncan Department of Mathematics, King's College, University of Aberdeen, Scotland

JOHN WILEY & SONS Sydney New York London

Copyright © 1968 by John Wiley & Sons, Ltd. All rights reserved. No part of this book may be reproduced by any means, nor transmitted, nor translated into a machine language without the written permission of the publisher. Library of Congress catalog card number 68-29702 SBN 471 22565 7 Cloth bound SBN 471 22566 5 Paper bound

Printed in Great Britain by Spottiswoode, Ballantyne & Co. Ltd., London and Colchester

PREFACE

This book is intended as a first course in con1plex analysis for students who take their tnathetnatics seriously. The prerequisites for reading this book consist of son1e basic real analysis (often as motivation) and a minitnal encounter with the language of n1odern mathematics. The text is suitable for students with a background of one year of analysis proper, that is one year beyond the usual preparatory informal courses on calculus and analytic geotnetry. There is sufficient material for a course varying from 40 to 60 lectures according to the background and needs of the students. Although the book is a development of several courses I have given to honours students of mathematics at Aberdeen University, I have tried to keep the exposition as simple as possible with the object of making some of the theory of complex analysis available to a fairly wide audience. Since there are already n1any books on complex analysis it is necessary to say a word about the special features of the present book. This can best be done by a sitnple illustration. The central theoren1 of complex analysis is the famous Cauchy theoren1 and it has long been considered a deep and difficult theoretn. This is certainly true of the most general case; but the theorem admits a straightforward proof for the case in which the given domain is starlike. This latter case is adequate for almost all the results of cotnplex function theory. Since I believe that the student has every right to demand a proper treatment of the version of the Cauchy theoren1 which is to be used for subsequent results, I have chosen to prove the Cauchy theoren1 only for the starlike case. This means that I am able to give complete (and straightforward) proofs of all the subsequent theoretns even though the theorems are not always the most general possible. I indicate how the theorems may be extended for the rare occasions on which this is necessary. This then describes the aitn of the book-to give an introductory course in complex analysis that is rigorous v

VI

Preface

and yet an1enable to the serious undergraduate student. I hope that the course is also enjoyable. I have freely borrowed from other sources the ideas that I consider to be the best and the simplest. Certain sources will be transparent to the discerning reader. For example Section 9.2 is wholly' inspired by the corresponding account in the more advanced book Theory of Analytic Functions of One or Several Variables, by Professor H. Cartan. On the other hand I think that occasionally some of the details are new as in parts of Chapters 5 and 10. It is a pleasure to record several personal acknowledgements. I owe much indirectly to Professor F. F. Bonsall; any lucidity of style that may occur in this book arises largely from his personal example as a supervisor and friend. Some of the work for this book was carried out while I was visiting Yale University where I was fortunate to have many fruitful conversations with Professor E. Lee Stout. I am grateful to my Aberdeen colleagues Eric E. Morrison and Alan J. White who read parts of the manuscript and gave encouragement and constructive criticisms. I am also indebted to the secretaries and typists of the mathematics departments both at Aberdeen and Yale who prepared various editions of the manuscript. Finally I wish to express my thanks to the editorial staff of John Wiley & Sons for their advice and encouragement.

Aberdeen University, 1968

JOHN DUNCAN

CONTENTS

1 METRIC SPACE PRELIMINARIES 1.1 1.2 1.3 1.4 1.5 I. 6

Set theoretic notation and terminology . Elementary properties of metric spaces . Continuous functions on metric spaces . Compactness Completeness Connectedness

1 3 13 19

28 30

2 THE COMPLEX NUMBERS 2.1 Definitions and notation 2.2 Domains in the complex plane 2.3 The extended complex plane .

35 43 50

3 CONTINUOUS AND DIFFERENTIABLE COMPLEX FUNCTIONS 3.1 3.2 3.3 3.4

Continuous complex functions Differentiable complex functions . The Cauchy-Riemann equations Harmonic functions of two real variables

54 60 67 71

4 POWER SERIES FUNCTIONS 4.1 4.2 4.3 4.4 4.5

Infinite series of complex numbers . Double sequences of complex numbers . Power series functions . The exponential function Branches-of-log . vii

75

79 84 90 95

...

Contents

VIti

5 ARCS, CONTOURS, AND INTEGRATION 5.1 5.2 5.3 5.4 5.5 5.6

101 105 108 114 119 122

Arcs . Oriented arcs Simple closed curves Oriented simple closed curves The Jordan curve theorem Contour integration

6 CAUCHY'S THEOREM FOR STARLIKE DOMAINS 131 135 138

6.1 Cauchy's theorem for triangular contours 6.2 Cauchy's theorem for starlike domains . 6.3 Applications 7 LOCAL ANALYSIS 7.1 7.2 7.3 7.4

Cauchy's integral formulae Taylor expansions The Laurent expansion. Isolated singularities

.

147 152 158 163

8 GLOBAL ANALYSIS 8.1 8.2 8.3 8.4 8.5 8.6 8. 7 8.8 8.9 8.10 8.11 8.12

Taylor expansions revisited . Properties of zeros Entire functions . Meromorphic functions Convergence in d(D) . Weierstrass expansions . Topological index Cauchy's residue theorem Mittag-Leftler expansions Zeros and poles revisited The open mapping theorem The maximum modulus principle

174 176 181 185 189 197 . 203 209

. 214 . 222 . 230

234

9 CONFORMAL MAPPING 9.I 9.2 9.3 9.4

Discussion of the Riemann mapping theorem . The automorphisms of a domain . Mappings of the boundary . Some illustrative mappings .

.

243

.

247

. 255 .

261

.

Contents

IX

10 ANALYTIC CONTINUATION 10.1 10.2 10.3 10.4

264 269

Direct analytic continuations . General analytic functions Complex analytic manifolds . The gamma and zeta functions

275

282

APPENDIX: RIEMANN-STIELTJES INTEGRATION

.

291

SUGGESTIONS FOR FURTHER STUDY

.

303 305

BIBLIOGRAPHY . INDEX OF SPECIAL SYMBOLS .

.

307

SUBJECT INDEX

.

309

1 METRIC SPACE PRELIMINARIES

1.1 Set theoretic notation and terminology The basic language of contemporary mathematics is the language of set theory. We shall assun1e that the reader is acquainted with the intuitive ideas of set theory. Since the notation of set theory varies from book to book we shall begin by listing briefly the notation which we shall employ in this book. We denote set n1en1bership byE; thus if E is a set and xis an element (or n1e1nber) of E, we write x E E. If x is not an element of E, we write x ¢ E. The set of all elements of E with property P is denoted by {x: x

E

E, P(x)}.

When there is no danger of confusion we shall write this set more simply as {x: P(x)}. When we wish to indicate that property P holds for each x in Ewe write P(x)

(x

E

E).

Singletons (i.e. sets consisting of exactly one element) are denoted by {a}. The empty set is denoted by 0. Set inclusion is denoted by c ; thus if A and Bare subsets of Ewe write A c B if every elen1ent of A is also an element of B. If A c B and B c A then A = B. If A c Band A # B we say that A is a proper subset of B. We use no special symbol for proper inclusion since it is rarely necessary to exclude the possibility of equality. Union, intersection and complementation are denoted by u, n, "respectively; thus if A and B are subsets of Ewe write Au B = {x: x E A or x E B} A nB = {x:xEAandxEB} A """ B = { x: x E A and x ¢ B}. 1

2

Elen~ents

of Con1plex Analysis

1.1

More generally if A is some index set and AA is a subset of E for each A E A we write

U {AA: A E A}

n {AA: A

E

A}

= =

{x: x {x: X

E E

AA for so~e A} AA (A E A)}.

If A and B are two non-empty sets we denote their Cartesian product by A x B; thus

Ax B

= {(x,y): xEA,yEB}.

When B = A we son1etin1es write A x A more briefly as A 2 • More generally we write An+ 1 for the Cartesian product of A with itself n times. The sets consisting of the integers, the positive integers (i.e. 1, 2, 3, ... ), the rationals and the reals are denoted respectively by Z, P, Q, R. The symbols Z, P, Q, R will be used exclusively for the above sets. We assume that the reader is acquainted with son1e of the basic properties of these sets. In particular we assume that the reader knows that the real numbers form a complete ordered field. Given non-empty sets X and Y, a function (or Inapping) from X to Y is a subset f of X x Y, such that (i) given x E X there is y E Y, such that (x, y) Ej (ii) (x, y) Ej, (x, z) Ej => y = z. We denote the n1apping syn1bolically by

f:

X->- Y.

As usual we writef(x) for the image of x under f GivenS c X, T c Y we write

f(S) = {f(x): x E S} J- 1 (T) = {x: x E X,f(x)

E

T}.

The set f(X) is called the range off We say that the n1apping f is onto* if f(X) = Y. We say that the mappingfis one-to-one if

f(x) = f(y)

=>

x

= y.

Iff is one-to-one the inverse function f- 1 : f(X) ->- X is defined by

j- 1 (y) =

X

if

X

= j(y).

The reader should be careful to distinguish the possible meanings of f- 1 at each occurrence. • We write 'onto' as one word since we regard it as a special mathematical term.

...

Contents

VIti

5 ARCS, CONTOURS, AND INTEGRATION 5.1 5.2 5.3 5.4 5.5 5.6

101 105 108 114 119 122

Arcs . Oriented arcs Simple closed curves Oriented simple closed curves The Jordan curve theorem Contour integration

6 CAUCHY'S THEOREM FOR STARLIKE DOMAINS 131 135 138

6.1 Cauchy's theorem for triangular contours 6.2 Cauchy's theorem for starlike domains . 6.3 Applications 7 LOCAL ANALYSIS 7.1 7.2 7.3 7.4

Cauchy's integral formulae Taylor expansions The Laurent expansion. Isolated singularities

.

147 152 158 163

8 GLOBAL ANALYSIS 8.1 8.2 8.3 8.4 8.5 8.6 8. 7 8.8 8.9 8.10 8.11 8.12

Taylor expansions revisited . Properties of zeros Entire functions . Meromorphic functions Convergence in d(D) . Weierstrass expansions . Topological index Cauchy's residue theorem Mittag-Leftler expansions Zeros and poles revisited The open mapping theorem The maximum modulus principle

174 176 181 185 189 197 . 203 209

. 214 . 222 . 230

234

9 CONFORMAL MAPPING 9.I 9.2 9.3 9.4

Discussion of the Riemann mapping theorem . The automorphisms of a domain . Mappings of the boundary . Some illustrative mappings .

.

243

.

247

. 255 .

261

1.2

Elernents of Con1plex Analysis

4

A nwtric space is a pair (X, d) where X is a non-etnpty set and dis a nwtric on X, i.e. dis a real valued function on X x X satisfying

(i) d(x, y) ~ 0 (x, y E X); d(x, y) = 0 iff x (ii) d(x, y) = d(y, x) (x, y E X), (iii) d(x, z) ~ d(x, y) + d(y, z) (x, y, z E X).

= y,

When no confusion is likely we denote a metric space by X for short. It is of course helpful to think of d(x, y) as the 'distance' from the point x to the pointy. For this reason, condition (iii) is referred to as the 'triangle inequality'. The student may also find it helpful on occasions to draw rough geometrical pictures to facilitate his understanding. The fundamental example of a metric space is obtained by taking X = R and (x, y E R). d(x, y) = lx - Yl It is a simple exercise to verify that dis indeed a metric on R. We say that dis the usual nzetric on R, and whenever we speak of the metric space R we mean that R has the usual metric. Every metric space gives rise to other tnetric spaces in the following manner. Let (X, d) be any tnetric space and let E be a non-empty subset of X. Let p be the restriction of the function d to E x E. It is trivially verified that pis a metric on E. We say that pis the relative nzetric induced onE by the metric d. Given any metric space X and any non-empty subset E of X, the set E with the relative tnetric will often simply be called the metric space E. The metric space E is often referred to as a subspace of E. \Ve shall subsequently see that the properties of subspaces n1ay be quite different from the properties of the original metric space. If Eisa non-en1pty subset of R, by the metric space E we n1ean the set E together with the relative metric induced on E by the usual metric on R. Another standard method for producing new n1etric spaces from old ones is by taking Cartesian products. We illustrate the tnethod by an example. Given n E P let X = Rn and define don Rn x Rn by

d((x, x 2 ,

••• ,

x.), (y., y 2 ,

••• ,

y.)) =

{~, (x,

_ y,)z} i.

The student may verify that dis a n1etric on Rn (only the triangle inequality needs any thought); we say that dis the usual n1etric on R 11 • Other examples of metrics on Rn may be found in Problem 1.2. The following metric space concepts are n1otivated by elementary geometry and the case of the tnetric space R. Let (X, d) be any metric space. Given p E X and e > 0, the set

{x: x

E

X, d(p, x) < e}

1.2

5

Metric Space Prelinzinaries

is called the open ball with centre p and radius E and is denoted by B(p, E). When we wish to specify which tnetric is involved we shall write Bd(p, E). The set {X:

X

X) ~

E X, d(p,

E}

is called the closed ball with centre p and radius E. A neighbourhood of p is any subset of X that contains son1e open ball \Vith centre p. A subset E of X is said to be bounded if it is contained in sotne ball (open or closed). Observe that B(p, E) is thus a bounded neighbourhood of p for each E > 0. Given a sequence {xn} in X we say that {xn} is convergent if there is x E X such that for every e > 0 there is N( E) E P such that

(n > N(e)). We call x a lirnit of the sequence {xn} and say that {xn} converges to x. We write this as Xn

--+ X

as

1l

--+ 00

or lim n-+ oo

Xn

=

X.

It is convenient at this point to make some simple observations about convergent sequences. For the metric space R the above definition coincides exactly with the classical definition of the convergence of a sequence of real nutnbers. For any metric space it is simple to verify that

lin1 Xn = n-+ oo

X

lim d(x, n-+ oo

Xn) =

0.

This indicates how convergence in the metric space R is fundamental to convergence in any 1netric space. If {xn} converges to x it is clear that any subsequence of {xn} also converges to x. In the metric space R 2 it is easy to see that {(xn, Yn)} converges to (x, y) if and only if {xn} converges to x and {Yn} converges to y; there is an obvious generalization to sequences in Rm for any m > 2. At this point we have still to determine how many limits a convergent sequence may have. We ought to suspect that a convergent sequence in any n1etric space can have only one limit. This is indeed true-but it requires to be proved. Proposition 1.1. Let (X, d) be a 1netric space and let {xn} be a sequence in X such that {xn} converges to x and toy. Then x = y. Proof We have

lim d(x, Xn) = 0 = lim d(y, Xn). n-+ oo n-+ oo

Elen1ents of Cornplex Analysis

6

Given E > 0 we thus have son1e N 1(E) d(x, Xn) < E

Si1nilarly, there is NiE)

E

E

1.2

P, such that

(n > N 1 (E)).

P, such that

d(y, Xn)
N2( E)).

Choose any integer n > max (N1(E), N 2(E)) and we have d(x, y) ~ d(x, Xn) + d(xm y) = d(x, Xn) + d(y, Xn)

< 2e. We have now shown that 0 ~ d(x, y) < 2e for every e > 0. It follows that d(x, y) = 0, and so x = y. We are now entitled to speak about the limit of a convergent sequence in a metric space. We consider next some special classes of subsets of a metric space. Given a metric space X, a subset 0 of X is said to be open if each point x of 0 has a neighbourhood that is contained in 0. Thus 0 is open in X if and only if for each x E 0 there is e > 0 such that B(x, E) c 0. Given A c X and p E X, we say that p is a cluster point of A if every neighbourhood of p contains a point of A other than p. Thus pis a cluster point of A if and only if for each e > 0 there is q E A n B(p, E) with q ¥= p. The set of all cluster points of A is denoted by A'. Given A c X we say that A is closed if it contains all its cluster points, i.e. if A' c A. Since both open and closed subsets of a metric space arc of particular significance we shall now prove several results about them. Proposition 1.2. The union of any farnily of open subsets of a nzetric space is open; the intersection of a finite fanzily of open subsets is open. Proof Let {0;.: A E 11} be a fatnily of open subsets of a n1ctric space and let 0 = U {O;.:A E 11}. Given p E 0 there is some A E A such that p E 0;.. Since 0;. is open there is E > 0 such that B(p, E) c 0 ).· Since 0;. c 0 we have B(p, E) c 0 and therefore 0 is open. Let 0 1 ,0 2 , •.• ,0n be open sets and let 0 = n{or:r = 1,2, ... ,n}. Given p E 0 we have p E Or (r = 1, 2, ... , n). Since each Or is open there is E"r > 0 such that

Let

E

=min

{Er:

(r

=

r = 1, 2, ... , n}. Then

E

1, 2, ... , n).

> 0 and (r

=

It follows that B(p, E) c 0 and so 0 is open.

l, 2, ... , n).

1.2

7

Metric Space Prelintinaries

We retnark at this point that Proposition 1.2 serves as the definition for the n1ore general objects known as topological spaces. More precisely a topology on a non-en1pty set X is defined to be a fan1ily .r of subsets of X that contains 0 and ..:\"" and is closed under the operations of arbitrary unions and finite intersections. A topological space is then a pair (X, :T) where X is a non-etnpty set and .r is a topology on X. Proposition 1.3. A subset F of a nzetric space is closed if and only if X" F rs open. Proof Let F be closed, so that F' c F. Let p E X"' F. Since p ¢ F' son1e neighbourhood of p n1ust be contained in X" F. Therefore X" Fis open. Let X" F be open and let p E X"' F. Then son1e neighbourhood of pis contained in X"' F, so that p cannot be a cluster point of F. This shows that F' n (X" F) = 0 and so F' c F, i.e. Fis closed.

Proposition 1.4. The union of a finite family of closed subsets of a n1etric space is closed,· the intersection of any family of closed sets is closed. Proof This follows easily fron1 Propositions 1.2 and 1.3, and the de Morgan laws for con1ple111entation of sets.

The next result gives a very useful characterization of closed sets in terms of convergent sequences. Proposition 1.5. Given a subset F of a nzetric space X, the following staten1ents are equivalent.

(i) F is closed. (ii) {xn} c F, X

=

lin1 Xn => X E F. n-+ co Proof (i) => (ii). Let F be closed and let {xn} be any convergent sequence such that {xn} c F and x = lin1 Xn. Suppose x EX" F. Given e > 0 we n-+ co n1ay choose N E P such that Xn E B(x, e) (n > N). This means that every neighbourhood of x contains points of {xn}, and so of F, other than x. Therefore x E F'. Since F is closed we have F' c F and sox E F. This is a contradiction and therefore x ¢ X"' F, i.e. x E F and so condition (ii) is satisfied. (ii) => (i). Let condition (ii) hold and let p E F'. Then every neighbourhood of p contains points of F. In particular, for each n E P we n1ay choose Pn

E

F n B(p,

!). We thus have {Pn}

c F and lim d(p, Pn)

n-+CO p. It now follows frotn condition (ii) that p fl

that lim Pn = n-+ co F' c F and so F is closed.

E

= 0 so

F. Thus

Eletnents of Con1plex Analysis

8

1.2

The basic examples of open sets and closed sets are given by open balls and closed balls. Let B(x, e) be an open ball in a metric space and let y E B(x, e). To show that B(x, e) is open we must produce some 8 > 0 such that B(y, S) c B(x, e). Let 8 = e - d(x, y) so that q > 0. Given p E B(y, o) we have d(x, p) ~ d(x, y) < d (x, y)

+ d(y, p)

+

8

= e.

Therefore B(y, 8) c B(x, e) and so B(x, e) is an open set. We leave the student to prove that a closed ball is a closed set. As a further example of closed sets we see from Proposition 1.5 that any singleton is a closed set. It follows from Proposition 1.4 that any finite subset of a metric space is closed. The unwary student may be tempted to think that any given subset of a n1etric space is either open or closed. In fact a subset of a metric space may be neither open nor closed-or it may be both open and closed! For example, in the n1etric space R it is easy to see that E

= {x: x E R, 0

~

x < 1}

is neither open nor closed. Indeed there is no neighbourhood of 0 contained in E, so that E is not open. Moreover 1 is a cluster point of E which does not belong to E. On the other hand it is trivial to verify that in any metric space X the set X itself is both open and closed. Suppose now that we give the above set E the relative n1etric p. Then the set E is both open and closed in the metric space (£, p) even though it is neither open nor closed in the metric space R. This situation is often a source of confusion to the student. Indeed much of the con1mon n1isunderstanding of metric spaces revolves around a misunderstanding of the relative metric. In subsequent chapters we shall frequently be working with relative n1etrics. It is therefore essential for the student to understand what happens to open sets, closed sets and convergent sequences when the relative metric comes into play. Proposition 1.6. Let (X, d) be a 111etric space and let E be a non-en1pty subset of X with relative n1etric p. A subset V of E is open in (E, p) iff it is of the fornz V = En U where U is open in (X, d). Proof. Given x

E

E we have BP(x, e)

= {y: y

£, p(x, y) < e} = {y:yEE,d(x,y) < e} =En Btt(x, e). E

Afetric Space Prelinzinaries

1.2

Let V be open in (E, p). For each x Brlx, E(x)) c V. It follows that V

E

9

V there is E(x) > 0 such that

= U {Bix, E(x)): x E V} = U {£ n Bd(x, E(x)): x E V} =En U

where U = U {Bd(x, E(x)): x E V}. Since each Bd(x, E(x)) is open in (X, d) it follows fron1 Proposition 1.2 that U is open in (X, d) and so V has the required fonn. Conversely let V = En U where U is open in (X, d). Given x E V we have x E U. Since U is open there is E > 0 such that Bix, E) c U. Therefore Br/x, E) = En Bd(x, E) cEnU=V.

Tlus shows that Vis open in (E, p) as required. Corollary. Let E be an open subset of a nzetric space (X, d). Then V c E is open in (E, p) iff it is open in (X, d). Proof Let V c E be open in (E, p). Then V = E n U, where U is open in (X, d). Since E is open in (X, d), we have V open in (X, d) by Proposition 1.2. Conversely, if V c E is open in (X, d), then V = En V is open in (£, p) as required.

Proposition 1.7. Let (X, d) be a nzetric space and let E be a non-en1pty subset of X with relative nzetric p. A subset F of E is closed in (E, p) iff it is of the fornz F = E n H, where 1-I is closed in (X, d). Proof This follows readily fron1 Propositions 1.3 and 1.6.

Corollary. Let E be a closed subset of a metric space (X, d). Then F c E is closed in (E, p) iff it is closed in (X, d). Proof Similar to the last corollary.

To illustrate the above two results let X

=

R and

E = {x: 0 ~ x ~ I} u {x: 2 < x < 3}.

Then V = {x: 0 ~ x ~ 1} is open in (E, p) but not in R; n1oreover F = {x: 2 < x < 3} is closed in (E, p) but not in R. Proposition 1.8. Let (X, d) be a n1etric space and let E be a non-enzpty subset of X with relative n1etric p. If {xn} is a convergent sequence in (E, p) then it is a convergent sequence in (X, d) (with the sanze linzit). If {xn} c E

1.2

Elenzents of Conzplex Analysis

10

is a convergent sequence in (X, d) then it is a convergent sequence in (E, p) iff its linzit is in E. Proof Let {xn} be a convergent sequence in (E, p). Then there is x E E such that lin1 p(x, Xn) = 0. Therefore lim d(x, Xn). = 0 and so {xn} conn-oo

n-oo

verges to x in (X, d). Let {xn} c E be a convergent sequence in (X, d) with limit x. Suppose that x E E. Then li1n p(x, Xn) = lim d(x, Xn) = 0

n-+ oo

n-oo

and so {xn} converges toxin (£. p). Suppose now that x ¢= E and that {xn} converges toy in (£. p). By the first part of this proposition we have that {xn} converges to y in (X, d). Proposition 1.1 now gives x = y. This contradiction con1pletes the proof.

To illustrate the above result, observe that the

sequence{~}

in R but fails to converge in the n1etric space {x: 0 < x

~

converges

1}.

We have already re1narked that a subset of a metric space 1nay be neither open nor closed. It is, however, possible, in a natural way, to associate with each subset of a n1etric space a corresponding open set and a corresponding closed set. This association will be useful in subsequent chapters. Let (X, d) be any n1etric space and let E c X. The interior of E is the set of all points of E that have so1ne neighbourhood contained in E; it is denoted* by E 0 • We show below that the interior of a set is open. The closure of E is the union of E and E'; it is denoted by E-. We say that a subset E of X is dense in X if E- = X. The boundary of E is the intersection of E- and (X"'- E)-; it is denoted by b(E). We show below that the closure of a set is closed and so the boundary of a set is also closed. Proposition 1.9. Let (X, d) be a 1netric space and let E c X.

(i) E 0 is the largest open set contained in E. (ii) E- is the snzallest closed set containing E. (iii) b(E) = b(X "'-E) = E- "'- £ 0 •

X

Proof (i) Given p E E 0 there is E > 0 such that B(p, t:) c E. Let E B(p, E). Since B(p, E) is open there is 0 > 0 such that B(x, o) c B(p, E).

* Tn some texts the interior of E is denoted by int E. We shall usc the symbol int for a difl'erent purpose in Chapter 5 and subsequently.

Metric Space Prelilninaries II Thus B(x, 8) c E and sox E £ 0. This shows that B(p, E) c E 0 and so E 0 is an open set. Let U be any open set such that U c E. Given x E U there is E > 0 such that B(x, E) c U c E. Thus x E E 0 and so U c E 0 • In other words E 0 is the largest open set contained in E. (ii) We observe that x E X""- E- iff son1e neighbourhood of x is contained in X""- E, i.e. iff x E (X""- E) 0 • It follows fron1 (i) above that X""- E- is open and so E- is closed by Proposition 1.3. Let F be any closed set such that F ::::> E. Then X""- F is open and X""- F c X""- E. It follows frotn (i) above that X""- F c (X""- E) 0 = X""- E-. Therefore F ::::> E- and so E- is the sn1allest closed set containing E. (iii) It is inunediate fron1 the definition of boundary that b(X"'- E) = (X""- E)- n (X""- (.X""- E))- = (.,\ ""- E)- n E- = b(E). Recall fron1 the proof of (ii) that 1.2

and so (X""- E)-

= X""- E 0 •

Therefore

b(E) = E-

n (X""- E 0 ) = E- ""- E 0 •

Observe fron1 the above result that E is open iff E = E 0 and E is closed {iff E = E-. To illustrate the above result further let X = R and E = x: 0 ~ x < I}. We n1ay easily check that E0

= {x: 0 < x < 1},

E-

= {x: 0

~

x

~

I},

b(E) = {0, 1}.

On the other hand if X is the Inetric space {x: 0 E = {X: 0 ~ X < 1}, then b(E)

~

x

~ I}

and

= {1}.

In other words, care is again required when working with the relative n1etric. PROBLEMS 1

1. Let X be any non-empty set. Define d on X x X by d(x, y)

1

if

= { o 1.f

X -:/:-

X=

y y.

Show that dis a metric on X (it is called the discrete nzetric). Which subsets of X are (i) open, (ii) closed, (iii) bounded? Which sequences in X converge?

Elen1ents of Con1plex Analysis

12

1.2

2. Let n E P and let X= Rn. Define d 1 and doo on X x X by n

d1((xh X2, · • ·, Xn), (yl, Y2, · · ·, Yn)) =

:2

r= 1

lxr -

Yrl

doo((x 1,x2, ... ,Xn),(yby2, ... ,yn)) = n1ax{!xr __:_ Yrl:r = 1,2, ... ,n}. Show that d1 and doo are metrics on Rn. Show also that Rn with the usual 1netric, (Rn, d1) and (Rn, doo) have the same open sets. Now let (Xn Pr) (r = I' 2, ... ' n) be metric spaces and let X = xl X x2 X ••• X Xn. Generalize the above results by defining analogous metrics on X.

3. Let S be any non-empty set and let 88R(S) be the set of all bounded real functions on S. Define don gou(S) x goR(S) by d(f, g) = sup {lf(s) - g(s)l: s E S}. Show that dis a metric on goR(S).

4. Show that any convergent sequence in a metric space is bounded.

5. Let E be a bounded subset of a metric space X. (i) Given p E X show that there is E > 0 such that E c B(p, E). (ii) Show that E- is bounded. 6. Let E be a subset of a metric space. (i) Given pEE' show that every neighbourhood of p contains an infinite subset of E. Show also that there is a sequence {pn} of distinct points of E such that p = lim Pn· n-+ oo

(ii) Show that E' is closed. Is (£')' = E'? (iii) Given p E E- show that there is a sequence {Pn} in E such that p = lim Pn- If {pn} is any convergent sequence such that {Pn} c E n-+oo

show that its limit is in E-.

7. Let E be a subset of a metric space. Show that E- is the intersection of all the closed sets that contain E, and E 0 is the union of all the open sets that arc contained in E. Show also that E- = Eu b(E),

E 0 = E ""- b(E).

8. Given subsets A, B of a n1etric space show that (i) (A 0 ) 0 = A 0 ; A c B => A 0 c B 0 ; (A n B) 0 = A 0 n B 0 , (ii) (A-)- = A-; A c B => A- c B-; (A u B)- = A- u B-.

9. Let Y be a subspace of a n1etric space X and let E c Y. Relate the interior, cJosurc, and boundary of E in Y to those in X.

1.3

Metric Space Prelilninaries

13

10. Given any open ball B(x, E) in a 1netric space detennine B(x, E)- and b(B(x, E)). Can b(B(x, E)) be en1pty?

11. In the n1etric spaceR detennine Q 0 , Q-, b(Q). Sho\v also that Q 2 is dense in R 2 (with the usual n1etric). Generalize this last result.

12. Let E be a bounded subset of R. Show that sup E and inf E belong toE-.

1.3 Continuous functions on ntetric spaces In this section we shall consider the itnportant concept of the continuity of tnappings between n1etric spaces and we shall show how it is related to the concepts of open sets, closed sets, and convergence. Let (X, d), ( Y, e) be n1etric spaces, let f be a function from X to Y, and let x E X. We say that f is continuous at x if for each € > 0 there is O(E) > 0 SUCh that

f(Bd(x, o{E))) c Be(f(x), E)

.

I.e.

d(X, y) < O(E)

=>

e(f(x),f(y))
0 let

o=

E.

~ d(x, y).

Then

d(x, y)


!f(x) - f(y)!
(ii). Let f be continuous at x and let {xn} converge to x. Given E > 0 there is 0 > 0 such that f(Bd(x, o)) c BeU(x), E). Since {xn} converges to x there is N E P such that

(n > N) and hence

(n > N). Therefore {f(xn)} converges to f(x) as required. (ii) ~ (i). Let condition (ii) hold and suppose that f is not continuous at x. Then there is sotne Eo > 0 such that for every o > 0 we have f(Bix, o)) ¢ Bc(f(x), Eo).

In particular, if we choose Ba

o = ! (n E P)

there is a corresponding

Xn

( n1) such that f(xn) ¢ Bc(f(x), Eo). 1t follows that {xn} converges to X,

ll

in X

and

(n

E

P)

so that {f(xn)} does not converge to f(x). This contradiction completes the proof.

1.3

Afetric Space Prelinzinaries

15

Proposition 1.11. Let (..r¥, d), ( Y, e) be nzetric spaces and let .f: ..rY ->- Y. The following statentents are equiz·alent.

f is continuous. f - 1 (F) is closed in X whenever F is closed in Y. f - 1 ( 0) is open in X whenever 0 is open in Y. Proof (i) => (ii). Let f be continuous and let F be closed in Y. Let {xn} c /- 1 (F) with lin1 Xn = x. Since f is continuous it follows frotn Pron-co position 1.10 that lim f(xn) = f(x). Since {f(xn)} c F and F is closed, it (i) (ii) (iii)

n-co follows froiD Proposition 1.5 thatf(x) E F and sox Ej- 1 (F). We conclude froiD Proposition 1.5 thatf- 1 (F) is closed, i.e. condition (ii) holds. (ii) => (iii). Let condition (ii) hold and let 0 be open in Y. By Proposition 1.3 Y"" 0 is closed in Y and so f- 1 ( Y "- 0) is closed in X by condition (ii). Since /- 1

cy"" o) = x ""/-

1

(0)

it follows that X""f- 1 (0) is closed in X and thereforef- 1 (0) is open in X by Proposition 1.3. Thus condition (iii) holds. (iii) => (i). Let condition (iii) hold and let x EX. Given E > 0 we have that Be(f(x), E) is open in Y and so by condition (iii) f- 1 (Be(f(x), E) is an open set in X which contains x. Hence there is 8 > 0 such that Bd(x, 8) c f- 1 (Be(f(x), E)).

Therefore f(Bd(x, 8)) c Be(f(x), E).

and sofis continuous at x. Since x was arbitrary we have thatfis continuous. The proof is complete. Continuity has the pleasant property of being preserved under composition. This statement is made precise in the next result. Proposition 1.12. Let X, Y, Z be n1etric spaces and let f: X--+ Y, g: Y--+ Z. Iff is continuous at x and g is continuous at f(x), then go f is continuous at x. Iff and g are continuous, then g of is continuous. Proof Let {xn} c X with lim Xn n-+co from Proposition 1.10 that

= x. Sincefis continuous at x it follows

lim f(xn) n-+co

= f(x).

Since g is continuous atf(x) it then follows that lim g(f(xn)) n-+co

= g(f(x)).

1.3

Elements of Complex Analysis

16

We conclude from Proposition 1.10 that g of is continuous at x. The final staten1ent of the proposition is now obvious.

Corollary. lffE Honz (X, Y), g

E

Honz (Y, Z), ~hen g ofE Hom (X, Z).

Proof. It is trivial to verify that go f maps X one-to-one onto Z. By the above result we have that go f is continuous. Since (go f) - l = f- 1 o g- 1 we also have that (g of) - 1 is continuous.

It is also important to study continuity in relation to the relative metric.

Proposition 1.13. Let (X, d), ( Y, e) be metric spaces, let f: X-+ Y, and let E be a non-enzpty subset of X with relative 1netric p. Iff is continuous on E then !IE is continuous on the n1etric space (E, p). If!IE is continuous on the metric space (E, p) and if E is open in (X, d), then f is continuous on E. Proof Let f be continuous on E and let x E E. Given E > 0 there is o > 0 such thatf(Bd(x, o)) c Be(f(x), E). Since B/)(x, o) c Bd(x, o) we have

f(B/)(x, o)) c f(Bd(x, o)) c Be(f(x), E). Since x was arbitrary this shows that fiE is continuous on the metric space (E, p). Let fiE be continuous on (E, p) and let E be open in (X, d). Given x E E there is o1 > 0 such that Bd(x, o1 ) c E, and so for 0 < o ~ o1 we have

Since fiE is continuous at x, given E > 0 there is o2 > 0 such that f(B/)(x, o2)) c BeU(x), E). Let o = min (ob o2) so that o > 0. We now have

and so f is continuous at x. Since x was arbitrary in E the proof is now complete. This last result will often appear tacitly in subsequent chapters. To illustrate the result, let a, bE R with a < b, let E = {x: x E R, a ~ x ~ b}, and let f be a continuous real function on the n1etric space E. In classical language the continuity at a and b is 'one-sided' continuity; we follow the classical notation by writing lim f(x) = f(a),

x-a+

lin1 f(x)

x-b-

=

f(b).

1.3

17

Afetric Space Prelinzinaries

Now let g: R ->- R be defined by g

( ) {10 X

=

if if

X E X¢

E E.

It is obvious that gjE is continuous on the tnetric space E. On the other hand g itself is not continuous on E since it clearly fails to be continuous at a and b. It is of course true that g is continuous on {x: a < x < b}. In subsequent chapters we shall frequently discuss sequences of functions with values in a rnetric space. For such sequences there are two cotnn1on concepts of convergence, narnely pointwise and uniforn1. More precisely, let X be any non-ernpty set, let E c X, and let ( Y, e) be a n1etric space. Let I andln (n E P) be functions frorn X to Y. We say that {In} converges to I pointll'ise on E if given x E E, E > 0 there is N(x, E) E P such that (n > N(x, E))

i.e. if for each x

E

E, {ln(x)} converges to l(x). We write

lirn

n-> co

In =I

We say that {In} converges to N(E) E P such that

(pointwise on E).

f

unifonnly on E if given

(n > N(E), x

E

E

> 0 there is

E).

We write lim

n-> co

In= I

(uniformly on E).

Roughly speaking, in uniforn1 convergence the same N satisfies the required condition for each of the points of E. Special interest occurs when X is itself a metric space and the functions In are continuous. We may then ask which types of convergence preserve continuity. Pointwise convergence fails to have this property. For example Y = R, let X= {x: x E R, 0 ~ x ~ 1},

In (x) =

xn

0 l(x) = { 1

(x

E

X, n E P)

if 0 if X

~X
0. Since {fn} converges to f uniformly on X, we 1nay certainly choose some N E P such that

Proof Let p

E

(x EX).

Since fN is continuous at p, we may now choose 8 > 0 such that

fN(Ba(p, 8)) c Be(JN(p), !€). If d(p, x) < 8 we now have

e(f(p),f(x)) ~ eU(p),fN(p)) + eUN(p),J:v(x)) + e(fN(x),f(x)) < -!€ + -1€ + j-E = €. This shows that f is continuous at p. Since p was arbitrary f is thus continuous as required. We shall further discuss continuity in relation to the topics to be introduced in §§1.4, 1.6.

PROBLEMS 1 13. Let (X, d), ( Y, e) be metric spaces and let f: X--:?- Y. Show that f is continuous as a mapping into the metric space Y if and only if it is continuous as a n1apping onto the n1etric space f(X). 14. Let (X, d) be a discrete 1netric space (i.e. d is the discrete n1etric on X as defined in Problem 1.1) and let ( Y, e) be any n1etric space. Which functions f: X ----7 Yare continuous? Which functions g: Y--:?- X are continuous? Give an exan1ple of a one-to-one function fro1n one n1etric space onto another which is continuous but is not a hon1eon1orphism.

15. Let (X, d) be a metric space, and Jet f and g be real functions on X. Define the functions f

+ g, fg, f

V

g, f

1\

g, lfl on X by

([+ g)(x) =f(x) +

g~Y)

(fg) (x) = f(x) g(x) (f V g) (x) = max (f(x), g(x)) (f 1\ g) (x) = min (f(x), g(x)) lfl (x) = lf(x)[. Iff and g are continuous show that the above functions are also continuous.

Afetric Space Prelinzinaries

1.4

19

16. Given.f: R---+R,g: R---+R define h: R2 ->-R by h(x, y) = (f(x), g(y)). Show that lz is continuous iff f and g are continuous.

17. Let X be a non-etnpty set and let {fn}, {gn} be sequences of real on X such that functions . lin1 fn = f

(unifonnly on X)

n ..... co

lin1 gn

n ..... co

=g

(unifonnly on X).

Show that

1i111 (Jn

n ..... co

+ g n)

=

f +g

(pointwise on X)

n ..... co

I i 111 (fn

(unifonnly on X)

V

n ..... co

g n)

=

f

V

g

(unifonnly on X) (uniforn1ly on X)

litn lfn I = IJI

(unifonnly on X).

n ..... co

Give an exatnple in which {fngn} fails to converge to fg uniforn1ly on X. 18. Let (X, d) be a 111etric space. Given p

fp(x)

=

d(x, p)

E

X let

(x EX).

If {pn} c X and lin1 Pn = p show that n ..... co

(unifonnly on X). 19. Given a n1etric space (X, d) define e on X x X by

d(x, y) . e(:\' y) = 1 + d( x, y) · Show that (X, e) is a n1etric space and that the identity 1napping is a homeomorphism of (X, d) with (X, e). Note that X= Be(x, 1) for each XE X. 20. Show that the relation of 'being hon1eon1orphic' is an equivalence relation on the class of all metric spaces.

1.4 Compactness In this section we shall discuss another special class of subsets of a metric space, namely compact sets. The concept of compactness is one of the most important ideas in topology. Many of the results that we obtain in

Eletnents of Con1plex Analysis

20

1.4

later chapters depend upon compactness argun1ents. We shall soon see that the technical significance of con1pactness revolves around the fact that it allows many problen1s to be treated by finiteness argutnents. There are several equivalent definitions of compactness. The following definition will be n1ost convenient for our purposes. Let (X, d) be a metric space and let K c X. We say that { U;.: A E A} is an open cover of Kif each U;. is open and K c U { U;.: A E A}. We say that K is cornpact if every open cover of K contains a finite subcover, i.e. if every open cover contains a finite nun1ber of sets whose union contains K. We may observe immediately that every finite subset of a metric space is compact, so that compactness Inay be regarded as a generalization of finiteness. We say that X is a cotnpact nzetric space if the set X itself is compact. We now have two possible concepts of cotnpactness for a subset K of X. K may be a compact subset of X or K may be a compact n1etric space with the relative metric. Happily the two concepts are equivalent and so there will be no need to distinguish between them. Proposition 1.15. Let (X, d) be a nzetric space and let K be a non-empty subset of X with relative nzetric p. Then K is con1pact in (X, d) if and only if (K, p) is a conzpact 1netric space. Proof Let Kbe a compact subset of(X, d) and let {V;.: A E A} be an open cover for Kin the metric space (K, p). It follows from Proposition 1.6 that for each A E A we may choose an open set U;. in (X, d) such that V;. = K n U;.. It is clear that { U;.: A E A} is now an open cover for Kin (X, d). Since K is compact in (X, d) there exist A1 , A2 , .•• , An E A such that

Kc

U {U;. 1 : j

= 1, 2, ... ,

n}.

It follows that

K c K n [U {U;.,:} = I, 2, ... , n}] = U {K n U;. 1 : j = 1, 2, ... , n} = U {V;. 1 : j = 1, 2, ... , n} so that {V;. 1 : } = 1, 2, ... , n} is an open cover forK in (K, p). Therefore (K, p) is a compact metric space. Suppose now that (K, p) is a compact metric space and let {U;.: A E A} be an open cover for Kin (X, d). By Proposition 1.6, each set K n U;. is open in (K, p) and hence {K n U;.: A E A} is an open cover forK in (K, p). Since (K, p) is compact there exist Ab A2 , ••• , An E A such that

K

c

U {K n U;.1 : j =

c

U {U;.,: j =

1, 2, ... , n} 1, 2, ... , n}.

21

Metric Space Prelin1inaries

1.4

Thus {U;.,: j = 1, 2, ... , n} is an open cover for K in (..-¥, d) and so K is con1pact in (X, d) as required. Cotnpact subsets of 1netric spaces have 1nany pleasant properties. The next result begins our study of these properties.

Proposition 1.16. A con1pact subset of a 111etric space is closed and boun-

ded; a closed subset of a co1npact set is co111pact. Proof Let K be a con1pact subset of a n1etric space (X, d) and let p EX"'-. K. For each x E Klet E(x) = -!d(x, p). We then have B(x, E(x))

n

B(p, E(x)) = 0

(x

E

K).

Clearly {B(x, E(x)): x E K} is an open cover of the con1pact set K and so has a finite subcover, say {B(x1, E(xJ): j

Let

o = n1in {E(x1): j

=

=

1, 2, ... , n}.

1, 2, ... , n} so that

o>

0. We now have

B(p, o) n K c B(p, o) n [ U{B(x1, E(x1)): j = 1, 2, ... , n}] c U{B(p, E(x1)) n B(x, E(x1)): j = 1, 2, ... , n} = 0.

It follows that B(p, o) c X"'-. K. Since p was arbitrary in X"'-. K this shows that X"'-. K is open and so K is closed. To see that K is bounded we observe that {B(x, 1): x E K} is an open cover of K and so has a finite subcover, say {B(x1, 1):j = 1, 2, ... , n}. Let = max{d(x1 , x 1):j = 1, 2, ... , n}. Given x E K there is some j such that x E B(x1, 1). Then

o

d(xb x) ~ d(xb x 1) < 0 + 1.

+

d(x;, x)

o

Therefore K c B(xb + 1) and so K is bounded. Suppose now that F is closed and F c K. We then have that X"'-. F is open. Let { U;.: A E A} be any open cover of F. Then {U;.: A E A} v {X"'-. F} is an open cover forK and so has a finite subcover since K is compact. If we exclude X"'-. F from the finite subcover (if necessary) the ren1aining sets must still cover F (since X"'-. F can cover no part of F). This proves that F is compact. In the final part of the above result we took F to be closed in X. There would be no difference if we took F to be closed in K. To see this we note that K is closed in X and so by Proposition 1.7 a subset of K is closed in (X, d) if and only if it is closed in (K, p).

1.4

Elements of Co1nplex Analysis

22

It is not true in general that every closed and bounded subset of a n1etric space is compact (sec Problcn1 1.25). On the other hand the above statetncnt is true for the n1ctric spaces Rn (n E P). This fact, which we prove below, is extretnely useful since it enables us to tell almost at a glance when a subset of Rn is compact.

Theorem 1.17. A subset ofRn is conzpact

iff it is closed and bounded.

Proof. In view of the last proposition we need only prove that a closed and bounded subset of Rn is compact. We shall give the proof for the case n = 2 (which is the 1nost significant for us); it will be clear that our proof can be adapted to deal with the general case. Let K be closed and bounded in R 2 • Then K is contained in some ball in R 2 and hence K is contained in sotne closed square S. (A closed square in R 2 is of course a subset of the form {(x, y): a ~ x ~ c, b ~ y ~ d} where a < c, b < d and c - a = d- b.) We shall show below that Sis compact and it will then follow from Proposition 1.16 that K is compact. Suppose that Sis not compact. Then there is an open cover { U;.: A E A} of S which contains no finite subcover. Suppose that S has sides length /. Join the mid points of opposite sides of Sand so divideS into four closed squares with sides length .Jxl. At least one of these four squares must fail to be covered by a finite subcollection of {U;.: A E A}, else S itself would be covered by a finite subcollection of { U;.: A E A}. Let S 1 be one such square and let its bottom left vertex be (a 1 , b1 ). We now repeat the process on S 1 and so obtain a closed square S 2 with sides length

2~ I and bottom left

vertex (a 2 , b 2 ) such that no finite subcollection of {U;.: A E A} covers S 2 • We now proceed by induction and so obtain for each n E P a closed square Sn with sides length in I and botton1 left vertex (an, bn) such that no finite subcollection of {U;.: A E A} covers Sn. It follows from the method of construction of the squares that the real sequences {an}, {bn} are monotonic and bounded, and so are convergent. Thus there exist a, b E R such that lim an

n- oo

= a,

lim bn =b.

n-oo

It follows that lim (an, bn) = (a, b).

n-oo

Since (an, bn) E S (n E P) and S is closed it follows from Proposition 1.5 that (a, b) E s. Since (a, b) E s there is Ao E A such that (a, b) E u).o· Since

I.4

Afetric Space Prelinzinaries

23

U;. 0 is open there is e > 0 such that B((a, b), e) c U;_ 0 • We tnay now choose nz E J> such that

Given (x, y)

E

Sm we now have

d((a, b), (x, y))

~

d((a, b), (am, bm)) + d((am, b111 ), (x, y))

,12.!

- Y, and let E be a non-en1pty subset of X. We say that/is unifornzly continuous onE if for every E > 0 there is o(E) > 0 such that (x E E).

Iff is uniformly continuous on X we say more sin1ply that/ is unifonnly continuous. The student should note the parallels between continuity and uniform continuity, and pointwise and uniform convergence. It is clear that unifonn continuity itnplies continuity, but the converse is false as is easily seen by taking

f(x)

=

x2

(x

E

R).

On the other hand, when X is compact the two concepts coincide. Proposition 1.19. Let (X, d), ( Y, e) be nwtric spaces with X conzpact, and let f: X----?- Y be continuous. Then f is unifonnly continuous.

Proof Let such that

€

> 0. Since f is continuous, given

X E

X there is o( €, x) > 0

The collection {Bix, o(E, x)): x EX} is clearly an open cover for the compact space X and so has a finite subcover, say {Bd(xh o(E, x 1)): j = I, 2, ... , n}. Let

so that 77( E) > 0. Given x, y E X with d(x, y) < 77( E), there is sotne j such that X E Bu(xh o(E, Xj)). Since 7)(€) ~ o(E, Xj) we have

d(y, x 1) Thus x, y

E

~

d(y,

X)

+ d(x,

x 1) < 2o(E, x 1).

Bixh 2o(E, x 1)) and so e(f(x),f(y))

~

e(f(x1),f(x))


0 choose S =

€

and then

d(x, y) < S =>

ldist (x, A) - dist (y, A) I
0 such that B(y, S) c B(x, e). Let 8 = e - d(x, y) so that q > 0. Given p E B(y, o) we have d(x, p) ~ d(x, y) < d (x, y)

+ d(y, p)

+

8

= e.

Therefore B(y, 8) c B(x, e) and so B(x, e) is an open set. We leave the student to prove that a closed ball is a closed set. As a further example of closed sets we see from Proposition 1.5 that any singleton is a closed set. It follows from Proposition 1.4 that any finite subset of a metric space is closed. The unwary student may be tempted to think that any given subset of a n1etric space is either open or closed. In fact a subset of a metric space may be neither open nor closed-or it may be both open and closed! For example, in the n1etric space R it is easy to see that E

= {x: x E R, 0

~

x < 1}

is neither open nor closed. Indeed there is no neighbourhood of 0 contained in E, so that E is not open. Moreover 1 is a cluster point of E which does not belong to E. On the other hand it is trivial to verify that in any metric space X the set X itself is both open and closed. Suppose now that we give the above set E the relative n1etric p. Then the set E is both open and closed in the metric space (£, p) even though it is neither open nor closed in the metric space R. This situation is often a source of confusion to the student. Indeed much of the con1mon n1isunderstanding of metric spaces revolves around a misunderstanding of the relative metric. In subsequent chapters we shall frequently be working with relative n1etrics. It is therefore essential for the student to understand what happens to open sets, closed sets and convergent sequences when the relative metric comes into play. Proposition 1.6. Let (X, d) be a 111etric space and let E be a non-en1pty subset of X with relative n1etric p. A subset V of E is open in (E, p) iff it is of the fornz V = En U where U is open in (X, d). Proof. Given x

E

E we have BP(x, e)

= {y: y

£, p(x, y) < e} = {y:yEE,d(x,y) < e} =En Btt(x, e). E

1. 5

29

Metric Space Prelirninaries

the cotnpleteness of R. This wilJ be illustrated in the proof below and in son1c of the problems that follow. Theorent 1.22. R'1 is conzplete for n

E

P, n

~

2.

Proof We shall give the proof for the case n = 2; it will be clear that our proof can be readily adapted to deal with the general case. Let {(xn, Yn)} be a Cauchy sequence in R 2 • Given nz, n E P it follows fron1 the definition of the usual tnetric d on R 2 that

lxm - Xnl ~ d((xm, Ym), (Xm Yn)) !Ym - Ynl ~ d((Xm, Ym), (xn, Yn)). It is now clear that {xn}, {Yn} are Cauchy sequences in R. Since R is conlplete there exist x, y E R such that

litn Xn =

n-+ oo

X,

litn Yn = y.

n-+ oo

It follows that n-+ oo

i.e. the sequence {(xn, Yn)} converges. Therefore R 2 is complete as required. PROBLEMS 1 32. Show that every discrete metric space is complete. 33. Let (X, d) be a complete metric space. Show that a non-empty subset E of X is closed iff E is a complete metric space (with the relative metric). 34. Let X, Y be metric spaces and let f be a continuous mapping from X onto Y. If X is complete does it follow that Y is complete? What iff is actually a homeomorphism? 35. Let (X1 , d1 ), .•• , (Xn, dn) be complete metric spaces and let X = xl X x2 X •.• X Xn. Let d be a nletric on X as in Problem 1.2. Show that (X, d) is complete. 36. Show that the metric space f!#R(S) of Problem 1.3 is complete. 37. Given any compact metric space X show that metric space.

~R(X)

is a complete

38. Given any metric space X let ~R(X) denote the set of bounded continuous real functions on X. Show that &a~R(X) is a closed subset of £?1R(X) and hence is a con1plete metric space with the relative metric (the usual metric for PJYCa( X)).

Ele1nents of Co1nplex Analysis

30

1.6

1.6

Connectedness

We still require one n1ore basic topological concept, nan1ely that of connectedness. Roughly speaking, a n1etric space is connected if it consists of a single 'piece'. For n1ost of this book we shall be working with connected metric spaces. Let (X, d) be a metric space. We say that X is disconnected if there exist non-empty open sets 0 1, 0 2 such that X= 0 1 u 0 2 and 0 1 n 02 = 0. We say that X is connected if it is not disconnected. In any metric space X the subsets 0 and X are both open and closed. It is easy to see from Proposition 1.3 that a metric space X is connected if and only if 0 and X are the only subsets of X that are both open and closed. We say that a non-empty subset E of X is connected if Eisa connected metric space (with the relative metric). In view of Proposition 1.6 we see that E is connected iff whenever 0 1, 0 2 are open sets in X withE c 0 1 u 0 2, En 0 1 n 0 2 = 0 then either En 0 1 = 0 orEn 0 2 = 0. If E is itself an open subset of X then we see that E is connected iff whenever 0 1 , 0 2 are open sets in X withE = 01 u 02, 01 n 0 2 = 0 then either 0 1 = 0 or 0 2 = 0. As a trivial example observe that any singleton is a connected subset in a metric space. The first result below shows how we may 'paste' together connected sets to form a larger connected set. Proposition 1.23. Let {E;_: A E Jl} be a fa1nily of connected subsets of a metric space X. Suppose there is A0 E J1 such that E;_ 0 n E;_ # 0 (A E Jl). Then E = U {E;_: A E .11} is connected.

Proof Let 0 1, 0 2 be open subsets of X such that En 0 1 n 0 2 = 0, E c 0 1 u 0 2. We shall show that either En 0 1 = 0 or En 0 2 = 0. Given A E J1 we must have E;_ c 0 1 orE;_ c 0 2. Otherwise E;_ n 0 1 =I= 0, E;_ n 0 2 # 0, E;_ c E c 0 1 u 0 2, and E;_ n 0 1 n 0 2 c En 0 1 n 0 2 = 0, i.e. E;_ is not connected. Suppose that E;_ 0 c 0 1 . We then have E;_ c 0 1 for each A E 11, for otherwise E;_ c 0 2 and so E;_ n E;_ 0 c En 0 1 n 0 2 = 0, which is a contradiction. It follows that E c 0 1. Thus En 0 1 = E and so En 0 2 = 0. Sin1ilarly if E;_ 0 c 0 2 we may show E c 0 2 and so En 0 1 = 0. We conclude that E n1ust be connected. This last result leads us to look for the largest possible connected subsets of a metric space. We define a conzponent of a n1etric space to be a maxin1al connected subset, i.e. a connected subset that is not properly contained in any other connected subset. In intuitive tenns the components of a metric space are its separate 'pieces'. We thus expect that every metric space can be split up into its separate 'pieces'. This is indeed the case as we now show.

1.6

Metric Space Prelinzinaries

31

Proposition 1.24. Every non-enzpty connected subset of a n-zetric space is

contained in exactly one con1ponent,· every nzetric space adn1its a unique decon1position into conzponents. Proof. Let E be a connected subset of a n1etric space X. Let A be the union of all those connected subsets of X that contain E. It follows fron1 Proposition 1.23 that A is connected. Suppose now that B is a connected subset of X and A c B. Then B contains E and so by the definition of A we have B c A. This shows that A is a n1axitnal connected subset, i.e. A is a cotnponent. Suppose now that Cis any cotnponent of X that contains E. Then Cis a connected set containing E and so C c A. Since A is connected and Cis a con1ponent we n1ust have C = A. This shows that E is contained in exactly one con1ponent. For each x E X we have that {x} is connected. By the above we thus have that each x belongs to exactly one con1ponent of X. This means that X is the union of its con1ponents. Further if A 1 , A 2 are two components of X then A 1 = A 2 or A 1 n A 2 = 0. This completes the proof. Thus far our only specific exatnples of connected sets have been singletons. We now produce some non-trivial examples of connected sets.

Proposition 1.25. Every interval of R is connected.

Proof Let I be any interval of R. (I may be bounded or unbounded and n1ay or may not include its finite end points.) Suppose that I is not connected. Then there exist open subsets Ob 0 2 of R such that In 0 1 # 0, In 0 2 # 0, I c 0 1 u 0 2, In 0 1 n 02 = 0. Choose a E In Ob bE In 0 2 • We have a # b. Suppose that a < band let C = SUp

{x:

X E

Ob

X

< b}.

It is clear that a ::::; c ::::; b. Since I is an interval and a, b E I we must have c E I c 01 u 02. Suppose that c E 0 1 • Since 0 1 is open there is E > 0 such that B(c, E) c 0 1 . In particular there is x E 0 1 such that c < x. If x ~ b we have c ::::; b ::::; X and sob E B(c, €) c ob which is impossible. On the other hand if x < b we have c < x < b. Since x E 0 1 this contradicts the definition of c. Suppose now that c E 0 2 • Since 0 2 is open there is E > 0 such that B(c, E) c 0 2. By the definition of c we 1nay choose x E 01 such that c - € < x ::::; c. Thus x E B(c, E) c 0 2. This is again a contradiction. We may derive similar contradictions if a > b. We therefore conclude that I is connected.

32

1.6

Elements of Con1plex Analysis

The student may observe that in the above proof we used only once the fact that I is an interval. We used it to obtain the property that if a, b E I and a ~ c ~ b then c E /. We show next that any connected subset of R has this property. This property is in fact equivalent to being an interval; but we shall not need this result and accordingly we shall not stop to prove it. Proposition 1.26. Let E be a connected subset of R. If a, b a ~ c ~ b then c E £.

E

E, a < b and

Proof Suppose that a < c < b and c ¢E. Let 0 1 = {x: x E R, x < c} and 0 2 = {x: x E R, x > c}. Then Oh 0 2 are open subsets of R with a E E (\ oh bEE(\ 02, E (\ 01 (\ 02 = 0. Since c ¢ E we also have E c 0 1 u 0 2 • We now have that E is not connected. This contradiction shows that c E E as required. We have now characterized all the connected subsets of R. We postpone till the next chapter a detailed discussion of the connected subsets of R 2 • We show next that the property of connectedness is preserved under continuous mappings. Proposition 1.27. Let X, Y be metric spaces with X connected and let f· X~ Y be continuous. Thenf(X) is a connected subset of Y.

Proof Suppose that f(X) is not connected. Then there exist open sets 0 1,0 2 of Y such thatf(X) n 0 1 # 0 ,f(X) n 0 2 # 0 ,f(X) c 0 1 u 0 2, f(X) n 0 1 n 0 2 = 0. Since f is continuous it follows from Proposition 1.11 thatf- 1( 0 1),f- 1( 0 2) are open in X. Moreover, we deduce from above that f- 1(01) # .0, f- 1(02) # 0, X= f- 1(01) u f- 1(02), f- 1(01) n f- 1 ( 0 2 ) = 0. This says that X is disconnected. We conclude that f(X) must therefore be connected. This last result corresponds to the intuitive notion that a continuous mapping cannot tear a single 'piece' into several 'pieces'. To illustrate the result let Y = R, and thenf(X) is a connected subset of R. Suppose that f(x 1) = p,f(x 2 ) = q with p < q. Given p ~ r ~ q it follows from Proposi~ tion 1.26 that r Ef(X), i.e. there exists x3 EX such that f(x 3 ) = r. In par~ ticular if we take X to be the n1etric space {x: x E R, a ~ x ~ b} we obtain the classical intermediate value theoretn. We conclude with a technical result that we shall wish to use in later chapters. Let X, Y be metric spaces and Jet f: X->- Y. We say that f is locally constant if for each x E .~¥ there is E(x) > 0 such that f is constant on B(x, E(x)).

1.6

f:

Afetric Space Prelilninaries

33

Proposition 1.28. Let X, Y be nzetric spaces with X connected, and let . .¥--+ Y be continuous and locally constant. Then f is constant.

Proof Let p EX, let y = f(p), and let E = {x: x E X,f(x) = y}. We shall show that E is both open and closed. Since X is connected it will follow that E = 0 orE= X. Since pEE we shall have E = X, so thatfis constant. Since {y} is closed in Y and f is continuous it follows fro1n Proposition 1.11 that E = f - 1 ({y}) is closed. Let x E E. Since f is locally constant there is E > 0 such that

f(z) = f(x) = y

(z

E

B(x, E)).

Thus B(x, E) c E and so E is open. The proof is now con1plete.

PROBLEMS 1 39. Which subsets of a discrete tnetric space are connected? 40. Show that a closed subset F of a 1netric space X is connected iff there exist non-en1pty closed sets F 1 , F 2 of X such that F = F 1 u F 2 , F 1 n F2 = 0. 41. A non-e1npty subset E of a n1etric space X is said to have property (C) if there exist non-e1npty open subsets Ob 0 2 of X such that 0 1 n 0 2 = 0, A c 0 1 u 0 2 • Is property (C) equivalent to being connected? 42. Let {En} be a sequence of connected subsets of a metric space such that En n En+ 1 =I= 0 (n E P). Show that U {En: n E P} is connected. 43. Given connected subsets A, B of R show that A x B is a connected subset of R 2 • State and prove son1e generalizations of this result. 44. Given a connected subset E of a metric space show that E- is connected. Deduce that the con1ponents of a metric space are closed. Is E 0 connected? 45. A metric space is said to be locally connected if each point has a connected neighbourhood. Show that the components of a locally connected metric space are open. Give an exa1nple of a metric space that is not locally connected. 46. Show that a non-empty subset of R is an interval iff it has the property stated in Proposition 1.26.

34

Elements of Complex Analysis

1.6

47. Describe a general open subset of R by decomposing it into its components.

48. Can a compact metric space have an infinite number of components? 49. What are the components of the metric space Q? 50. Show that any continuous function from a connected metric space to a discrete metric space is constant.

2 THE COMPLEX NUMBERS

2.1

Definitions and notation

In this chapter we shall discuss son1e of the fundan1ental algebraic and topological properties of the co1nplex nu1nbers. We assun1e, as in Chapter I, that the student is acquainted with the basic properties of the real nu1nbers. In particular we recall that the real nu1nbers R fonn a field under the operations of addition and n1ultiplication. Let us now consider R2 . The algebraic structure on R2 that is usually encountered first by the student is the vector space (or linear space) structure defined by the operations of addition and scalar Inultiplication, nan1ely,

(a, b) + (e, d) = (a + e, b + d) t\(a, b) = (t\a, t\b)

(a, b, e, dE R) (t\, a, b E R).

Under these operations R2 becon1es a two-dimensional real vector space. We now define a multiplication * on R 2 in such a way that R 2 becon1es a field under the operations of + and *· We define (a, b)

* (e, d) =

(ae - bd, ad

+ be)

(a, b, e, dE R).

Proposition 2.1. R 2 is a field under the operations of+ and*· Proof We already know fro1n the vector space structure of R 2 that R 2 is a con1mutative group under +, in which the zero elen1ent is (0, 0). We show next that R 2 """{(0, 0)} is a con1mutative group under *· Using the algebraic properties of R we obtain the following relations for all ele1nents of R 2 """{(0, 0)}. (a, b)* (e, d) = (ae - bd, ad+ be) = (ea - db, da + eb) = (e, d)* (a, b) 35

Elenzents of Conzplex Analysis

36

2.1

[(a, b)* (c, d)]* (e,f) = (ac - bd, ad + be)* (e,f) = (ace- bde- adf- be/, acf- bdf + ade +bee) = (a, b)* (ce - df, cf + de) = (a, b) * [(c, d) * (e, f)] (a, b)* (1, 0) = (a, b) -b ) a (1, 0) = (a, b)* ( a2 + b2' a2 + b2 . We thus see that R 2 ""- {(0, 0)} is a cotnmutative group under *· It remains to show that* is distributive over +. For all elements of R 2 we have

(a, b)

* [(c, d) + (e,f)] = = =

(a, b)

* (c +

e, d +f) (ac + ae - bd - hf, ad [(a, b) * (c, d)] + [(a, b)

+ af + be + be) * (e, f)]

and so the proof is complete. The above field is called the cornplex field and we shal1 denote it by C. We shall also refer to Cas the conzplex nun1bers or the cornplex plane. The student should realize that R 2 and C are precisely the same set, namely, all ordered pairs of real nun1bers. When we wish to emphasize the vector space structure on this set, we use R 2 ; when we wish to en1phasize the field structure we use C. Since the subject of this book is conzplex analysis we shall almost always use C rather than R 2 . Our next step is to show how the real field R can be enzbedded in the complex field C. ln other words we shall show that the field C contains an isomorphic copy of the field R. We can then follow the usual practice of identifying R with its isomorphic copy in C. We define h: R-+ C by

h(a)

=

(a, 0)

(a

E

R).

l t is clear that lz maps R one-to-one into C and that

h(a + b) h(ab)

= (a + b, 0) = (a, 0) + (b, 0) = h(a) + h(b) = (ab, 0) = (a, 0) * (b, 0) = h(a) * h(b).

This shows that the set {(a, 0): a E R} with the operations + and * is an isomorphic copy of the real field. Fro1n now on we shall feel free to regard R as a subset of C by identifying any real nlllnber a with the ordered pair (a, 0). Furthermore we shall now discard the sytnbol * for n1ultiplication and follow the standard practice of writing sin1ply (a, b)(c, d) for (a, b)* (c, d). The complex field has three special cleinents, nan1ely (0, 0), (1, 0) and (0, 1). The clement (0, 0) is the zero elcn1ent of the additive structure of C and the element (I, 0) is the unit ele1nent of the nntltiplicative structure of

2.1

37

The Cornplex Nwnbers

C. Under the above identification these eletncnts are denoted by 0 and 1 respectively. The cletnent (0, I) is denoted by i and has the property

= (0,

i2

1)(0, 1)

= (- I, 0) = -

1.

By extending the real field to the cotnplex field we have thus produced a concrete 'square root of tninus one'. We also have that (a, b)

= (a, 0) + (0, b) = (a, 0) + (0, I)(b, 0) = a + ib.

Fron1 now on we tnay denote a cotnplex nutnber by (a, b) or a + ib according to which seen1s n1ore appropriate to a given context. More often than not we shall usc the fonn a + ib. It is usually n1ost convenient to denote a cotnplex nutnber by a single letter, say a = a + ib. We then say that a is the real part of a and b the inzaginary part of a. We write a =

Rca,

b = In1 a.

We define the conzplex conjugate of a by

a = a - ib. Observe that

1

Rea = 2 (a

The n1apping a-+ a (a

E

+ a),

hn a

=

/a - a).-

1 2

C) is easily seen to have the following properties.

a+fJ=a+P af1 = aP = a = a. Geometrically the n1apping a -+ a is sitnply reflection in the real axis {(a, 0): a E R}. The perpendicular axis {(0, b): bE R} is called the intaginary axis. Complex numbers of the form (0, b) are called pure ilnaginary. Note that a is real if and only if a = a, and a is pure imaginary if and only if a = -a.

Given a complex nutnber a = a

Ia!

=

+ ib we define the 1nodulus of a by (a2 + b2)!.

Geotnetrically the tnodulus of a cotnplex nun1ber is its distance fron1 the origin. It is obvious frotn the definition that

Ia!

=

Ia!,

a!

~

Ia!,

It is also obvious that IRe

jin1 aj

~

Ia!.

38

Elenzents l?( C01nple.x Analysis

2.1

On the other hand we easily verify the inequality [af ~ [Rea[

+

[Im a[.

We give below three further sin1ple properties of the modulus function. We single out these three properties since they are significant for generalizations of R and C (see Problen1 2.6). Proposition 2.2 The nzodulus function has the following properties. (i) [a[ = 0 if and only if a = 0. (ii) [at?[= [alit?[ (a,f3EC). (iii) Ia + ~~ ~ lal + lt?l (a,~

E

C).

Proof (i) Let a = a + ib. We have lal = 0 iff a 2 + b 2 = 0. Since a and b are real, this occurs iff a = b = 0, i.e. a = 0. (ii) la~[ 2 = (a~)(at?) = aat?P = [al 2 1~1 2 • Since the 1nodulus is non-negative we conclude that lat?l = lall~l(iii) Making use of several earlier properties we obtain

Ia

+ t?[ 2

=

=

= ~

=

(a + t?)(a + P) aii. + t?P + a"P + ~a lal 2 + [~[ 2 + 2 Re (a"P) lal 2 + lt?l 2 + 2[a"fil lal 2 + 1~[ 2 + 2[allt?l.

Since the n1odul us is non-negative we conclude that

Ia + ~~

~ Ia[

+

lt?l.

Roughly speaking the argwnent of a non-zero co1nplex nun1ber a is the angle t between the positive real axis and the line segn1ent joining 0 to a. This gives the equations Rea

cos

t

Im a

.

= lal , Sin

t

= lal ·

It is well known that there are an infinite nun1ber of real nun1bers t which satisfy the above equations and that any two such nun1bers differ by a rnultiple of 27T. We shall therefore find it convenient to define a whole family of argument functions. In order to do this we recall the following property of the cos and sin functions. Given A E Rand given a, bE R with a 2 + b 2 = 1 there is exactly one real number t such that cos t = a, sin t = b, and A < I ~ A + 21r. Given 0 E R we define the function arg 0 : C "'- {0} ->- R by

arg 0 (a) = t

2.1

The Contplex Nwnbers

39

where

.

Rea

cost= Tal'

Sill

lin a t = jaj '

8-

< t

1T

~

8

+

1T.

Observe that the range of arg 0 is {t: 8 - 1r < t ~ 8 + 1r}. We call arg0 the principal L'alue of the argutnent and denote it n1ore sitnply by arg. For any 8 E Rand any a E C "'- {0} we have

a

=

jaj{cos arg0 (a)

+ i sin arg0 (a)}.

Further, arg 0 is constant on the ray {ra: r > 0} detennined by a. This is clear since Re (ra) Rea jraj = Taf' Conversely suppose that arg 0 (a)

!raj

lin a

=Tal'

= arg 0 ({3). Then

Rea Tal

=

Re f3 lBf

and so

f3 = ~~~ Im "· It follows that f3 = ~~~ "· We have thus sho\VD that arg 8 (a) = arg 0 ((3) if and only if f3 = ra for sotne Re {3

=

~~~

I Ill (ra)

Re ex, and similarly Im

r > 0. We define N 8 = {0}

U

{a: arg 8 (a) = 8

It follows fron1 the above that if a N0

E

+ 1r}.

N 8 , a =f. 0 then

= {ra: r

~

0}.

We write N 0 n1ore sin1ply as N. Observe that N is sitnply the negative real axis (including the origin)*. We turn no\V to the tnetric structure on C. Since the set C is the same as the set R 2 it follows that we have a n1etric on C given by

p(a If z

= a + ib,

w

+

ib, c

+ id) =

[(a - c) 2

+ (b

- d) 2 ]!.

= c + id then the tnetric is given more succinctly by p(z, w) = jz - l-vj.

The n1etric space C is complete (Theoren1 1.22). A subset of C is conlpact if and only if it is closed and bounded (Theoren1 1.17). For obvious geometrical reasons we shall refer to balls in C as discs. The following important notation will be observed throughout this book. We make no claims as to the general acceptance or otherwise of our notation, but it will be essential for the student of this book to fatniliarize *Note that we also use N for a positive integer. The context will make clear which meaning is intended.

2.1

Elenzents of Conzple.x Analysis

40

hitnself with our notation. In the list that follows, a and {3 denote complex nutnbers, r and R positive real nutnbers with r < R.

Ll(a, r) = Ll'(a, r) = .J(a, r) = C(a,r)= V(a, r)= Y'(a, r) = A(a; R, r) = A(a;R,r) = (a, {3) = (a, f3] =

{z: z

E

C, jz - al < r}.

Ll(a, r) ""'{a} {z :

z

E

C, Iz - a I ~ r}

{z: zEC,

{z: {z:

z E C,

C, {z: z E C, {z: zEC, {( 1 - t )a {(I - t )a [a, {3) = { ( 1 - t )a [a, {3] = {(I - t)a ZE

lzlzlz-

al = r} al > r} al ~ r} r < jz - al < R} r ~ jz- al ~ R} + t{3: t E R, 0 < t + t f3 : t E R, 0 < t + t f3 : t E R, 0 ~ t + t/3: t E R, 0 ~ t

< I} ~ I} < I} ~

I}.

We conclude this section with a technical result on compact sets that we shall need in Chapter 6.

Proposition 2.3. Let {Kn} be a sequence of non-e1npty con1pact subsets of C such that Kn + 1 c Kn (n E P). Then K = Kn: n E P} is 1zon-ernpty. If, further, lim diam Kn = 0, then there is a E C such that

n{

n-+ co

(i) K = {a} (ii) given e > 0 there is N

Kn

C

E

P such that (n > N).

.d(a, e)

Proof: For each n E P, Kn is compact and so is closed by Proposition 1.16. It follows that C""' Kn is open in C. We now have fron1 Proposition 1.6 that K 1 ""'Kn = K 1 n (C""' Kn) is open in the tnetric space K 1 • We note from Proposition I.I5 that K 1 is a con1pact tnetric space. Suppose that K is en1pty. Then

U {K1 ""'Kn:

11 E

P}

=

K1 ""'

K=

K1.

This n1eans that {K1 ""' Kn: 11 E P} is an open cover of the con1pact tnetric space K 1 and so has a finite subcover, say {K1 ""' Knr: r = 1, 2, ... , 1n}. Let p = n1ax {nr: r = 1, 2, ... , nz}. Since Kn+ 1 c Kn (n E P) it follows simply that

This gives the contradiction that KP

=

0. 'vVe conclude that K #- 0.

2.1

Tlze C01nplex Nunzbers

Suppose now that lin1 dian1 (K11 )

41

0. Since K is non-empty we 1nay

=

n-+ oo

choose son1e a

E

K. Given f3

E

K we have a, [3

E

Kn (ll (n

E

E

P) and so

P).

It follows that ja - ,Bj = 0 and so f3 = a. This proves that K = {a}. Given E > 0 there is N E P such that

(n > N). Given n > N, z

E

K 11 we then have

I= -

aj

~ dian1

(Kn)
N) as required. PROBLEMS 2

1. Show that the natural ordering ~ on R cannot be extended to C in such a \Vay that C becon1es an ordered field. (Hint: consider the possibilities i ~ 1, i ~ 1.) 2. (i) Use the n1ultiplication table 1

l

1

1

l

l

l

-1

to show that the con1plex field is isomorphic with the field (under n1atrix addition and multiplication) of all 2 x 2 real matrices of the form

_: ~)·

(

(ii) Show that there is no multiplication * for which the vector space R3 becomes a real division algebra. (Hint: suppose there is such a nlultiplication *· Show that there is an isomorphic copy of the division algebra consisting of certain 3 x 3 realtnatrices. Use the fact that every such n1atrix has a real eigenvalue to obtain a contradiction.) (iii) Let * be defined on R4 by (ab a2, a3, a4) * (bh b2, b3, b4) = (a1b1 - a2 b2

a 1b3 - a 2b4

-

+

a3b3 - a4b4, a1b2 a3b1 + a4b2, a1b4

+ a2b1 + a3b4 - a4b3, + a2b3 - G3b2 + a4b1).

Show that (R 4 , +, *) is a skew field, i.e. has all the properties of a field except commutativity of multiplication.

42

Elements of Complex Analysis

2.1

3. Let {an}, {.8n} be complex sequences with lim an = a, lim .Bn = ,8. n-+oo

n-+oo

Show that

+ .Bn)

(i) lim (an n-+ oo

=

a

(ii) lim (Aan) = Aa (A

E

n-+ oo

+ ,8 C)

n-+ oo

(iv) lim ~n n-+ oo t-Jn

=~ tJ

(ftn, ,8 E C "{0})

using (a) Proposition 2.2 (b) the real and imaginary part of the sequences and the analogous results for real sequences.

4. Given z,

E

C "{0} (r = 1, 2, ... , n) show that

with equality if and only if arg (zr) is constant. Show also that

lzn - Ztl

~

n-1

L

r=l

lzr+l - z,j

and find necessary and sufficient conditions for equality.

z = x + iy let llzll = !xl + jyj. Show that llz1 + z2ll ~ llztll + llz2ll, llz1z2ll ~ llz1llllz211 Cz1, Z2 EC) lzl ~ llzll ~ v2!zl (z EC).

5. Given (i) (ii)

6. Let F denote either R or C, and let X be a vector space over F. We say that a real function 11·11 on X is a norm if (i) llxll ;;?; O(xE X), llxll = 0 iff x = 0 (ii) ll"xll = jAjllxll (x EX, A EF) (iii) llx + Yll ~ llxll + IIYII (x, Y EX). If d(x, y) = llx - Yll (x, y EX) show that dis a metric on X. If (X, d) is complete we say that (X, 11·11) is a Banach space over F. If further X is a linear algebra over F and (iv) llxyll ~

llxiiiiYII (x, Y EX)

we then say that X is a Banach algebra over F. If X is a Banach algebra and {xn}, {Yn} are sequences in X with lim Xn = x, lim Yn = y, show that n-+oo

(a) lim

n-+ oo

(X n

+ Yn) = X + Y

(b) lim (Axn)

= AX (A E F)

(c) lim (XnYn)

=

n-+ oo

n-+ oo

xy.

n-+oo

2.2

The C01np!ex Nwnbers

43

7. (i) Let 11·11 be defined on Rn by ll(xb X2, · · ·,

Xn)ll = (

i

r=l

)!·

lxrl 2

Show that (R't, 11·11) is a Banach space over C (with the usual vector space operations). (ii) Let 11·11 be defined on en by ll(zi, z2, ... , zn)ll = 1nax {lzrl: r

= I, 2, ... , n}.

Show that (Cn, 11·11) is a Banach algebra over C (with n1ultiplication defined coordinate-wise).

8. Show that C(O, 1) = {cos t

+ i sin t:

t

E

[0, 21r]}.

9. Let 0 E R. For which con1plex ntunbers zb .:2 is it true that (i) arg 8 (z1z2) = arg 0 (z 1) + arg 8 (z2) (ii) arg 8 (zl + z2) = arg 8 (z 1 ) + arg8 (z 2).

10. Sketch the sets of con1plex nutnbers specified by each of the following conditions.

(i) lzl >I, larg(z)l

0 it also follows that z ~ 0. This contradiction shows that 1 is a star centre and so C "- N is a starlike domain. Clearly, i, - i E C "- 1V and 0 E (i, - i]. Since 0 EN, this shows that C "- N is not convex. Further examples of non-convex starlike domains are illustrated in Figure 2.1. Not ali dotnains are starlike. For example it is easy to see that

Figure 2.1

the 'punctured' disc Ll'(a, r) is open. It is also straightforward to verify that Ll'(a, r) is polygonally connected. Thus Ll'(a, r) is a domain. Given f3 E Ll'(a, r) let y = f3 - a. Then a + y, a - y E Ll'(a, r) and evidently a E [a + y, a - y]. This shows that no point of Ll'(a, r) is a star centre and so Ll'(a, r) is not starlike. In the same way we may verify that C "-{a}, V(a, r), and A(a; R, r) are non-starlike domains. These examples of nonstarlike domains will occur frequently in subsequent chapters. The student should now have some idea of how to produce increasingly complicated examples of domains. One such construction is given in Problem 2.20. The problem of describing a general dotnain in C can be sitnplificd slightly in that it is enough to study bounded don1ains. To sec this we need first a sitnple letnma.

Lemma 2.6. Let f: C ->- C be defined by f(z)

=

1 _;

lzl.

Then f is a honzeonwrphisnz of C with Ll(O, 1).

2.2

The Con1plex Nwnbers

47

Proo.f We have

lf(z) I = 1

lzl

+ Iz I
0. Since f is continuous at ::: there is o > 0 such that d(z, H') < o in1plies lf(z) - f(w)l < E. Since IRef(z) - Ref(w)l ~ lf(z) - /(lV)I it follows that d(::, w) < o in1plies IRef(z) - Re.f(H')I < E. This shows that Ref is continuous at z. Since= was arbitrary in £, Ref is thus continuous. A similar argument shows that I n1f is continuous. Suppose now that Ref and Im f arc both continuous. Given z E £, E > 0 we tnay choose > 0 such that d(=, w) < itnplies

o

o

IRef(.:) - Ref(w) I < jE,

1In1f(z) - Imf(w)l
.. E C), f + g,fg E rc(£).

Proof Let zn, z

E

E with lim Zn n-+ co

litn f(zn)

n-+ co

= f(z),

=

E

0. Since {fn} converges to f uniformly on X, we 1nay certainly choose some N E P such that

Proof Let p

E

(x EX).

Since fN is continuous at p, we may now choose 8 > 0 such that

fN(Ba(p, 8)) c Be(JN(p), !€). If d(p, x) < 8 we now have

e(f(p),f(x)) ~ eU(p),fN(p)) + eUN(p),J:v(x)) + e(fN(x),f(x)) < -!€ + -1€ + j-E = €. This shows that f is continuous at p. Since p was arbitrary f is thus continuous as required. We shall further discuss continuity in relation to the topics to be introduced in §§1.4, 1.6.

PROBLEMS 1 13. Let (X, d), ( Y, e) be metric spaces and let f: X--:?- Y. Show that f is continuous as a mapping into the metric space Y if and only if it is continuous as a n1apping onto the n1etric space f(X). 14. Let (X, d) be a discrete 1netric space (i.e. d is the discrete n1etric on X as defined in Problem 1.1) and let ( Y, e) be any n1etric space. Which functions f: X ----7 Yare continuous? Which functions g: Y--:?- X are continuous? Give an exan1ple of a one-to-one function fro1n one n1etric space onto another which is continuous but is not a hon1eon1orphism.

15. Let (X, d) be a metric space, and Jet f and g be real functions on X. Define the functions f

+ g, fg, f

V

g, f

1\

g, lfl on X by

([+ g)(x) =f(x) +

g~Y)

(fg) (x) = f(x) g(x) (f V g) (x) = max (f(x), g(x)) (f 1\ g) (x) = min (f(x), g(x)) lfl (x) = lf(x)[. Iff and g are continuous show that the above functions are also continuous.

3.1

Continuous and Differentiable Conzplex Functions

57

There is a difficulty in trying to assign a convention for the n1eaning of oo + oo or oo. 0, and we shall return to this point shortly when we consider the rational functions. The n1ain purpose in n1aking the above definitions is to obtain notational brevity. For exatnple we tnay now define the function j: coo -+ coo by

=

j(z)

!z

Our conventions then in1ply that j(O) = oo, j(oo) = 0.

Lentma 3.3. j is an isornetry of coo witlr itse(f Proof If=

E

C, = =1- 0 then

d *(J·(o), J.(7)) = (

~

I

+

and d*(j( oo ), j(z)) = d*(O, j(z)) z =1- 0, w =1- 0 then

d*(j(z), j(1r))

1 z

= ( I+

Finally, d*(j(O), j(oo))

I

rt

-

( l

lzl + lzl')• -

d*(O -)

,"

= d*(j(O), j(j(z))) = d*( oo, z). If z,

wE C,

1 lV

rn~

lz- wl + lzl2)l(l +

_! 2)! - (1 + w = d*(z, w).

!w!2)!

= 1 = d*(O, oo). The proof is complete.

Proposition 3.4. Let E be a non-en1pty subset of coo and let f be a function fronz E into C or coo. If z E E n C then f is continuous at z with respect to d* iff it is continuous with respect to the usual111etric. If oo E E,f is continuous at oo (with respect to d*) ifff o j is continuous at 0 with respect to the usual 111etric. Proof Recall frotn the final paragraph on page 52 that the n1etric spaces (En C, d*), (En C, p) have precisely the same open sets. It follows from Proposition 1.11 that the identity mapping is a hon1eon1orphisn1 of (En C, d*) with (E n C, p). The first part of the proposition is now itnmediate from Proposition 1.12. Suppose no\v that f is continuous at oo. By Lemma 3.3 and Proposition 1.12 it follows that/ o j is continuous at 0 with respect to d* and so, by the above, with respect to the usual tnetric. Conversely, iff o j is continuous with respect to the usual metric then it is continuous with respect to d*. Arguing as above we see that f = f o j o j is continuous at oo with respect to d*.

58

3.1

Elements of Con1plex Analysis

The above result indicates that continuous functions on subsets of ceo can be effectively described in terms of the usual Jnetric on C. The result will also serve as a motivation for subsequent definitions involving the point at infinity. Note in passing that a continuous c01nplex valued function on ceo is auto1natically bounded by Proposition 1.20. A rational function is a function q: C ~ceo of the form q = PdP2 where p 1 , p 2 arc polyno1nial functions such that {z: p 1 (z)

= 0} n {z: p 2 (z) = 0} =

If p 3 , p 4 , are other polynomial functions such that q P1P4 = P2P3 and so deg (pi)

+ deg (p4) =

0.

= p 3 /p 4 then we have

deg (P1P4) = dcg (p2p3) = deg (p2) + deg (p3).

Therefore

and so we may now define the degree of q by

whenever q

=

pdp 2 • We now extend q to

ceo as follows: if

n

Pl(z)

=

_2: CXrZr, (an "/= 0), r=O

we define q(oo)

m

P2(z) =

2

fJrzr,

(fJm =1- 0),

r=O

= ;~. (This is again independent of the choice of poly-

nomials p 1 , p 2 with q = pdp 2 .) In particular q(oo) = 0 if dcg (q) < 0 and q(oo) = if deg (q) > 0. Since j is continuous on C "'- {0} the function l/p 2 is a continuous complex function whenever p 2 (z) =1- 0 and hence q is continuous at those points of C where p 2 (z) =1- 0. If piz) = 0, then p 1 (z) =1- 0 and we easily show that

Finally it is easy to sec that q is continuous at oo. We have thus shown that every rational function is a continuous function fron1 ceo into ceo. We conclude this section by discussing the continuity of the tnodulus and argument functions. The continuity of the argu1ncnt functions will be of particular significance for our discussion of logarithmic functions in the next chapter.

Continuous and Differentiable C01nplex Functions

3.1

59

Proposition 3.5. (i) The function z---+ lzl is continuous on C. (ii) Giren 8 E R, arg 0 is continuous on C ""- N 0 and is not continuous at each point of N 0 ""- {0}.

Proof (i) This follows in11nediately fron1 the inequality

llzl -

Jwll ~ lz - wl

(z,

wE

C).

=E C ""- N 0

and let arg0 (z) = t. Choose any E > 0 such that (f - E, f + E) C (8 - 7T, e + 7T). Jt follOWS fr0111 the propertieS Of COS and sin that there is 1] > 0 such that Is- tl < E whenever SE(B- 7T, e + 7T) and Icos s - cos t I < YJ, Isin s - sin t I < YJ· Since C ""- N 0 is an open set (see Exan1ple 2.4) there is o1 > 0 such that LJ(z, 1 ) c C ""- N 0 • Let = Inin (ol, -tlziYJ) so that 0 > 0. Given H' E LJ(z, o) let argo (w) = s. Then (ii) Let

o

o

cos s - cos t 1

1

=

Re

Re z

11'

lwl - Tzf jwl)

Re w (lzl

~

2jw-

lzl

+ I wl

(Re w- Re z)

lwllzl

zl

< YJ· Sin1ilarly, we tnay show that Jsin s - sin tl < 'YJ· It then follows that larg0 (w) - arg 0 (z)j < E. This shows that arg 0 is continuous at each point of C ""- }l0 • Given f3 E N 0 ""- {0}, let

f3n

M+ i sin ( 0 + " -

= lf31 {cos ( 0 + " -

M}

We then have lin1 f3n

n-+ co

= (3,

lin1 argo (f3n) = lin1

n-+co

n-+co

(e +

7T -

!)n = e+ 7T.

Now let Yn

= lf31 {cos ( 0 - " +

~) + i sin ( 0 -

"

+

M}

Then lim Yn

n-+ co

= (3,

lim arg0 (yn) n-+ co

= lim

n-+ co

(e -

1r

!)

+ 11 =

8-

1r.

It follows fron1 Proposition 1.10 that arg0 is not continuous at (3. The proof is complete.

3.2

Elenzents of Con1ple.x Analysis

60

PROBLEMS 3

1. (i) Let E be a compact metric space. Define

IIIII =

sup {1/(z)l: z

E

I . II on rri(E) by

£}.

Show that (11(£), I . II) is a Banach algebra (see Problem 2.6). (ii) Let E be an arbitrary n1etric space and let .t!J~'C(E) denote the set of bounded continuous complex functions on E. Define II . II on f!IYt'(E) by

IIIII = Show that (l!il'C(E),

sup {1/(z)l: z

E

E}.

II . II) is a Banach algebra.

oarguments.

2. Prove Proposition 3.2 by e -

3. If p, q arc polynoinial functions show that deg (p + q) deg (pq)

= =

max (deg (p), deg (q)) deg (p) deg (q).

Do the san1c equations hold if p, q arc rational functions?

4. Let a, {3, y, 8 E C with o:8 - f3y # 0. Let f be defined on f(z)

coo

by

= az + fJ yz + 8

(f is called a honwgraphy or fractional linear transfornzatioll). Show that

/E Honz (Coo, C

00 ).

5. Complete the details of the following sketch proof of the continuity of arg0 on C"' N 0 • Suppose arg0 is not continuous at z E C"' N 8 • Choose a sequence {zn} in C"' N 0 such that li1n Zn = z and no subsequence of

oo {arg0 (zn)} converges to arg 0 (z). Since {arg0 (zn)} c (8- n, 8 + n] son1e subsequence converges to a point t E [8 - n, 8 + n]. Consider cos and sin of this subsequence and show that n-+

cost = cos arg0 (z),

sin t = sin arg 0 (z).

Hence deduce a contradiction.

3.2

Differentiable contplcx functions

In this section we define differentiability for complex functions and deduce some elcn1cntary facts about differentiable functions. Let E be a non-empty subset of C and let f: E ->-C. As far as the text is concerned we shall define the concept of con1plcx differentiability only at interior points of E. (The history of con1plcx analysis has shown that this is the most fruitful definition to consider; a In ore general definition is given in

Continuous and Differentiable Con1plex Functions

3.2

61

Problen1 3.8.) Suppose then that a E E and that Ll(a, r) c E for some r > 0. \Ve say that f is differentiable at a if there exists y E C such that the following condition holds.

(l) Given

E

> 0 there is S such that 0 < S < r and f(a

+ h)

- f(a)

-----:------'----'- - y
0 such that Ll(a, r) c E and.fis differentiable at each point of Ll(a, r). This is a n1uch stronger condition than sin1ply asking that f be differentiable at a; it also turns out to be a tnuch n1ore interesting condition. Given a nonetnpty open set 0 of C (or of C 00 ) We say thatj: 0-+ Cis analytic Oil 0 if it is differentiable at each point of 0. Thus f is analytic on 0 if and only if it is regular at each point of 0. Since differentiability is a local property of a function we tnay just as well study .f separately on each con1ponent of 0. In other words we are reduced to studying functions that arc analytic on a dotnain D of C (or of C 00 ) . We sha11 denote by d(D) the set of a11 functions that are analytic on D. The n1ain purpose of this book is to perfonn a detailed study of s/(D). Iff is analytic on C we say that f is an entire function. Given f E d(D) the derivatire off is the function z-+ f'(z) (z E D), denoted of course by f'. As usual the higher derivatives off are defined by induction. Thus iff' E s/(D) than .f" is defined to be the derivative off'. We say thatfis infinitely differentiable on D if all the higher derivatives off exist on D. One of the astonishing facts that we shall prove about functions in ._r:Y(D) is that they are auton1atically infinitely differentiable on D. Given f: D-+ C we say that a E D is a singularity off iff is not regular at a. We say that a E D is an isolated singularity off iff is differentiable at all points of son1e neighbourhood of a except at the point a itself. Observe that an isolated singularity is indeed a special kind of singularity. Several retnarks are in order here on the matter of tenninology. Some text books use the word holonzorphic instead of analytic, and some use the word integral instead of entire. The phrase 'an analytic function' appears to have many different n1eanings in the literature on con1plex analysis. We shall avoid this phrase whenever possible and use the n1ore precise ternlinology 'f is analytic on D' or 'f E .»/(D)'. The above definition of a singularity may seem harsh in view of the fact (Problem 3.11) that a can be a singularity of a function f even iff is differentiable at a. On the other hand it is natural in that it contradicts the n1ore itnportant condition of regularity at a point, and as such it will avoid unnecessary pathological situations in subsequent chapters. Proposition 3.7. Given any donzain D of C, d(D) is a subalgebra of ~(D). Given J, g E .9/(D), ;\ E C we have

(;\f)'

=

;\f',

(f

+ g)'

= f'

+ g'

(fg)'

= f'g + fg'.

Proof This is a straightforward exercise on any of the conditions (1), (2) or (3).

64

Elen1ents of C01nplex Analysis

3.2

Proposition 3.8. Let D 1 , D 2 be don wins of C, let f: D 1 -+ C, g: D2 -+ C and let f( D1 ) c D 2 •

(i) Iff E s1(D 1 ), g E.>'/( D 2 ) then g of E s/( D 1) and (go f)' = (g' o f)f'. {ii) Iff E ~(D 1 ), g E .#(D2 ), with g'(z) =f:. 0 (z E D 2 ) and g o f(z) = z (z

then f

E

s/(D 1 ) and f'

E

D 1)

= I /g' of

Proof (i) Let a E D 1 • Then f(a) E D 2 and so there is r > 0 with .d(f(a), r) c D 2 • Since g is differentiable atf(a) it follows fron1 condition (2) that there is g: .tJ (f( a), r) -+ C such that

g(f(a)

+ k)

g(f(a))

=

+ k{g'(f(a)) +

g(j(a)

+

(k

k)}

E

.d(O, r))

where g is continuous at /(a) and g(j(a)) = 0. Since f is continuous at a there is 8 > 0 such that .d(a, 8) c D 1 andf(.d(a, 8)) c .d(f(a), r). It follows that for each hE .d(O, 8) we have

g(f(a

+

h)) = g(J(a))

+ {J(a +

h) - f(a)}{g'(f(a))

+

g(f(a

+ h))}.

Sincefis differentiable at a there is '1: .d(a, 8)-+ C such that

f( a

+

h)

=

f(a)

+ h{J'(a) + YJ(a +

where YJ is continuous at a and YJ(a)

go J(a

+

h)

=

go J(a)

+

(hE .d(O, 8))

h)}

= 0. We now have

h{g' 0 J(a) .J'(a)

+

'(a

+

h)}

(hE .d(O, 8))

where

'(a + h) = f'(a) · g(f(a + h)) + g' o f(a). YJ(a + h)

+

g(j(a

+

h)). YJ(a

+ h)

(h

E

LJ (0, 8)).

It is clear that 'is continuous at a and '(a) = 0. By condition (2) we conclude that go f is differentiable at a with (go f)'(a) = (g' o f(a)f')(a). Since a was any point of D 1 this completes the proof of (i). (ii) Observe that f is one-to-one on D 1 since f(z 1) = /(z 2 ) implies z 1 = g(f(z 1 )) = g(f(z 2 )) = z 2 . Let a E D 1 • Then /{a) E D 2 and there is r > 0 with .d(f{a), r) c D 2 • Since f is continuous at a there is 8 > 0 such that .d(a, 8) c D 1 and f(.d(a, 8)) c .d(f(a), r). Let {hn} c L1'(0, 8) with lim hn = 0. Then /(a + hn) E Ll'(f(a), r) and lin1 /(a + hn) = f(a) by n-+co

n-co

Proposition 1.1 0. Since g is differentiable at f(a), condition (3) gives I . g(f(a + hn)) - g(f(a)) _ '(f( )) tmco /(a + I-z 11 ) - ;·(a ) - g a . n-

3.2

Continuous and Di.lferentiable Co111plex Functions

65

Since g'(f(a)) =I 0 we have litn f(a

+ hn)

n--+ ro

hn

- f(a) = , 1 g (J(a))

By condition (3) f is thus differentiable at a with

Since a was any point of D 1 , the proof is cotnplete. We can now see that s/(D) is well supplied with eletnents for any doBlain D of C. Indeed it is trivial to verify that 1 and u are entire functions with 1' = 0, u' = 1. It follows frotn Proposition 3.7 that any polynotnial function is an entire function. Thus d(D) contains all polynon1ial functions (restricted to D). It is equally trivial to verify that j is analytic on C "'{0} with J''(z)

=-

1

(z

,..2

E

C "'{0}).

£.

Now let q be a rational function, say q in C such that

= p 1 . (j o p 2 ). If D is any don1ain

{z: P2(z) = 0} n D = 0 then qlv E .s 0 there is 8 > 0 with (a - 8, a + 8) c I and f(a

+ h) - f(a) h

y
-

C and let {an} c D with lim an = a n-. co

E

D. If each an is a

singularity off show that a is a singularity off. Give an example in which f is differentiable at each an and at a.

13. Usc each of conditions (1), (2), (3) to prove Proposition 3.7. Which proof is most elegant?

3.3

3.3

Continuous and Differentiable Contplex Functions

67

The Cauchy-Rientann equations

1n this section we shall consider what the statetnent 'f E d(D)' n1eans in terms of the associated functions u = Ref, v = Itnf We have already shown that f is continuous if and only if u and v are continuous. In other words a continuous con1plex function can be obtained by piecing together arbitrary continuous real functions u and v. On the other hand we shall see that the functions u and v nu1st be very intirnately related if we are to obtain a differentiable function. \Ve begin by recalling son1e real analysis.

Let U be an open subset of R 2 and let g: U--->- R. We denote by

~~· ~~

the partial derivatives of g (if they exist) with respect to x, y respectively. Thus for exatnple if g has a partial derivative with respect to x at (a, b) then

og ( b) - a ox '

=

1" g(a 1n1

+ h, b)

- g(a, b)

h

h-+0

·

Recall that g is said to be (real) differentiable at (a, b) E U if there exist A, B E R such that for (x, y) in son1e neighbourhood of (a, b) we have

g(x, y) = g(a, b)

+ (x

- a){A

+ e 1 (x, y)} + (y

- b){B

+ e2 (x, y)}

where e 1 , e 2 are continuous at (a, b) with e 1 (a,

b) = e 2 (a, b)

= 0.

It then follows that g has partial derivatives with respect to x and y at (a, b) and

og ox (a, b)

= A,

og

oy (a,

b) = B.

Theorem 3.9. Let f: D ---+ C with f = u + iv and let a E D. Then f is differentiable at a if and only if u, v are differentiable at a and satisfy the Cauchy-Rie1nann equations at a, i.e.

ou

ov

ox (a) = oy (a),

ou

ov

oy

ox

-(a) = --(a).

Thus f E d(D) if and only if u, v are differentiable and satisfy the CauchyRientmm equations on D. Proof Let a = (a, b) and choose r > 0 such that Ll(a, r) c D. Suppose that f is differentiable at a. Then by condition (2) there is ~: Ll(a, r) -7 C such that f(z)

= f(a) + (z - a){f'(a) +

~(z)}

(z

E

Ll(a, r))

Ele1nents of Con1plex Analysis

68

3.3

where ~ is continuous at a with ~(a) = 0. Let ~ = g + iYJ and z = (x, y). On taking real parts of the above equation we obtain

u(x, y) = u(a, b)

+ (x + (y

- a){Ref'(a) + .g(x, y)} - b){ -Itnf'(a) - IJ(X, y)}.

Since g, TJ are continuous at (a, b) with u is differentiable at (a, b) with

~~(a, b)=

Ref'(a),

g(a, b) = TJ(a, b) = 0, it follows that

~;(a, b) = -

Jmf'(a).

Similarly, by taking itnaginary parts we see that vis differentiable at (a, b) with

~;(a, b) =

In1j'(a),

~;(a, b) =

Ref'(a).

It follows that u and v satisfy the Cauchy-Riemann equations at a. Suppose conversely that u and v are differentiable at a and satisfy the Cauchy-Riemann equations at a. It follows that there exist real functions Er (r = 1, 2, 3, 4) in some neighbourhood of (a, b) such that

au (a, b) u(x, y) - u(a, b) = (x - a) { ox

+

E1

(x, y) }

+ (y v(x, y) - v(a, b)

=

(x -

a){~~ (a, b)+ 1(L1(a, r)). Moreover v E .Yt(Ll(a, r)) by Proposition 3.12 so that v is an harmonic conjugate of u. Suppose now that v1 is any harmonic conjugate of u. Thcnf1 = u + il 1 E s:I(Ll(a, r)) and so f- / 1 E s>/(LJ( a, r)). Since f- f 1 = i(v - t' 1) it follows from the Cauchy-Ric1nann equations that 1

0 ox (v - v1)

=

8 oy (v - v1)

=0

so that v - v1 is constant on Ll(a, r) as required.

3.4

Continuous and Differentiable Conzplex Functions

73

Exaruplc 3.14. Let D = L1 '(0, 2) and let

u(x, y) Then u

E

log (x 2 + y 2)

=

£(D) and u has

110

((x, y) ED).

hannonic conjugate on D.

Proof It is routine to verify that u E .Ye(D). Suppose that u has an harnlonic conjugate v on D. Define g: [0, 27T] ---+ R by g(t) It follows that g g'(t)

= v(cos t, sin t)

E ~R([O,

(t

E

[0, 27T ]).

27T]) with g(27T) = g(O). Moreover

= :.~(cost, sin t)( -sin t) + :; (cost, sin t)(cos t).

Since u and v satisfy the Cauchy-Riexnann equations we have g'(t)

=-

~u(cost,sint)(-sint) + ux ~u(cost,sint)(cost) uy

2 sin 2 t 2 cos 2 t cos 2 t + sin 2 t + cos 2 t + sin 2 t = 2 (t E (0, 21r)).

It follows that g(t)

=

g(O)

+ 2t

(t

E

[0, 27T])

and so g(27T) =/=- g(O). This contradiction shows that u cannot have an hannonic conjugate on D. PROBLEMS 3

24. Given that f = u + iv E d(D) is infinitely differentiable on D show that u, v have partial derivatives of all orders on D. 25. Show that the following functions are harn1onic on C and find harmonic conjugates for each. (i) u(x, y) = x2 - y 2 + 2x. (ii) u(x, y) = ex cosy. (iii) u(x, y) = sin x cosh y. 26. Show that the Laplace equation corresponds to

o2f 8z 8z = 0 ·

27. Let f E d(D) with f(z) =/=- 0 (zED). Show that log assuming that f is suitably differentiable.

Ill

E

.Ye(D),

1.4

Elenzents of Conzplex Analysis

24

We now consider a stronger concept of continuity. Let (X, d), ( Y, e) be n1etric spaces, letf: X->- Y, and let E be a non-en1pty subset of X. We say that/is unifornzly continuous onE if for every E > 0 there is o(E) > 0 such that (x E E).

Iff is uniformly continuous on X we say more sin1ply that/ is unifonnly continuous. The student should note the parallels between continuity and uniform continuity, and pointwise and uniform convergence. It is clear that unifonn continuity itnplies continuity, but the converse is false as is easily seen by taking

f(x)

=

x2

(x

E

R).

On the other hand, when X is compact the two concepts coincide. Proposition 1.19. Let (X, d), ( Y, e) be nwtric spaces with X conzpact, and let f: X----?- Y be continuous. Then f is unifonnly continuous.

Proof Let such that

€

> 0. Since f is continuous, given

X E

X there is o( €, x) > 0

The collection {Bix, o(E, x)): x EX} is clearly an open cover for the compact space X and so has a finite subcover, say {Bd(xh o(E, x 1)): j = I, 2, ... , n}. Let

so that 77( E) > 0. Given x, y E X with d(x, y) < 77( E), there is sotne j such that X E Bu(xh o(E, Xj)). Since 7)(€) ~ o(E, Xj) we have

d(y, x 1) Thus x, y

E

~

d(y,

X)

+ d(x,

x 1) < 2o(E, x 1).

Bixh 2o(E, x 1)) and so e(f(x),f(y))

~

e(f(x1),f(x))


GO

We say that 2 an is absolutely convergent if 2 lanl is convergent. Given that cp: P ---7 P is one-to-one and onto we say that 2 act> is a rearrangeJnent of 2 an-

2

Proposition 4.2. If an is absolutely convergent then it is contergent. If act> is a rearrangen1ent of an then act> is absolutely convergent and

2

2

GO

L

2

GO

act>(n) =

n=l

2

an·

n=l

Proof Let Tn = jail + la2l + · · · + lanl (n E P). Since 2 a 11 is absolutely convergent {Tn} is convergent and therefore Cauchy. Since

ISm - Snj ~

ITm - Tnl

4.1

Power Series Functions

77

it follows that {Sn} is Cauchy and therefore convergent since C is a con1plete n1etric space. Since IRe an! ~ !an!, !In1 an! ~ !ani (n E P) it follows fron1 the C0111parison test for real series that L Re an and L In1 an are absolutely convergent. It now follows fron1 the theory of real series that L Re aCJ? and I lin aCJ? arc absolutely convergent with co

co

co

" Re aCJ?(n) = " Re

L n=l

L I 111 aCJ?(n) = n=l L lin a

an,

L..

co

n=l

11 •

n=l

Since

(n

E

P)

L

it follows fron1 the con1parison test for real series that acp is absolutely convergent, and so also convergent. Using Proposition 4.1 we now obtain co

L

co

aCJ?(n)

n=l

=

L

co

Re acp(n)

n=l

+i L

co

=

L

In1 acp(n)

n=l

co

Re an

n=l

+

i

L

In1 an

n=l

In the theory of absolutely convergent real series the n1ain tools are the cotnparison test and the nth root test (or the weaker ratio test). The same is true for con1plex series. We recall first the definition of the limit superior of a bounded sequence of real nun1bers. Given a bounded real sequence {an} the lilnit superior of {an}, denoted by lim an, is that real nutnber a with n-. co

the following properties: (i) Given

E

> 0 there is N an
N).

E

> 0 there is a subsequence {ana} of {an} such that ana(n) > a -

E

(n

If {an} is not bounded above we write lim an

n-. co

= +oo.

E

P).

78

4.1

Elements of Complex Analysis

Proposition 4.3. (i) If L an is absolutely convergent and if there is N E P such that I,Bnl ~ !ani (n > N) then L ,Bn is absolutely convergent and so convergent. (ii) Let A = lim n-+oo

diverges

lanl 11 n. Then L an is absolutely convergent if A

< I and

if A > I.

Proof. (i) This follows from the comparison test for real series and

Proposition 4.2. (ii) Suppose A < I and choose € > 0 such that t = A follows that there is N E P such that

+ € < I. It

(n > N)

and so (n > N).

L

It follows from the comparison test for real series that !ani is convergent. Suppose now that A > I and choose € = A - I. Then there is a subsequence {aCT} of {an} such that Ia

CT(n) 11/CT(n) > A -

- 1

(n

€ -

E

P)

and so

It follows that {an} does not converge to zero and hence converge.

L an

cannot

PROBLEMS 4 1. Given z E C discuss the convergence of the series

L arg (z + n). 1 +in 2. Given 0 < a < b show that 00

f(a) =

L na +1 zn. 1

n~l na + in 3

show that f is continuous on (2, 3).

b

converges iff 2a - b > I. If

(2 < a < 3)

Power Series Functions

4.1 3. Given a 11 ,

/3

79

n

11

= 2 ar.

E C (n E P) let An

Show that

r=l

N-1

N

L

arf3r

2

=

(f3r - f3r + l)Ar

r=l

r=l

~ ! {cos nt +

Discuss the convergence of n

=1

+

f3NA N·

i sin nt}.

11

4. Given a, {3. y, :: E C where y =1- 0, -1, -2, ... , discuss the convergence of

L a(a +

1) ... (a + 11 - 1){3({3 + 1) ... ({3 y(y + 1) ... (y + 11 - I)

+ 11

-

1) zn. n!

2:

llxnll 5. Let B be a Banach space (see Problern 2.6). If {xn} c Band converges show that 2 -"n converges. Let X be a con1pact tnetric space, let B = 0'(X) and let/E B with II/II < I. Show that 1 + 2fn converges. \Vhat is the sun1? Suppose now that B is a Banach algebra and y E B is such that lin1 n-+ co

llrn l 11n
0 there is N(e) E P, such that

(n1 > N( e), 11 > N( e)). If {am.n} also converges to fLit is sin1ple to show that fL lirn m,n-+ co

.

am n

= ,\,

We write

= ,\,

It is sotnetimes helpful to visualize a double sequence as a doubly infinite array as indicated below. al, 1

al, 2

al, 3

a2.1

a2, 2

a2, 3

a3,1

a3,2

a3,3

4.1 Elements of Complex Analysis 80 It follows trivially from the definition that if {am, n} converges to .\ then lim an, n = .\. In pictorial terms this says that the dexter diagonal conn ... oo

verges to the limit of the double sequence. Our interest is essentially confined to double sequences which arise in special ways.

Proposition 4.4. Given {an}, {,Bn} c C let m

Sm, n =

n

L a, + r=l L ,8,.

r=l

Then {Sm,n} converges if and only if both 00

00

L a, + r=l L ,8,.

lim Sm,n = m,n-+oo

La" L ,8, converge, in which case

r=l

La" L

Proof. Suppose that ,8, converge to .\, p. respectively. Given e > 0 there is m1 (e) E P such that

Similarly, there is n1 (e) E P such that

Choose N(E) = max (m 1 (E), n 1 (E)) and we have

+ fL)I < E so that {Sm. n} converges to .\ + p.. ISm,n- (.\

(m > N(E), n > N(E))

Suppose now that {Sm,n} converges to S. Given E > 0 there is N(E) such that

E

P

(m > N(E), n > N(E)). Fix n > N(E). Then for n1 1 > m 2 > N(E) we have ml

L

r=m 2 + 1

a, = ISm~on - Sm2.nl
0 choose S =

€

and then

d(x, y) < S =>

ldist (x, A) - dist (y, A) I
0 such that !Ami ~ M(mE P). Given E > 0 we may choose N(E) E P such that E

IBn - ILl < 2M

(n > N(e)) E

lAm - Aj < 2{l + IILI)

(m > N{e)).

We now have !Sm,n - AIL!
N{e), n > N{e)),

E

and so {Sm. n} converges to AIL. Observe that the above result sums the double array {a,,B8 } by rectangles. It is trivial to verify that the same sum is obtained by the row or column method. The situation is much more complicated with regard to summing by triangles. This latter method is very significant for the next section of this chapter and warrants special definition. Given complex sequences {an}, {,Bn} (n = 0, I, 2, ... ) we define their convolution to be the sequence {yn} defined by n

'Yn =

2

r=O

(n=0,1,2, ... )

a,,Bn -r

and we write {yn} ={an}* {,Bn}. Proposition 4.6. Let

2 am 2 ,Bn be absolutely convergent and let {yn} =

2 'Yn is absolutely convergent and

{an} * {,Bn}. Then

00

00

00

2 'Yn = n=O 2 an n=O 2 ,Bn•

n=O

Proof Since n

2

r=O it follows that

IYrl ~

n

2

r=O

n

!a,j

2

r=O

I,B,I ~

oo

2

oo

r=O

la,l

2

ro::O

I,Brl

{,t lr,l} is monotonic and bounded. Thus 2

Yn is abso-

n

lutely convergent and so convergent. Let An = n

Cn =

2

r=O

n

y, Dn =

2

r=O

n

!a,j, En =

2

r=O

oo

j,B,j, A =

2

r=O

2

r=O a,, p. =

n

a,, Bn = oo

2

r=O

have

IC2n

-

IC2n

A2nB2nl + IA2nB2n - Ap.l ~ IDnEn- D2nE2nl + IA2nB2n- Ap.j.

Ap.j ~

-

2

,Bn

r=O ,8,. We then

Power Series Functions

4.1

83

Since {A 2 nB2 n} converges to Af-t and since {DnEn} converges it follows that lim IC2 n

n-+ oo

Af-t!

-

= 0.

Since any subsequence of a convergent sequence converges to the limit of the sequence we conclude that 00

2 Yn

n=O

= lim

n-+oo

C2n

= Af-t.

PROBLEMS 4 6. (i) Let a

n1n

(m, n E P).

=----.,.

(m

m,n

+ in)2

Show that lim (lim

m-+ oo n-+ oo

= 0 = lim (lim

am, n)

n-+ oo m-+ oo

am, n)

but that {am. n} does not converge. (ii) Let a m,n

=

(-l)m+n(_!_ m

i)

+n

(m, n E P).

Show that {am. n} converges to 0 but that for each fixed mE P the sequence {am. n} fails to converge.

1

7. Let {a,,,} be such that the double sequence {~

J Ia,, ,I} 1

converges.

Show that lim

i

~ s

m,n-+ oo T = 1 =1

aT.

s = T~= ( s~= aT. s) = s~= ( T~= = lim 2 aT,s• 1

n-+oo

1

1

1

aT,

s)

T+s~n

(This shows that the double array has the same sum by rectangles, rows, columns, or triangles.) 8. For which z do the following series converge? ~

L -oo

n ~ log ( 1 + In I) ~ z 'L _ 00 -oo n2 +n+z ' L

.

2z Z

2

-n

2

9. Let

an=(

( -l)n

n

+ IY

(n=0,1,2, ... )

2

and let {yn} = {an}* {an}. Show that

2 an converges while 2 Yn diverges.

84

Elements of Complex Analysis

4.3

10. (i) Let L 1{Z +) denote the set of all complex sequences an (n = 0, 1, 2, ... ) such that Iani converges. Show that L 1{Z +) is a complex linear algebra with the operations defined by

2:

A{an} {an} + {,Bn} {an} . {,Bn}

= {Aan} = {an + ,Bn} = {an} * {,Bn} ·

If 0 there is N E P such that ISnl < series into two parts and deduce the result.)

4.4

(n > N). Now split the

£

4.4 The exponential function We shall now examine the so-called standard or elementary functions. We shall assume that the reader has a working knowledge of the corresponding real functions. Aside from the polynomial functions the most important standard function is the exponential function. We define exp: C ~ C by exp (z) = 1

+

co

zn

L n.- 1

(z

C).

E

n= 1

Proposition 4.11. The function exp is an entire function with the following properties: (i) exp' = exp; (ii) exp (z1 + z2 ) = exp (z 1) exp (z 2 ) (zr, z2 E C); (iii) exp (z) = ex{ cosy + i sin y} (z = x + iy E C).

Proof. (i) It is well known from real analysis that (n !) 11n ~ oo as n ~ oo so that the power series for exp has infinite radius of convergence. It thus follows from Theorem 4.10 that exp E d(C) and moreover exp' (z)

co

=

L

nzn-1

n.1

n=1

(ii) Given a

E

= exp (z)

(z

It follows from Propositions 3. 7 and 3.8 that f =

exp (z) exp {a - z)

We now have/'

. t.e.

E

E

C).

C).

d(C) and

+ exp (z){ -exp (a

- z)}

= 0 and so by Proposition 3.10 f(z) = /(0) = exp (a) exp (z) exp (a - z)

Given Zr, z 2 E C let z

as required.

E

C define f on C by

f(z) = exp {z) exp {a - z)

f'(z)

(z

= Zr,

a

= exp {a)

(z

E

C), (z E C).

= z 1 + z2 and we obtain

(z

E

C).

Power Series Functions

4.4

91

(iii) Using (ii) we now obtain for z = x + iy exp (z) = exp (x

= ex

+ iy) =

exp (x) exp (iy)

{1 + ~ (iy?n} n. n =1

v2m

00

eX

=

ex{cosy + isiny}.

1

y2n + 1

00

+ m2:= 1 (- 1)In c2 111)f• + i n2=0 (- 1)n (2n +

=

{

}

1) f

.

Corollary. The function exp has no zeros.

Proof It follows frotn (ii) that exp (z) exp ( -z)

= exp (0) =

(z

1

E

C).

and therefore exp cannot have any zeros. Since the function exp is an extension of the usual exponential function on the real line we shall often write exp (z) n1ore briefly as ez. The occurrences of the exponential function in con1plex analysis are too numerous to tnention. It is thus itnportant to have as clear a geometrical picture as possible of how the exponential function behaves on C. Given.{: C---+ C \Ve say that y is a period off if

f(z

+ y) = f(z)

(z E C).

It is easy to see that such a function/repeats itself in strips of width jyj. If y is a period off so is 2y, for

f(z + 2y) = f(z + y) = f(z)

(z

E

C).

More generally we have that ny is a period ofjfor each n E Z. We say that y is a fundmnen tal period off if y is a period off, y =I= 0, and no proper subtnultiple of y is a period off If y is a fundan1ental period off so also

.

IS -y.

Proposition 4.12. The function exp has fundantental periods 27Ti, - 27Ti and these only. Given e E R, exp nzaps the strip {z: e - 7T < Im z ~ e + 7T} one-to-one onto C ""- {0}, and the do1nain {z: 8 - 7T < Itn z < 8 + 7r} oneto-one on to C ""- N 0 •

Proof We have by Proposition 4.11 (ii) and (iii) exp (z + 27Ti) = exp (z) exp (27Ti)

=

=

exp (z) {cos 27T + i sin 27T} (z E C) exp (z)

92 so that

27Ti

is a period of exp. If y is any period of exp we have exp {y)

If y

4.4

Elentents of Conzp/ex Analysis

=

a

+

= exp {y + 0) = exp (0) = 1.

ib we then have ea {cos b

+

i sin b}

=

I.

On taking the n1odulus of both sides we obtain ea = I and thus a = 0. We then have cos b = 1, sin b = 0 so that y is an integral multiple of27Ti. This proves the first statement. Let z 1 = x 1 + iy 1 , z 2 = x 2 + iy 2 be such that exp (z 1) = exp (z 2 ). It follows that [exp (z 1 )! = [exp (z 2 )[, i.e. ex1 = ex2 so that x 1 = x 2 • We now have cos y 1

+

i sin Y1

=

cos Y2

+

i sin Y2·

It follows that exp is one-to-one on any strip {z:

e-

7T

< lin z

~

e + 7T}.

We have already observed that exp n1aps C into C ""- {0}. Given wE C "'-. {0} let z = log !wl + i arg 0 (w). Then 8 - 7T < lm z ~ 8 + 7T and exp (z)

=

= =

exp (log [w!) exp {i arg 0 (w)) Iwl {cos (arg0 (w)) + i sin (arg0 {w))} lV.

Therefore exp maps the strip {z: 8 - 7T < ln1 z ~ 8 + 7T} one-to-one onto C ""- {0}. For the final part it is sufficient to verify that exp maps the line + 7T} onto N 0 "'-. {0}. {z: Im z =

e

As a further illustration of the exponential function the student should verify that exp maps the seg1nent [a + ic, b + ic] onto the segment [eaetc, ebetc]. We may now obtain the image of rectangles under the exponential mapping. The situation is described pictorially in Figure 4.1.

··· .

.

... •

Imz=7T ----a 'ic ::· ·.·.·. ··.·.·.·... .·. :- .·.·. · · b •i c ·.

a+id ·

b ic

e e

o ic •

0

e e . b~id

Im z = - 1r

Figure 4.1

0

ee

·o id •

Power Series Functions

4.4

93

There are several standard functions related to the exponential function. \Ve define the complex functions sin, co~, sinh, cosh by sin (z) =

~i {exp (iz) - cxp (- iz)} i{exp (i.:) + cxp (- iz)}

(z

E

C)

(z E C) (z E C) (z E C).

cos(.:) = sinh (z) = -!-{exp (z) - exp (- z)} cosh (z) = j{exp (z) + exp ( -z)}

It is clear fron1 Propositions 3. 7 and 3.8 that each of these functions is an entire function. By considering their power series expansions we n1ay easily see that the above functions agree on the real line with the associated real functions. We set out below son1e of the sin1ple properties of these functions leaving the proofs as exercises. The student should fatniliarize hin1self with these properties since they will be required for 1nany subsequent problen1s. (i) sin' = cos, cos' = -sin, sinh' = cosh, cosh' = sinh. (ii) cos 2 + sin 2 = cosh 2 - sinh 2 = 1. (iii) sin (z 1 + z 2 ) = sin (z 1 ) cos (z 2 ) + cos (z 1) sin (z 2 ) cos (z 1 + z 2 ) = cos (z 1 ) cos (z 2 ) - sin (z 1) sin (z 2 ). (iv) jsin (z)l 2 = sin 2 x + sinh 2 y, Ieos (z)j 2 = cos 2 x + sinh 2 y. (v) sin, cos have funda1nental period 21r; sinh, cosh have fundan1ental period 21Ti. (vi) The zeros of sin, cos, sinh, cosh consist respectively of the sets {n11: n E Z}, {(n + t)7T: n E Z}, {in11: n E Z}, {i(n + -!)11: n E Z}. We shall feel free to speak of other such con1plex functions as Sill

tan=-, cos

sech

=

1

1 COSl

,

and to state son1e of their obvious properties. For exatnple it is clear that tan E..#( D) where D = C ""'- {(u + -!)11: 11 E Z}. PROBLEMS 4

16. Show that exp is the only entire function f such that f(R) c R and f(I) = e, Find all entire functions satisfying ( *). Find a continuous nowhere differentiable function f satisfying ( *). (Hint. Functions of z are helpful for the latter part.)

94

4.4

Elenzents of Cotnplex Analysis

17. Let g: C-+ C be defined by

g(z) = {

exp (

_!)z

0

if

~

larg (z)l

4'

7T

lzl > 0,

otherwise.

Where is g differentiable? Where is g regular?

18. Use Proposition 4.9 to prove Proposition 4.11 (ii). 19. At which points do the following series converge? On which sets do they converge uniforn1ly? co

co

co

2: e-nz2,

2: e-nz,

2: 2nze- nz2,

2:

e-nz2,

""

L... e -

n2z2

.

-co

-co

0

0

0

co

co

20. Establish the following results.

= i sinh (z), sinh (iz) = i sin (z), = cosh (z), cosh (iz) = cos (z) (z E C). (ii) sin (z) = sin (z), cos (z) = cos (z) (z E C). (iii) lim e-Ysin(x + iy) = !{sinx + icosx}. (i) sin (iz) cos (iz)

y- +co

(iv)

li1n tan (x

21. Let

+ iy)

=

D 1 = {z: IRe

i.

zl

< ;}

{z: lim zl < 1}, D 4 = C ""-{(-co, -1]

D2

= {z: 0
0 and a E R then ta is defined by ea log t. This motivates the following definition. Given ,\ E c, 8 E R we define p~: C "-. {0} ~ C by p~

If we take ,\

=

(z) = exp (,\loge (z))

(z -:F 0).

1 in the above definition we obtain for any 8 that p~

(z) = z

(z -:F 0).

More generally, for any integer n E Z we may easily show that

p: (z) = zn

(z 1: 0).

For the case ,\ = -!we have (pj (z)) 2 = exp {! loge {z)

+ ! loge (z)} =

z

(z -:F 0)

4.5

Power Series Functions

99

so that each function p~ is a 'square root' function. Moreover for each ==I= 0 P~+2n(z) = exp {-!-logo+2n (z)} = exp {-!loge (z) + 1ri} = - p~ (z). This coincides with the intuitive notion of 'positive' and, negative' square roots. On the other hand when we take A to be con1plex our intuition breaks down in the face of forn1ulae such as

pb(i) = exp (i log (i)) = exp (;; i) = e-•1 2 • In line with our earlier practice we write p~ n1orc sin1ply as p'A. If A = f-L + iv and t > 0 then

p'A(t) = exp {(f-L + iv) log t} = t 1t{cos (v log t) + i sin (v log t)}. We shall so1netin1es \Vrite t'\ for p"A(t) (t > 0) to show its close relation with the corresponding real function.

Proposition 4.19. Given A, f-L

E

C,

eE R we hare

(i) p~+Jl = p~p~ (ii) p~ E d(C ""- Ne) and (p~)' = Ap~ - 1 • Proof (i) p~+ 11 (z)

exp {(A + f-L) loge (z)} = exp {A loge (z)} exp {f-L loge (z)} =

=

p~(z)

pHz).

(ii) It follows frotn Propositions 4.16 and 3.8 that p~ is analytic on C ""- Ne and (p~)'(z) = exp' (A loge (z))A log~ (z) A

= - p~(z) z

=

;\p~-I(z).

If ;\ is an integer then p~ is in fact analytic on C ""- {0}. If A is not an integer we may use the argun1ent of Proposition 3.5 to show that p~ is not continuous at any point of Ne ""- {0}, and hence in this case each point of Ne ""- {0} is a singularity of p~. It may also be verified (see Problen1 4.28) that given a E C ""- Ne we may represent p~ by a power series on son1e disc about a. In particular using the technique of Exan1ple 4.18 we 1nay verify that oo (z - 1)n p'A(z) = 1 + L A( A - 1) ... (A - n + 1) (z E Ll(l, 1)). 1 nal

n.

100

Elements of Complex Analysis

4.5

PROBLEMS 4

24. Given n E P show that n log ( 1 + where l/n(z)j

~

~)

=

z + fn(z)

(z E ..J(O, !n))

lzl 2 (z E Lf(O, !n)). Deduce that n

lim n-+co

(1 + n~)n = exp (z)

(z E C)

and show that the convergence is uniform on any compact subset of C. 25. Let D be a domain in C with 0 ¢ D. We say that x: D continuous argument on D if xis continuous and x(z)

= arg (z)

-~

R is a

(mod 21r).

Show that there is a branch-of-log on D iff there is a continuous argument on D. 26. Given zh z 2 E C "- Ne find necessary and sufficient conditions for loge (z1z2)

= loge (z1) + loge (z2).

27. Let D be a domain in C such that for some R > 0

C(O, R) c D c C "- {0}. Show that there is no branch-of-log on D.

28. (i) Letfbe a branch-of-log on D. Given L1(a, r) c D represent/ on L1(a, r) by a power series. (ii) Given a E C "- Ne represent p~ by a power series on some disc about a. What is the largest such disc? 29. Given A = p,

+ iv show that IP~(z)j =

!zltt e-v argo.

30. Discuss the convergence of the following series. co

L

n=O

P~nn(z),

5 ARCS, CONTOURS, AND INTEGRATION

The n1ain purpose of this chapter is to develop the n1achinery of integration along arcs and contours in the con1plex plane. Every Inatheinatical theory of integration that is not trivial involves a considerable amount of technical detail, and this chapter will prove no exception. We have atten1pted son1e streainlining of the treatinent of integration by relegating to the Appendix the technical details of the Rien1ann-Stieltjes integral. Consequently, the larger part of the chapter is concerned with the topological aspects of arcs and contours in the con1plex plane. This study is of interest in itself apart fron1 applications to con1plex analysis. We include a sin1ple proof of the Jordan curve theoren1 for a special class of simple closed curves.

5.1 Arcs An intuitive idea of an arc in the complex plane is given by a picture that looks like a piece of bent wire. As a first step in formalizing this picture we n1ight regard an arc, r say, as the range of a continuous complex function defined on the interval [0, 1]. If we do not wish the arc to intersect itself we n1ust take the continuous function to be one-to-one. (Indeed without this restriction the curve* could fill out a square!) Unless we wish to label r with exactly one such function we are obliged to consider all the continuous functions on [0, 1] that are one-to-one and have range r. Since [0, 1] is a compact metric space it follows fron1 Proposition 1.18 that the above functions are in fact the hon1eomorphis1ns of [0, 1] with r. We are now ready for our fonnal definition. Recall that given any two metric spaces X and Y, Honz (X, Y) denotes the set of all homeomorphis1ns of X with Y.

* See, for example, G.

F. Simmons, Introduction to Topology and Modern Analysis, (McGraw-Hill, 1963) Appendix 2.

101

I

102

Elernents of Con1plex Analysis

r

5.1

An arc is a subset of C that is homeotnorphic with the closed interval [0, 1]. In other words, a subset of of C is an arc if and only if the set Hon1 ([0, 1], T) is not etnpty. We say that an arc is represented by each elen1ent of Horn ([0, I], T). We also say that each element y of Ho1n ([0, 1], T) is a representative function for Note that a subset r of C is an arc if and only if it is the range of a continuous function that is one-to-one on [0, 1]. In particular, given a, f3 E C, a i= {3, [a, {3] is an arc. To see this we sin1ply take

r

r

r. .

y(t)

=

{1 - t)a + tf3

(t

E

(0, 1]).

As a further example take

y(t)

=

a

+ r exp (rrit)

(t

E

[0, 1])

and we obtain a semi-circular arc. The student should have no difficulty in producing many other simple exan1ples. We begin our study of arcs by analysing the set Horn ([0, 1], T). It turns out that there is a very sirnple description of this set. In what follows, the set Honz ([0, 1], [0, 1]) will be of special in1portance; we denote it by f/J for short. The lemma below shows that the elements of f/J are (strictly) monotonic functions. In view of Proposition 1.18, f/J thus consists of the continuous strictly monotonic functions on [0, 1] that have range [0, 1]. (In fact, f/J consists of the strictly n1onotonic functions on [0, 1] that have range [0, 1]; see Problem 5.1.)

Lcmn1a 5.1. Each elenzent of cJ> is a nwnotonic function. Proof Let r:p E cJ> and suppose r:p(O) < r:p(1). Suppose there is t E {0, 1) such that r:p(t) < r:p(O). We then have r:p(O) E [r:p(t), r:p(1)]. Since r:p is continuous on [t, I] it follows fron1 the intermediate value theoren1 that there is u E (t, 1) with r:p(u) = r:p(O). This contradicts the fact that r:p is one-to-one and so we must have r:p(t) > r:p(O) (t E (0, 1)). A sirnilar argument shows that r:p(t) < r:p( 1) ( t E (0, 1)). Suppose now that 0 < s < t < 1 and r:p(s) > r:p(t). We then have r:p(s) E (r:p(t), r:p(l)). Arguing as above we obtain v E (t, 1) such that r:p(v) = r:p(s). This is again a contradiction and so r:p(s) < r:p(t). We have thus shown that r:p is strictly increasing if r:p(O) < r:p(l). A sirnilar argun1ent shows that r:p is strictly decreasing if r:p(O) > r:p(l ). The proof is con1plete. Given r:p E cJ> it follows frorn the above lernrna that r:p(O) = 0 or r:p(O) = 1. It is then clear that r:p is increasing if and only if r:p(O) = 0, and r:p is de-

Arcs, Contours, and Integration

5.1

creasing if and only if .

((f

=

>.E C).

(jefC(r)). o

y _)y'- =

J.o f(y(l 1

t))( -y'(l - t)) dt

I/

f(y(u))y'(u) du = -

(r = I, 2, ... , n) form a direction-preserving decomposition

Proof. We may choose a representative function y for r such that y(O) is the first point of F 1 • Lett, (r = 0, I, ... , n) be such that t 0 = 0 and y(t,) is the last point of r, (r = I, 2, ... , n). Then r, may be represented by rl[t,_ 1 ,t,1 (r = I, 2, ... , n). Using Theorem 5.9 (i) we now obtain

l f = J. f 1

r

o

0

y dy

r

=

L:= J.t,t, n

=

L:n

1

r= 1

l

r,

f

0

y dy

-1

f.

For the next property it is necessary to introduce the concept of the length of r. A subdivision of [0, I] is a set of the form P = {t,: r = 0, I, ... , n} where 0 = 10 < t 1 < · · · < tn = I. We denote by fJJ by the set of

5.6

Arcs, Contours, and Integration

127

all subdivisions of [0, 1]. Given such a subdivision P of [0, 1] we define n

2

T(y, P) =

/y(tr) - YCtr-1)/

r= 1

= sup {T(y, P): P E .9'}.

V(y)

Since y E 8&ir-([O, 1]) we certainly have 0 ~ V(y) < + oo. Suppose that Tis an arc and that y 1 is any representative function for T. Then y 1 = yo fP for son1e fP E cJ> +. Given P E .9' with P = {tr: r = 0, I, ... , 11} we easily see that {+ we also have {ffJ- 1 (tr): r fP- 1 (t) it now follows that

Since fP

o

ffJ- 1 E

t =

V(y) = sup {T(y o fP o fP- I, P): P E .9'} ~ sup {T(y o fP, P): P E .9'}

=

V(yl).

We have thus shown that V(y 1 ) = V(y) for any function y 1 that represents r. (Observe that we have in particular given a proof of Problem 5.5.) We leave the reader to verify that this is also true in the contour case. We tnay nO\V define the length of r by

= V(yl)

lr

r.

where y 1 is any representative function for By the above reasoning the length of is thus well defined. Choosing the given representative function \Ve then obtain

r

lr

=

J: lrl

The proof of this last state1nent is given in the Appendix. Given a non-empty subset S of C and given f: S--* C we shall use the following abbreviated notation: sup s

CI 5.

l/1

= sup

{lf(z)/: z E S}.

J/ ~ s~p l/1 JJ = J: y)y' ~ J: If rllr'l (/E ~(T)).

lr

(f o

Proof

~

sup {1/(y(t))l: t

= lr sup

r

Given f

E

o

E

[0, I]}

J: lr'l

If/.

rc(D) recall that g

E

si(D) is a primitive off if g' =f.

5.6

Elements of Complex Analysis

128

Cl 6. Iff E

~(D)

with primitive g and ifF c D,

= g(y(l)) - g(y(O)).

fJ

In particular, if r is a simple closed contour, fJ=O.

Proof

JJ = J:

J:

(f o y)y' =

(g'

y)y'

o

=

J:

(goy)'

g(y(l)) - g(y(O)).

=

To illustrate CI 6 suppose first that r is an arc with first point z0 and last point z 1 • Then for n = 0, I, 2, ... , we have

Jrr zn dz

=

n

1

+

1

{zn + 1 - z 1

n + 1}

o

•

If r is a contour and p is any polynomial function then we have

Moreover, if the contour

r is such that 0 ¢ r we have

Jrr zn~1 dz On the other hand if we take

r

=

1

Let fn

E ~(I)

(n

E

lim In = f

~(T)

E

P).

O

w)w'

= 2TTi

rJo 1 = 2TTi. 1

P) and let

n-o oo

ThenjE

(n

to be C(O, 1) then we obtain

rJc !z dz = Jof (j CI 7.

0

(uniformly on I}.

and

Jrf! =

lim

n-ooo

fIn· r

Proof It follows from Proposition 1.14 that f E 0 there is N E P such that

sup lfn - fl ~ e r

(n > N).

5.6

Arcs, Contours, and Integration

129

Using CI 1, 2 and 5 we now obtain

L1• - J/ - L

(f. -f)

~ lr sup

r

~

lr

ffn - fl

£

(n > N).

It is now clear that

Jrf I=

lim

n-+ oo

JrrIn·

Remark. CI 7 asserts that f ~ frf is a continuous mapping from the metric space CC( I) to C.

PROBLEMS 5

34. Evaluate the following integrals.

f.

[O,a]

Re z dz,

f.

J

lm z dz,

C(O,l)

zn dz.

C(O,l)

35. Show that the function z ~ z log (z) has a primitive on C "\_Nand evaluate

f.

z log (z) dz.

[0 ,i]

36. Let/Ed(D) be such that/' E rt'(D) andf(D) c C "\_ N. Show that

J/'tf= 0 for any simple closed contour

rc

D.

37. Letfbe a power series function on LJ(a, p). Show that

for every simple closed contour

r

c LJ(a, p).

38. Let r be a simple closed contour starred with respect to a, and let {rn} c (0, 1) be such that rn+l ~ rn (n E P) and lim rn = 1. For each n-+ oo n E P let Yn(t) = rn y(t)

+ (1

- rn)a

(t E [0, 1]).

5.6

Elements of Complex Analysis

130

Show that each Yn represents a starlike simple closed contour, K = T* "'- int ( T 1 ) show that

Jrf I=

lim n-.oo

f I Jrn

rn say. If

{fE ~(K)).

39. Show that . IIm

R-.

+ 00

J. C(O.R)

z dz

z

3

+

I

= 0'

lim R-.+oo

J. [-R,-R+t)

z2 exp (z) dz Z + I

= 0.

40. Develop an elementary integration theory for arcs and contours that consist of unions of line segments and arcs of circles. Such a theory is in fact adequate for almost all complex function theory; but the ideas of this chapter provide an easy introduction to certain topics on the topology of the complex plane and also a useful application of the RiemannStieltjes integral.

6 CAUCHY'S THEOREM FOR STARLIKE DOMAINS

Cauchy's theoren1 (in son1e fonn) is the n1ost in1portant single theoren1 in con1plex analysis. Ahnost every deep result that we shall prove depends in one way or another on Cauchy's theoren1. Roughly speaking the theore1n states that iff is analytic on a do1nain containing a sin1ple closed contour rand its Jordan interior then

fJ=O. No atte1npt is n1ade in the present book to prove the n1ost general form of the theoren1. We content ourselves by proving a version of the theoren1 sufficient for most applications. The version that we consider has the merit of ad1nitting a straightforward proof, and also illustrating the spirit of more general proofs. Geon1etrical language is occasionally used in this chapter to facilitate understanding, but the student who so wishes may express everything in this chapter in purely analytical language.

6.1

Cauchy's theorem for triangular contours

In this section we shall prove Cauchy's theorem for the case in which the sin1ple closed contour is triangular, i.e. as in Example 5.5. Tllis preliminary case adn1its an elegant proof that is due to Pringshein1. The proof gives remarkably deep penetration into what is involved in any version of the theorem, namely (a) the completeness of C (b) CI 1-6 (c) the topological nature of the given sin1ple closed contour in relation to the domain in which the given function is analytic. By a triangle in the con1plex plane we mean the convex hull (see Problem 2.14) of three distinct points that do not lie on a straight line. Thus 131

6.1

Elements of Complex Analysis

132

given cx1, cx 2, cxa E C, not all on the same straight line, the associated triangle is given by

T = {"lcxl + "2«2 + "acxa: "h "2' "a ~ 0, "1 + "2 + "a = 1}. It is straightforward to verify the obvious fact that. b(T) = [cxh cx 2] U [cx 2, «a] U [cxa, ext].

We have already seen (Example 5.5) that b(T) is a simple closed contour; we shall denote it by F(T). Given cx 1, cx 2, «a (in that order) we agree to give T(T) the direction determined by the subarc [cx1, cx 2]. Thus if /E ~(F(T)) then

For the purposes of this section it is irrelevant whether F(T) has the positive or negative direction since of course

Jr 1 = o

r(T)

J

-r(T)

1 = o.

Theorem 6.1. Given a triangle Tin a domain D,

f f

JnT>

{fE d(D)).

= 0

Proof. Let T have vertices cxh cx 2, cxa (in that order) and sides of length a, b, c. Given f E d(D) let

A=

r

JnT>

1.

We split T into four smaller triangles as follows. Let

f3t = !(cx2

+ «a),

f12 = !(cxa

+ «t),

f3a = !(ext

+ cx2)

and let Ti (r = 1, 2, 3, 4) be the triangles with vertices (cxh f3a, {3 2), (cx2, f3h f3a), (cxa, {32, f3t), {f3h f32, f3a) respectively (see Figure 6.1).

Figure 6.1

Cauchy's Theorenz for Starlike Domains

6.1

133

By CI 4 we have

By CI 3 we have

r t+rj j

t=o [Dk.DJl

UJJ.Dkl

(j, k = 1, 2, 3; j :;6 k).

It follows that

Jr t=

r

r(T)

±Jr

= 1 r(Tr)

f.

There is some r in {1, 2, 3, 4} such that

A~

4

r

Jrcrr>

1

for otherwise we should have the contradiction

r

Jrcr>

1

f .

Since Tk+ 1 c Tk and Tk+ 1 has sides length 2-a, 2-b, 2-c, it follows from the principle of induction that there is a sequence of triangles {Tn} such that (i) Tn+1 c Tn (n E P), (ii) Tn has sides length 2- na, 2- nb, 2- nc (n (iii) A

~

4n

J

r

f

(n

E

P).

E

P),

6.1

Elements of Complex Analysis

134

It is readily verified that diam (Tn)

~

2-n(a + b + c)

(n EP)

so that lim diam (Tn) = 0. By Proposition 2.3, n{Tn: n E P} consists of a n-+ oo

single point, a say, and then a E T c D. Since Dis open there is p < 0 with Ll(a, p) c D. Since/is differentiable at a there is 7J: Ll(a, p) ~ C such that

/(z) =/(a)

+ (z

- a){f'(a)

where 7J is continuous at a with 7J(a) Ll'(a, p) and

+ 7J(z)} (z E Ll(a, p)) = 0. Since f is continuous

7J(Z) = f(z) - f(a) -/'(a) z-a

on

(z E Ll'(a, p))

it follows that 7J is continuous on Ll(a, p). Given 0 < 8 ~ p and

E

> 0 there is 8 such that

(z E Ll(a, S)).

Appealing to Proposition 2.3 again we may choose m E P such that Tm c Ll(a, S). Using CI 2 and 6 we obtain

r

Jn~

r {f(a) + (z - a)f'(a)} dz = r (z - a)7J(Z) dz. JnTm>

j(z) dz =

Jn~

+

r

Jn~

(z - a)7J(Z) dz

Finally by CI 5 we have 0

~A ~

r

r

I ~ 4m (z - a)l](Z) dz JnTm> JnTm> ~ 4mlnTm> sup {[z - a[[17(z)[: z E F{Tm)}

~

4m

4m2-m(a + b + c)2-m(a + b + c)E

= (a + b + c) 2 E. Since E was an arbitrary positive number we conclude that A = 0 and so the proof is complete. It is possible to use a combinatorial argument to show that Cauchy's theorem holds for polygonal contours r such that F* c D (although the technical details are by no means simple). Given an arbitrary simple closed contour r with T* c D (note the tacit assumption of the Jordan curve

6.2

Cauchy's Theore111 for Starlike Donwins

I35

thcoren1) and given e > 0, it is possible to show that there is a polygonal contour T 1 such that Tt c D and

f f- f I r1

r

0 it follows from the uniform continuity off on K that there is sl > 0 such that lf(z) - f(w)l
0 we have

L"' e-x• cos (2bx) dx = e-•• L"' e-x• dx, L"' e-x• sin (2bx) dx = e-•• J: ex• dx. Proof. It is well known from real analysis that each of the above integrals converges absolutely. Given n E P let be the boundary of the rectangle with vertices 0, n, n + ib, ib with direction determined by the subarc [0, n]

rn

(see Figure 6.4.).

.---------...1

mib

ib

0

rn•

n

Figure 6.4

Let f(z) = exp (- z 2 ) (z E C). Since f is an entire function Theorem 6.3 gtves (n

By CI 4 we have

J. n e-x2 dx + 0

f.

[n, n + tb)

f +

E

J.o e-.z)} dz

h'(>.) =

0

=

Therefore frf

{zf(.\z)} dz

().. E (0, 1)).

= h(l) = h(O) = 0.

4. Let D = {z: Re z > 0} and let/E ~(D-) be analytic on D. If there is M > 0 such that

lf(z)l ~

J

show that ~oo f(iy) dy

M lZf2

(zEn- 11 V(O, 1))

= 0.

S. Let 8b 82 be such that -1r < 81 < 0 < 82 < TT, and let the domain D = {z: 81 < arg (z) < 82 }. Let /E

(w - z)

n

dw

(z

E

12 E

P

L1(a, r)).

Then hE d(L1(a, r)) and

, l

lz (z) =

C(a,r)

(W

ng(w) )n+ 1 dw -

(z

E

L1(a, r)).

Z

Proof Let z0 E L1(a, r). Choose p < r such that z0 z 1 E L1(a, p). Given wE C(a, r) we have

1

1

(w - z 1)n

(w - z 0)n

E

L1(a, p) and let

(w - z 0 )n - (w - z 1 )n (w - z 0 )n(w - Z1)n

(z1 - Zo) {w- zo)n-1 (w - z 0 )n(w - z 1)n

+ (w-

zo)n-2(w- z1)

+ (lV z ~ r 1

+ ...

-

z 1) n - 1}

- z 0 1 n(2r)n-1.

(r _ p)2n

Since g is continuous on the compact set C(a, r), it is bounded, say lg[ ~ M. It follows from CI 5 that

[lz(z1) - lz(zo)[

=

l {( g~v) C(a,r)

lV

)" - ( g~w) )"} dw Z1 lV Zo

Mn(2r)n- 1 ~ 27Tr (r - p ) 2 n [zl - zol· This holds in particular for z1 E L1(z 0 , p - [z 0 I) and so it is clear that h is continuous at z0 •

7.1

Elements of Complex Analysis

150 We now have

i

h(z1) - h(zo) = Z1 -

f

J= 1 Jcca,r>

Zo

n~~~{

(w -

Zo)

.

(w -

1 dw.

Z1)

Applying the above continuity result to each integral we see that h is differentiable at z 0 and

Ln

I

h(z0 )=

i

1=1

C(a,r>

- f -

( _g(w))n+ 1 dw W

Zo

d

ng(w)

Jcca,r>

(w -

Zo)n + 1

W.

Since z 0 was any point of Ll(a, r), the proof is now complete. Theorem 7.3 (Cauchy). Let f E d(Lt(a, R)). Then f is infinitely differentiable on Ll(a, R), and for n E P, 0 < r < R, f (Z ) Proof Let z

E

=~ 2 . 7Tl

i

d (W f(w) )n+1 W - Z

C(a,r>

(z E Ll(a, r)).

Ll(a, r), where 0 < r < R. By Theorem 7.1 f(z) = ~ 27Tl

f

f(w) dw.

JC(a,r) W

-

Z

Sincefis continuous on C(a, r) it follows from Lemma 7.2 that I f (z) = -2 . 7Tl 1

J. C(a.r>

(

f(w) W -

Z

) 2 dw.

We now proceed by induction. Suppose that f has k derivatives on Ll(a, r) and that

k! f (z) = -2 . 7Tl

J. C(a,r)

(W

d )k +1 w Z

f(w) -

(z ELl(a, r)).

It follows from Lemma 7.2 thatj is analytic on Ll(a, r) and f () z

= (k + I)! 27Tl.

J. C(a,r)

(W

f(w) -

z

)k+2

d

w

(z ELl(a, r)).

The result of the theorem now follows by induction. The above theorem marks the beginning of an increasing divergence between the results of real and complex analysis. Recall that a real differentiable function on (- I, I) need not be twice differentiable at any point

Local Analysis

7.1

151

of ( -1, 1). In fact let g be a continuous real function on ( -1, 1) that is nowhere differentiable,* and let

=

f(x)

r

(xE(-1, 1)).

g(t) dt

By the fundatnental theoren1 of calculus f is differentiable on ( -1, 1) and f' = g, so that f is nowhere twice differentiable. This situation cannot occur for functions that are analytic on a disc. In other words, the condition that a function be analytic on a disc is n1uch n1ore restrictive than n1ight be indicated by experience with real analysis. PROBLEMS 7

1. Let f

E

1) at a and let g(z) (z - a)mf(z) (z E Ll(a, R)). Then

=

k'(a) # 0, then h(a) Res (f; a) = k'(a)·

Res (f· a) = '

1 (m - 1)!

g(m- l)(a)

·

Proof. (i) We have 00

(z - a)f(z)

=

(z - a)

2

an(Z - a)n

n= -1

and therefore Res (f; a) = a_ 1 = lim (z - a)f(z). Now letfbe of the form

i

as indicated. Then

.

h(z)

Res {f: a) = !1~ (k(z) - k(a))/(z - a) h(a) = k'(a)· (ii) If a is a pole of order m then f is of the fonn 00

2

f(z) =

an(z - a)n

n= -m

and so 00

g(z) =

2 n=O

an-m(Z - a)n.

58

3.1

Elements of Con1plex Analysis

The above result indicates that continuous functions on subsets of ceo can be effectively described in terms of the usual Jnetric on C. The result will also serve as a motivation for subsequent definitions involving the point at infinity. Note in passing that a continuous c01nplex valued function on ceo is auto1natically bounded by Proposition 1.20. A rational function is a function q: C ~ceo of the form q = PdP2 where p 1 , p 2 arc polyno1nial functions such that {z: p 1 (z)

= 0} n {z: p 2 (z) = 0} =

If p 3 , p 4 , are other polynomial functions such that q P1P4 = P2P3 and so deg (pi)

+ deg (p4) =

0.

= p 3 /p 4 then we have

deg (P1P4) = dcg (p2p3) = deg (p2) + deg (p3).

Therefore

and so we may now define the degree of q by

whenever q

=

pdp 2 • We now extend q to

ceo as follows: if

n

Pl(z)

=

_2: CXrZr, (an "/= 0), r=O

we define q(oo)

m

P2(z) =

2

fJrzr,

(fJm =1- 0),

r=O

= ;~. (This is again independent of the choice of poly-

nomials p 1 , p 2 with q = pdp 2 .) In particular q(oo) = 0 if dcg (q) < 0 and q(oo) = if deg (q) > 0. Since j is continuous on C "'- {0} the function l/p 2 is a continuous complex function whenever p 2 (z) =1- 0 and hence q is continuous at those points of C where p 2 (z) =1- 0. If piz) = 0, then p 1 (z) =1- 0 and we easily show that

Finally it is easy to sec that q is continuous at oo. We have thus shown that every rational function is a continuous function fron1 ceo into ceo. We conclude this section by discussing the continuity of the tnodulus and argument functions. The continuity of the argu1ncnt functions will be of particular significance for our discussion of logarithmic functions in the next chapter.

7.4

Local Analysis

169

frotn which the coefficients am tnay be detennined inductively. In particular al = -1, a 2 = 112. It is not at all obvious frotn ( *) that a 2 n + 1 = 0 (n E P). To see this we use a sitnple trick. Since (z

E

L1'(0, 21r))

we have ...

C()

----,------- n=O 2 an( -zY

(z

exp ( -z) - 1

E

L1'(0, 21r))

and therefore C()

- z =

2

2a2n + 1 z2n + 1

(z

E

L1'(0, 21r)).

n=O

The uniqueness of the Taylor expansion now gives a 1 = -t and a 2 n + 1 = 0 (n E P). Let Bn = n!an (n = 0, 1, 2, ... ). It is clear frotn (*) that each Bn is a rational nutnber. The nutnbers Bn are called the Bernouilli nunzbers* and they n1ake several interesting appearances in analysis (see, for example, Problen1s 7.25, 7.26, 8.47). It is easily verified that the Bernouilli numbers satisfy the following equations.

z

exp(z)- 1

= ~ Bn zn n=o

nil ('~)Bj = 0

J=O

n!

(z E L1'(0, 21r))

(n = 2, 3, ... ).

}

To con1plete this section it remains to study the case of an isolated essential singularity. We already know that a function cannot be bounded on any neighbourhood of an isolated essential singularity, and neither can the function converge to oo as z approaches the singularity. In fact the behaviour of the function near the singularity is rather surprising, as we shall see below.

Theorem 7.14 (Casorati-Weierstrass). Let f E d(.d'(a, R)) have an isolated essential singularity at a. Thenf(L1'(a, o)) is dense in Cfor 0 < o ~ R, i.e. given Z E C, € > 0, there is WE LJ'(a, o) such tlzat lf(w) - zl < €. Proof Suppose the result is false. Then there is 0 < Eo > 0 such that (wE .d'(a, o0 )). lf(w) - zol ~ Eo

* The student should

o

0

~

R, z 0

E

C,

note that some authors define the numbers ( -t)n + 1 (2n) !a 2 n as the 'Bernouilli numbers'.

170

7.4

Elenzents of Conzplex Analysis

We tnay therefore define g on Ll'(a, S0 ) by I

g(w)

=

(wE Ll'(a, S0 )).

f(w) - z 0

Evidently g is analytic and bounded by 1/Eo on ..d'(a, S0 ). It follows from Proposition 7.9 that g has a unique extension to a function g that is analytic on Ll(a, S0 ). If g(a) =I= 0, then f- z 0 1 has a removable singularity at a and sof has a removable singularity at a. If g(a) = 0, it follows from Theorem 7.11 (ii) that f- z0 1 has a pole at a and so f has a pole at a. Both these possibilities lead to a contradiction and therefore the theorem is established. Corollary. Iff E d(v(O, R)) has an isolated essential singularity at oo, then f(\1(0, S)) is dense in C for S ~ R. Proof Let S ~ R. Since fo j has an isolated essential singularity at 0 it follows from the theorem above that

Picard established a n1uch stronger result than the above theorem, namely that eachf(Ll'(a, S)) is either the whole complex plane or the whole complex plane except for one point. We shall not attempt to prove Picard's theorem in this book, but we give an illustrative exan1ple below. Picard's theorem is a 1nost striking result and its original publication stimulated much research in complex analysis. Observe that the Casorati-Weierstrass theorem may be expressed as follows. Given {3 E C there is a sequence {zn} in Ll'(a, S) such that lim

n-.oo

Zn

=

a,

This gives son1e indication of the nature of the discontinuity at a. Example 7.15. Iff(z) = exp

G) (z

E

C "'-. {0}) thenf(LI'(O, 3))

=

C "'-. {0}

for each S > 0. Proof Clearly f has an isolated essential singularity at 0, and f never takes the value 0. Let S > 0 and let {3 E C "- {0}. Given r > 0 we have

r

f(refO) = exp ( l c-to ) =

er1

cos 0

e--;1

sino.

7.4

Local Analysis

171

\Ve wish to cho osc r, 0 such that 1

e;-

cos

o

= LBI '

- !r sin 0 =arg (f3)

(n1od 21r).

Let

rn = {(arg (f3)

+

2111r) 2

+

(log

lf31) 2 } -l/2

and then choose On so that rn, On satisfy the equations above. We then have f(rn ewn) = f3 (11 E P) and if 11 is suft1ciently large rn eton E L1'(0, S). This con1pletes the proof. The student tnay care to visualize the surface in R 3 given by

PROBLEMS 7

18. Ifjhas an isolated singularity at a with Laurent coefficients an, show that li1n n-+ co

Ia_ n [ 11 = n

0.

19. Iff has a pole of order n at oo show that S(f; oo) is a polynomial function of degree

11.

20. Letfhave a pole of order 111 at a and letp be a polynomial function of degree n. Show that p of has a pole of order 11111 at a. 21. Show that an isolated singularity off cannot be a pole of exp of Hence or otherwise show that an isolated singularity is re1novable if and only if Ref or Imf is bounded above or below on so1ne neighbourhood of the singularity.

22. Let the functions below be defined on L1'(0, 1). Show that each function has a pole at 0, and detennine the order and residue in each case.

. z2 + 1 (I) f(z) = z2(z4 + 1) (ii) f(z) = z- 11. Then lin1 f3n = a and lim f(,Bn) = oo. This shows that f

n-oo

n-.oo

cannot be differentiable at a, so that a is a singularity off Then a is a nonisolated singularity off This contradicts the hypothesis thatf E vii(D) and so P 1 has no cluster points in D. (ii), (iii) These parts follow exactly as in Theorem 8.3. Recall that whenever we wish to assign values of a function at its poles we adopt the convention

f(a) = 00 (a E P1). This n1eans that meromorphic functions are continuous as functions from DtoC We are already familiar with one class of n1eromorphic functions. If D is any domain in C or if D = coo then each rational function belongs to vfi(D). This is clear from the fundamental theorem of algebra. Indeed if q is a rational function then there exist k E C, ah ... , am E C (not necessarily distinct), and ,8 1 , . . . , f3n E C (not necessarily distinct) such that 00 •

q(z) = k (z - a 1) . . . (z - am) (z - {31) ... (z - ,Bn) where {a 1 , . . . , am} n {f3b ... , ,Bn} = 0. It is a well-known fact of algebra that the set of rational functions forms a field. The san1e is true of the tneromorphic functions, but care is needed in defining the sum and product of two meromorphic functions. Letf E Jlt(D) and let a ED. By arguing as in Theorem 8.1 we easily see that there is 111 E Z and f 1 E JII(D) such that

f(z)

=

(z - a)"'f1 (z)

(zED)

where f 1 is regular at a with f 1 (a) =P 0. Similarly, if g n E Z and g 1 E ./K( D) such that

(zED)

E

Jlt(D) there is

8.4

Global Analysis

187

where g 1 is regular at ex with g 1 (ex) =P 0. It fo1lows readily that we tnay now define f + g and fg by (f

+ g)(ex)

lim (.f(z)

=

(fg)(ex) =

+ g(z))

lirn (f(z)g(z)). z-+a

If ex ¢= P1 u Pg the above definitions agree with the usual pointwise definitions. It is now routine (though tedious) to verify that .A(D) is a con1plex linear algebra. Theorem 8. 12. ult(D) is a field.

Proof In view of the above remarks it will be sufficient to show that fE ~/lt(D) itnplies 1/.(E "'II( D). GivenfE . Jt(D) it is clear that 1/fis analytic on D ""- {P1 u Z 1}. By Theorcn1 7.11 (i) 1/.(is differentiable at each point of P 1 (and has a zero there). By Theoren1 7.11 (ii) 1// has a pole at each point of Z 1 . Thus the only singularities of I/.( are poles and so 1/.f E ult( D) as required. The student n1ay recall that the field of rational functions is the field of quotients obtained frorn the integral dornain of polynotnial functions. This is si1nply to say that the rational functions are of the fonn p 1 /p 2 where pr, p 2 are polynotnial functions. (We have of course the convention that P1/P2 = P31P4 if P1P4 = P2P3; we tnay thus suppose, as in our original definition of rational function, that pr, p 2 have no con1n1on zero.) If D is any dotnain in C it is clear that {fig: f, g

E

S'f(D)} c .A( D).

In fact the above sets coincide so that Jlt(D) is the field of quotients obtained fro1n the integral domain d(D). We shall prove this result in §6 for the case D = C. The result fails spectacularly for the case D = coo. Indeed we have seen fron1 Liouville's theorem that d(C 00 ) consists of the constant functions. Tllis 1neans that d(C 00 ) is its own field of quotients. On the other hand we have already remarked that ..4't(C 00 ) contains all the rational functions. We show below that uh'(C 00 ) actually consists of the rational functions. We need first a useful technicalletnma. Recall that iff: D -+ C has an isolated singularity at ex, then the principal part off at ex is denoted by S{f; ex). n

Lemma 8.13. Let f

E

is regular at each ex 1 (j

A( D) and let exr, ... , exn

= 1, ... , n).

E

P1 . Then f -

L

S(f; ex1)

f=l

Proof Recall from Chapter 7 that each S(.f; ex1) is analytic on C ""- {ex 1}, and f- S(f; a 1) is regular at ex 1• This proves the ,result if n = 1. If n > 1

Elen1ents of Con1plex Analysis

188

8.4

and / 1 = f - S(f; a 1) then / 1 is regular at a 1 • Since S(f; a 1 ) is regular at a 2 it follows that S(/1 ; a 2 ) = S(f; a 2) and so

Since / 1 - S(/1 ; a 2) is regular at a 1 and a 2 this proves the result if n = 2. We continue in the fashion and the proof is compl~ted in a finite number of steps. Observe in passing that if P, = {ab ... ' an} then there is g E .PI( D) such that n

f

=

g

+ 2:

S(f; a 1).

1=1

This remark will be generalized in §9. Theorem 8.14. vlt(C 00 ) consists of the rational functions.

Proof It is clearly sufficient to show that each/ in .Af(C 00 ) is a rational function. Either f is regular at oo or f has a pole at oo. In either case there is R > 0 such that f is analytic on \7(0, R). By Theorem 8.11 (ii) f has a finite number of poles in the compact set Lf(O, R), say a 1 , . . • , an· Let g

= f-

n

L S(f; a

1),

so that g is an entire function by Lemma 8.13. More-

J=1

over g - S(g; oo) is regular at oo. Since S(g; oo) is a polynomial function we have g - S(g; oo) E s1(C 00 ) and so g - S(g; oo) is constant by the corollary to Liouville's theoren1. It follows that f is a rational function as required. It is convenient at this point to remark on an obvious analogue of Theoretn 8.10 for merom orphic functions on C. Suppose that f E vK(C) with Z 1 = {ab ... , am}, P1 = {{31, ... , f3n} where a1 is a zero of order J-LJ and {31 is a pole of order vi. Then there is an entire function g such that

(z

E

C).

The proof proceeds by the technique of Theorem 8.10. This result also will be generalized in §6. Some of the results of this section (e.g. Lemma 8.13) apply equally well to functions all of whose singularities arc isolated. We shall concentrate our attention in the text on the more interesting class of meromorphic functions and relegate such extensions to the problems.

8.5

Global Analysis

189

PROBLEMS 8 12. (i) If f(z)

=

sin

G)

cosec

G)

(z E C) determine Z 1, P1 and their sets

of cluster points. (ii) If/(z)

=

cosec

(t ~ z)

(z E C) show that

f

is meromorphic on

Ll(O, 1) but not on C.

13. Letf: D--+ C have isolated singularities at each point of E let a be a cluster point of E in D.

c

D and

(i) If each point of E is a ren1ovablc singularity show that f tnay be differentiable at a. (ii) If each point of E is an essential singularity show that f cannot be differentiable at a.

14. Let .f(D) denote the set of functions f: D--+ C all of whose singularities are non-retnovable isolated singularities. (i) State and prove the analogue of Theoretn 8.11 for J(D). (ii) Show that .f(D) is not a field. (iii) State and prove the analogue of Lemma 8.13 for J(D). (iv) Describe a general element of f(C 00 ) .

15. Let f be a rational function whose zeros and poles are all of even order. Show that there is a rational function g such that f = g 2 •

16. (i) Letfbe a rational function whose zeros and poles are all simple. 1 1

Suppose the zeros are ab ... , an and the poles are-, ... , - , where al

an

la 1l #

1 (j = 1, ... , n). Show that If[ is constant on C(O, 1). (ii) Given any rational function f with no poles on C(O, 1), show that there is a rational function g with no poles on Ll(O, I) such that jgj = Ill on C(O, 1).

8.5 Convergence in d(D) In this section we shall construct a metric on the set d(D). It turns out to be natural to define the metric on the larger set ~(D) and then induce the metric on d(D). We show that convergence in this metric is equivalent to uniform convergence on the compact subsets of D. We show further that the metric space ~(D) is complete. It then follows via a converse of Cauchy's theorem that d(D) is a closed subset of the metric space ~(D). This means that if fn E d(D) (n E P) and if fn --+ f uniformly on each coolpact subset of D then f E d(D). Moreover for each k E P we even have

Elen1en1s of Co1nplex Analysis

190

8.5

f~k>

-)- j unifonnly on each con1pact subset of D. This result is cxtren1ely itnportant in con1plcx analysis; two sin1plc applications arc given in this section and further significant applications appear in §6 and §9. At the end of this section we discuss (without proof) some of the deeper topological properties of the 1netric space d(D). Let K be any cotnpact subset of C. If

d(f, g) = sup {lf(z) - g(z)l: z E K} ·(J, g E rt'(K)) recall that dis a metric (the usual metric) on ri(K). The compactness of K ensures that the functions in f(?(K) are bounded and hence dis finite valued. Now let D be any domain in C. The functions in Cff(D) are not all bounded (proof?) and a little care is needed to construct a useful metric on rc(D). Let {Kn} be a fixed con1pact exhaustion of D (cf. Proposition 2.7). Thus each Kn is compact, D = U{Kn: n E P} and given any compact subset K of D there is n E P such that K c Kn. Let dn denote the usual metric on ri(Kn). Given f, g E ri(D), n E P let dn(J, g) (J, ) an ' g = 1 + dn(f, g) where, for simplicity of notation, we do not distinguish f, g from f!Kn' gjKn respectively. Observe that 0 ~ an(J, g) ~ 1 (n E P) so that the series 2: 2 -nan(f, g) converges by the comparison test. We may thus define a on rt'(D) X r!(D) by 00

a(j, g) =

2:

2- nan(f, g).

n=l

Proposition 8.15. (CC(D), a) is a 1ne1ric space.

Proof Given f, g E t(}(D) it follows easily fron1 the definition of a that 0 ~ a(f, g) < + oo, a(f, g) = a(g,f) and a(!, f) = 0. If a(f, g) = 0 then an(f, g) = 0 (n E P), and so f(z) = g(z) (z E Kn, 11 E P). Since D = U{Kn: n E P} it follows that f = g. Finally, suppose f, g, lz E f(?( D). Since dn is a metric on r!( Kn) we have Given r, s, t ~

dn(J, g) 0 with I

~ ~

dn(f, h) + dn(h, g) (n E P). r + s it is easy to check that

I ----

I+t

~

r

I+r

+

s

I+s

.

It follows that and hence Therefore a

a(f, g) ~ a(f, h) + a(h, g). is a metric on CC(D).

8.5

Global Analysis

191

Corollary. s:/(D) is a 1netric space with the relative 111etric a. The above 1netric a is rather con1plicated but convergence in (rc(D), a) has a siinple characterization. Recall that convergence in (??(K), d) is equivalent to unifonu convergence on K. Proposition 8.16. Let fn staten~ents are equivalent. (i) lin1 a(J:Hf) = 0.

E

rt'(D) (n

E

P), f

(ii) For each con1pact subset K of D, fn

~f

E

qf(D). Then the following

un((onnly on K.

Proof (i) => (ii). Let a(/11 , / ) ~ 0 and let K be any con1pact subset of D. Choose n1 E P such that K c Km. Since a(J,fn) ~ 0 it follows that am(J,fn) ~ 0 and hence dm(J,fn) ~ 0. Since K c Km we conclude that fn ~ f unifonuly on K. (ii) => (i). Letfn---+ funifonnly on each con1pact subset of D. Since each Km is con1pact we deduce in particular that dm(J,fn)---+ 0 and so am(f,fn)---+ 0 for each 111 E P. Let E > 0. Since am ~ 1 (111 E P) we n1ay choose 111 0 E P such that 00

L

(11 E P). 2 -mam(j,Jn) < !E m=mo+ 1 Since am(f,fn)---+ 0 for 111 = 1, 2, ... , 111 0 , we n1ay now choose that am(f,fn)


11 0 ,

11 0 E

P such

111 = 1, 2, ... , 111 0 ).

For n > n0 we now have mo

a(f,fn)
-

C and let {an} c D with lim an = a n-. co

E

D. If each an is a

singularity off show that a is a singularity off. Give an example in which f is differentiable at each an and at a.

13. Usc each of conditions (1), (2), (3) to prove Proposition 3.7. Which proof is most elegant?

8.5

Global Analysis

193

Theoren1 8.18 (Morera). Let fE f(}(D) and suppose that for each a E D there is Ll(a, R) c D such that .L.f = Ofor every silnple closed contour in Ll(a, R). Then f E d(D).

r

Proof. Given a ED and the corresponding Ll(a:, R) we have frf

= 0 for

every triangular contour in Ll(a:, R). It fol1ows fron1 the argutnent employed in Theoren1 6.2 thatfhas a pritnitive on Ll(a:, R), say g. Since g is analytic on Ll(a:, R) it fo11ows frotn Theoretn 7.3 that f = g' is analytic on Ll(a:, R). Since a was any point of D we conclude that/ E d(D).

s/(D) (n E P) be such that fn-+ f unifonnly on each colllpact subset of D. Then jE ._0/(D) and for each k E P, J~k>-+ J unifonnly on each co1npact subset of D. Theoren1 8.19. Let fn

E

Proof It follows as in Proposition 8.17 that/ E f(}(D). Given a ED there is R > 0 such that Ll(a:, R) c D. Let r be any sitnple closed contour in Ll(a:, R). Since fn is analytic on L1(a:, R), Cauchy's theoretn gives frfn = 0 (n E P). Since r is a con1pact subset of D we have fn-+ f uniforn1ly on r and then CI 7 gives

Jr!

=

litn n-+oo

Jrf In = o.

Therefore f E s/(D) by Theorem 8.18. Let k E P and let r = tR. For each z

f (7) -

=

k!

21Tl.

r( ) _ ~ Z - 2 . 1Tl

Jn

For each z

E

i i

C(a,R)

C(a, R)

E

Ll(a:, R) Theoren1 7.3 gives

(W

f(w) d ' )k + 1 l1 - Z

( lV

fn(w) d , )k +1 H • - z

Ll(a:, r) we now have

IJ(z) -

J~k>(z)l = ~

i

f(w) - ~~~~? dw 21Tl c (w - z)

(2)k+1

k! ~ 2- 21TR R

1T

sup

C(a,R)

If- fnl·

Since fn-+ f uniformly on C(a:, R) it follows that f~k>-+ J uniformly on J(a:, r). If K is any compact subset of D then {Ll(a:, r): a E K} is an open cover of Kand so has a finite subcover, say{Ll(a:1, r 1):j =I, ... , 111}. Since J~k>-+ J uniformly on each Ll(a:b r1), it follows easily that f~k>-+ j uniformly on K. Tllis cotnpletes the proof.

8.5

Ele1nents of Conzplex Analysis

194

Corollary. (s:I(D), a) is a conzplete 111etric space. be Cauchy in (.r:I(D), a). Then{./~} is Cauchy in (rt'(D), a) and so by Proposition 8.17 there is /E rc'(D) such that aU,fn) ~_.. 0. By Proposition 8.16 and the above theorem IE ._..o/(D). Since aU, In)--:;.. 0 the proof is cotnplete. ·

Proo_f

Let{/~}

The above theoretn will be a key tool in the next section. For the present we consider applications to the Riemann zeta function and the infinite Laplace transform. Recall that n-z = exp ( -z log n) so that ln-zl = n-x where x = Re z. It follows from the comparison test that the series n -z converges absolutely for Re z > I. The Rienzann zetafimction is defined by

L

(Rez>l). n=1

Example 8.20. The Rienzann zeta fimction is analytic on {z: Re z > 1}.

Proof Given n E P we define In on C by n

ln(z)

=

L J-z

(z E C).

j=1

It is then clear that each In is an entire function. Given S > 0 and z such that Re z ~ 1 + S, we have co

l~(z)

- ln(z)j

=

L: J-z

00

L: J-1-6

~

f=n+1

J=n+1

and so In ~ ~ uniforn1ly on {z: Re z ~ 1 + S}. Since S was arbitrary it follows easily that In~~ uniformly on every con1pact subset of the do1nain {z: Re z > 1}. By Theoren1 8.19 ~ is analytic on {z: Re z > 1}. Moreover the derivatives of ~ are given by

Y(z) = ~ ':> ~ (- I )k (log !z)k . n =1

(Rez > 1, k~E P) .

n~

We shall show in §10.4 how the Riemann zeta function n1ay be extended to a meromorphic function on C. Letlbe a cotnplex function defined on {t: t ~ 0} and letlbe Rien1ann integrable on any bounded subinterval of {t: t ~ 0}. We say that f is of exponential type with index .:\ if there is M > 0 such that

rl(t) I ~ MeAt

(t

~

0).

Given such I and Rc z > A, the infinite integral

8.5

Global Analysis

195

converges absolutely, i.e. J~ le-z~f(t)ldt converges as R sided) infinite Laplace tran~fonn off is defined by F(z)

=

J."' c-''f(t) dt

-->-

+oo. The (one-

(Re ;; > ,\).

The Laplace transfonn is a useful tool in the theory of differential equations and elsewhere.

Example 8.21. Iff is of exponential type with index ,\, the Laplace transfonn off is analytic on {z: Re z > ,\}. Proof For each n

E

P let (;; E

C).

By Exa~nple 7.5 each Fn is an entire function. Given Re z ~ ,\ + o we have

o>

0 and z such that

IF(z) - Fn(z)l

Therefore Fn -7 F uniformly on {z: Re z ~ ,\ + o} and it follows as in the last example that F is analytic on {z: Re z > ,\}. The student may observe that Theorem 8.19 leads to an alternative proof that a power series function is analytic on its open disc of convergence. In fact we need only recall from Proposition 4.8 that a power series converges uniformly on closed subdiscs of the disc of convergence. Since polynomial functions are entire, the result follows from Theoren1 8.19. The result may be expressed in different terms as follows. Given any open disc LJ the polynon1ial functions are dense in (d(LJ), a). This leads us to ask for which domains D the polynon1ial functions are dense in (d(D), a). A famous theoren1 of Runge asserts that the polynon1ial functions are dense in (d(D), a) provided that D is sinzply connected, i.e. CCX) "' D is connected. Recall from Problem 2.27 that every starlike domain is simply connected. Runge's theorem leads very easily to an extension of Cauchy's theorem to the case of a sin1ply connected d01nain D. In

Elenzents of Conzplex Analysis

I 96

8.5

fact givenfE.S'i(D) and any simple closed contour r c D we may choose a sequence of polynomial functions {Pn} that converge to f uniformly on r. It follows from CI 6, 7 that lim J Pn = 0. . Jf = n-+oo r

r

(A proof of Runge's theorem is given in Saks and Zygmund, Analytic Functions (second English edition) Warsaw 1965.) Finally we mention one further deep result concerning the metric space (.9/(D), a). If {fn} is a sequence in sf( D) that is uniformly bounded on each compact subset of D, then some subsequence of {fn} converges in (d(D), a). {Observe the analogy with the result that every bounded real sequence has a convergent subsequence.) This remarkable result is called Mantel's theorem. It is a key tool in the standard proofs of the Riemann mapping theorem (cf. page 243). PROBLEMS 8

17. (i) Let D be any domain inC and let H 00 (D) denote the set of bounded functions that are analytic on D. Show that H 00 (D) is a complex Banach algebra with the supremum norm (cf. Problem 2.6). (ii) Let D be any bounded don1ain in C and let An denote the set of functions in ~(D-) that are analytic on D. Show that Av is a complex Banach algebra with the supremun1 norn1. (A.1 is usually called the disc algebra. For further remarks on Hoo(LJ(O, 1)) and Aa see page 303.)

18. Let D = {z: -1 < Imz 0 with

fm jj(t)j

dt

~

(n-1, 11

M

E

P).

If

F(z) = _1. Joo 2rrl

_

J(t) .. dt

00

t -

(z

.L.

E

C"' R)

show that F is analytic on C "' R and that lin1 y-oO+

{F(x + iy) - F(x - iy)}

=

f(x)

(x

E

R).

Evaluate Fin the case in which f(t) = 1 1 2 + t

22. Recall that .Yf'(D) denotes the set of hannonic functions on D. Show that Yl'(D) is a closed subset of (~(D), a). 23. Show that

(1 - 21-z)l;;(z)

=

2oo ( - 1)nz + n==1

1

(Re z > 1).

1l

24. Let f be of exponential type with index ,\ and let F be the Laplace transforn1 off Sho\v that

£C•l(z) =

f' (-

t)• e -z'j(t) dt

(Re z > A).

Iff is such that f' is of exponential type with index A, find the Laplace transform off' in terms of F. 25. Find the Laplace transforms of the following functions. (i) f(t) = eat (t ;::: 0, a E C) (ii) J(t) = tn (t ;::: 0, 1l = 0, 1, 2, ... ) (iii) f(t) = sin (t + a) (t ;::: 0, a E R). 26. Show that the polynomial functions are not dense in (d(L1'(0, 1)), a). Generalize this result.

8.6 Weierstrass expansions In this section we derive the Weierstrass expansions of entire functions and n1eromorphic functions on C. These expansions are given in terms of infinite products in which the zeros and poles appear in the expected

Ele1nents of Con1plex Analysis

68

3.3

where ~ is continuous at a with ~(a) = 0. Let ~ = g + iYJ and z = (x, y). On taking real parts of the above equation we obtain

u(x, y) = u(a, b)

+ (x + (y

- a){Ref'(a) + .g(x, y)} - b){ -Itnf'(a) - IJ(X, y)}.

Since g, TJ are continuous at (a, b) with u is differentiable at (a, b) with

~~(a, b)=

Ref'(a),

g(a, b) = TJ(a, b) = 0, it follows that

~;(a, b) = -

Jmf'(a).

Similarly, by taking itnaginary parts we see that vis differentiable at (a, b) with

~;(a, b) =

In1j'(a),

~;(a, b) =

Ref'(a).

It follows that u and v satisfy the Cauchy-Riemann equations at a. Suppose conversely that u and v are differentiable at a and satisfy the Cauchy-Riemann equations at a. It follows that there exist real functions Er (r = 1, 2, 3, 4) in some neighbourhood of (a, b) such that

au (a, b) u(x, y) - u(a, b) = (x - a) { ox

+

E1

(x, y) }

+ (y v(x, y) - v(a, b)

=

(x -

a){~~ (a, b)+ n0 ). It follows that lf(z)[ ~ -![Hn 0 (z)l so thatf(z) =1- 0 as required. The infinite products to be considered below arc constructed from the Weierstrass prinzary factors. These latter are defined by

E 0 (z) = 1 - z

En(z)

=

z2 zn) (1 - z) exp ( z + 2 + · · · +

n

(n

We need first two sintple lemn1as. Lemma8.23.j£n(z)- II::::; 3jzjn+I (zELl(O,j-),nEP).

E

P).

200

8.6

Elements of Complex Analysis

Proof. Given z

E ~(0,

1- z

I) we have

= exp (log {I - z)) = exp - 2: z"' -) 00

(

"' •1

and thus

= exp -

En.(z)

(

n

zl) .. L: -. J ... n+l } 00

If lzl ~ -!then

It is elementary to verify that lexp (w)-

II

~

(we C)

lwl exp (lwl)

and it now follows that for z E .J{O, !) IEn.(z)-

II

~

lzln+l exp {!zl"'+ 1)

~

3lzl"'+ 1 •

Lemma 8.24. Given {P.n.} c P there exists {Pn.} c P such that converges. Proof Choose p. > lo! 2 (n

Therefore P.n(!)"" test.

~

L: P.n.(!)""

+ log p..) (n E P) and it follows that

log (p.n.)

+ Pn log(!)

(!)"' (n

P) and the result follows from the comparison

E

~

n log(!).

We are now ready for the key result of this section. Recall from Theorem 8.3 that the zeros of an entire function are at most countable. If there are an infinite number of zeros, say {an.}, we then have lim a"' = oo, since n-+ao

the set of zeros has no cluster point in C. We may thus assume that the zeros are indexed such that Ia"'+ 1 1 ~ Ian. I (n E P). We show now that any such sequence can be the set of zeros of an entire function. Theorem 8.25 (Weierstrass). Let {an} c C ""- {0} be such that (n

E

P)

and lim an = oo, and let {P.n} be any sequence of positive integers. Then n-+ oo

there is an entire function g such that Zg = {an}, where a"' is a zero of order ILn·

8.6

Global Analysis

201 Proof Let K be any con1pact subset of C. Since an~ oo we n1ay choose n 0 E P such that (n >

11 0 ,

z

E

K).

Let Pn be as in Len11na 8.24. We have by Len11na 8.23

1~-ziP"+ 3(l)P" an 2 2

(anz)

£--1~3Pn

Using Lemma 8.24 we now see that

L JLn

(n > n0 , z

EP.(:.) - I

E

K).

converges uni-

fonnly on K. Since K was any con1pact subset of C it follows fron1 Proposition 8.22 that if (z

E

C)

then g is an entire function with Z 9 = {an}· It is also clear fron1 Proposition 8.22 that an is a zero of order fLn· Ren-1ark. If we wish also to have a zero of order kat 0 we simply take g 1 (z) = zkg(z) (z E C).

Theorem 8.26 (Weierstrass). Iff is an entire function, there is an entire function h such that

f(z)

= zk exp (h(z))

.Q {£ •.(:.)

r

(z

E

C)

where k ~ 0, {an} = Z 1 "'- {0}, an being a zero of order Jl-n, and Pn is chosen as in Lenuna 8.24. Proof By Theorem 8.25 and the remark following, we may choose an entire function g such that Z 9 = Z 1 , each zero of g having the same order as the corresponding zero off It follows thatffg is an entire function with no zeros. By Theorem 8.9 there is an entire function h such that ffg = exp o h. With g as in Theorem 8.25 the result now follows.

Theorem 8.27 (Weierstrass). Iff E Jt(C) there exist entire functions g, h such that f

=

gjh.

Proof Since f

E ~(C)

it follows from Theorem 8.11 that the set P 1 is at most countable and has no cluster points in C. By Theorem 8.25 we may choose an entire function h such that Zh = P" each zero of h being of the same order as the corresponding pole off It follows that !if is an entire function, say g. Then/= gjh as required.

8.6

Ele1nents of Complex Analysis

202

Observe that the above entire function g has the property that Z 9 = Z 1, corresponding zeros being of the same order. Note also frotn Theorem 8.25 that each f E .AI(C) can be represented as the quotient of two infinite product expansions. It should be understood that the results of this section arc essentially existence theorcn1s. Given an entire function f with Z 1 = {an} we may readily obtain the Weierstrass expansion off nzodulo an entire function with no zeros. The determination of the exact expansion is generally quite difficult. One such example will be given later in Example 8.38. PROBLEMS 8 27. Given {an} c C "'- {0} let Pn = a 1 a 2

converges iff L log

an (n

•..

E

P). Show that {pn}

a" converges, in which case }~~ Pn = exp (.~, log a.)·

28. For which z E C do the following infinite products converge to a non-zero complex number?

(i)

fr (I + ~)

n=l

11

(ii)

n=

(

ro

n

I +

.. k

2)

(iii)

2

fr n sin (~). n

n=l

11

1

29. Let {un} c si(D) be such that 2 lunl converges uniformly on each compact subset of D and I + un(z) =I= 0 (z E D). If ro

n (I

+ lln(z))

(zED)

f'(z) = ~ un'(z) n=l l + un(z) f(z)

(=ED)

f(z) =

n=l

show that

the series converging uniformly on each compact subset of D. What =I= 0 ? modifications are required if

z,

30. Let {an}, {~n} c C with

lim

n-.ro

an = lim

n-.ro

~n = 00 and

am =f.

~n

z,

= {an}, (1n, 11 E P), and let {f.tn}, {vn} c P. Construct f E ..4/(C) with P 1 = {,Bn}, an being a zero of order fln, ~n being a pole of order vn. Comment on the case when lin1 (an - ~n) = 0.

31. Given 0 < r < I show that there is Mr > 0 with

(wE Lf(O, r)). Derive the analogues of Theorems 8.25, 26, 27 when C is replaced by Ll(O, 1). Can you generalize the argtuncnt to an arbitrary don1ain D of C?

8.7

Global Analysis

203

8. 7 Topological index In this section we define the topological index of a simple closed contour with respect to a point not on the contour. Roughly speaking the topological index specifics how often a curve winds round a point and in which direction. In fact the topological index leads to a very useful definition of the positive direction on a starlike sitnple closed contour. As such the topological index sitnplifies the statetncnt of several subsequent thcoretns in this chapter. We consider first a weakened concept of the pritnitive of a function analytic on a dotnain. Let denote, as usual, an oriented piecewise sn1ooth arc or a simple closed contour with representative y (y being suitably differentiable). Let f E s/(D) where c D. We say that g E 't:([O, 1]) is a prilnitive off along if for each T E [0, 1] there is a pritnitive F-r off in sotne neighbourhood of y( T) such that g(t) = Fly(t)) fort in son1e neighbourhood ofT. Iff has a prin1itive on D in the usual sense, i.e. if there is FE s/(D) with F' = J, thenfhas a pritnitive along In fact we sitnply take g = F o y. The concept of a pritnitive along r is thus a weaker concept than the usual concept of a pritnitive. We know that a given f E si(D) need not have a prin1itive on D (see Example 4.17 and Proposition 4.15). On the other hand f always has a prin1itive along any in D.

r

r

r

r.

r

Theorem 8.28. If fE d(D) and

r,

r

c D, then

(i) f has a prinzitive along g say; (ii) the set of all prinzitives off along (iii) rf = g(l) - g(O).

f

r is given by {g + cl : c E C};

Proof Note that the caseD = Cis trivial sincefthen has a primitive on D by Theorem 6.2. (i) Let E = dist (F, C"' D) so that E > 0 by Proposition 1.21. Since y is continuous on the con1pact set [0, 1] it is uniformly continuous on [0, 1] by Proposition 1.19. Hence there is > 0 such that

o

x, y Choose

11 E

E

[0, 1],1x -

P such that! < 11

Yl

< 8

=>

lr(x) - y(y)j
0 with

(x

E

R).

Since f is continuous on R it follows by the con1parison test that integrable on R and so

J

lin1

f(x) dx =

oo

R-+

_ 00

+ oo

JR

f

is

f(x) dx.

-R

Let T(R) be the sen1i-circular contour described above, with the positive direction. If all the poles off in the upper half plane lie inside T(R 0 ) then

l

f =

i

21Ti

r(R)

Res (/;

aJ

(R > R 0 ).

j=l

Since deg (f) ~ -2 there is K > 0 and R1 > R 0 with

lf(z)l ~

K

TZf2

If rl(R) denotes the arc {R efO: ()

E

[0, 7T]} it follows fronl CI 5 that

and so lim R-+

+ oo

l

f= 0.

Ft(R)

We now conclude that

J-oo~ f(x) dx = ""

lim R-++oo

l

r(R)

f

= 21Ti

±

Res (f; ai).

j=l

212

8.8

Elements of Complex Analysis

Example 8.33. Given a > 0 we have

J.

xsinx d

eo

2

x +a

o

2

x

1T

-a

= -2 e .

Proof. Let g

(z)

=

z exp (iz) z2 + a2

(z E C).

so that g e vlt(C) with Pg = {ia, - ia}. The pole ia in the upper half plane is simple {Theorem 7.11) and so by Proposition 7.12 we have R es (g: za 0

=

)

ia exp (i 2a) . 2 za

= z~ e -a .

Integrating g round the contour T(R) of the above proposition we obtain g [ Jr(R)

=

1Ti

e-a

(R > a).

For R > 2a we have

f

Jr1

~

g

= ~

~

f." fexp (iR e )iR e - a o f." e-R dO R2 R2- a2 o

R• R

18

2

18 l

dO

sinO

J. 6 J. 6

n/2

e- R

0

sin 8

d0

n/2

e -8 dO

o

37T(l -e -R) -- R so that

It follows that •

hm

R-++eo

JR -RX

X etx 2 2

+a

dx

= 1rie-a.

On taking imaginary parts we conclude that

l

eo X sin X 2 2 0 X

+a

d X -- ~Z I'lffi

JR X sin X 2 2 R-++eo -RX

+a

d X --

Further examples are given in the problems below.

1T -

2

e

-a

.

3.4

Continuous and Differentiable Conzplex Functions

73

Exaruplc 3.14. Let D = L1 '(0, 2) and let

u(x, y) Then u

E

log (x 2 + y 2)

=

£(D) and u has

110

((x, y) ED).

hannonic conjugate on D.

Proof It is routine to verify that u E .Ye(D). Suppose that u has an harnlonic conjugate v on D. Define g: [0, 27T] ---+ R by g(t) It follows that g g'(t)

= v(cos t, sin t)

E ~R([O,

(t

E

[0, 27T ]).

27T]) with g(27T) = g(O). Moreover

= :.~(cost, sin t)( -sin t) + :; (cost, sin t)(cos t).

Since u and v satisfy the Cauchy-Riexnann equations we have g'(t)

=-

~u(cost,sint)(-sint) + ux ~u(cost,sint)(cost) uy

2 sin 2 t 2 cos 2 t cos 2 t + sin 2 t + cos 2 t + sin 2 t = 2 (t E (0, 21r)).

It follows that g(t)

=

g(O)

+ 2t

(t

E

[0, 27T])

and so g(27T) =/=- g(O). This contradiction shows that u cannot have an hannonic conjugate on D. PROBLEMS 3

24. Given that f = u + iv E d(D) is infinitely differentiable on D show that u, v have partial derivatives of all orders on D. 25. Show that the following functions are harn1onic on C and find harmonic conjugates for each. (i) u(x, y) = x2 - y 2 + 2x. (ii) u(x, y) = ex cosy. (iii) u(x, y) = sin x cosh y. 26. Show that the Laplace equation corresponds to

o2f 8z 8z = 0 ·

27. Let f E d(D) with f(z) =/=- 0 (zED). Show that log assuming that f is suitably differentiable.

Ill

E

.Ye(D),

Elements of Con1plex Analysis

214

1 41. Let f(z) = -: exp (iz) (z

..

8.9

C). Integrate f round the contour gtven tn 0

E

Exan1ple 8.31. Let r -:... 0 +, R

l

+ oo and deduce that

--?

oo

sin x dx

o

•

=

7T.

2

X

Use a similar argutnent to show that

l

oo

0

sinx

x(x 2 + a 2 )

dx- _!!__(I- e-a) • 2a 2

(a > 0).

42. Establish the following results. (i)

l"' /og+X\ o

dx

= 0.

x d (n..) looo (1 log + x = -4. sinh ax d x = t an (l a) (111... ) looo SID . h 7T

x2)2

2

7TX

(-7T e and let h(x)

= axex

- 1

(x

E

[0, I]).

Then h E CC n([O, 1]), h(O) = - 1, h(I) = e - 1 > 0. By the intermediate value theoren1 there is t E (0, 1) such that h(t) = 0. This con1pletes the proof. Theoren1 8.44 (Hurwitz). Let {fn} c .s.I(D) converge to funifonnly on each conzpact subset of D, and let r be a starlike sinzp/e closed contour with T* c D. Iff has no zeros on r and k zeros on int (T) then, for son1e n 0 E P,fn has k zeros on int (T) (n > n 0 ), zeros being counted according to their 1nultiplicity.

Proof Observe from Theoren1 8.I9 that f E d(D). Since f has no zeros on the con1pact set r, it follows from Proposition I.20 that e = inf {1/(z)j:

Since In ~ f uniformly on

r

lfn(z) - f(z)l
0.

there is llo E

~

E

p such that

lf(z)j

(z E

r, 11

> llo).

The conditions of Rouche's theorem are now satisfied with g Therefore fn has k zeros on int (T) for n > n0 •

= fn - f

Corollary. Iff has no zeros on int ( T) neither has In for n > n0 • The above Hurwitz theoren1 gives a good deal of information about the zeros off in terms of the zeros of fn· A little more analysis provides even more detailed information. Theorem 8.45. Let {fn}, f be as in the above theorem with f :/= 0. (i) If an E z,n with lim an = a E D, then a E z,. n--+ oo

(ii) Given a E z, there is a, E z," such that ~im a, J

]-+00

= a.

Proof (i) (This part requires only that fn E CC(D).) Choose R > 0 such that ~(a, R) c D. Since an---? a there is m 0 E P with an E ~(a, R) (n > m 0 ). Let e > 0. Sincefn---? /uniformly on ~(a, R) there is n 0 > 1n 0 such that

lf(z) - fn(z)j < !-e

(z

E

Ll(a, R), n > n 0 ).

226

Elernents of Conzplex Analysis

8.10

l n particular (n > llo).

Since f is continuous at a there is n 1

E

P such that

For n > n1ax (n 0 , n 1 ) we now have

1/(a)l

=

[/(a) - fn(an)[

~ [J(a) - f(an)l


0 such that a is the only zero off on J(a, R). Suppose a is a zero of order k. Given 0 < r < R it follows from Hurwitz's theorem that there is n(r) E P such that fn has k zeros on .Ll(a, r) for n > n(r). In particular chooser; = R/} (j E P). For eachj E P we may choose some zero a; of some fn 1 with ai E Ll(a, r;). We then have ~im a; = a as required. J-+ OCI

The next result is a simple extension of the principle of the argument. Iff E v'II(D) has n zeros, counted according to multiplicity, on some compact subset of D, we shall here denote these zeros by a 1 , ••• , an. Thus if a is a zero of order k then a will appear k times in a 1 , .•• , an· A similar convention applies for poles.

Proposition 8.46. Let f

JII(D), g E d(D) and let r be a starlike sinrple closed contour, with positive direction, such that T* c D. Suppose f has no zeros and no poles on rand has zeros on int (T) at ab .. . , an and poles on int ( T) at f3h ... , f3m· Then

1 -2 . 7Tl

E

1 J' r

g f. =

:L n

i=l

g(aj) -

:L m

g(f3;).

i=l

Proof We employ the same method as that used for Theoren1 8.41. It is

only necessary to verify that

In general it is not possible to apply the above result directly to the case in which g is a logarithmic function. In the case when r is a circle such an application would lead formally to the Jensen formula (Theorem 8.48) which relates the size of log Ill on a circle to the zeros and poles inside the circle. We choose a ditrerent approach to derive the result, beginning with a simple le1nma.

8.10

227

Global Analysis

Lemma 8.47.

f" Proof Given

a E

II+

log

-

gives

=

.fc f

.ct(.1(0, !

) ). 1 1

and so Cauchy's theorem

0. Therefore

l

log(l ~ aew) iewde e

2.n

o

and so

.1(0, 1)).

+ az)

0 it follows thatfE =

E

.1(0, l) let

1 f(z) = -: log (I

Since log 1

(a

ac 10 1 dO= 0

l

2.7t

0

log

II + a ewl de =

=

0

0

as required.

Theorem 8.48 (Jensen). Let f E vf((Ll(O, R)) be regular at 0 with f(O) =1= 0. Given 0 < r < R let f have no zeros and no poles on C(O, r), and let the zeros off on Ll(O, r) be ab . . . , an, and the poles (3b ... , f3m· Then 1

2

1T

l2:t log lf(rew)l de 0

n

= log

r

lf(O)I + j~1 log ~

r

m

- j~ 1 log lf3il.

Proof Define F on Ll(O, R) by

f3m f3m

Z

f(z)

• J(O)•

Clearly FE .Af'(Ll(O, R)) and F has no zeros and no poles on Lf(O, r). Hence there is t such that r < t < R and F is analytic on Ll(O, t) and has no zeros on Ll(O, t). By Proposition 6.4 there is G E d(Ll(O, t)) such that exp (G(z)) = F(z)

(z E Ll(O, t)).

Since F(O) = 1 we have G(O) = 2k1ri for some k E Z. By subtracting a constant from G if necessary we may suppose that G(O) = 0. It follows that the function z--+ G(z) is analytic on Ll(O, t) and Cauchy's theorem then z g1ves

l

C(O,r)

G(z) dz = 0. Z

8.10

Elements of Complex Analysis

228 Thus

J.

2.n

G(ret 0 ) dB = 0

0

and so

f" Re G(re

18)

d9

= 0.

Since IF(z)l = exp (Re G(z)) we have Re G(z) = log IF(z)l and thus

2~ f" log l/(re ")1 d9 = 1

+

i 211T J.

i=l

2

.n

log

t

1/(0)I + 1 ~ log 1.811 ~ 1 1 log la1l

log la 1

-

~

retel dB -

0

i=l

1 21T

J.

2

.n

log 1,81

-

ret 8 1 dB.

0

It is clear that

Lf"

log

la1 - re 18 l d9

+ 2~

=log r

f"

log l -

~e

18

d9

=log r by Lemma 8.47, and similarly

2~

f"

log

1.81 - re 18 l d9

= log r.

The theorem now follows.

Corollary. Iff E s1(Ll(O, R)) then log

l/(0)1 ..;; 2~

f"

log

l/(re1")1 d9

with equality if and only iff has no zeros on Ll(O, r). PROBLEMS 8 49. Use Rouche's theorem to prove the fundamental theorem of algebra.

SO. Show that four of the roots of the equation z 5 to A(O; 2, f).

+

15z + I = 0 belong

51. Let f(z) = tan z - az (z E C) and let rn be the boundary of the square with vertices + n1T + n1Ti. Show that/has 2n + 1 zeros on int (Tn) provided that

8.10

Global Analysis

229

Show that the zeros are real if a is real. 52. Show that the equation z = exp (z) has an infinite nu1nber of roots and that they belong to {z: Re z < -I}. 53. Given r > 0 show that there is n(r)

E

P such that all the roots of

zn I+z+-+· .. +-=0 2! ll! z2

belong to v(O, r) for n > n(r). 00

54. Let f(z) =

I

n=o

anzn (z

E

Ll(O, I)), let 0 < r < I, let n1(r) be the nu1n-

her of zeros off on Ll(O, r), counted according to tnultiplicity and let ft =

inf {!.f(z)!: z E C(O, r)}.

Show that ft

~

!ao! +

!at!

+ · · · + !am 0 such that LJ(f(a), o) c f( U). This will show that f( U) is open and hence that f is an open mapptng. By Theorem 8.1 there is k ~ I and g

E

s;-/(D) such that

f(z) - f(a) = (z - a)kg(z)

(zED)

where g(a) =f. 0. Chooser > 0 such that Lf(a, r) c U, g(z) =f. 0 (z and let 8

= inf {[(z - a)kg(z)[:

E

J(a, r)),

z E C(a, r)}.

Since C(a, r) is compact it follows from Proposition 1.20 that 8 > 0. Given {3 E LJ(j(a), o) We have

f(z) - f3

=

(z - a)kg(z)

+ f(a)

- f3

(zED).

Since

[f(a) - {3[ < 8

~

[(z - a)kg(z)[

(z

E

C(a, r))

it follows from Rouche's theorem that the equation f(z) = f3 has k roots on LJ(a, o). Thus each point of LJ(f(a), o) is the image under f of at least one point of LJ(a, r) c U, i.e. LJ(f(a), o) c f( U) as required.

Corollary. f(D) is a don1ain.

Proof By the theorenlf(D) is open and by Proposition 1.27 f(D) is also connected, i.e. f(D) is a domain.

8.11

Global Analysis

231

The above theorcn1 is son1cti1nes called the open 111apping theorenr or the interior 1napping princriJle. Using the idea in the proof we can relate when a function is locally one-to-one to the behaviour of the derivative. Proposition 8.50. Giren f E ._c:/(D) and a E D, f is one-to-one on so1ne neighbourhood of a ({and only if f'(a) =I 0.

Proof Suppose thatf'(cx) =I 0. In the notation of the proof of the last theoren1 we have k = 1 so thatfis one-to-one on Ll(a,r). Suppose now thatjis one-to-one on Ll(a, R) c D and thatf'(a) = 0. If r = -!R it follows frotn the proof of the last theorcrn that for each f3 E Ll(f(cx), S) the roots off(::) = f3 on Ll(cx, r) arc repeated. Thus

f'(z)

=

0

Sinccfis continuous there is t such that 0 < t

f'(z) = 0

~

rand

(z ELl(a, t))

and sojis constant on Ll(a, t). This contradicts the hypothesis thatjis oneto-one on Ll(a, R) and so we n1ust havef'(cx) =I 0. Observe that the above result is essentially a local result. For exan1ple if D = C and f = exp then f' has no zeros on D. Given a E C, exp is one-to-one on Ll(a, R) if and only if 0 < R ~ 7T. In particular exp is certainly not one-to-one on D. On the other hand, for any f E d(D), iff is one-to-one on D it does follow fron1 the above proposition thatf'(z) =I 0 (z E D). This leads to a considerable strengthening of Proposition 3.8 (ii). The result below is son1etin1es called the inverse 1napping theore1n. Theorem 8.51. Iff E d(D) is one-to-one on D then f- 1

E

d(f(D)).

Proof Let U be open in D. By the open mapping theoren1 we have f(U) = (f- 1) - 1 (U) open in f(D). It follows from Proposition 1.11 that f - 1 is continuous on f(D). We also have f'(z) =I 0 (z E D) by Proposition 8.50 and the result now follows frotn Proposition 3.8 (ii). Given f E d(D) it is often desirable to be able to detennine explicitly what the domainf(D) looks like. Up till now we have n1ainly used ad hoc techniques to determine f(D). The result below allows for a more systematic tnethod. Theorem 8.52. Let f E d(D) and let y represent a starlike sinzple closed contour r, with positive direction, such that T* c D. Iff o y represents a starlike silnple closed contour T 1 , with positive direction, then f 1naps int ( T) one-to-one onto int ( T 1).

Elenwnls of Conrplex Analysis

232

8.11

Proof Given f3 E C"" T 1 the ntunber of roots on int (T) of the equation f(z) = f3 is given by I f'(z) N(f3) = 2ni r f(z) - f3 dz

l 1J l

=

I 2ni

=

2ni r 1

I

1

y)y' y - {3

(/' o

0

o

dw

{3

W -

Therefore

N(f3) = { I 0

if {3 if f3

E E

int ( T 1), ext (T1 ).

It follows that

int (T1 ) c /(int (T)) c T 1 *. To complete the proof it is now sufficient to show thatj(int (T)) c int(T1 ). Suppose conversely that there is a E int (F) such that f(a) E T 1 • Since int (T) is open andfis an open n1apping, it follows that there is r > 0 such that LJ(f(a), r) c f(int (T)). But f(a) E b(ext (T1 )) and so LJ(f(a), r) must contain points of ext (T1 ). This contradicts the fact established above that f(int (T)) next (T1 ) = 0, and so the proof is complete. The theoren1 may not lead us directly to f(D). In some cases it is necessary to proceed via the 'squeezing principle' as we illustrate below. The theorem has extensions to the cases in which the contours need not be starlike, but the theorem in its present form suffices for most applications. Example 8.53./f D = {z: - ; < Rc z < ;} then sin (D)= C"'{x: x ~- 1 orx ~I}. Proof Recall that sin is an entire function with sin (x + iy) = sin X cosh y + i cos X sinh y. Let be the boundary of the rectangle with vertices

rn

Tr -,

2

Tr

2

+

•

rn,

Tr

-- + 2

(cf. Figure 8.6). Then sin is one-to-one on

[-I, 1], (;, ; the ellipse

+ in]

onto [I, cosh 11], [;

u2 cosh n

-.__,....+ 2 in the upper half plane, and

Tr

•

111

'

rn

2

and Olaps [ -1-n,

+ in, -; + in]

-!n]

onto

onto the arc of

v2 = I sinh 2 n

[-I+ in, - ;] onto [-cosh n, -I].

8.11

233

Global Analysis _rr+-m

-7T+tn 2

2

i sinh n

"l

!!

2

2

L ___~

-cosh n

-1

1

coshn

Figure 8.6

By Theoren1 8.52. sin n1aps the inside of T 11 one-to-one onto the inside of the setni-ellipse in Figure 8.6. Since n was arbitrary it follows that sin maps{=: - ; < Rc z < ;, lm z > 0} one-to-one onto {z: Im z > 0}. A similar argument shows that sin maps { z: - ; < Re z < ;, Im z < 0}

one-to-one onto {z: In1 z < 0}. The required result follows. PROBLEMS 8 59. Let f!(D) denote the set of all open tnapping fron1 D into C. (i) Show that @(D) is not closed under sun1s or products. (ii) Show that f E ti(D) itnplies / 11 E l!J(D) (n E P). (iii) Iff E l!J(D) has no zeros show that I If E l!J(D). (iv) Given/ E l!J(D 1 ), g E (!)( D 2 ) with g( D 2 ) c D 1 show that/ o g E l!J( D 2 ). (v) Give an exatnple ofjE (!)(D) that is not continuous on D. 60. LetfE s/(D) be one-to-one on D, and let r be a starlike contour in D with star centre a. If y represents r with the positive direction show that f o y represents a simple closed contour T 1 such that n(T1 ; f(a)) = 1. 61. Determine f(D) in each of the following cases. (i) f(z) = cosh (z), D = {z: 0 < Im z < 1r}. (ii) f(z)

(iii) f(z)

= tan (z), D = {z: =

z +

~'

-

- ; < Re z < ;}-

D = LJ'(O, I).

(iv) f(z) = log z, D = LJ(I, I). (v) f(z) = log (1 - z 2 ), D = {z: Itn z > 0}. 62. Let f be as in part (ii) above. Show that

f

_ 1

(z) =

J.

[O,z]

I

dw

+ W2 =

1 1 + iz 2.1 log 1 - IZ

Obtain analogous results for inverse functions of sin, cos.

234 8.12

Elenzents of Co1nplex Analysis

8.12

The rnaxin1un1 rnodulus principle

In this final section we study the behaviour of 1/1 whenjEs'I(D). The basic result is the n1axin1un1 modulus principle which asserts that if f E .rl( D) is not constant then Ill has no absolute maxin1um point on D. This fan1ous result admits several different proofs, and could easily have been proved at the beginning of this chapter. We have chosen to delay this section until now since the results are best seen in the light of the open Ina pping theoren1. We use the tnaximUJn modulus principle to derive some deeper results. The first concerns a sharpening of the bound of a function and the second concerns the behaviour on the boundary of D when D is an unbounded dotnain. As a final application we derive the corresponding maximum tnodulus principle for harn1onic functions.

Theorem 8.54 (Maximum Modulus Principle). (i) IfjE d(D) is not constant then Ill has no absolute maxilnum points on D. (ii) If D is bounded and g E CC(D -) is analytic and not constant on D then the absolute nzaxilnwn points of lgl belong to b(D). Proof (i) Suppose that a is an absolute maximum point for l/1, i.e. lf(z) ~ lf(a)l(z ED). By the open mapping theorem f(D) is open and so there is r > 0 such that LJ(f(a), r) c f(D). We may clearly choose f3 E LJ(f(a), r) such that lfJI > If(a) I. Since fJ Ej(D) this contradicts the hypothesis that a is an absolute maximum point for l/1. This proves (i). (ii) If D is bounded then D- is closed and bounded, and so compact. Since lgl is a continuous real function on D- it has at least one absolute n1aximun1 point on D-. Suppose a is an absolute Inaximum point for lgl and a ED. We then have in particular lg(z)l ~ lg(a)l

(z E

D)

so that a is an absolute maximun1 point for lgl on D. This contradicts part (i) and so the absolute maximun1 points of lgltnust therefore belong to D- x D = b(D).

Corollary. (i) Iff has no zeros on D then Ill has no absolute nzininzunz points on D. (ii) If g has no zeros on D- then the absolute nzininzunz points of jgl belong to b( D). Proof Apply the theoren1 to the functions

Iff and Ifg.

The maxi1num rnodulus principle 1nay be expressed in geometrical terms as follows. Given f E s/( D) the graph of I/I is a subset of R 3 -it may be

8.12

235

Global Analysis

thought of as the landscape of [/I over D. The first part of the theoretn says that there arc no tnountain tops. In fact there are not even local hills for we would then have an absolute tnaxin1un1 point for 1!1 restricted to some open disc. A zero off is obviously an absolute 1ninimun1 point for lfl, but ifjhas no zeros on D then the graph of.fhas no hollows over D. The second part of the theorcn1 says that the highest part of the graph always occurs at the edge. Moreover if the function has no zeros then the lowest point also occurs at the edge. The student should illustrate these ideas by drawing the landscapes associated with exp, sin, and log. It is worthwhile to consider one of the other standard proofs of the 1naxin1un1 n1odulus principle. The student should realize that it is sufficient to show that if .f E .ci(Ll(a, R)) is such that If I has an absolute tnaxin1un1 point at a then .f is constant. If f(a) = 0, this last statctnent is trivial. Suppose that f(a) # 0. By the Corollary to Theorcn1 7.1 we have

f(a)

= -

1

27T

f.2:r .f(a + rciO) dO

(0 < r < R).

0

Let

Then h E ~{[0, 27T]) and since a is an absolute tnaxin1un1 point for If! it follows that h(O) ~ 1 (0 E [0, 27T]). Since

S:'

{I - h(B)} dB =

Re

J:' {

1 - f(a

/~a;e'")} dB

= 0

it follows by integration theory that h( 8) = 1 ( 8 E [0, 27T ]). Since r was any point of (0, R) it follows that Ref is constant on Ll(a, R). We conclude frotn Problem 3.20 that f is constant on Ll(a, R) as required. The maxin1un1 modulus principle leads to a study of the growth of 1/1 on circles of increasing radius. More precisely suppose thatjEd(Ll(a, R)) is not constant and define

Ma(r) = sup {jf(z)j: z E C(a, r)}. Since f is continuous on the cotnpact set C(a, r) we have

Ma(r) = max {lf(z)l: z E C(a, r)}. Moreover by the maximum modulus principle we tnay equally write

Ma(r) = tnax {lf(z)l: z E Lf(a, r)}.

8.12

Elenzenls of Cotnplex Analysis

236

It follows that M a is an increasing function. In fact (Problem 8.64) it is strictly increasing and continuous. Further properties of Nla arc given in the problcn1s at the end of the section. By way of illustration observe that if f(z) = zn then Ma(r) = (r + lal)n, and if f(z) = cxp (z) then Ma(r) = exp (r + Rca). The next result is usually called Schwarz's Lemma but we have dignified it as a theorem. Observe that the result gives a sharpening for the bound of a function. We shall give a significant application of the result in Lemma 9.10.

Theorem 8.55 (Schwarz). Let f 1/(z)l ~ M (z E LJ(a, R)). Then

1/(z)l

M

~ R

E

s/(LJ(a, R)) be such that f(a)

(z E LJ(a,

lz - «I

with equality if and only if f(z) = A(z - a) (z Proof Since f(a) = 0 there is g f(z) = (z - a)

E

E

=

0 and

R))

LJ(a, R)) where IAI = lvl/ R.

d(LJ(a, R)) such that

(z E LJ(a, R)).

g(z)

Given 0 < r < R we have

lg(z)l

~

M

M

z -

(z

=-

r

a

E

C(a, r)).

It follows from the maximum modulus principle that

lg(z)l ~ Given f3

E

M

(z

r

E

J(a, r)).

LJ(a, R) we have M lg(f3)1 ~r

and therefore lg(f3)1

(1{3 - a I < r < R)

~ ~- This shows that

1/(z)l

~

M

R lz -

ex!

(z

E

LJ(a, R)).

If equality holds at any point of LJ(a, R) then lgl has an absolute maximum point on LJ(a, R) and so must be constant. The re1naining details of the proof are trivial. The second part of the maximum modulus principle may be phrased as follows: if D is a bounded domain and iff E rt'( D-) is analytic on D with 1/1 ~ M on b(D) then 1/1 ~ M on D. We now consider the situation in

8.12

Global Analysis

237

which D is an unbounded don1ain. For this case the theoren1 general as we sec in Exatnple 8.56 below, and it is necessary to son1c condition on the rate at which Ill grows towards infinity. results of this nature were obtained by Phragn1en and Lindelof, shall sin1ply give one illustrative theorcn1. Example 8.56. Let D

Then Ill

= ( z: -~ < Im z < ;) and let

l(z) = exp (exp (z)) (z E C). 1 on b(D) while I is unbounded on D.

~

Proof Clearly b(D)

=

(x + i;: xER) and

!(x + i ;) Thus Ill

fails in in1pose Several but we

~

=

lexp ( + ix)l

~

(x

I

E

R).

1 on b(D) but it is obvious that I is not bounded on R c D.

Theorem 8.57 {Phragmcn-LindclOf). Let D

= { z:

~

-;

Re z < ; } and

let IE C(J(D -) be analytic on D. Suppose there exist M, a > 0, .\ < I such that ll(z)l ~ exp (a eAIRe zl)(z ED). ll(z)l ~ M (z E b(D)), Then ll(z)l ~ M (zED). Proof Choose p, > 0 with A < p, < I. Given E > 0 let ge(z) = exp (- Ecosh (p,z)) so that each g( is an entire function. For z = x

(z E C)

+ iy ED-

Re {2 cosh (p,z)) = cosh (p,x) cos (p,y) ~ where I> = cos

(IL ;) > 0, since 0 < IL
0 and A < fL it follows that lim l(z)ge(z) = 0.

lxl-+ + oo

Given

a

E D we may choose R > Re

ll(z)ge(z)l ~ M

a

such that

(lxl > R)

(zED).

Ele111ents of Conzplex Analysis

238

1t foiiows that

1.!:_~€

I~

8.12

M on the boundary of the rectangle with vertices

+ R + i ;. By the n1aximum n1odulus principle

lfg(l

~ M on all of the

rectangle and in particular 1/(a)g/a)l ~ M. Since a was any point of D this shows that lfg(l ~ M on D. Since lin1 g((z)

(-+0+

we conclude that l/1

~

=

1

(z

E

D)

M on D as required.

Recall from Chapter 3 that there is a close relation between functions analytic on a don1ain and functions harmonic on a domain. We show finally how the 1naximun1 modulus principle for si(D) leads to a corresponding principle for .YE(D). Theorem 8.58. (i) If u E .Yt(D) is not constant then u has no absolute n1axin111111 nor nzininzunz points on D. (ii) If D is bounded and u is a continuous real function on D-, harmonic on D, then the absolute 1naxinuun and rninimun1 points for u belong to b(D).

Proof In view of the method of proof for Theorem 8.53 it will be sufficient to show that if u E .YE(D) is not constant then u is an open 1napping from D into R. Given a ED there is R > 0 such that L1(a, R) c D. By Proposition 3.13 there is v E :ff(D) such that f = u + iv is analytic on L1(a, R). By the open n1apping theorem there is o > 0 such that L1(f(a), o) c f(L1(a, R)). It follows that

(u(a) -

o, u(a) + o)

c u (L1(a, R)) c u(D).

This proves that u is an open mapping as required. PROBLEMS 8 63. Let D be a bounded domain and letfn E rt'(D-) be analytic on D for each 11 E P. If {/n} converges uniforn1ly on b(D) show that {fn} converges uniformly on D-. (flint. Usc the maxinn1m 1nodulus principle to show that {fn} is Cauchy in the n1etric space rt( D-).)

64. GivcnjE si(L1(a, R)) show that the function Ma is strictly increasing and continuous. (Hint. For the continuity part derive contradictions from the suppositions sup {Ma(r): r < r 0 } < i\1a(r0 ) < inf {Nla(r): r > r 0 }.)

8.12

239

Global Analysis

65. (Hadatnard·s three circle theoren1.) Given fE.r:f(iJ'(a, R)) and 0 < r 1 ~ r ~ r 2 < R show that

(Hint. Apply the maxin1un1 n1odulus principle to the function z-+ zP(f(z)q) (p, q E Z, q > 0). Choose AE R such that r/'Ma(r 1 ) = r/. Ma(r 2 ) and then

choose Pn, qn so that

lin1 Pn co q,1

=

A.) If g(u) = log Ma(eu) show that g is a

n-+

conre.Y function i.e.

g(tu1 + (1 - t)u 2)

66. Given f

E

~

tg(u 1)

+ (1

- t)g(u 2)

(t

E

(0, 1)).

s1(C), AE (0, 1) let I1() r

=

M 0 (Ar) M 0(r)

(r > 0).

Show that h is strictly decreasing and that litn h(r) = 0 unless f is a polyT-+

+ 00

non1ial function. What is the lin1it iff is a polynotnial function? 67. Given f

E

d(iJ(a, R)) with f(a) = 0, use Schwarz's lemn1a to show

that the function r-+ Ma(r) is increasing. Show also that the function is

r

strictly increasing if it is not constant.

68. Let f E .9/(iJ(O, 1)) be such that f(iJ(O, 1)) c iJ(O, 1). If f(a) = a, f(fJ) = f3 where a, fJ E iJ(O, 1), a =/= fJ, show thatf(z) = z (z E iJ(O, 1)), i.e. if f fixes two distinct points of iJ(O, 1) then it fixes all the points of iJ(O, 1). (Hint. Consider the following functions defined on iJ(O, 1.)

+a)

z h(z) = f ( 1 - az ' 69. Let f

E ~(J(O,

g(z)

h(z)- a = 1 - ah(z).

R)) be analytic on iJ(O, R) and let

A(r)

= max {Ref(z):

z

E

C(O, r)}.

(i) Show that A is strictly increasing and continuous. (ii) If f(O) = 0 show that

2r

M 0 (r) ~ R _ r A(R) (Hint. Let g(z)

lemma.)

=

(0 < r < R).

2A(~~z~ f(z) (z E LI(O,

R)) and apply Schwarz's

240

Elements of Complex Analysis

8.12

(iii) Show that

Mo(r)

~

2' A(R) R-r

+ RR-r + r lf(O)I

{0 < r < R).

70. Give an alternative proof of Theorem 8. 58 by considering the function expo f where f E d(Ll(a, R)).

9 CONFORMAL MAPPING

The ain1 of the last chapter was to study the fundamental properties of functions that are analytic or n1eron1orphic on a don1ain D. The aim of this chapter is to study the existence of functions in .s1(D) with special properties. In particular we are interested in functions in .s:I(D) that are one-to-one. As a prelitninary n1otivation we consider some ideas from applied n1athetnatics. The flow of air past an aerofoil (wing) is clearly of great importance in the design of an aircraft, but the associated problems are complicated because of the shape of the aerofoil. If the aerofoil has uniform cross-section then the original three-dimensional problem can be reduced to a problem in two real din1ensions-or one complex dimension. The next step is to find a change of variable that will transform the aero foil to a simpler shape -and in this context the simplest shape is a disc. In fact in this case it is n1ore appropriate to transform the outside of the aerofoil (where the flow occurs) to the outside of a disc (see Problem 9.19 for an example). The original problen1 about the aerofoil n1ay now be transformed to a corresponding problem about the disc and this latter problem is usually much n1ore amenable. Having solved the problem for the disc we then transform back again to solve the problem for the aerofoil. We consider next the properties that should hold for this change of variable. We wish to transforn1 two real variables, x, y say, to two new real variables, u, v say; or equivalently we wish to transform one complex variable z to a new complex variable w, say by the relation w = f(z). The mapping fshould be one-to-one so that we can transform the problem back and forth between the aerofoil and the disc. Since differential equations figure prominently in the aerodynamics of the problem, both u and v ought to be suitably differentiable in terms of x andy (rather than merely continuous). The potential theory aspects of aerodynamics make it desirable that the mapping should preserve angles between curves; thus if two curves intersect 241

242

Elen1ents of Con-1plex Analysis

9.1

at right angles in the x, y plane the transforn1ed curves should intersect at right angles in the u, v plane. The appropriate condition for this property is that f = u + iv should be analytic and such that f' has no zeros (sec Probletn 9.1 ). Recall fro In Proposition 8.50 that a necessary condition for f to be one-to-one is that f' have no zeros. Alternatively we may recall fro1n real analysis that the tnapping (x, y) ~ (u, v) is one-to-one on some neighbourhood of a point provided the Jacobian function

ou ov ox ox

-J=

ou oy

ov oy

does not vanish at that point. Iff= u + iv is analytic then the CauchyRiemann equations give J =

(ou) ox

2

+ ( ov) 2 =

lf'l 2

ox and we see again the necessity of the condition that f' have no zeros. We thus see in the context of aerodynamics the relevance of investigating the existence of functions in .s1(D) that are one-to-one and have a specified domain as their range. It is now natural for the mathematician to pose several general questions about analytic mappings between domains. The first four questions below are suggested by the above example. The final question arises in connection with potential theory and the existence of harmonic functions with specified behaviour on the boundary of a domain; it is also suggested by Cauchy's integral formula. I. Given domains D, D 1 , does there exist f E d(D) which maps D one-to-one onto D 1 ? 2. If so, what is the set of all such n1appings? 3. What special features arise in the case D 1 = D? 4. Can the above functions be extended to be continuous on D- and one-to-one on b(D)? 5. Given a continuous function g on b(D), when can g be extended to a continuous function on D- that is analytic on D? In their full generality the above questions arc beyond the scope of this book. In §I we discuss, without proof, part of the answer to question I. In the process we discuss an equivalence relation on the set of all domains that is finer (i.e. induces Inore equivalence classes) than the equivalence relation induced by hon1con1orphisn1. Questions 2 and 3 are answered in §2 and some part answers to questions 4 and 5 are given in §3. In the final section we give some illustrative mappings.

Power Series Functions

4.1

83

Since {A 2 nB2 n} converges to Af-t and since {DnEn} converges it follows that lim IC2 n

n-+ oo

Af-t!

-

= 0.

Since any subsequence of a convergent sequence converges to the limit of the sequence we conclude that 00

2 Yn

n=O

= lim

n-+oo

C2n

= Af-t.

PROBLEMS 4 6. (i) Let a

n1n

(m, n E P).

=----.,.

(m

m,n

+ in)2

Show that lim (lim

m-+ oo n-+ oo

= 0 = lim (lim

am, n)

n-+ oo m-+ oo

am, n)

but that {am. n} does not converge. (ii) Let a m,n

=

(-l)m+n(_!_ m

i)

+n

(m, n E P).

Show that {am. n} converges to 0 but that for each fixed mE P the sequence {am. n} fails to converge.

1

7. Let {a,,,} be such that the double sequence {~

J Ia,, ,I} 1

converges.

Show that lim

i

~ s

m,n-+ oo T = 1 =1

aT.

s = T~= ( s~= aT. s) = s~= ( T~= = lim 2 aT,s• 1

n-+oo

1

1

1

aT,

s)

T+s~n

(This shows that the double array has the same sum by rectangles, rows, columns, or triangles.) 8. For which z do the following series converge? ~

L -oo

n ~ log ( 1 + In I) ~ z 'L _ 00 -oo n2 +n+z ' L

.

2z Z

2

-n

2

9. Let

an=(

( -l)n

n

+ IY

(n=0,1,2, ... )

2

and let {yn} = {an}* {an}. Show that

2 an converges while 2 Yn diverges.

244

Elenzents of Co1nplex Analysis

9.1

definition of what we n1ean by a domain D having n holes. In fact we have already discussed the case n = 0 in connection with Runge's theorem (see page 195). Recall that a domain D is sinzply connected if cro ""- D is connected, and recall that any starlike domain is simply connected. A domain D is said to be doubly connected if coo ""- D has 2 components. This corresponds to the domain D having one hole; for· example D = A(O; 2, 1) or D = LJ'(O, I). More generally, a domain D is said to have connectivity n if cro ""- D has n components. Note that we have not yet exhausted all the possibilities since coo ""- D may have an infinite number of components. The connectivity of a domain is preserved under homeomorphisms and in fact any two domains of connectivity n are homeomorphic; but we shall not prove these statements. The most important case is the sin1ply connected case (which includes in particular the starlike case). Although all simply connected domains are hon1eon1orphic they are not all isotnorphic. ln fact C is not isomorphic with LJ(O, 1). Otherwise we should have an entire function f such that Ill < I on C andfis one-to-one on C. This contradicts Liouville's theorem and so there are at least two equivalence classes amongst the simply connected dotnains. It is a surprising fact that there are only two such equivalence classes. This is a simple consequence of the famous Riemann mapping theorem stated below. Every sinzply connected don1ain D of C, other than C itself, is isomorphic with LJ(O, I). More precisely the theoren1 asserts that given a E D there is exactly one mapping fE I(D, Ll(O, I)) such that f(a) = 0 andf'(a) > 0. We shall not attempt to prove the theorem since it involves more topological machinery than we have developed. (The proof would be well within our grasp had we first proved Runge's theorc1n and Mantel's theorem (see page I96).) The Riemann mapping theorem implies that, up to isomorphism, there are only two simply connected don1ains in C, namely C and LJ(O, 1). For many problems this means that if we can deal with the cases D = C and D = LJ(O, 1) then we can (at least in theory) deal with the case of any simply connected domain. For this reason particular attention is given in advanced texts to the case D = LJ(O, 1). We shall return to this point in a moment. It must be clearly understood that the Rien1ann n1apping theorem is an existence theorem and gives little clue as to how we might determine an actual isomorphism of L1(0, 1) with a given sin1ply connected domain other than C. In practice one usually consults a 'dictionary' of conformal mappings, or resorts to nun1erical techniques.

Confornwl A1apping

9.1

245

The results for doubly connected dotnains arc n1ore complicated. Every doubly connected dotnain is isotnorphic with some annulus centred on the origin; but A(O; R2 , R 1 ) is isomorphic with A(O; r2 , r1 ) if and only if R 2 / R 1 = r2 /r 1 • It follows that the equivalence classes for doubly connected don1ains are in one-to-one correspondence with the 1nen1bers of [0, 1). The results for higher connectivity are even more complicated. \Ve turn now to tnore algebraic considerations. Proposition 9.2. Given rp E I(D 1 , D 2 ) the 1napping f isonwrphisrn of d( D 2 ) with d(D 1).

~f o

rp is an algebraic

Proof Let Tf = f o rp (f E d(D 2 )). Then Tf = Tg itnplies f = Tf o rp - 1 = Tg o rp- 1 = g, so that Tis one-to-one. Given hE d(D 1 ) we have h o rp- 1 E d(D 2 ) and T(h o rp- 1 ) = h, so that T tnaps d(D 2 ) onto .s1(D 1 ). Given f, g E ._d( D 2 ), ,\ E C we have T(,\f) = ,\f o rp = ,\Tf T(f + g) = (f + g) o rp = f o rp + g o rp = Tf + Tg T(fg) = fg o rp = (f o rp)(g o rp) = TjTg. The proof is con1plete. Corollary. If rp of d(D).

E

A(D) the nwppingf ~ f

o

rp is an algebraic auto1norphisn1

If the domain D 1 has a sitnpler structure than D 2 we may first solve problems for the algebra d(D 1 ) and then transfer the results to .91(D 2 ) via the above algebraic isomorphisn1. In particular if D 2 is a simply connected domain, D 2 =f. C, problems on d(D 2) can sometimes be solved by considering the algebra d(Ll(O, 1)). The above proposition is essentially routine and trivial. The converse below is much more exciting and by no means trivial. We need to recall some algebraic terminology. A non-en1pty subset M of d(D) is an ideal iff, gEM, ,\ E C impy f + g, ,\fE M and if fh EM whenever fE M, h E d(D). An ideal M is a 1naxirnal ideal if M =f. .#(D) and if .#(D) is the only ideal that properly contains M. Theorem 9.3. Let T be an algebraic isonwrphisnz between .s1(D 2 ) and d(D 1 ). Ifu is the identity 1napping on D 2 , then Tu: D 1 ~ D 2 is an isomorphisnz.

Proof Given f E .#(D 2 ), w Ef(D) iff f - wl has a zero on D, i.e. iff f - wl has no multiplicative inverse in .s1(D 2 ). We show now that Tf has the same range as f Since Tis a non-zero algebraic isomorphism we have Tl = 1. Iff, g E .#(D 2) with (f- wl)g = 1 then T(f- wl)Tg = Tl and so (Tf- wl)Tg = 1, i.e. Tf- wl has a multiplicative inverse. A similar

246

Elentents of Conzplex Analysis

9.1

r-

1 shows that f argutncnt applied to wl has a nniltiplicative inverse if Tf- w1 has. It follows fron1 above that/ and Tf have the same range. In particular if cp = Tu then cp has the same range as u, namely D 2. Since cp E d(D 1 ) it is now sufficient to show that cp is one-to-one. Suppose that cp is not one-to-one so that there exist z 1 , z 2 E Db z 1 =I z2 with cp(z1 ) = cp(z2) = w. Let M = {f: f E d(D 2), f(w) = 0}. It is trivial to verify that M is an ideal of d(D 2) and clearly M f:; .91(D 2 ). Suppose J is an ideal of d(D 2 ) that contains M strictly. Then there is g E J such that g(w) f:; 0. Since g - g(w)1 EM c J it follows that g(w)1 E J and so 1 E J. Since J is an ideal we have J = d(D 2) and so M is a maximal ideal. Moreover M = {(u- w1)f:/Ed(D 2 )}, since f(w) = 0 iff there is g Ed(D 2 ) with f(z) = (z - w)g(z) (zED). Let N = {Tf:/E M}. Since Tis an algebraic isomorphism it follows that N is an ideal, and it is clear that N is in fact a maximal ideal since M is a maximal ideal. On the other hand

N = {(Tu - wT1)Tg: g E ..#(D 2 )} = {(cp - w1)h: hE d(D 1)} so that N is contained in the proper ideal {/: f E d(D 1 ), f(z 1 ) = 0}. But N is strictly contained in this ideal for if g(z) = z - z 1 (z E D 1 ) then g ¢: N. Therefore N is not a maximal ideal. Tills contradiction shows that cp must be one-to-one as required. This last theorem and the previous proposition show that two domains are isomorphic if and only if the associated algebras of analytic functions are isomorphic. This suggests that one might obtain an algebraic proof of the Riemann mapping theorem. Unfortunately no simple proof has yet been obtained along these lines. PROBLEMS 9

1. An oriented arc r is regular at z 0E r if r is smooth in some neighbourhood of z0 and if there is some representative function y such that Zo = y(to), y'(to) # 0. Let rl, r2 be oriented arcs that intersect at Zo and are regular at Zo. The angle between r1 and r2 at Zo, < ( rb r2; Zo), is defined up to equivalence modulo 2rr by

< (Fb F2; Zo)

=arg y~(t1) -

arg y~(t2)

where Zo = Y1(t 1) = Y2Ct2).

(i) Show that the definition is independent of the choice of those representative functions Yb y 2 such that y~(t 1 ) f:; 0, y~(t 2 ) f:; 0. (ii) LetjE .9/(D) and let r be an oriented arc in D regular at z 0 where f'(zo) # 0. Show that/( F) contains an oriented arc regular atf(z0 ).

9.2

Confonnal Mapping

247

(iii) Let.fE. 0 we have

lji(z)l2 = x x

+ i(y + i(y +

1) 2 = x2 + y2 + 1 - 2y < 1 1) x 2 + y 2 + 1 + 2y

so thatf(II+) c Ll(O, 1). Given a

E

Ll(O, 1) let

fJ = i 1 + a. 1- a Then f(fJ) = a and

Im f3

=

Re ( 1

a)

+ a. 1 -

1- a 1- a

=

1 - Ia 12 II - al 2

> 0

so that f3 E II+. Therefore f maps II+ onto Ll(O, 1) and the proof is complete. Lemma 9.10. H(Ll(O, 1); 0) is the group of rotations

e=

{au: a

E

C(O, 1)}.

Proof Clearly B c H(Ll(O, I); 0). Given f E H(Ll(O, I); 0) we have f E d(Ll(O, 1)), f(O) = 0 and lf(z)l < 1 (z E Ll(O, 1)). By Schwarz's lemma we have

lf(z)l ~ lzl The same argument applied to f

-l

(z E L1(0, 1)). gives

(wE Ll(O, 1)). Given z E Ll(O, I) there is wE Ll(O, I) such that z = f- 1 (w) and so

lf(z)l ~ lzl = lf- 1(w)l ~ lwl

=

lf(z)l

(z

E

L1(0, 1)).

It follows from the maximum modulus principle that the mapping

z ~f(z) is constant and hence there is a E C(O, 1) with f(z) = az z (z

E

Ll(O, 1)). The proof is complete.

Theorem 9.11. A(II +) consists of the mappings of the form

f(z) = az + f3 yz

where a,

+

fJ, y, S E R, aS - fJy > 0.

S

(z

E

II+)

9.2

Confornzal Mapping

253

Proof Let G be the set of 1nappings specified in the staten1ent of the theoren1. Given f E G it is clear that f is analytic and one-to-one on II+. As an hon1ography f n1aps R u {oo} one-to-one onto R u {oo} and so the dotnain f(II+) is contained in II+ u II-. Therefore f(II+) is contained in the con1ponent II+ or the con1ponent II-. Since

. Tn1.((r)

=

aS - {3y Y2 + 02 > 0

it follows that f(II+) c II+. Since f tnaps coo onto coo it follows that f(II-) c II- and then f(II+) = II+. We have thus shown that G c A(II+). It is straightforward to verify that G is a subgroup of A(II+) that is transitive on II+. In view of Proposition 9.6 the proof will now be conlplete when we show that 1-!(II+; i) c G. Letfbe the iso1norphisn1 of II+ with LJ(O, I) given in Letntna 9.9. Since f(i) = 0 it follows frotn Proposition 9.4 (ii) that the group H(II+; i) is isotnorphic with the group H(LJ(O, I); 0) under the tnapping if;---+ .fo if; of- 1 . Len11na 9.10 now gives

H(II+;i)

=

{f- 1 ox of:

XE

B}.

By sitnple tnanipulation we verify that any function g of H(II+; i) is of the fonn

e

=COS

a(,..)-

2+

. e Sill

2

. e + cos e 2 2

b.:.

-z Sin

Therefore H(II + ; i) c G as required. Assuming the Riemann mapping theoren1 we may now determine A(D) for any sin1ply connected domain D, other than C. Indeed since II+ is sin1ply connected Proposition 9.4(iii) describes A(D) in terms of A(II+). In practice this may not lead directly to the neatest description of A(D). The student may appreciate this by trying the case D = LJ(O, 1). We give below an alternative technique for detern1ining A(Ll(O, 1)). The essence of the proof is to characterize the homographies that are hon1eon1orphisms of L1 (0, 1) with itself. Theorem 9.12. A(LJ(O, 1)) consists of the 1nappings of the fonn

f(z) where

=

e E R and A. E Ll(O, 1).

ete z + _A. 1 + ;\z

(z E .:1(0, 1))

4.3 Power Series Functions 81 Proof. It is well known from real analysis that lim n 11 n = 1. It follows n-+ oo that lim jnanl 11 n = litn lanl 11 n n-+ oo

so that

p

n-+ oo

is also the radius of convergence for the power series function 00

2:

g(z) =

n=1

nan(Z- a)n- 1 •

Let z E Ll(a, p) with lz - al = r and choose R such that r < R < p. Given hE C with Ihi = 8, 0 < 8 ~ !(R - r) we have a + hE L1(a, p). For each n E P we have

(z- a+ h)hn - (z- a)n- n(z- a)n-1

= t

~

=

Since lanRnl

2

~ (n),.n-t8t-2

lhl

t

t=2

~

i= (n)(za)n-tht-1 t

lhl n(n

- I) 2

i (n - 2),.n-t8t-2 2 t -

t=2

JhJ n(n; I) (r +

w-•.

2: an(z ~

- a)n converges on C(a, R) there is M > 0 such that M (n E P). It now follows that

f(z + h) - f(z) - g (z) ;.........;._-----::-----'h

~ = L., an {(z - a - h)n - (z - a)n - n(z - a)n-1} n=1 h ~ ~ ~ IhI n(n - 1) (r + 8)n- 2 n=l R

2

~~~11 n(n Since r + 8 ~ r + t (R - r) =

!)(' ~

l r2·

< R the above series converges and we conclude that f is differentiable at z with f'(z) = g(z). Since z was arbitrary we have thus shown that/is differentiable on L1(a, p) with 00

f'(z) =

2:

nan(Z - a)n - l

(z E Ll{a, p)).

n=l

Suppose now thatfis differentiable k times on Ll(a, p) with 00

J(z) =

2: n=k

4

n(n- 1) ... (n- k

+

l)an(z- a)n-k

(z EL1(a, p)).

9.3

Conjorn1al Mapping

255

In this section we have detennined the three groups A(C), A(C 00 ), A(JI+). The group A(C) is son1etin1es called the 'az + {3' group. The group A(C 00 ) is easily seen to be ison1orphic with the group SL(2, C) of all 2 x 2 complex n1atrices with detern1inant 1. Sin1ilarly the group A(JI+) is isotnorphic with the group SL(2, R) of all 2 x 2 real n1atrices with detern1inant 1. The groups SL(2, C), SL(2, R) are of fundan1ental in1portance in the theory of Lie Groups.

Ren1ark. We state without proof the interesting fact that for every doubly connected don1ain D, the group A(D) is ison1orphic with the group S0(2, R) of all 2 x 2 real orthogonal matrices with detenninant 1. PROBLEMS 9 7. Let D be a don1ain such that A(D) is transitive on D. (i) Given a, f3 E D show that the groups H(D; a), H(D; {3) are isotnorphic. (ii) If D is ison1orphic with D 1 , show that A(D 1 ) is transitive on D 1 •

8. Show that A(C) has an infinite number of transitive subgroups. Consider the same problem for A(.d(O, 1)) and A(C 00 ) . 9. Show that f E A(C 00 ) iff f is a homeomorphism of of whose singularities are isolated.

ceo with itself, all

10. Sho\V that an homography is uniquely determined by its values at three distinct points. 11. Determine the fixed points of an homography J, i.e. the points p such that f(p) = p. Iff has distinct fixed points a, f3 E C show that f is of the form f(z) - a = A z - a. f( z) - f3 z - f3

What happens in the other cases? 12. Every function in A(D) admits a unique extension to an homography if D = C, .d(O, 1), JI+. If D is isomorphic with .d(O, 1) show that D has the above property iff every isomorphistn of D with .d(O, 1) is the restriction of an homography. Deduce that D has this property iff it is a disc or a half plane.

9.3

Mappings of the boundary

In this section we examine questions 4 and 5 of the introduction. The questions are difficult in general and we shall not attempt to prove many results. We begin by discussing question 4.

256

9.3

Elenzents of Complex Analysis

Let Db D2 be domains and lctfbe an isomorphism of D 1 with D 2 • We should like to know iff can be extended to a homeomorphistn of D- 1 with D- 2 , i.e. iff has an extension to D- 1 that is continuous on D- 1 and tnaps b(D 1 ) one-to-one onto b(D 2 ). If D 1 , D 2 are bounded and if b(D 1 ), b(D 2 ) are sitnple closed curves then the answer. is yes, although we shall not give the proof. Thus, for example, every isomorphism of Ll(O, I) with the semi-disc Ll(O, I) n n+ adtnits a continuous extension to .J(O, I). It is not possible to remove the condition that b(D 1 ), b(D 2 ) be simple closed curves. For example it is not possible to extend an isomorphism of L1(0, I) with D = Ll(O, I)"'- [0, I] to be a homeomorphism of Lf(O, I) with D- = Lf(O, I). Otherwise we \vould have the contradiction that C(O, 1) is homeon1orphic with C(O, I) u [0, 1] (sec Problem 5.32). If we remove the condition that f be one-to-one the situation becomes much n1ore complicated. We shall illustrate by considering the case of the domain L1(0, 1). GivenfE 0'(Ll(O, 1)) we say thatfhas property (E) iff can be extended to a continuous function Jon Lf(O, I). Iff has property (E) the extension is clearly unique. Iff has property (E) then f is continuous on the compact set Lf(O, I), and so /is bounded on Lf(O, I) and/is bounded on L1(0, I). In order for fto have property (E) the condition thatfbe bounded is necessary but not sufficient, as the following example shows. Example 9.13. If

(~ ~

f(z) = cxp

(z

:)

E

Ll(O, I))

tlren f is analytic and bounded on Ll(O, 1) but does not lzal'e property (E). Proof Clearly f

E

..7/(L1(0, I)). Given z

1)

z + Re ( z - I

and so lf(z)l < I. Suppose .J(O, I). We then have

f

=

x2

=

+

x

+ y2

(x - I )2

iy

.d(O, I),

I 0

1 - 2r cos u

(0 < r < 1, u E R).

(ii) This may be established by real integration. Alternatively we may employ the residue theorem to obtain

J

2n

=

Pr(u) du

f.

0

1 - r2

C(O, 1)

=

r(z + :z)1 + r2

1-

dz -. IZ

! J.

(1 - r2) dz i c (1 - rz)(z - r)

= 27T. (iii) Given 0 < 8 < 1rj2, 0 < r < 1, we have

lui

sup {P,(u):

~

8}

= 1 - 21r -cosr 8 + r2 2

and it follows that lim sup {Pr(u):

r-+1-

lui

~ 8}

= 0.

Theorem 9.15. Given a continuous complex function f on C(O, 1), the following statements are equivalent. (i) f has a unique extension to a function and analytic on L1(0, 1).

(ii)

J:x e'•BJ(e"') dO = 0

(n

E

J that is continuous on J(O, 1)

P).

Proof. Let condition (i) hold. It follows from Problem 6.2 that

f.

znf(z) dz

=

0

(n

= 0, 1,2, ... )

C(O,l)

and therefore

fx

e'"BJ(e' 8) dO

=0

(n E P).

Let condition (ii) hold. Extend/to a function/ on J(O, I) by defining/ on L1(0, 1) by

/(z) =

~ J. 27Tl

f(w) dw

C(O,l) W -

Z

(z E L1(0, 1)).

4.3

Power Series Functions

89

part (possibly void) of its boundary, or for C ""-{ex}. Moreover it follows easily frotn the work above that the series converges absolutely at each point of the open annulus and unifonnly on any closed subannulus. Furthennore h is infinitely differentiable on the open annulus and the series n1ay be differentiated 'tenn by tenn'. Functions of tllis type will appear frequently in Chapter 7. We shall call such functions quasi power series functions. In particular it will turn out that any function in d(C ""-{ex}) is a quasi power series function. PROBLEMS 4 11. Show that the following series have radius of convergence 1. ~

~

.... n £.

'

Show that the first converges nowhere on C(O, 1), the second converges on C(O, 1) ""- {1}, and the third converges on C(O, 1).

12. (i) Show that

L zn does not converge unifonnly on L1(0, 1). -n+ 1

(ii) Show that

£.

.... n

rived series

L n(n + 1) converges unifonnly on L1(0, 1) while the de-

L ~11 does not converge uniformly on L1(0, 1).

13. If L exn(z - ex)n has radius of convergence p, what is the radius of convergence of the following power series?

14. Which power series functions are differentiable at oo? 15. Suppose that converges. If

L exnzn

has radius of convergence 1 and that

L an

00

f(z) =

L

(z E L1(0, 1))

exnzn

n=O

show that 00

lin1 f(x) = x-+1-

L exn

(Abel).

n=O

(Hint. Show first that it is sufficient to consider the case in which 00

L

exn = 0. Let Sn = ex 0

+ ex 1 + · · · + exn and show that

n=O 00

f(z) = (I - z)

L Snzn

n=O

(z E L1(0, 1)).

260

Elements of Complex Analysis

9.3

For this S we may by Lemma 9.14(iii) choose r0 < I such that E

(r 0 < r < 1). sup {Pr(t - 8): It - 81 ~ S} < 4 M It follows that lim/ (r eft) = f( eft) (uniformly in t). r-+1-

It is obvious that J is continuous on Ll(O, 1). Given euJ E C(O, I) let rn eftn ~ eW with 'n eftn E Ll(O, 1). Then 'n ~ 1 and eftn ~ e19 • Given E > 0 it follows from (*) that there is n 1 E P such that

Since f is continuous on C(O, 1) we may choose n 2 E P such that 1/(eftn) -/(e19)1

n2 ).

If n 0 = max (nh n 2 ) we have

(n > no) so that /(rn ettn) ~ /(ei 9). This shows that f is continuous at each point of C(O, 1) and so f is continuous on .J(O, I) as required. If g is any extension off that is continuous on Lf(O, 1) and analytic on Ll(O, 1), then Problem 7.1 . gtves

g(z)

=

/(z)

(z

E

.d(O, 1))

and so g = j~ The proof is complete. PROBLEMS 9

13. Let f

E

d(Ll(O, 1)).

(i) Iff has property (E) and g is a primitive off, show that g has property (E). (ii) Iff has property (E) show that/' need not have property (E).

14. (i) If g(z)

= (z -

1) exp

G~ D(z

E

LI(O, 1)), show that g has pro-

perty (E). Observe that g has no extension that is analytic on Ll(O, I for any E > 0. (ii) If g(z)

=

0}. Since 0 < " < 2 it is easily verified that p"A is an ison1orphism of D 1 with D. The proof is complete. Note that f does not even have a continuous extension to LT(O, 1) in this case; but this is no surprise for fis unbounded. Example 9.18. Let D

= L1(0, I) n n+ and let

1 + 2iz + z 2 f( z ) = - .z ----1 - 2iz + z 2

(zED-).

Then f is an isonwrphisnz of D with L1(0, I) and f 1naps b(D) one-to-one onto C(O, 1), the senzi-circle in the upper half-plane being 1napped onto itself Proof We easily check that f is the composition of the homography g used in Exatnple 9. I 7, the mapping z ---7- z 2 and the homography of Lemma 9.9. The homography g is an isomorphism of D with the quadrant {z: Re z > 0, I m z > 0} (proof?) and the mapping z ---7- z 2 is an isomorphism of this quadrant with n+. It follows from Lemma 9.9 thatf is an isomorphism of D with LJ(O, 1). The i1nage of the boundary is easily traced under each of the three mappings. Note that the behaviour at the boundary is very satisfactory in this example.

PROBLEMS 9

J 9. If f(z)

= Hz

+ ~)

(z E C "{0}) show that f is an isomorphism of

L1(0, I) with C ""- [ -1, 1]. Detennine the inverse n1apping. I f - IE C(a, r)

and 1 E L1(a, r) show thatfis an ison1orphisn1 of Y'(a, r) with the outside of an acrofoil (Joukowski's aerofoil).

9.4

Conformal Mapping

263

20. Let D = { z: 0 < Im z < ; } and let f(z) = tan2

m

(zED).

Show thatfis an isomorphism of D with L1(0, I)" [ -1, 0] and use it to produce an isomorphism of L1(0, 1) with L1(0, 1)" [ -1, 0]. Discuss the behaviour at the boundary in both cases. (Cf. Problem 5.32.)

10 ANALYTIC CONTINUATION

In the previous chapter we considered briefly the problem of extending a function fin s/(D) to a continuous function on D-. In this chapter we are concerned to extend f to a function that is analytic on a domain larger than D. This leads to the concept of a complete analytic function and thence to con1plex analytic manifolds and Riemann surfaces.

10.1

Direct analytic continuations

We consider here the first step of extending a function fin .sd(D) to a function analytic on a larger domain. We shall see in particular that this is not always possible. A function elernent is an ordered pair (f, D) where D is a domain in C and/ E s/(D). Thus if (/1 , D 1), ( /2 , D 2 ) are two function elements we have ( /1 , D1) = (/2 , D 2 ) if and only of D1 = D 2 and / 1 = / 2 • Given a function element (f, D) we say that a function elen1ent (/1 , D 1 ) is a direct analytic continuation of (f, D) if D n D 1 =/:. 0 and

f(z) = f1(z)

(zED

n

D 1).

It is evident that (f, D) is then a direct analytic continuation of (fb D 1 ). Every function element (f, D) admits direct analytic continuations. We simply take any domain D 1 c D and / 1 = fln 1 • It is then clear that ( /1 , D 1 ) is a direct analytic continuation of (f, D). We say that such direct analytic continuations are trivial and all other direct analytic continuations are proper. As a simple example of a proper direct analytic continuation take D = .:1(0, I) and 00

f(z)

=

L

zn

(z

E

Ll(O, 1)).

n=O

Evidently (f, Ll(O, 1)) is a function elen1ent. Now let D 1

/1(z)

=

1 _ z 1

(.: 264

E

C ""{1}).

= C ""{1} and

Power Series Functions

4.4

91

(iii) Using (ii) we now obtain for z = x + iy exp (z) = exp (x

= ex

+ iy) =

exp (x) exp (iy)

{1 + ~ (iy?n} n. n =1

v2m

00

eX

=

ex{cosy + isiny}.

1

y2n + 1

00

+ m2:= 1 (- 1)In c2 111)f• + i n2=0 (- 1)n (2n +

=

{

}

1) f

.

Corollary. The function exp has no zeros.

Proof It follows frotn (ii) that exp (z) exp ( -z)

= exp (0) =

(z

1

E

C).

and therefore exp cannot have any zeros. Since the function exp is an extension of the usual exponential function on the real line we shall often write exp (z) n1ore briefly as ez. The occurrences of the exponential function in con1plex analysis are too numerous to tnention. It is thus itnportant to have as clear a geometrical picture as possible of how the exponential function behaves on C. Given.{: C---+ C \Ve say that y is a period off if

f(z

+ y) = f(z)

(z E C).

It is easy to see that such a function/repeats itself in strips of width jyj. If y is a period off so is 2y, for

f(z + 2y) = f(z + y) = f(z)

(z

E

C).

More generally we have that ny is a period ofjfor each n E Z. We say that y is a fundmnen tal period off if y is a period off, y =I= 0, and no proper subtnultiple of y is a period off If y is a fundan1ental period off so also

.

IS -y.

Proposition 4.12. The function exp has fundantental periods 27Ti, - 27Ti and these only. Given e E R, exp nzaps the strip {z: e - 7T < Im z ~ e + 7T} one-to-one onto C ""- {0}, and the do1nain {z: 8 - 7T < Itn z < 8 + 7r} oneto-one on to C ""- N 0 •

Proof We have by Proposition 4.11 (ii) and (iii) exp (z + 27Ti) = exp (z) exp (27Ti)

=

=

exp (z) {cos 27T + i sin 27T} (z E C) exp (z)

266

Elen1ents of Co1nplex Analysis

IO.I

We 1nay thus define f 2 for Re z > - 2 by f(z + 2) f2(z) = z(z + I)

(Re z > -2).

Arguing as above we sec that this gives a direct analytic continuation (f2, D 2) of(/, D), where D 2 = {z: Re z > - 2} '\,. {0, -I}. Evidently fh f 2 agree for Re z > -I by Proposition 8.I (or direct verification). Continuing in this way we n1ay extend f to the whole of C and the extended function will be n1eromorphic on C. Note that this method will clearly work for many kinds of functional relationships. Suppose now that f E ..W'(L1(0, I)). Given a E LJ'(O, I), f is regular at a and so in son1e neighbourhood of a we have by Taylor's theorem 00

f(z)

=

2

n=O

an(z - a)n.

Suppose the above power series has radius of convergence Pa· Then Pa ~ I - lal by Proposition 8.2. If Pa > I - lal and we define 00

fa(z) =

2

n=O

an(z - a)n

(z

E

LJ(a, Pa))

then (fa, L1(a, Pa)) is evidently a proper direct analytic continuation of (f, L1(0, I)). If Pa = I - Ia I we pick another point f3 E LJ'(O, 1) and hope to obtain a proper direct analytic continuation starting from {3. It may happen that Pa = I - Ia I for each a E L1(0, I). This situation prompts the following definition and exmnple. Given a function element (f, D) we say that b(D) is a natural boundary for (f, D) if(/, D) adrnits no proper direct analytic continuation. A domain D is said to be a do1nain of holonwrphy if there is a function elen1ent (/, D) for which b(D) is a natural boundary. Example 10.2. If (z

E

L1(0, I))

n=O then C(O, 1) is a natural boundary for (J, L1(0, 1)) and so L1(0, 1) is a don1ain of holonwrphy. Proof It is clear that f E ..W'(LJ(O, I)). Suppose that (J, L1(0, 1)) admits a proper direct analytic continuation. Thenf must have a continuous extension to some subarc of C(O, I) (why?). Any such arc must contain a point of the form e10 where 0 = 2prrf2q and p, q E P. We shall show below that lim lf(re 10)j T-+

1-

=

+ oo.

Analytic Continuation

10.1

267

This contradiction will then con1plete the proof.

If= = re 10 we have q-1

L

f(=) =

c:o

z

2:

+

2"

r 2 " exp

(2n-q+

prri)

1

n=q

n=O c:o

2:

= g(z) +

,.211.

n=q

It is clear that g is bounded on L1(0, 1). Let c:o

h(r)

Suppose that lin1 h(r)

=

L r 2" n=q

(0 < r < 1).

= a. Then for each N E P we have

T-+ 1-

(0 < ,. < 1) and therefore N

+

q+N

1-

li1n r-+1-

L

n=q

This contradiction shows that lin1 h(r) = T-+

r211 ~ a.

+ oo and therefore

1-

lin1 lf(re 18 )l = +oo, T-+

1-

as required. Let us now return to our original question about extending a given function in d(D) to a larger domain. If(/, D) has a proper direct analytic continuation (/r, D 1) we Inay of course extend f to a function g on the don1ain D u D 1 by taking

g(z) = {

f(z)

if zED,

/1(z) if z E D1 • It is then clear that g is analytic on D u D 1. It would seem that we could now take a proper direct analytic continuation (/2, D 2 ) of (/r, D 1) and so build an extension off on the dotnain D u D 1 u D 2 • To get such an extension we n1ust have / 2 agreeing with f on D n D 2. This n1ay in fact fail as the next example shows. Example 10.3. Let D 1 = L1(1, 1), D 2 = L1(e< 2n°13 , 1), D 3 = L1(e< 4 n 1>13 , 1), / 1 =log lv1,/2 = logtnlv2,/3 = log1rlv 3. Then (jj+ 1, D1 + 1 ) is a direct analytic continuation of(!;·, Di) (j = 1, 2) but / 1 (z) i= / 3 (z) (z E D 1 n D3 ).

268

Elen1ents of Con1plex Analysis

10.1

Proof It is clear that each (jj, Di) is a function eletnent. Let A = {rc 0 such that L:l(a, rA) c DA and (fA, L:l(a, rA)) E f. We have already seen in Proposition 10.5 that (f, L:l(a, r)) can be linked to (fA, L:l(a, rA)) by a chain of function elements of the fonn (g, L1(f3, p)). We shall show below that this chain can be chosen such that the points f3 have rational real and imaginary parts and each p is rational. The set of all finite collections of such discs forms a countable set. Thus there is at tnost a countable nun1ber of ways of fonning an analytic continuation of (f, L:l(a, r)) of the fonn (fA, L:l(a, r,,)). fn other words f has at tnost a countable nun1ber of 'values' at a. Suppose that {(It, L1(f3h p1): j = 1, 2, ... , n} are the links of a chain joining U: L1( u, r)) to (fA, L:l( a, rA)). Let tI =

111 in

E

(a, pI - a, b, pI - b)

where a = dist (f3b L:l(a, r)), b = dist (f3I, L1({3 2 , p2 )). We have !I > 0 since L:l(a, r) n L1(f3I, PI) f= 0 and L:l{fti, PI) n Ll(f32, P2) t= 0. Recall that given

94

4.4

Elenzents of Cotnplex Analysis

17. Let g: C-+ C be defined by

g(z) = {

exp (

_!)z

0

if

~

larg (z)l

4'

7T

lzl > 0,

otherwise.

Where is g differentiable? Where is g regular?

18. Use Proposition 4.9 to prove Proposition 4.11 (ii). 19. At which points do the following series converge? On which sets do they converge uniforn1ly? co

co

co

2: e-nz2,

2: e-nz,

2: 2nze- nz2,

2:

e-nz2,

""

L... e -

n2z2

.

-co

-co

0

0

0

co

co

20. Establish the following results.

= i sinh (z), sinh (iz) = i sin (z), = cosh (z), cosh (iz) = cos (z) (z E C). (ii) sin (z) = sin (z), cos (z) = cos (z) (z E C). (iii) lim e-Ysin(x + iy) = !{sinx + icosx}. (i) sin (iz) cos (iz)

y- +co

(iv)

li1n tan (x

21. Let

+ iy)

=

D 1 = {z: IRe

i.

zl

< ;}

{z: lim zl < 1}, D 4 = C ""-{(-co, -1]

D2

= {z: 0
0. Since d(p, q) = 0 we tnay choose a polygonal line joining p to q such that lr < p. It follows c ~(a, p). We n1ay now choose a chain of function elements that {(fj, Di):j = 1, 2, ... , n} linking (f, ~(a, r)) to (g, ~(a, t)) such that each Di c ~(a, p). This 1neans that eachjj is sin1ply the restriction of fto Di and so f and g agree on ~(a, p), i.e. p = q. It is clear from the definition of d that d(p, q) = d( q, p) (p, q E Mr). It ren1ains to prove the triangle inequality. Let p, q, r be any points of Me. Let be a polygonal line joining p to r and let T' be a polygonal line joining r to q. Then u T' is a polygonal line joining p to q and so we have

r

r

r

d(p, q) ~ Ir This is true for any such pair

d(p, q)

~

r,

u

r ~ Ir

r

+ Ir.

T' and so

inf {lr: r joins p to r} + inf {lr,: T' joins r to q}

= d(p, r)

+ d(r, q).

The proof is now con1plete. Our next obvious step is to investigate the properties of the metric space (Mh d). Is it complete, or connected, or compact? In answering such questions for a metric space it is often helpful to have a clear picture of what the open balls look like. In fact it is usually enough to know what the open balls look like when the radius is sn1all enough. The next result shows that they sin1ply look like open discs in the complex plane.

278

Elements of Conzplex Analysis

10.3

Theoren1 10.8. Let f be a co1nplete analytic function and let p E Mr with p = (a, .f, L1 (a, r)). For 0 < € < r, B(p, e) is isonwtric with L1(a, E).

Proof Let q E B(p, €) and let q = ({3, g, L1(f3, t)). We have Ia - f31 ~ d(p, q) < € so that f3 E L1(a, t:). Since d(p, q) < E we may choose a polygonal line r joining p to q such that lr < €, It follows that r c L1(a, t:) and so by the argun1ent used in Theorem 10.7, (g, L1(f3, t)) is a direct analytic continuation of(j, L1(a, r)). Thus ({3, g, L1(f3, t)) = (f3,f, L1(a, r)) and so B(p, t:) c {({3,/, L1(a, r)): f3 E L1(a, t:)}.

For any q = (f3,f, L1(a, r)) we may join p to q by the line segment [a, {3] so that d(p, q) = Ia - f31. This means that

= {({3,/, L1(a, r)): f3 EL1(a, t:)} and that the n1apping (f3,f, L1(a, r)) -7 f3 is an isometry of B(p, t:) with B(p, t:)

L1(a, t:). We note further that M, is a connected metric space. This is readily proved from the definition of connectedness; alternatively we may use the method of Theorem 2.I since we can join points of Mr by polygonal lines. The metric space Mr is never compact and need not be complete (see Problem 1O.I3). Before proceeding further with the general theory it is helpful to form some intuitive geometrical pictures of Mr for various complete analytic functions f. If f is determined by a function f meromorphic on C then M, simply looks like C (minus the poles of f). If f is determined by the function element (f, L1(0, I)) of Example 10.2, then M, simply looks like L1(0, I). Suppose now that f is determined by the function element (p!, L1(1, 1)). It is not difficult to see that the function elements of f of the form (f, L1(a, r)) are simply (pZ, L1(a, Ia!)) where a E C ""- {0} and 8 = arg (a) + 2nrr. For any disc Ll(a, Ia!) there are essentially only two different functions, namely PZ and p~ + 2 n. Thus each point a of C ""- {0} is associated with two points of M,. This suggests that M, should look something like two copies of C ""- {0}. Since Mr is connected these copies must so1nehow be joined together. To sec how we should pcrforn1 the joining let us note how points of M, arc joined (in the sense introduced above). To join (I,p!, LI(I, I)) to (I, P~n' L1(l, I)) we n1ust take a polygonal line which goes round the origin once in the positive sense. If we go round the origin once more in the positive sense we come to the point (I, p!," L1(I, 1)) i.e. (l,p!, L1(1, 1)); in other words we return to our starting point. A similar situation will obtain for circuits starting from any point of M 1• This sug-

I0.3

Analytic Continuation

279

gcsts the following construction. Take two copies of C ""'{0} placed one above the other and cut then1 along the positive real axis. Take the lower tlap of the botton1 plane and paste it to the upper flap of the top plane; now take the lower flap of the top plane and paste it to the upper flap of the botto1n plane. This last operation is of course itnpossible to realize, but we have at least fonncd son1e idea as to what Mr looks like in this case. Renzark. The above space Afr is of course not con1pact. If however we add on two other points corresponding roughly to 0 and oo then it is possible to turn this extended space into a cotnpact connected tnetric space (by a 111ethod akin to that used for Ceo). It is this extended Space which is usually referred to as the Rienzann swface associated with the function pt. It is a re1narkable fact that the extended space is actually honleOnlorphic with Ceo. Suppose now that f = log. We tnay fonn a picture of Mr by a tnethod sitnilar to the one used above. Only, in this case we n1ust take a copy of C ""'{0} for each 11 E Z and paste then1 all together suitably to fonn an infinite continuous spiral staircase. The student n1ay anntse hitnself by producing tnodels for Mr in n1any other cases (see Problen1 10.12). To return to the general case, Theoren1 10.8 tells us that we tnay regard Afr as being n1ade up of lots of open balls each looking like a disc in C. Let us exatnine this situation a little n1ore closely. Given p E M 1 let CfJv be the hon1eon1orphistn of B(p, e) with Ll(a, e) given in Theoren1 I 0.8. Thus

({3

E

Ll(a, e)).

\Ve say that the tnappings cpp (p E Me) fonu a fatnily of coordinate n1appings for Mr. There are clearly 1nany other fan1ilies of homeotnorphistns of the balls B(p, e) with open discs inC, but we shall use only the above tnappings. It is significant to know how the coordinate tnappings con1pare on overlapping balls. We n1ake the con1parison as follows. Let p, q E Mr with associated coordinate 1nappings cpp: B(p, e)--+ C, CfJv: B(q, o)--+ C and B(p, e) n B(q, o) =1- 0. If

Upq = cpp(B(p, e) n B(q, o)) then Upq. is a connected open subset of L1(a, e) and we 1nay define the tnapping ifipq: Upq--+ C by ,/, 'f'pq = cpq 0 C{Yp-1 . The tnappings ~'pq are called coordinate transfonnations. In this case it is trivial to verify that each ifipq is sin1ply a restriction of the identity Inapping. In particular, the coordinate transforn1ations are conforn1al 1nappings.

280

Ele1nents of Complex Analysis

10.3

We have now developed sufficient motivation to introduce the concept of a 1nanifold. The student should understand that we shall only be considering very special cases of 1nanifolds; our definitions are by no means the tnost general possible. For our purposes we shall define a nzanifold (of con1plex din1ension 1) to be a connected metric space M such that for each p EM there is a homeon1orphis1n (/)p of so1ne B(p, e) with an open disc in C. The n1appings { 0 we 1nay choose S > 0 such that the first integral on the right is less than E/2. With this fixed we tnay now choose N 0 such that the second tenn on the right is less than E/2 whenever N > N 0 • This completes the proof.

o

We shall now show how the formula in Theorem 10.12 enables us to extend ~ to the whole plane. Let us define P and Q for Re z > 1 by

1 1

P(z) =

Idt

t

o e -

=J oo

Q(z)

1z-1

1

1z-1

t dt. e - 1

Using the methods employed for r we may easily show that the formula for Q(z) actually defines an entire function. Recall from Example 7.13 that

t = e - 1 t

Bntn

co

2:-, 1l.

(t

E

[0, 1])

n=O

the convergence being uniform on [0, 1]. We deduce for Re z > 2 that P(z)

=

co

L

n=O

_

11

B 1n+z-2 n

,

/l,

0

dt

~ Bn n = o 1l !(n + z - I)

_ 1 _ _!_ - z - I 2z

+

~ B2n , n = 1 (2n) !(2n + z - I)

Using the fact that

lim

n-. co

B

l/2n 2n

(2n)!

1 < 1 21T

=-

4.5

Power Series Functions

99

so that each function p~ is a 'square root' function. Moreover for each ==I= 0 P~+2n(z) = exp {-!-logo+2n (z)} = exp {-!loge (z) + 1ri} = - p~ (z). This coincides with the intuitive notion of 'positive' and, negative' square roots. On the other hand when we take A to be con1plex our intuition breaks down in the face of forn1ulae such as

pb(i) = exp (i log (i)) = exp (;; i) = e-•1 2 • In line with our earlier practice we write p~ n1orc sin1ply as p'A. If A = f-L + iv and t > 0 then

p'A(t) = exp {(f-L + iv) log t} = t 1t{cos (v log t) + i sin (v log t)}. We shall so1netin1es \Vrite t'\ for p"A(t) (t > 0) to show its close relation with the corresponding real function.

Proposition 4.19. Given A, f-L

E

C,

eE R we hare

(i) p~+Jl = p~p~ (ii) p~ E d(C ""- Ne) and (p~)' = Ap~ - 1 • Proof (i) p~+ 11 (z)

exp {(A + f-L) loge (z)} = exp {A loge (z)} exp {f-L loge (z)} =

=

p~(z)

pHz).

(ii) It follows frotn Propositions 4.16 and 3.8 that p~ is analytic on C ""- Ne and (p~)'(z) = exp' (A loge (z))A log~ (z) A

= - p~(z) z

=

;\p~-I(z).

If ;\ is an integer then p~ is in fact analytic on C ""- {0}. If A is not an integer we may use the argun1ent of Proposition 3.5 to show that p~ is not continuous at any point of Ne ""- {0}, and hence in this case each point of Ne ""- {0} is a singularity of p~. It may also be verified (see Problen1 4.28) that given a E C ""- Ne we may represent p~ by a power series on son1e disc about a. In particular using the technique of Exan1ple 4.18 we 1nay verify that oo (z - 1)n p'A(z) = 1 + L A( A - 1) ... (A - n + 1) (z E Ll(l, 1)). 1 nal

n.

Analytic Continuation

10.4

289

\Ve turn finally to the question of the zeros of l;. The distribution of these zeros turns out to be of considerable significance in the theory of nutnbers. It is clear fron1 our earlier considerations that has zeros at -2, -4, - 6, .... 1t was shown by Rietnann that satisfies the equation

s

s

(z

E

C).

s

This equation enables us to deduce the behaviour of for Re z < 1- frotn its behaviour for Re ~ > -!-. In particular it rnay be used to prove that the only zeros of~ outside the strip 0 ~ Re z ~ I occur at -2, -4, -6, .... It is known that ~ has an infinite nutnber of zeros in the strip 0 ~ Re z ~ 1, that none lie on Re ~ = 0 or Re ~ = 1, and that an infinite number lie on the line Re z = 1-. There is a conjecture of long standing that all the zeros in this strip lie on the line Re z = t· The conjecture has not yet been proved or disproved and is known as the Rienwm1 lrypothesis. PROBLEMS 10

16. Show that r has exactly one absolute 1ninin1un1 point in {x: x > 0} and that it lies in (1, 2). 17. Let

F(-) - r(-) .t:.

-

.t:.

oo

-

(

l)n

"' - -~ n~o 11 !(z + 11)

(z E C) .

Show that F is an entire function and that (Rez>1).

18. Show that

r(z) = r(z)

(z

E

C)

and deduce that

lr(iy)l = 1rj(y sinh (1ry))

(y E R "'{0}).

19. Let f-L: P-+ R be defined by f-L(l) = 1 and f.L(ll) = {

( -1Y if n is the product of r distinct prime factors, 0

otherwise.

Show that for Re z > 1

1 s(z) s(z) s(2z)

=

I

fL(11)

n= 1

liZ

~

lfL(l:)l.

n= 1

n~

290

10.4

Elements of Complex Analysis

20. Let d(n) denote the number of divisors of n and let ak(n) denote the sum of the kth powers of all the divisors of n. Show that

(~(z)) 2 = ~(z)~(z

- k)

=

I I

n=-1

n=l

d~) n

ak~n) n

(Re z > 1) (Re z > max (1, k

+

1)).

100

Elements of Complex Analysis

4.5

PROBLEMS 4

24. Given n E P show that n log ( 1 + where l/n(z)j

~

~)

=

z + fn(z)

(z E ..J(O, !n))

lzl 2 (z E Lf(O, !n)). Deduce that n

lim n-+co

(1 + n~)n = exp (z)

(z E C)

and show that the convergence is uniform on any compact subset of C. 25. Let D be a domain in C with 0 ¢ D. We say that x: D continuous argument on D if xis continuous and x(z)

= arg (z)

-~

R is a

(mod 21r).

Show that there is a branch-of-log on D iff there is a continuous argument on D. 26. Given zh z 2 E C "- Ne find necessary and sufficient conditions for loge (z1z2)

= loge (z1) + loge (z2).

27. Let D be a domain in C such that for some R > 0

C(O, R) c D c C "- {0}. Show that there is no branch-of-log on D.

28. (i) Letfbe a branch-of-log on D. Given L1(a, r) c D represent/ on L1(a, r) by a power series. (ii) Given a E C "- Ne represent p~ by a power series on some disc about a. What is the largest such disc? 29. Given A = p,

+ iv show that IP~(z)j =

!zltt e-v argo.

30. Discuss the convergence of the following series. co

L

n=O

P~nn(z),

292

A.l

Elen1ents of Con1plex Analysis

linear co1nbinations of monotonic functions belong to .%'i"'([a, b]). We then show in Proposition 2 that we obtain all the functions of .~i'([a, b]) in this way. Proposition 1. .%'i''([a, b]) is a linear space.

Proof Let f, g

E

afi/([a, b]) and let P

T(f + g, P)

~ ~

E

f!J. Then

+ T(g, P) V(f; a, b) + V(g; a, b).

T(f, P)

It follows that

V(f + g; a, b)

~

V(f; a, b)

+ V(g; a, b)

and so f + g E &ai"'([a, b ]). Given a E C we easily see that V(af; a, b) = la!V(f; a, b) so that afEYJ-•F'([a, b]). This completes the proof. We now make son1e simple observations. Since {a, b} is a subdivision of [a, b], iff E afi/([a, b]) we have

lf(b) - f(a)[

~

V(f; a, b).

Suppose now that [c, d] c [a, b]. Given P E f!J[c, d] we may clearly extend P to be a subdivision P' of [a, b] by adding the points a and b. It follows that for each K E f!J[c, d]

[f(c) - f(a)l + T(f, P) + [f(d) - f(b)l ~ T(f, P') ~ V(f; a, b) Therefore /l[c. dJ

E

YJ:I'([c, d]) and V(f; c, d)

~

V(f; a, b).

If a < x < b we have in particular that [a, x]. Further if x ~ y ~ b we have that

V(f; a, x)

~

f

is of bounded variation on

V(f; a, y).

We need one other observation before the next result. Given/ E 88-i/'([a, b]) and c E (a, b) we have

V(f; a, b) = V(f; a, c) To sec this Jet P 1 have

E

.tJt'[a, c], P 2

E

+ V(f; c, b).

:Y'[c, b] so that P 1 u P 2

and this gives

V(f; a, c)

+

V(f; c, b)

~

V(f; a, b).

E

&[a, b]. We then

Rien1a1m-St ie!tjes Integration

A.2

293

Conversely, given P E &'[a, b] we 1nay adjoin c to P (if necessary) to fonn P*. Evidently there exist P 1 E .~[a, c] and P 2 E &[c, b] such that P* = P1 u P2. Then

T(f, P) ~ T(f, P*) = T(f, P 1) + T(j, P 2) ~ V(f; a, c) + V(f; c, b) and so it follows that

V(f; a, b)

~

V(f; a, c)

+

V(f; c, b).

Proposition 2. Given f E 91i"([a, b]) there exist four non-negative nondecreasing ji1nctions fk such that

f

=

fl -

/2 +

i(f3 - !4).

Proof It follows easily from the inequalities IRe zj ~ 1=1, IIn1 zj ~ jzj that Ref, linfE 91i '([a, b]). It is thus sufficient to show that if f E @i/'([a, b]) is real valued then there exist non-negative non-decreasing functions g, h such that f = g - h. We 1nay choose M > 0 such that f(a) + A1 ~ 0. Define g and lz on [a, b]) by g(x) = f(a) h(x) = f(a)

+ +

M M

+ +

V(f; a, x) V(f; a, x) - f(x).

It is then clear that g is non-negative and non-decreasing and f We have h(a) = M > 0 and if a ~ x < y ~ b then

=

g - h.

h(y) - h(x) = V(f; a, y) - V(f; a, x) - (f(y) - f(x)) = V(f; x, y) - (f(y) - f(x)) ~ 0. This shows that h is both non-decreasing and non-negative. The student shoud observe that the decon1position of a real function of bounded variation into the difference of two non-negative non-decreasing functions is not unique. For example iff= g - h is one such decomposition and if M > 0 then f = (g + M) - (h + M) is another such decoinposition. We shall return to this remark in the next section.

A.2

The Riemann-Stieltjes integral

We begin by defining the integral for a special case and we then go on to define the integral for the case we wish to consider. Let a, bE R with a < band let/be a bounded real function on [a, b] and g a non-decreasing

Elements of Complex Analysis

294

function on [a, b]. Let P = {t,:j We define the norm of P by

v(P)

= 0,

= max {t 1

-

A.2

1, ... , n} be an element of gJ[a, b].

=

t1 _ 1 : j

1, ... , n}.

For j = 1, ... , n we define M1

m1 For each P

E

= sup {f(t): t1 _ 1 < t < t1} = inf {f(t): t1 _ 1 < t < t1}.

9 we define n

S(f, g; P) =

L M {g(t 1

1) -

g(t1 -t)}

i=l n

s(f, g; P) =

L: m1{g(t1)

-

g(t1 -1)}.

1=1

We define the upper and lower Riemann-Stieltjes integrals off with respect tog by

r

fdg = inf {S(f, g; P): p E 91'}

J_ f dg = sup {s(f, g; P): P

E

91'}

respectively. If these two integrals are equal we say that f is RiemannStieltjes integrable with respect to g. We then denote their common value by

J!dg or J:fdg or J:!(t)dg(t) and call this the Riemann-Stieltjes integral off with respect to g.

Theorem 3. Iff is a continuous real function on [a, b] and g is nondecreasing on [a, b] then f is Riemann-Stieltjes integrable with respect to g. If his another non-decreasing function on [a, b] then

J:!d(g

+h)=

J:!dg

+

J>dh.

Proof Since f is continuous on the compact set [a, b ], f is bounded, say Ill ~ M. Given P E 9 we easily verify that - M{g(b) - g(a)} s{f, g; P)

~

~

S(j, g; P) M{g(b) - g(a)}.

A.2

295

Riemann-Stieltjes Integration

It follows that

r

-M{g(b)- g(a)} ,;;

J_ fdg ,;; M{g(b) -

fdg

g(a)}.

It is clear that for any P E 9

s(f, g; P)

~

S(f, g; P).

Given P1, P2 E 9, let P 3 be the subdivision of [a, b] formed from P1 u P 2. We now verify that

s(f, g; P1) ~ s(f, g; P3) ~ S(f, g; P 3) ~ S(f, g; P2) and it follows from this that

r r 1 dg

,;;

1 dg.

Since f is continuous on the compact set [a, b], it is uniformly continuous. Given £ > 0 there is thus S > 0 such that

tb t2 E [a, b], !t1 - t2l < S =>

!f(tl) - j(t2)!
0 show that I/fEPJJf/([a, h]) with b) V(f; a, b) V(I/.11 J; a, ~ (inf (/)) 2

3. Given f

E 86'i~"'([a,

b]) let

IIIII =

+ 2 V(f;

lf(a)l

a, b).

Show that 86'1/'([a, b]) is a Banach algebra with respect to 2.6).

II ·II (see Problem

4. Give exan1ples of the following situations. (i) f E ~'C([a, b]), f tt 88i~"'([a, b]). (ii) f E 86'f/([a, b ]), f ¢ CC([a, b ]). 5. Given f E ~'C([a, b]) n 86'i""([a, b]) show that the mapping x---+ V(f; a, x) is continuous on [a, b ]. 6. Given f E 86'i""([a, b]) show that f has one-sided limits at each point of [a, b]. Let E be the set of points in [a, b] for which one of the following inequalities holds: lim f(t h-0+

+ h)

=1- f(t),

lim f(t h-0+

li1n f(t - h) =1- f(t), h-0+

+ h)

=1-

lin1 f(t - h). h-0+

Show that E is at most countable. It follows that f is continuous except on a countable set. 7. Let g be a step function on [a, b ], i.e. there is {t 1 : j = 0, ... , n} E .9[a, b] such that g is constant on each (t 1 _ 1 , t1), say f3h and g(t 1) = y 1• Show that g E .98r([a, b]) and determine V(g; a, x) for x E [a, b ]. Show that any bounded function/is Rien1ann-Stieltjes integral with respect to g. Evaluate f~ f dg. 8. Let fbi~

E

f f f,

re([a, b]),

(A,f,

+

AI, A2 E

A.j2) dg,

d(A,g,

+

.\2g2)

C, gb g2

=

A,

=

A,

f/ f f,

1

E

.18i''([a, b]). Show that

dg,

+

A2

J: /

dg 1

dg,

+

A2

f /1

dg2.

2

ElenJents of Con1plex Analysis

104

5.1

To illustrate the above definitions, note that any arc of the form [a, ,8] (a, ,8 E C, a =1= ,8) is smooth. Now let be the sen1i-circular arc represented by

r

y(t) = a

+ r exp (7Tit)

(t E·[O, 1]).

Then y is differen tiablc on (0, 1) and

y'(t) = 7Tir cxp (7Tit)

(t

E

(0, 1))

so that r is smooth. To obtain a sitnplc example of a piecewise smooth arc let ab a 2 , a 3 be distinct points of C such that a 3 ¢: {ta 1 + (1 - t)a 2 : t E R}. Then [a 1 , a 2 ] U [a 2 , a 3 ]

is easily shown to be a piecewise stnooth arc.

PROBLEMS 5 1. Let cp be a strictly monotonic real function on [0, 1] with range [0, 1]. Show that cp E C/>. 2. Give examples of functions in C/> that are differentiable except at (i) one point (ii) n points (iii) a countable infinity of points. 3. Use the Weierstrass continuous nowhere differentiable function to produce a representative function y for an arc such that y is nowhere differentiable. 4. If y(O)

= i

and y(l)

=

1 sin (

~) +

i exp (1) (IE (0, 1]), show that y

represents an arc which is not rectifiable.

r

r

5. Let be an arc and let y be a representative function for which is of bounded variation. Use Lcn1n1a 5.2 to show that every representative function for r is of bounded variation. 6. Let r be a sn1ooth arc. Use Problcn1 5.2 to produce a representative function for r that is not differentiable. 7. Let

r

be the arc [0, 1]

u [1, 1 + i].

1 - (t - I) 2 { p.(t) = 1 + i(t - I ) 2

Usc the function

(t (t

E

[0, 1])

E

(I, 2])

to show that Tis stnooth. More generally let T be any piecewise stnooth arc with representative function y which has one-sided derivatives at the points where y is not differentiable. Show that r is SI1100th.

SUGGESTIONS FOR FURTHER STUDY

In the course of the book we have attempted to introduce the student to the basic ideas of con1plex analysis. We hope that the student's appetite has been whetted to read deeper and wider in this subject. Almost every theorem that we have proved in this book was known in son1e form to analysts at the end of the 19th century; but complex analysis has not stood still since 1900 and this century has seen n1any developments in the theory. We give here some descriptive ren1arks about a few of these developments. It is in order to begin by mentioning some of the more elementary aspects of the dassical theory that we have omitted. Apart from the gamma and zeta functions we have not discussed many of the special functions of complex analysis. In particular we have neglected the rich theory of doubly periodic functions (or elliptic functions). Other standard functions such as those of Legendre and Bessel appear only in veiled references in the problems. We have also omitted the elementary theory of asymptotic expansions. The student may recall that several theorems were discussed in the text without proof, e.g. Runge's theorem, Mantel's theorem and the Riemann mapping theorem. He may now wish to take up these topics in some of the references suggested in the bibliography e.g. Ahlfors, Cartan, Hille, Saks and Zygmund. He may also wish to study some advanced topics such as the Nevanlinna theory of meromorphic functions, analytic nun1ber theory, cluster set theory, and Riemann surfaces. At several points in the book we attempted to show the value of interrelating the algebraic and analytic aspects of a problem. This is one of the themes that has received much attention in several branches of a large subject known as functional analysis. To give a simple illustration let A be the disc algebra (see Problem 8.17), i.e. the complex Banach algebra of all complex functions that are continuous on Lf(O, 1) and analytic on Ll(O, 1). The disc algebra is dearly a dosed subalgebra of ~(Lf(O, 1)). Wermer has proved the interesting fact that A is actually a maximal closed subalgebra of l{f(Lf(O, 1)), i.e. that l{f(Lf(O, 1)) is the only closed subalgebra t_hat properly contains A. Conversely certain maximal dosed suba1gebras of~(Ll(O, 1)) admit properties of an analytic nature. Now let B be the algebra H 00 (L1(0, 1)) (see Problem 8.17), i.e. the complex Banach algebra of all complex functions that are analytic and bounded on Ll(O, 1). We may regard A as a closed subalgebra of B by identifying A with those functions in B that have property (E) of 303

304

Ele1nents of Cornplex Analysis

Chapter 9. The algebra B is much larger than the algebra A and its properties are n1uch more complicated. It follows from a theorem of Gelfand (for general commutative Banach algebras) that B can be represented as a subalgebra of the continuous functions on a certain compact topological space, called the carrier space, which contains a homeomorphic image of the open disc LJ(O, 1). It follows from a theorem of Carleson that the image of the disc LJ(O, 1) is even dense in the carrier space (this remarkable result is called the corona theorem). The carrier space consists of the set of maximal ideals of B endowed with a suitable topology. Certain analytic problems are intimately related to properties of this set of maximal ideals. For example, a sequence {zn} in LJ(O, 1) is called an interpolation sequence if for each bounded sequence {wn} of complex nun1bers there exists f E B such that f(zn) = Wn (n E P ). The characterization of such sequences is intimately related to the corona theorem. Many topics of similar nature are being investigated at the present time (Carleson's corona theoren1 dates from 1963) in a subject known as function algebras. In elementary complex analysis we have been studying functions of the form f: D ->-C. We can generalize in two directions. In the first place we may allow our functions to take their values in a more general space than the complex numbers. This programme has been carried through in functional analysis where one considers functions with values in a complex Banach algebra, for example. It is quite simple to define analytic in this context and it is then possible to obtain extensions of the classical theory. For example, Liouville's theorem extends in the expected manner, and as such it formed a significant tool in Gelfand's original development of the theory of commutative Banach algebras. This generalized complex analysis also forms a basic tool for the standard treatment of the topic known as spectral theory. We n1ay also generalize in the opposite direction and consider complex valued functions of several complex variables. The standard definition of analytic in this case is given in terms of power series expansions in several complex variables. Although some theorems generalize (for example the important maximum modulus principle) the student should be forewarned that the theory of functions of several complex variables is quite different from the theory for a single variable. For example, in two or more variables an isolated singularity is automatically removable. Moreover there are quite innocent-looking open connected subjects of C 2 such that every function that is analytic on the set admits an extension to a function that is analytic on a strictly larger open connected set. We have indicated in this book that the technical difficulties in obtaining the most general theorems for functions of a single complex variable arc usually of a topological nature. These difficulties arc multiplied for several complex variables, not least by the fact that it is not possible (for the common mortal) to visualize more than three real dimensions. I ndccd it is not possible to understand even the statements of some of the deeper theorems for several complex variables without a knowledge of some of the language of algebraic topology. Advances in mathematical research and teaching have brought elementary complex function theory within the scope of an ever-widening audience, and it is to be hoped that the same may be done in the future for functions of several complex variables.

BIBLIOGRAPHY

This selected bibliography is in three parts. The first consists of collateral reading. Ahnost all the books in this section contain n1uch more material than the present book. The second section consists of advanced topics in con1plex analysis. The final section gives a few references to generalizations. I have n1ade no attempt to give a complete bibliography, and the selection reflects n1y own personal tastes.

A. Collateral Ahlfors, L. V., Complex Analysis, (2nd ed.), McGraw-Hill, New York, (1966). Boas, R. P., Jr., Entire Fnnctions, Academic Press, London, (1954). Cartan, H., Elementary Theory of Analytic Fnnctions of One or Several Variables, (translated from the French), Hermann, Paris, (1963). Gleason, A. M., Fundamentals of Abstract Analysis, Addison-Wesley, Reading, Mass., (1966). Hille, E., Analytic Function Theory, 2 volumes, Blaisdell, New York, (1963). Markushevich, A. I., Theory of Functions of a Complex Variable, 3 volumes, (revised English edition), Prentice-Hall, London, (1965). Nehari, Z., lntrodnction to Complex Analysis, Allyn and Bacon, Boston, (1961). Rudin, W., Real and Complex Analysis, McGraw-Hill, New York, (1966). Saks, S. and Zygmund, A., Analytic Fnnctions (2nd English ed.), Polish Scientific Publishers, Warsaw, (1965). Titchmarsh, E. C., The Theory of Functions, (2nd ed.) Oxford University Press, Oxford, (1939).

B. Advanced Collingwood, E. F. and Lohwater, A. J., The Theory ofC/nster Sets, Cambridge University Press, Cambridge, (1966). Hayman, W. F., Meromorphic Functions, Clarendon Press, Oxford, (1964). Heins, M., Selected Topics in The Classical Theory of Functions of a Complex Variable, Holt, Rinehart and Winston, London, (1962). Springer, G., Introduction to Riemann Swfaces, Addison- Wesley, Reading, Mass., (1957). 305

306

Elements of Complex Analysis

Titchmarsh, E. C., The Theory of the Riemann Zeta-Function, Oxford University Press, Oxford, (1951). Whyburn, G. T., Topological Analysis, (revised ed.), Princeton, New Jersey, (1964).

C. Generalizations Gunning, R. C. and Rossi, H., Analytic Functions of Several Complex Variables, Prentice-Hall, London, (1965). Hoffman, K., Banach Spaces of Analytic Functions, Prentice-Hall, London (1962).

Simmons, G. F., Introduction to Topology and Modern Analysis, McGraw-Hill, New York (1963). Seminars on Analytic Functions, Institute for Advanced Study, Princeton University, New Jersey, Vol. II, Seminar V, (Analytic Functions as Related to Banach Algebras), (1958).

INDEX OF SPECIAL SYMBOLS

d(D) 63 A(D) 243 AD 196 A(a; R, r), A(a; R, r) 40 arge 38 (a, {3), (a, {3], [a, {3), [a, {3] 40 81'i/'([a, b]) 66 Bn 169 B(p, €) S b(E) 10 36

Hom (X, Y) 13 l(D1, D2) 243

51 ct'(E) 55 C(a, r) 40

int {F) 120 j 57 lim 5 lim 77 loge 97 lr 127 JI(D) 185 Ma(r) 235 Mt 276 Ne 39 n(F; p) 205 w 109

deg 56

p

2

diam 26 dist 25 d* 52 J(a, r), J(a, r), .1'(a, r) tff 181 £ 0 , E- 10 exp 90 ext (F) 120 f 269 !IE 3 f/J 102 f/J +' f/J- 103 r 283 r 101 r* 122 [y] 144 .7t'(D) 71 H 00 (D) 196 H(D ,· z0) 249

P1

185 98

c coo

p~

40

n+ ntp ' 113

251

lJf+, tp_

115

Q 2 R 2 Res(/; a) 164 [/ 181 S(f; a) 164 u

z z, ~

56 2 177 194

56 0'" 1

1

Sr f

11·11

307

124 42

5.2

I 07

Arcs, Contours, and Integration

Then y 1naps [0, 2] one-to-one onto r. It is now sufficient to show that y is continuous. It is obvious that y is continuous except possibly at t = 1. Since li1n y(t) = lin1 y 1 (t) = z1 t-+1-

t-1-

liin y(t) t-+1+

= li1n

y 2 (t)

t-+1+

=

Z1

it follows that y is indeed continuous at t = 1. We have seen a sin1ple exa1nple of this situation in the last section for the case F 1 = [a 1 , a 2 ], r2 = [a2, a3] where a3 does not lie on the infinite line detennined by a1 and a 2 • Consider now the opposite process. Given an oriented arc we say that oriented arcs Fr (r = I, 2, ... , n) fonn a direction-preserving decmnposition of r if

r,

r

u {rr :

(i) = r = I ' 2, ... ' II} (ii) the first point of r1 is the first point of r (iii) the first point of Fr + 1 is the last point of Fr for r = I, ... , n - I.

Figure 5.2

In other words the 'pieces' rr are joined 'head to tail' as in Figure 5.2. As an exan1ple suppose that r is represented by y and let tr be such that 0

=

t1