Lecture Notes on Complex Analysis

van F r a n c i s Lecture Wilde Notes on COMPLEX AN ALYS i s Imperial College Press Lecture Notes on COMPLEX A...

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van F r a n c i s

Lecture

Wilde

Notes

on

COMPLEX AN ALYS i s Imperial College Press

Lecture

Notes

on

COMPLEX ANALYSIS

This page is intentionally left blank

Lecture

Notes

on

COMPLEX AN ALYS i s

van F r a n c i s W i l d e

Imperial College Press

Published by Imperial College Press 57 Shelton Street Covent Garden London WC2H 9HE Distributed by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

LECTURE NOTES ON COMPLEX ANALYSIS Copyright © 2006 by Imperial College Press All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 1-86094-642-9 ISBN 1-86094-643-7 (pbk)

Printed in Singapore by World Scientific Printers (S) Pte Ltd

To Erica


Preface

This text forms what is often referred to as "a first course in complex analysis". It is a slight enhancement of lecture notes first presented to undergraduate students in the Mathematics Department of Bedford College, University of London, as part of the Mathematics BSc. degree, and then given for many years in the Mathematics Department of King's College, London. During this time they have been continually revised, reorganized and rewritten. The aim was to provide a rigorous and largely self-contained but extremely gentle introduction to the basics of complex analysis. The audience for the course comprised not only single subject mathematics BSc. and MSci. students but also a number of final year joint honours students as well as postgraduate students who missed out on the subject in their undergraduate programme. There are a number of core topics (such as Cauchy's theorem, the Taylor and Laurent series, singularities and the residue theorem) which simply must be offered to any student of complex analysis. However, quite a bit of preparation is required, so these important results unavoidably tend to appear in rather rapid succession towards the end the course. This leaves very little room for extra topics, especially if they are particularly complicated or involve a lot of additional machinery. The presentation here is for the benefit of the student audience. There has been no quest for ultimate generality nor economy of delivery. Nowadays, it seems that many students do not get to see an account of metric spaces, so this aspect of complex analysis has been presented in quite some detail (in Chapter 3). It is then but a small step for the student wishing to go on to study metric spaces in general. The exponential and trigonometric functions are defined via their power series expansions in Chapter 5, so a certain amount of manoeuvring is required to extract

vii

viii

Lecture Notes on Complex

Analysis

those properties familiar from calculus—for example, the appearance of the number IT is carefully explained. Those for whom this is familiar territory can quickly press on. Most of the core results are contained in Chapters 8-12. The next two chapters, covering the maximum modulus principle and Mobius transformations have been moved around a bit over the years. For example, the maximum modulus principle (Chapter 13) could be discussed anytime after having dealt with Cauchy's integral formulae (Chapter 8). The treatment of Mobius transformations (Chapter 14) is essentially a stand-alone topic so could fit in almost anywhere. It might well be read after Chapter 8 so as to provide a little variety before embarking on the study of the Laurent expansion in Chapter 9. It has to be admitted that the final section of Chapter 13 (on Hadamard's Theorem), possibly Chapter 15 (on harmonic functions) and Chapter 16 (on local properties of analytic functions) could be considered a bit of a luxury. In practice, they were all usually squeezed out because of lack of time. Most of the rest of the material in these notes just about fits into a one semester course. The majority of students embarking on this subject will have studied calculus and will usually have also been exposed to some real analysis. Nevertheless, experience has shown that the odd reminder does not go amiss and so an appendix containing some pertinent facts from real analysis has been included. These are all consequences of the completeness property of R (so tend not to be very carefully covered in calculus courses—or else are deemed obvious). A number of text books were consulted during the preparation of these notes and these are listed in the bibliography. No claim is made here regarding originality. As an undergraduate student in the Mathematics Department of Imperial College, it was my privilege to be taught analysis by M. C. Austin and Professor Ch. Pommerenke. Their lectures could only be described as both a joy and an inspiration. It is a pleasure to acknowledge my indebtedness to them both. /. F. Wilde

Contents

Preface 1.

Complex Numbers 1.1 1.2 1.3 1.4 1.5 1.6

2.

3.

vii 1

Informal Introduction Complex Plane Properties of the Modulus The Argument of a Complex Number Formal Construction of Complex Numbers The Riemann Sphere and the Extended Complex Plane . .

1 2 4 8 12 14

Sequences and Series

17

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

17 17 18 21 23 24 25 26

Complex Sequences Subsequences Convergence of Sequences Cauchy Sequences Complex Series Absolute Convergence n t h -Root Test Ratio Test

Metric Space Properties of the Complex Plane

29

3.1 3.2 3.3 3.4 3.5

29 32 34 36 38

Open Discs and Interior Points Closed Sets Limit Points Closure of a Set Boundary of a Set ix

x

4.

5.

6.


3.6 Cantor's Theorem 3.7 Compact Sets 3.8 Polygons and Paths in C 3.9 Connectedness 3.10 Domains

40 41 49 51 56

Analytic Functions

59

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

59 59 61 66 70 73 74 77

Complex-Valued Functions Continuous Functions Complex Differentiable Functions Cauchy-Riemann Equations Analytic Functions Power Series The Derived Series Identity Theorem for Power Series

The Complex Exponential and Trigonometric Functions

79

5.1 The Functions exp z, sin z and cos z 5.2 Complex Hyperbolic Functions 5.3 Properties of exp z 5.4 Properties of sin z and cos z 5.5 Addition Formulae 5.6 The Appearance of TV 5.7 Inverse Trigonometric Functions 5.8 More on exp z and the Zeros of sin z and cos z 5.9 The Argument Revisited 5.10 Arg z is Continuous in the Cut-Plane

79 80 80 83 84 86 89 91 92 94

The Complex Logarithm

97

6.1 6.2 6.3 6.4 7.

Analysis

Introduction The Complex Logarithm and its Properties Complex Powers Branches of the Logarithm

97 98 100 103

Complex Integration

111

7.1 Paths and Contours 7.2 The Length of a Contour 7.3 Integration along a Contour

Ill 113 115

Contents

8.

9.

10.

11.

7.4 Basic Estimate 7.5 Fundamental Theorem of Calculus 7.6 Primitives

120 121 123

Cauchy's Theorem

127

8.1 Cauchy's Theorem for a Triangle 8.2 Cauchy's Theorem for Star-Domains 8.3 Deformation Lemma 8.4 Cauchy's Integral Formula 8.5 Taylor Series Expansion 8.6 Cauchy's Integral Formulae for Derivatives 8.7 Morera's Theorem 8.8 Cauchy's Inequality and Liouville's Theorem 8.9 Identity Theorem 8.10 Preservation of Angles

127 133 136 138 139 142 145 146 149 154

The Laurent Expansion

157

9.1 Laurent Expansion 9.2 Uniqueness of the Laurent Expansion

157 163

Singularities and Meromorphic Functions

167

10.1 Isolated Singularities 10.2 Behaviour near an Isolated Singularity 10.3 Behaviour as \z\ —> oo 10.4 Casorati-Weierstrass Theorem

167 169 172 174

Theory of Residues 11.1 Residues 11.2 Winding Number (Index) 11.3 Cauchy's Residue Theorem

12.

The Argument Principle 12.1 Zeros and Poles 12.2 Argument Principle 12.3 Rouche's Theorem 12.4 Open Mapping Theorem

13.

xi

Maximum Modulus Principle

175 175 177 179 185 185 187 189 193 195

xii


Analysis

13.1 Mean Value Property 13.2 Maximum Modulus Principle 13.3 Minimum Modulus Principle 13.4 Functions on the Unit Disc 13.5 Hadamard's Theorem and the Three Lines Lemma 14.

Mobius Transformations 14.1 Special Transformations 14.2 Inversion 14.3 Mobius Transformations 14.4 Mobius Transformations in the Extended Complex Plane

15.

Harmonic Functions 15.1 Harmonic Functions 15.2 Local Existence of a Harmonic Conjugate 15.3 Maximum and Minimum Principle

16.

Local Properties of Analytic Functions 16.1 Local Uniform Convergence 16.2 Hurwitz's Theorem 16.3 Vitali's Theorem

Appendix A A.l A. 2 A.3 A.4 A.5 A.6 A.7 A.8 A.9

Some Results from Real Analysis

Completeness of R Bolzano-Weierstrass Theorem Comparison Test for Convergence of Series Dirichlet's Test Alternating Series Test Continuous Functions on [a, b] Attain their Bounds Intermediate Value Theorem Rolle's Theorem Mean Value Theorem

195 196 200 201 204 207 207 209 210 . 215 219 219 220 221 223 223 226 229 231 231 233 235 235 236 236 238 238 239

Bibliography

241

Index

243

Chapter 1

Complex Numbers

1.1

Informal Introduction

What is a complex number? It is any number of the form z = x + iy, where x and y are real numbers and i obeys i2 = —1. Of course, there is no real number whose square is negative, and so i is not a real number. Accordingly, x is called the real part of z, denoted Hez, and y = Imz is called the imaginary part. (Notice that Imz is y and not iy.) Complex numbers are declared equal if and only if they have the same real and imaginary parts; if z\ = x\ + iyi and z^ = £2 + it/2, then z\ = z^ if and only if both x\ = X2 and 2/1 =2/2- We write 0 for 0 + iO. Addition and multiplication are as one would expect; •Zl + 22 = (Xl + X2) + i (j/1 + J/2)

z\ Z2 = On + iyi) (x2 + iyi) = xix2 + iy\x2 + iy\iy2 + £1*2/2 = (xix2 - 2/12/2) + i (2/1X2 + Z12/2) • If z = x + iy, then — z = — x — iy. Suppose that z = x + iy, and z ^ 0. Then at least one of x or y is non-zero. In fact, z ^ 0 if and only if x2 + y2 > 0. We have 1 z

1 x + iy

x — iy x — iy (x + iy)(x — iy) x2 + y2 x j y 2 x +y2 x2 + y2

, 1 x , 1 —y so that Re - = — 5 ^ and Im - = — ^ ^. z xz + 2/ z x1 + 2/ By definition, complex conjugation changes the sign of the imaginary part, that is, the complex conjugate of z = x + iy is defined to be the l


Analysis

complex number z = x — iy. Notice that z z = x2 + y2 and that Re z = x = Proposition 1.1

and Im z = y = . y 2 2z For any complex numbers z\, z2, we have

(i) (ii) (iii)

z\ + z2 = ~z[ + z^, £ i z 2 = zT zi, T\ = z\.

(iv)

lfz2^0,

then (-) J} a n d applying the first part, the result follows. Let Co = 0 and Q = Cj-i + Zj, for j = 1 , . . . , n and let P be the polygon [Co, Ci] U ' ' ' U [Cn-i, Cn], where [u>, z] denotes the straight line segment from w to z. Then the equality in question is the statement that the distance between the initial and final points of P is equal to the sum of the lengths of its segments. This can only happen if the polygon stretches out into a straight line (and does not turn back on itself).

8


Example 1.3

For given zi,z2,z3

Analysis

&C, what is the set

{w € C : w = az\ + (3z2 + 7Z3, some a,/?,7 <E [0,1] with a + /3 + 7 = 1} ? In fact, this set is the triangle (including its interior) with vertices zi, Z2, zz • To see this, first notice that we can write w = az\ + (3z2 + 723 = (1 - 7)((1 - A*)2i + ^ 2 ) + 7^3 where LI = (3/(a + /3) (assuming that the denominator is not zero).] Now, the set { £ : C = (1 — \i)z\ + fiz2, 0 < fi < 1} is just the line segment from z\ to z2. Let (^ = (1— [i)zi+ fiz2 be some point on this line segment. The set Lfj, = { w : w = (1 — 7 ) ^ + 72:3, 0 < 7 < 1} is the line segment from £M to Z3. As Li varies between 0 and 1, so £M varies along the line segment from z\ to z2 and the LMs fill out the triangle. 1.4

The Argument of a Complex Number

We have agreed that a complex number can be usefully pictured as a point in the plane. Now, we can use polar coordinates rather than cartesian coordinates, giving the correspondences (assuming z ^ O ) z = x + iy

(x,y)

(r,6),

where r = ^Jx1 + y2 = \z\, and where 8 is given by the pair of equations . COS V =

x

x

Re z

— =

-r—r =

-7-7-

r

\z\

\z\

and • a

V V = 7—7 r \z\

sin 0 = -

=

Imz \z\

-7—7- .

The value of 6 is determined only to within additive multiples of 2ir, that is, if 6 satisfies both cos# = x/r and sin# = y/r then so does 0 + 2kn, for any k € Z. Moreover, these are the only possibilities: if ip also satisfies cos ip = x/r and sin ip = y/r then ip = 8 + 2rnr for some suitable n £ Z. The angle 8 is called the argument of z, denoted arg z. Note that according to the above discussion, arg z is not well-defined. One could call arg z an argument of z, i.e., a solution to cos# = x/r and sin# = y/r, or one could define arg z to be the set of all such solutions, arg z = { 8,8±2ir, 8±4n, • • •},

Complex

Numbers

'(x,y)

Fig. 1.2

z = x + iy

Cartesian and polar coordinates.

where 6 is any solution. We prefer the first idea, even though it is something of a nuisance. By convention, we can pick on a particular choice. There is a unique solution 0 satisfying — TT < 6 < TT; this choice of 6 is called the principal value of the argument of the complex number z and is denoted by Argz. Thus, for any z ^ 0, Argz is well-defined and is uniquely determined by the requirement that Argz = 6 € (—TT, TT] and cos# = x/r and sin# = y/r. For example, Argz = 0 for any real number x with x > 0. If a; is real and x < 0, then Argx = TT. Also Argi = n/2, Arg(—1) = TT, Aig(-i) = —TT/2.

Fig. 1.3

Arg(l + i) = f, A r g ( - 1 + t) = 3f, A r g ( - 1 - i) =

-%.

If z\ — ri(cosî + i sin Si) and zi = ^(cosfo +isin^2) then, using the

10


Analysis

standard trigonometric formulae, we find that Z1Z2 = rir2(cos(6»i +9?) + i sin(6>i + 62)) and so 9\ + ^ is a possible choice of argument for the product z\ Z?, for any choices 9\ and 9i of arguments for Z\ and z C given by cf>: a — i > (a, 0) is a yieW isomorphism ofM. onto F. Proof. Prom the definitions, we see that (a, 0) + (6,0) = (a+b, 0) and that (a, 0) • (6,0) = (o6,0), for any (a,0),(6,0) G F. Furthermore, the additive inverse of (a, 0) is (—a, 0) € F and, if a ^ 0, the multiplicative inverse is ( l / a , 0 ) G F. It follows that F is a subfield of C. Next, w e n o t e t h a t 0 ( a + 6) = (a+6,0) = (a,0) + (6,0) = 0 ( a ) + 0(6) and (ab) = (ab,0) = (a,0) • (6,0) = 0(a) • 0(6), 0(0) = (0,0) and 0(1) = (1,0) and so 0 is a homomorphism with respect to both operations + and •. Finally, we observe that (a, 0) = 4>(a) and so 0 maps R onto F, and if 4>(a) = 0(6), then (a, 0) = (6,0) and therefore a = 6. Hence 0 maps R one-one onto F and is a field isomorphism. • This means that F and R are "the same", that is, R can be embedded in C as F. This is just the formal proof that the "real line" is still the "real line" when we consider it as the x-axis of the complex plane. This is not an entirely vacuous statement because we are also considering the additive and multiplicative structures involved. (The plane is more naturally considered as a linear space, so that addition is natural but multiplication is a little special. In fact, it can be shown that R™ (with n > 1) can be given a multiplication making it into a field only for n = 2, in which case the multiplication is as above.) Now, any (a, 6) G C can be written as (a, 6) = (a, 0) + (0,6) = (a, 0) + (0,1) • (6,0) = 4>{a) + (0,1)- 0(6) = 4>{a) + i 0(6)

= a + ib where we have dropped the isomorphism notation 0 by writing (j>(x) as just x, for any x G R. Also, we have set i = (0,1), and we have dropped the •, denoting multiplication merely by juxtaposition, as usual. Thus, with this new streamlined notation, any complex number has the form a+ib, with a, b G R, and where i satisfies i2 = (0,1) • (0,1) = (-1,0) = 0 ( - l ) = - 1 , i.e., i2 = —1. We have therefore given substance to the hopeful but vague idea of "a number of the form x + iy, with x, y G R, and with i2 = —1", and have recovered our original formulae for addition and multiplication. The complex numbers are well-defined—they form a field and contain the set of real numbers R as a subfield.

14


Analysis

W h a t ' s going on? We might worry about simply asserting that i2 = — 1 without having said precisely what i was in the first place. However, it turns out to be all right. We can just go ahead and write any complex number as a + ib, where i2 = — 1 and not worry. It can all be justified. (But we still need to sort out square roots.) R e m a r k 1.5 T h e set of real numbers has a notion of order, defined in terms of positivity. For any real number x, precisely one of the following three statements is true; x = 0, x > 0, —x > 0. A property of positivity is t h a t if x > 0 and y > 0, then xy > 0. It follows t h a t 1 > 0. (To see this, first we note t h a t , clearly, 1 ^ 0 (otherwise, x = 1 x would be 0 for all i £ l ) . Hence either 1 > 0 or —1 > 0, but not both. If — 1 > 0 were true, then we would have ( - 1 ) ( - 1 ) > 0. B u t ( - 1 ) ( - 1 ) = l 2 = 1 so t h a t also 1 > 0. B o t h — 1 > 0 and 1 > 0 is not allowed, so we conclude t h a t — 1 > 0 is false and therefore 1 > 0.) Is there such a notion for complex numbers which extends t h a t for the real numbers? If this were possible then, for example, we would have either % > 0 or — i > 0 (since i ^ 0). If i > 0 were true, we would have — 1 = i2 > 0, which is false. Hence — i > 0 must be true. B u t then, again, this would imply —1 = (—i)2 > 0, which is false. We must concede t h a t there is no generalization of positivity extending from the real numbers to the set of complex numbers. For real numbers x and y, the inequality x > y is just a way of writing x — y > 0. This latter does not make sense, in general, for complex numbers, so it follows t h a t inequalities, such as z > C, do not make sense for complex numbers. W h a t ' s going o n ? Try as we might, we cannot make (useful) sense of inequalities between complex numbers.

1.6

T h e R i e m a n n Sphere and the E x t e n d e d C o m p l e x Plane

Let S2 denote the sphere {(x,y,z) eR3 : x2+y2 + z2 = 1} in K 3 and let N denote t h e point ( 0 , 0 , 1 ) , t h e "north pole" of S2. Think of C as the plane {(x,y,z) : z = 0 } , containing the equator of S2. Then given any point P in this plane, the straight line through P and N cuts the sphere S2 in a unique point, P', say. As P varies over the plane, the corresponding point P' varies over S2 \ {N}. This sets up a one-one correspondence between C

and S2 \ {N}.

Complex

Numbers

15

We note t h a t points far from t h e origin in C are m a p p e d into points near the n o r t h pole (and points close to the origin are m a p p e d into points close to the south pole of S2, i.e., the point (0,0, —1)). Notice too t h a t if (Pn) is a sequence of points in C which converges to some point P in C, t h e n t h e images P'n of Pn converge in S2 t o the image P' of P. We also see t h a t if (zn) is a sequence in C such t h a t \zn\ —> oo, then the sequence (P^) of their images converges to N in S2 (and vice versa). T h e point N is called the "point at infinity". T h e extended complex plane, Coo, is defined t o b e C together with one additional element, t h a t is, C ^ = C U {oo}, where {oo} is a singleton set with oo 0 C. It does not m a t t e r what oo actually is, as long as it is not already a member of the set C. For example, we could take oo to be 0 , which is certainly not a complex number. (Note t h a t a and { a } are different mathematical objects, so, in particular, 0 is not the same as { 0 }. Indeed, the objects { 0 } , { 0 , { 0 } } and { 0 , { 0 , { 0 } } } are different, as different as the numbers 1,2,3.) W h a t ' s going on? The issue is how to augment a given set to give it just one new element. That is, given a set A, how does one construct a new set B such that B \ A is a singleton set? In the case above, A = C and B = Coo is the set we seek. If it does not matter what the new element is, as in the case here, then the explicit construction above is just one of many possibilities. There is then a one-one correspondence between Coo and S2 given by oo N together with the correspondence between C and S2 \ {N}, as introduced above. T h e extended complex plane, Coo, viewed in this way is referred to as the Riemann sphere. This gives a sensible realization of "infinity". For example, the mapping z I—> 1/z is not a priori denned at z = 0 G C. However, if we consider the extended complex plane, or the Riemann sphere, then in addition to the mapping z — i > 1/z for z G C \ { 0 }, we can define 0 — i > N and N H-» 0, (or, in more suggestive notation, 0 — i > oo (1/0 = oo) and O O H O ( l / o o = 0)). This defines the m a p 2 H 1/Z as a mapping from Coo —> Coo- This construction is reasonable in t h a t if zn —> z, then 1/zn —> 1/z even if z or any zn is equal to 0 or to oo. T h e point here is to notice t h a t by studying Coo, rather t h a n just C, we can sometimes handle singularities just as ordinary points—after all, one point on a sphere is much the same as any other. In real analysis, one considers the limits x —> oo and x —> —oo. Whilst it must be stressed at the outset t h a t this is just shorthand symbolism, nevertheless, it does invoke a kind of image of two infinities—one positive

16


Analysis

and the other negative. In view of the picture of complex numbers as points in the plane, one might wonder if it might be worth considering some kind of collection of "complex infinities", each being somewhere off in some given direction (perhaps corresponding to some "end of the rainbow" at the "end" of the ray r (cos 6 + i sin 6) as r becomes very large). The view of C as being wrapped around a sphere, as developed above, suggests that we can bundle all these "infinities" into just a single "point at infinity", namely, the north pole. It should be stressed that whilst C is a field (so one can do arithmetic), this is no longer true of Coo- There is no attempt to assign any meaning whatsoever to expressions such as oo + oo or 0 x oo. The operations of addition and multiplication are simply not directly applicable when oo is involved.

Chapter 2


2.1

Complex Sequences

A sequence of complex numbers is a collection of elements of C labelled by integers; for example, ai, 02,03, To be more precise, a sequence of complex numbers is a map a from the natural numbers N into C. If we write an for the value a(n), then we recover the intuitive notion above. The important thing about a sequence is that there is a notion of "further along", e.g., the term a-no is "further along" the sequence than, say, aio6Of course, this property is inherited from the ordering within N; an is further along the sequence than am if and only if n > m. We denote the sequence a\,a^,... by (a„) or (a„)„ej,e It is often very convenient to allow a sequence to begin with ao rather than with a\. (We can still express this in terms of a map from N into C by considering the sequence (bn) where b : N —> C is the map b : n 1—» o„_i, n € N.) The notation (o„)" = 0 or (o„)„>o might be appropriate here.

2.2

Subsequences

A subsequence of a sequence (an) is any sequence got by removing terms from the original sequence (o„). For example, ai, 03, as, 0,7,... (where " . . . " means "and all further terms with an odd index") is a subsequence of the sequence 01,02,03,.... More formally, we define subsequences via mappings on N as follows. Suppose that <j> : N —> C is a given sequence in C Let ip : N —» N be any given map such that ip(n) > tp(m) whenever n > m (i.e., V preserves the order in N). Then ^ » o ^ : N - + C i s a sequence of complex numbers; <j>oi/j maps n into <j>(tp(n)) € C, for n € N. Any sequence of this form is said to be a subsequence of <j>. 17

18


Analysis

In our example above, suppose that 4>(n) = an. Let ip : N —> N be the map ip : k \-* 2k — 1, for fc e N (note that ip preserves the order in N). Then the sequence <j> o ip is n H-> <j){ip{n)) = 0(2n — 1) = ain-\ = a^(n)> i-e-> the subsequence a^,(i),a^(2),av(3)' • • • = a ij a 3v°5, Often the image ip(k) of k £ N under •;/> is denoted by n^, in which case the subsequence (/>oip is denoted by (a„fc)keN (or (anjt)k>o if we start with k = 0 rather than with fc = l ) .

2.3

Convergence of Sequences

We have a useful notion of distance between two complex numbers, so we can use this to define convergence in C. Definition 2.1 The sequence (an) of complex numbers converges to £ in C if for any given e > 0 there is N € N such that \an — (\ < e whenever n > N. £ is the limit of the sequence. We signify that (an) converges to ( by writing o„ —> £, as n —> oo, or limnôo o n = CRemarks 2.1 (1) One expects that the smaller the given e, the larger TV will need to be. (2) This definition looks exactly the same, typographically, as that of the convergence of a sequence of real numbers. In fact, if £ is real and each an is real, then this reduces to the definition for real numbers. In other words, this is a generalization of the notion for real sequences to complex ones. (3) The value \an — C\ is the distance between an and £, and so (an) converges to C if and only if the distances between £ and the various terms an become smaller than any preassigned positive value provided we go far enough "along the sequence". This last part is usefully paraphrased by saying that for any given e > 0, the distance between ( and an is "eventually" smaller than e. Our first result confirms that any subsequence of a convergent sequence also converges, and to the same limit as the original sequence. Proposition 2.1 Suppose an —> ( as n —> oo and (ank) is a subsequence of (an). Then ank —> ( as k —> oo.


19

Proof.

Let e > 0 be given. Since (an) converges to C> there is some such that n > N implies that \an — CI < e- Now, (rik) is a strictly increasing sequence of integers and so there is K such that UK > N. But then nfc > n ^ > N whenever k > K. It follows that \anic — C| < £ whenever k > K, i.e., by definition, (ank) converges to C,. D JVGN

As a consequence, we can say that a sequence with two convergent subsequences, but with different limits, cannot converge (for if it did, every convergent subsequence would have the same limit, namely, the limit of the original sequence). For example, the sequence (an) = ((—1)") does not converge, since a\, 0,3, as, • • • and 02,04, a^,... are convergent subsequences with limits —1 and 1, respectively. What does it mean to say that a given sequence does not converge? The sequence (bn) does not converge if it is never eventually close to any point in C; that is, for any given C G C, (bn) is not eventually close to £. This means that there some £0 > 0 such that it is false that (bn) is eventually within £0 of C- This, in turn, means that no matter what we choose for N, there is always some n > N such that \bn — £| > £0. In particular, there is n\ such that \bni — £| > £0. Similarly, there is ri2 > n\ such that \bn2 — C| > £ o, and so on, giving an increasing sequence n\ < n-i < "3 < • • • such that \bnk — (\ > £Q for k = 1, 2 , . . . . Hence, for any given C £ C, there is some £0 > 0 and some subsequence (bnk) with the property that \bnh — C| > £0, k € N. The subsequence and £0 will, in general, depend on the chosen point (. The next theorem tells us that the convergence of a sequence of complex numbers is equivalent to that of its real and imaginary parts, with the real and imaginary parts of the limit being the limits, respectively, of the real and imaginary parts of the sequence. Theorem 2.1 The sequence of complex numbers (a n ) converges to C if and only if both Rea„ —> ReC and I m a n —> ImC, as n —+ 00. Proof. Suppose an = an + i(5n and C = C + iv a n d that an —> C a s n —> 00. Let £ > 0 be given. Then there is some JV G N such that \an ~ CI < £> whenever n > N. The inequalities and

|Rea„ - ReC| = |Re(a„ - ()\ < \an - Cj | I m a n - I m C | = |Im(o„ - ()\ < \an - (\

show that both |Rea n — Re£| < e and |Ima n — ImC| < £ whenever n > N, that is, Re an —> Re C and Im an —> Im C as n —*• 00.

20


Analysis

For the converse, suppose that both Re an —» Re C and Im an —»Im C as n —• oo. Let e > 0 be given. Then there is TVi € N such that |Rea n — Re£| < e/2 whenever n > N\ and there is N2 £ N such that |Ima„ — Im£| < e/2 whenever n > A^2- Let N = Ni + N2 (or max{JVi, A^}, it will work just as well). Then n > N implies that €

I On - CI < |Re a„ - Re C| + |Im a„ - Im C| < ^+'o

£

= e

and the result follows.

•

Remark 2.2 A somewhat streamlined proof of this last result can be given as follows. We begin by noting that the convergence of (a„) to C is equivalent to that of the real sequence (\an — C|) to 0. This is equivalent to that of (\an — £| ) to 0, i.e., it is equivalent to the convergence of the real sequence ((a n — £) 2 + (/3n — rj)2) to 0 in R. But this, in turn, is equivalent to the convergence of both (an) to £ (in R) and (/?„) to r\ (in R), and the proof is complete. Theorem 2.2 (i) (ii) (iii) (iv)

Suppose that zn —> z and (n —> C as n —> oo. Then

a zn + b Cn —> a z + b £, for any a,b £ C; zn Cn —> z C as n —> oo; i/Cn 7^ 0 /or a/Z n and C 7^ 0, *^ en VCn -* 1/C as n ~~> ° ° ; if Qn^Q for all n and C 7^ 0, i/ien 2n/Cn —> z/C as n —> 00.

Proof. We can prove these statements directly or we could appeal to the corresponding familiar statements for real sequences. For example, we can prove (iii) directly as follows. Let e > 0 be given. We wish to show that eventually |1/C„ - 1/C| < e. Now, |1/C„ - 1/CI = |C - 0", we deduce that there is some N\ € N such that |C — Cn| < \ |C|> whenever n > N\. Hence ICI = IC - Cn + Cn| < IC " Cn| + ICn| < \ |C| + ICn|

whenever n > N\, and therefore (rearranging) \ \C\ < |Cn| whenever n > Ni. Let K = min{± |C|, |Ci|, • • •, |6v x |}- Then K > 0, since each of the Ni + 1 terms is positive, and |C«| > « for all n e N. Hence l/(|Cn| ICI) < l / ( « ICI) for all n. •

Sequences and

21

Series

We are now in a position to piece things together. Let e > 0 be given. Then there is N G N such that n> N implies that

|Cn-CI<e«|C|. Note that the right hand side is strictly positive. Hence 1 /c

/Cnl

IC»I ICI

N, as required. Next, we shall prove (ii) via the familiar real versions. To this end, let zn = xn + iyn, z = x + iy, („ = £„ + irjn and C = f + iv where each xn,yn, £n,Vn,x,y,£,T) belongs to R. Then ZnCn = (Xn + iyn)(tn

+ «7n) = Xn€n ~ VriVn + i(XnVn

+ 2/nfn)-

We are told that zn —> z and (n —* C a n d so xn —> a;, j / n —> y, £„ —» £ and f?n —> »7- By "real analysis", we conclude that xn£n — ynT]n —* %€ — VV = Re zC, and xnrjn+yn£n —> xr?+j/£ = Im z£. In other words, Re z„Cn ~~* R-e ZC and Im z„C„ —> I m zC> a n d therefore znCn -» z CPart (i) can be verified in a similar way (or directly), as can (iv). We also note that (iv) follows from (ii) and (iii). •

2.4

Cauchy Sequences

It makes sense to talk about Cauchy sequences in C, they are defined just as for the real follows. Definition 2.2 The sequence (zn) is said to be a Cauchy sequence in C if for any given e > 0 there is N G N such that \zn — zm\ < e whenever both m,n > N. Proposition 2.2 only if both (Rezn) Proof.

The sequence (zn) is a Cauchy sequence in C if and and (Imz n ) are Cauchy sequences in R.

The proof follows from the observations that

Re zn -Rezm=

Re(z„ - zm)

and

Im zn -lmzm

= Im(zn -

zm),

22


Analysis

SO t h a t

\Rezn - Rezm\

~\ > < \zn - zm\ < \Rezn - Rezm\ + \lmzn |Imzn-Imzm| J

—lmzm\.

The left hand inequalities in each pair imply that both (Rezn) and (lmzn) are Cauchy sequences if (zn) is, whereas the right hand inequality implies that ( zn) is a Cauchy sequence whenever both (Rezn) and (Imz n ) are. (For given e > 0 there is iVi such that \Rez„ — R e z m | < e/2 whenever m,n > Ni. Similarly, there is TV2 such that |Imz„ — I m z m | < e/2 whenever m,n> TV2. Putting TV = max{TVi, TV2}, we find that \z„ — zm\ < \Rezn — Rez m \ + |Imz n — Imz m \ < e whenever m,n > TV.) D The set of complex numbers C, in common with the set of real numbers R, possesses the property that Cauchy sequences necessarily converge (the completeness property). In fact, this important property is inherited from the real number system, as we show next. (It does not hold, for example, for Q, the set of (real) rationals.) Theorem 2.3 A sequence of complex numbers (zn) converges in C if and only if it is a Cauchy sequence. Proof. Suppose that zn —• z as n —> oo. We shall show that (zn) is a Cauchy sequence. Let e > 0 be given. Then there is TV e N such that \zn — z\ < e/2 whenever n > TV. Hence, for any m, n > TV, we have that \zn ~ zm\ = \{z„ -z) + (z< \zn ~ z\ +

zm)\

\Z-Zm\

< e

and so we see that (zn) is a Cauchy sequence, as claimed. Conversely, suppose that (zn) is a Cauchy sequence in C. Then both (Rez„) and (Imz„) are Cauchy sequences of real numbers. It is a basic property of the real number system R. that any Cauchy sequence of real numbers has a limit (R is complete). Hence there is some a in R and some (5 in R such that Rez„ —> a and Imzn —> (3 as n —• oo. But then zn —+ a + i/3 as n —> oo. •


2.5

23

Complex Series

We can set up a theory of complex series just as is done for their real counterpart. A series is simply a limit of a sequence of partial sums. Of course, it is necessary to worry about the existence of any such limit, that is, we must pay attention to the convergence, or otherwise, of the sequence of partial sums. Definition 2.3 Suppose that WQ,WI,W2, • •. is a sequence of complex numbers. The series Y^kô Wk 1S s a ^ *° converge provided that the sequence of partial sums Sn = Y?k=oWk converges in C as n —* oo; otherwise, it is said to diverge. If Sn —> S as n —> oo, then we write Y^kLo Wk f° r &> t n e limit of the series. By considering the real and imaginary parts, we see that the series Yl'kLo Wk converges if and only if both series Sfelo ^ e Wk an( ^ SfcLo ^ m Wk converge. If this is the case, then oo

oo

oo

"^Wk = ^ReiUfe + zyÎmwfc. k=o fc=o fe=o Indeed, for any n G N, Sn = Re Sn + i Im Sn and Sn converges if and only both Re Sn and Im Sn converge. Writing Wk — Uk +ivk, we see that Re Sn = R e E L o w * : = Y^k=ouk and> similarly, Im5„ = YX=oVk- Therefore Sn converges if and only if both J^felo uk and J 3 ^ . 0 Ufe converge; in which case oo

Wk =

53

oo

Uk+i

H

oo

X!Vk'

fc=0 fe=0 fc=0

as claimed. Remark 2.3 If J2T=o Wk converges, then Sn = Y^k=o Wk converges and so is a Cauchy sequence; that is, for any given e > 0, there is TV £ N such that |5„ — Sm\ < e whenever m,n > N. In particular, if m = n — 1 and n > N + 1 (so that m > N), then \Sn — 5„_i| < e. But Sn — Sn-i = wn, and so \wn\ < e whenever n > N + 1; that is wn —> 0 as n —> oo. This observation tells us that if it is false that wn —> 0, as n —> oo, then 2^fcLo Wk does not converge. Notice that wn —» 0 as n —• oo is a necessary but not a sufficient condition for the convergence of Yl'kLo Wk- Indeed, taking Wk = l/(fc + 1), k = 0 , 1 , 2 , . . . provides a counterexample (familiar from real analysis).

24


Analysis

Proposition 2.3 Suppose that both the series J2T=o wk and Y^T=o Cfc are convergent. Then, for any a, b G C, the series YH*Lo(.awk + HAO converges and oo

oo

^2(awk+b(k)

=

oo

aJ2wk+b^2Ck.

fc=0 fc=0 fc=0

In other words, convergent series can be added term by term, and multiplication by any complex number can be performed termwise. Proof. 2.6

This follows from £ £ = o a w k + b ^ k = a £ £ = 0 wk + b££=0Cfc•

•

Absolute Convergence

The following definition of absolute convergence is just as for real series. Definition 2.4 The series Y^kLo wk ls s a ^ t o converge absolutely if the series Yl'kLo \wk\ (with real non-negative terms) converges. The series Y^k=o Wk ls s a ^ ^° converge conditionally if it converges but does not converge absolutely. ~

jfc

Example 2.1

The series \ J -rr r converges conditionally. (The real k + fc=o ^ ^ and imaginary parts are convergent, by the alternating series test.) Proposition 2.4

Every absolutely convergent series is convergent.

Proof. Suppose that Yl'kLo Wk ls absolutely convergent. Then Y^LQ \wk\ is convergent and its partial sums Tn = X^fc=o \wk\ form a Cauchy sequence. The generalized triangle inequality |]Cfc=m+i ^ 1 - 23fc=m+i l^fcl implies that the partial sums Sn = Y^k=owk form a Cauchy sequence in C and therefore converge. • It turns out that absolutely convergent (power) series play a central role in complex analysis, so any indications as to whether or not a given series is absolutely convergent are welcome. The n th -root test is one such. To discuss this, we recall some terminology. For the moment we consider only real sequences. Let (an) be a sequence of real numbers which is bounded from above; that is, there is some M G R such that an < M for all n. For each k G N, set 6k = sup„ >fc an. Since an < M, it follows that Ak = {an : n > k} is a bounded set of real numbers and so Bk is well-defined. Furthermore, since

25


M is an upper bound for Ak, it follows that (3k <M. Next, we observe that Ak+i C Ak and so any upper bound for Ak is certainly an upper bound for Afc+1. Hence (3k is an upper bound for Ak+i and therefore fik+i < At, since /3fc+i = supAfc+i. We see that ((3k)keN is a decreasing (i.e., non-increasing) sequence of real numbers. It follows that either (/3k) is bounded from below, in which case there is some (3 G R such that (3k —> /? as n —> oo (in fact, /? = inffc /3fc) or else ((3k) is not bounded from below and so neither is (an). In the first case, we call (3 the limit superior of the sequence (an) and write (3 = limsupa„ 71—>00

or, alternatively,

(3 = l i m a n . n

If ((3k) is not bounded from below, we write limsupa n = —oo. Similarly, if (an) is not bounded from above (so that no (3k is defined), then we indicate this by writing limsupa„ = oo. Note that these last two "equalities" are no such thing, but simply convenient and suggestive shorthand pieces of notation. Examples 2.2 (1) If an = i then limsup„ an = 0. (2) If an = (—1)™ then l i m s u p n a „ = 1. (3) If an = 2" then limsup„o;„ = oo (i.e., (an) is not bounded from above). (4) If an = —n + (—l) n n then limsup n an = 0 (but notice that (an) is not bounded from below). Remark 2.4 The value of l i m s u p n a n is unchanged by the alteration of the values of any finite number of the ans. Indeed, suppose that (a'n) is another sequence in R such that a'n = an for all n > N. Then (3'k = sup„ >fc a'n = sup„ >fc an = fik for all k > N and so the sequences (f3'k) and (Pk) either both converge to the same limit or are both unbounded from below. 2.7

n t h - R o o t Test

We can now discuss the so-called n t h -Root Test for absolute convergence of a complex series. Theorem 2.4 ( n t h - R o o t Test) Let (w„)n>o be a given sequence of complex numbers and suppose that limsup n G N |twn| ' " = L. Then the series

26


Analysis

Yl^Lo wn converges absolutely if L < 1 and diverges if L > 1. Proof. For any k e N, let j3k = sup n > fe \wn\1/n. Suppose first that L < 1 and let r € E be such that L < r < 1. Since /3fc —> L, it follows that for all sufficiently large k, we have r > /?fc > L. In other words, there i s i f e N such that \wn\ ' n < r whenever n > K. Taking n t h powers, we deduce that \wn\ < r" whenever n> K. It follows that the series X^^°=o \Wn\ converges by comparison with the convergent geometric series l £ L 0 r n . Suppose now that L > 1. Then /?£ > L > 1 for all k. It follows from the definition of the supremum that for any given 5 > 0, there is some m > k such that \wm\ 'm > sup n>fe \wn\ — 5. In particular, if we take S to be L — l, then it follows that, for any k, there is some m> k such that

\wm\1/m >/3k-6>L-S

= L-(L-l)

= l.

But then \wm\ > 1 for such m and it is false that \wn\ —> 0 as n —-> oo. We conclude that Y^=o wn cannot be convergent. • Remark 2.5 Note that the theorem says nothing about the situation when L = 1. This is because the information that L = 1 is simply insufficient to determine the convergence or otherwise of the series. For example, if wo = 1 and wn = 1/n, for n > 1, then \wn\ ' n = 1/n 1 '" —> 1 as n —> oo and L = 1. The series Y^^Li V n i s n ° t convergent. However, putting vn = w„, we still have limsup \vn\ ' " = 1 but now Y^Lovn is absolutely convergent.

2.8

Ratio Test

We can recover the Ratio Test as a corollary. Theorem 2.5 (Ratio Test) Suppose that wn ^ 0, for all n = 0,1,... and that the ratios \wn+i\ /\wn\ —> L, as n —> oo. Then if L < 1 i/ie series X^^Low™ converges absolutely, and if L > I, it diverges. Proof. To simplify the discussion, we introduce a harmless adjustment to the series Y^=own- Indeed, if K ^ 0 and vn = nwn, then X ^ L o ^ " converges absolutely (or diverges) if and only if the same is true of the series £ ^ L 0 u n -

27


For any given e > 0, there is some J V g N such that L_£ N + l, I |_ P™

K]

— I

_

= fJ-N+1/ |«>jv+i| so

K-ll I I

|^n|

K

lîV+2| |^JV+l[ I ' ''

K»-l|

|tu„_i| |ion_2|

I

I

\wN+2\ "'

N 4-1

|UJV+I|

=

nN+1.

N+l A4

JV+1

/^N

|WAT+I|

Suppose L < 1. Then we may choose L < \x < 1 and we may suppose that e > 0 is so small that L + e < /it. From (*) and (**), we see that

K | < nn for all n > N + 1 and therefore limsup„ \vn\ < fj, < 1. Hence the series IC^lo Un a n d consequently 2^Lo '"'n converges absolutely, by the n th -root test. On the other hand, if L > 1, then we may choose 1 < fi < L and we may suppose that e > 0 is so small that /z < L — e. But then (*) and (**) imply that

K l > /i" for all n > N + 1 and so limsup n \vn\ 'n > /J, > I and we conclude that D^Lo vn a n d therefore Y^=o wn diverges. D As before, the series with wn = \/n for n > 1 (and WQ = 1, say) or with wn = 1/n 2 for n > 1 (and WQ ~ 1) show that nothing can be said when L = l.


Chapter 3

Metric Space Properties of the Complex Plane

3.1

Open Discs and Interior Points

In this chapter we introduce some terminology and discuss some properties of various types of subsets of C The ideas readily extend to the more general setting of metric spaces. Definition 3.1 The open disc with centre z0 and radius r > 0 is the set D(z0,r) = {z€ 0 }. Every point w € S is an interior point; indeed, for any w € S, D(w,r) C S whenever 0 < r < Imw. (If £ G D(w,r), then Im£ = Imw + Im(£ — w) > Imw — |£ — w\ > l m w - r > 0 . ) (2) Let S = { z : Imz > 0 } . Then w is an interior point of S whenever Imw > 0, as above. If Imw = 0, then w is not an interior point of S, since D(w,r) will always contain points z with l m z < 0 n o matter 29

30


Analysis

what r > 0 is chosen (for example, the point w — ir/2) and so cannot be contained in S. (3) The set { z : Re z = 0 } contains no interior points. (It is a line, and a line cannot contain discs.) More formally, let w £ S = {z : R e z = 0 }. Then, for any r > 0, the disc D(w,r) contains the point w + ir/2 which does not belong to S, so there is no disc, D(w,r), wholly contained in S. Therefore w is not an interior point of S. (4) Every point in C is an interior point of C. (For any given point w £ C, any r > 0 of your choice will be such that D(w,r) C C So w is an interior point of C.) Definition 3.3 A set G C C is called open if for any point w £ G there is some r > 0 such that D(w, r) C G. In other words, G is open if and only if it consists entirely of interior points. Examples 3.2 (1) C is open. (2) G = { z : Im z > 0 } is open. (3) A = { z : Re z > 0 } is not open. (For example, w = i £ A, but it is impossible to find r > 0 such that D(i, r) C A. Indeed, the disc D(i, r) contains the point — r/2 + i ^ A.) (4) 0 is open. This is a rather peculiar situation. If 0 were not open, then there would be some point w € 0 such that, for any r > 0, the disc D(w, r) contained some point not in 0 . Since there is no such point w, we conclude that 0 cannot fail to be open, i.e., it is open essentially by default. Whilst this result may appear a little bizarre, it is nonetheless very convenient, since it means that the intersection of two open sets is also open, even if they have no points in common—see below. First let us show that the so-called open discs D(z, r) really are open. Proposition 3.1

For any z £ C and any r > 0, the set D(z,r)

is open.

Proof. Let w £ D(z,r) be given. Then \w - z\ < r. Let p > 0 be such that 0 < p < r — \w — z\ (note that the right hand side is positive). Then \w — z\ < r — p. We claim that the disc D(w,p) is contained in D(z,r).

Metric Space Properties of the Complex

Fig. 3.1

To see this, let C € D(w,p).

The set D(z,r)

Plane

31

is open.

Then we have

|C-2| = |(C-«;) + («;-z)| < | £ — iu| + |i« — ;z|,by the triangle inequality, < P + (r ~ p) = r, as required. Since w € D(z,r) open.

is arbitrary, we conclude that D(z,r)

is •

Remark 3.1 This proposition justifies the terminology "open disc". The set {z : \z — w\ < r } consists of the "inside" (interior) together with the circumference (boundary) of the disc, and is called the closed disc, denoted by D(w,r). We shall discuss closed sets shortly, as well as this "overbar" notation. Theorem 3.1 (i) Let { G Q } Q G / be an arbitrary collection of open sets in C, indexed by the set I. Then the union [jaeI Ga is an open set. (ii) Let G\,..., Gm be any finite collection of open sets in C. Then their intersection G\ n Gi D • • • f~l Gm is an open set. Proof. Let {Ga} be given. If they are all empty, then so is their union and we know that 0 is open. Otherwise, let £ G | J a Ga. Then there is some ao e I such that (, G Gao. By hypothesis, Gao is open and so there is some r > 0 such that D(C,r) C Gao. But then it follows that D((,r) C |J Q Ga and so 1JQ Ga is open.

32


Analysis

Suppose now that G\,..., Gm are open sets in C. If D^Li Cj = 0 ) t n e n we are done, since 0 is open. Otherwise, let w £ flj=i Gj- Then w £ Gj for each j = 1,2,..., m. By hypothesis, each Gj is open and so there is some rj > 0 such D(w, rj) C GJ, j = 1,2,..., m. Let r = m i n { r i , r ^ , . . . , r m } . Then r > 0 and clearly £>(w,r) C D(w,rj) C Gj, j = 1,2,... ,m. Hence D(w, r) C P|^=i Gj and we conclude that the intersection G\ n • • • n G m is open. D Remark 3.2 These are basic and crucially important properties of open sets. Abstractions of these are the building blocks for a topological space. Remark 3.3 Note that it is false in general that an arbitrary intersection of open sets is open. For example, for each n £ N, let Gn be the open disc -D(0,1/n) = { z : \z\ < 1/n } . Then each Gn is open, but their intersection is n~=i 0 such that D(0,r) C {0}). Remark 3.4 We can express convergence of sequences using discs. The sequence (zn) converges to z in C provided that for any given e > 0 it is true that zn is eventually within e of z, i.e., there is some N £ N such that n > N implies that \zn — z\ < e. This is the same as saying that zn £ D(z,e), whenever n > N. Thus, zn —> z if and only if (zn) is eventually in any given disc D(z, e), e > 0 (centred on z). The mental image of a convergent sequence in C is one of a collection of dots (representing the points of the sequence) eventually moving inside (and staying inside) any given disc around the limit.

3.2

Closed Sets

Next, we define closed sets. Definition 3.4 A set F in C is closed if its complement C \ F is open. In other words, closed sets are precisely the complements of open sets, and vice versa. Examples 3.3 (1) The empty set 0 is closed, because its complement, C, is open. The whole complex plane, C, is closed, because its complement, 0 , is open. (2) The set F = { z : Im z > 0 } is closed. To see this, we just have to show that its complement is open. But if w £ F, then we must have


33

Plane

Imui < 0. Putting r = \ \Lva.w\ we see that D(w,r) deduce that C \ F is open, i.e., F is closed.

C C \ F and we

The following is a very useful characterization of closed sets. Proposition 3.2 The non-empty set F in C is closed if and only if every sequence in the set F which converges in C has its limit in F; i.e., F is closed if and only if whenever (zn) is a sequence in F such that zn —> z, for some z £ C, as n —> oo, then it is true that z £ F. Proof. Suppose that F is closed and suppose that (an) is a given sequence in F such that an —> z, as n —> oo. We must show that z £ F. Suppose, on the contrary, that z £ F. Then z belongs to the open set C\F. Hence there is some r > 0 such that D(z,r) C C \ F. In particular, this means that D(z,r) D F = 0 . However, an —• z means that an £ D(z,r) for all sufficiently large n. This is a contradiction and we conclude that z £ F, as required. For the converse, suppose that an —-> z with an £ F implies that z £ F. We must show that C \ F is open. If F = C, there is nothing to prove, so suppose that F ^ C . Let z £ C \ F. We claim that there is r > 0 such that D(z, r) C C \ F . If this were not true, then, no matter what choice of r > 0 we make, we will have D(z,r) n F ^ 0 . In particular, for each n £ N, we would have D(z, £) D F ^ 0 . So suppose a„ e D(z, £) D F . Then an £ F and \an — z| < ^, for all n £ N. It follows that z, as n —> oo. By hypothesis, this means that z £ F, a contradiction. We conclude that there is indeed some r > 0 such that D(z, r) C C \ F . Therefore C \ F is open and so F is closed. • Remark 3.5 So the set F fails to be closed if there exists some sequence of points in F which converges to a point not belonging to F . In fact, we could therefore define a set F to be closed if (an) in F and an —> z in C implies that z £ F. Example 3.4 The disc D(w, r) is not closed. To see this, we can take, for example, a sequence of points converging radially outwards towards a point on the circumference of the disc. Specifically, consider the sequence zn = w + (1 — £)r, n £ N. Then zn £ D(w, r), for each n, but zn —> u; + r which does not belong to D(w,r). Remark 3.6 A given set A C C need be neither open nor closed. For example, the set S = {z : 0 < Rez < 1} is not open, nor is it closed. Indeed, 1 £ S, but there is no r > 0 such that D(l,r) C S, so 1 is not an

34


Analysis

interior point, and therefore S is not open. On the other hand, the sequence *n = ~, n £ N, belongs to S and converges to 0 ^ S. So S is not closed. So there are sets which are neither open nor closed, but is it possible to have sets which are both open and closed? We have seen that C and 0 have this property. It is no use trying to find further examples: we will see shortly that these are the only possibilities.

3.3

Limit Points

Closed sets can also be characterized in terms of limit points, which we now consider. Definition 3.5 The point w is said to be a limit point of a given subset A of C if every disc D(w, r), r > 0, contains some point of A other than w. Limit points are also called cluster points or accumulation points. Note that the point w may or may not actually belong to A. If D'(w, r) denotes the punctured disc {z : 0 < \z — w\ < r }, then the definition says that w is a limit point of A if D'(w, r) n A ^ 0 for all r > 0. The idea is that there should be points of A not equal to, but arbitrarily close to w. By setting r = - for each n in N we see that w is a limit point of A if and only if there is some sequence (an) of points in A, all different from w, such that an —> w, as n —> oo. Examples 3.5 (1) If G is a non-empty open set, then every point of G is a limit point of G. Indeed, if w € G, then there is some p > 0 such that D(w, p) C G. Hence, for any r > 0,

D'(w,r)nGDD'{w,R)^0 where R = min{r, p}. (2) The point w — i is a limit point of the unit disc D(0,1). To see this, let zt = i - it = (1 - t)i for 0 < t < 1. Then zt e D(0,1) and \zt -i\=t. It follows that zt ^ i and zt £ D'(i, r) whenever 0 < t < r. Hence i is a limit point of the unit disc. In fact, every point on the circumference of the disc is a limit point of the disc (take a sequence inside the disc which converges radially outwards to the point of the circumference). The set of limit points of the disc D(£, r) is the set { z : \z — C\ < r }.


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35

(3) If A contains a finite number of points, then it has no limit points. This is clear if A — { a }, a singleton set. Otherwise, for any w e t , let p = min{ \w — a\ : a £ A and a ^ w}. Then p > 0 and the punctured disc D'{w, p) contains no points of A, so w cannot be a limit point of A. Proposition 3.3 limit points.

The set F is closed if and only if F contains all its

Proof. Suppose first that F is closed, i.e., Fc, the complement of F, is open. Suppose that C is a limit point of F. We must show that £ G F. If this were not the case, then we would have Q £ Fc. But then this would mean that there was some r > 0 such that £>(£,r) C Fc (because Fc is open, by hypothesis). In particular, D(C,r)f~)F — 0 . But for ( to be a limit point of F, we must have that -D'(Ci r) n F ^ 0 , which is a contradiction. We conclude that £ £ F, as required. Next, suppose that F contains all its limit points. We must show that Fc is open. If Fc = 0 , then we are done. So suppose that Fc ^ 0 and let w £ Fc. We wish to prove that there is some r > 0 such that D(w,r) C Fc. If this were not true, then no disc D(w,r) would be wholly contained in Fc. This is just the statement that D(w, r) D F ^ 0 for every r > 0. But w $. F, by hypothesis, and so we may say that D'(w,r) ^ 0 for all r > 0. This means that w is a limit point of F and so must belong to F, if F is assumed to contain all its limit points. We have a contradiction. We conclude that, indeed, Fc is open and so F is closed. • Remark 3.7 The statement that a set A contains all its limit points is equivalent to the statement that whenever (an) is a convergent sequence in A then its limit is also in A. To see this, we first note that if w is a limit point of A, then it is the limit of some sequence (an) in A and so the first statement above follows from the second. Suppose now that (o n ) is a sequence in A and that an —> w. If w = ak for some k, then w £ A (because ak £ A). On the other hand, if an ^ w for all n, then this means that w is a limit point of A. Hence the second statement follows from the first. We therefore have three equivalent statements, namely: • the complement of F is open; • F contains the limit of any of its convergent sequences (that is, if (zn) belongs to F and zn —> z for z £ C, then z actually also belongs to F); • F contains all its limit points.

36

Lecture Notes on Complex Analysis

What's going on? The equivalence of these statements means that a closed set in C can be defined (equivalently) in several ways. Which formulation one chooses amounts largely to personal preference. To each definition adopted, one then has two propositions of the form "F is closed if and only if ...", where "..." denotes either of the other two equivalent statements. It turns out that the most appropriate definition is the one we have given, namely that in terms of the complement being open. This generalizes to metric spaces and also to the theory of topological spaces, where the use of sequences is generally not the best approach to take.

3.4

Closure of a Set

Definition 3.6

For any A CC, the closure of A is the set

A = { z : there is some sequence (an) in A such that an —> z } . In words, A consists of all those points of C which are the limits of some sequence from A. In particular, for any a £ A, we can set an = a, for all n £ N. Then, trivially, an —> a and so a £ A and therefore A C A. Furthermore, if A is closed, then (an) in A and an —> z implies that z belongs to A. Therefore A = A if A is closed. Remark 3.8 Alternatively, one could define the closure of a set A to be the set A itself together with all its limit points. This follows because any point w £ A is a limit point of A if and only if w is the limit of a sequence of points of A. Example 3.6 What is the closure, K, say, of the disc D(w,r)? We know that D(w,r) C K. Let C, £ K. Then, by definition, there is some sequence (zn) in D(w,r) such that zn —> C> as n —* oo. In particular, \zn —w\ C — w and therefore \zn — w\ —> |C — w\, as n -> oo. Since \zn — w\ < r for each n G N, it follows that \C — w\ < r (the real sequence (\zn — w\) lies in the interval [0,r], so, therefore, must its limit). We deduce that {z :\z — w\ C as n —• oo. We must show that C € A. By definition of A, this follows if we can show that there is some sequence (a„) in A with a„ -> C _ To see this, we note that for each n, zn S A, which means that zn is itself the limit of some sequence of points from A. In particular, any such sequence is eventually within 1/n of zn, that is, there is certainly some element of A within distance 1/n of zn. Let an be one such element of A; an € A and \zn — an\ < 1/n. Doing this for each n yields a sequence (an) of elements of A. We claim that an —> (, as n —> oo. Let e > 0 be given. Then there is iV0 6 N such that n > NQ implies that \zn — C\ < e/2. Let N = max{./Vo, 2/e}, so that, in particular, 1/n < e/2 whenever n> N. We have \an ~ CI < I On - Zn| + |Zn ~ CI

< i + | e,

whenever n>

N0,

< \e + \e = e whenever n > N. It follows that an —> C giving ( s i that A is closed. Remark 3.9 set D(w,r).

and we conclude •

This result justifies the terminology "closed disc" for the

Remark 3.10 The above result implies that the set A together with its limit points is closed. This means that this "enlarged" set has no new limit points. Any limit point of A is already a limit point of A. This can be seen directly as follows. Let C be a limit point of A and let r > 0 be given. Then there is some w € A with w ^ C a n d |C — w\ < \r- Now, w £ A, and so for any given p > 0 there is some a e A with a ^ w such that \w — a\ < p. In particular, we may take p = |C — w\ so that a ^ C (it is closer to w than C is) so that |C — a\ < \( — w\ + \w — a\ < r and it follows that C is a limit point of A.

38


Analysis

Since a set is closed if and only if it contains its limit points, we see that the above discussion constitutes, in fact, an alternative proof that the closure of any set is closed.

3.5

Boundary of a Set

Next, we define the boundary of a set. It is usually quite clear from a diagram what the boundary of a set is, but we still need a formal definition. Definition 3.7

The boundary of a set A C C is the set

dA = { z : for all r > 0, D(z, r) n A ^ 0 and D(z, r) n (C \ A) =£ 0 }. In other words, a point z belongs to the boundary of the set A provided every open disc around z contains both points of A and points not in A. The next result is a useful characterization of the boundary of a set. Proposition 3.5 For given A C z and bn —> z, as n —=> oo. Proof. First suppose that z £ dA. Then, for any n e N, the sets D(z, - ) n A and D(z, - ) n (C \ A) are non-empty. Let an be any point of D(z, i ) fl A and let bn be any point of D(z, £) D (C \ A). In this way, we have constructed sequences (a„) in A and (bn) in C \ A. Furthermore, \z — an\ < - since an € D(z, A), and so an —> z as n —> oo. Similarly, \z ~ bn\ < i , and so bn —* z as n —* oo. Now suppose that there are sequences (o n ), (bn) such that an £ A, bn $ A, for all n and an —> z and 6„ —> z, as n —> oo. Let r > 0 be given. We must show that both D(z, r)(lA and D(z, r) n (C \ yl) are non-empty. But a„ converges to z and so (a„) is eventually in D(z,r). This implies that D(z, r)P\A^0. Similarly, bn —> z and therefore (bn) is eventually in D(z,r) a,nd so D(z,r)n(C\ A) ^0. D From this result, we deduce that

8A = Example 3.7

AnC\A.

The sets D(w,r) and D(w,r) have the same boundary, dD(w, r) = dD(w,r) = {z : \z -w\ = r }.


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Example 3.8 Let A be the horizontal strip A = {z : —1 Imz. But zn $. A implies that |Imz„| > 1 and the imaginary part of any limit of zn would also have to satisfy this inequality.) Similarly, if |Imz| > 1, then no sequence from A can converge to z. The following is the counterpart of theorem 3.1 for closed sets—but notice that now the union must be finite whereas it is the intersection which can be arbitrary. Theorem 3.2 (i) For any finite collection F\,..., Fm of closed subsets of C, the union Fi U F2 U • • • U Fm is a closed set. (ii) Let {Fa}aei be an arbitrary collection of closed sets in C, indexed by the set I. Then the intersection f]aeIFa is a closed set. Proof. This follows from theorem 3.1 by taking complements. Indeed, if i*i,..., Fm are closed, then the sets C \ Fi,..., C \ Fm are open. Hence, by theorem 3.1, C \ (Fi U • • • U Fm) = (C \ Fi) n • • • n (C \ Fm) is open and we deduce that JFI U • • • U Fm is closed. For any collection {Fa} of subsets of C, we have

C\(f]Fa)=[jC\Fa. a

a

Now, if each Fa is closed, each (C \ Fa) is open and so, by theorem 3.1, \Ja C \ Fa is open. Hence C \ (f\a Fa) is open and so f] Fa is closed. •

40

3.6


Analysis

Cantor's Theorem

Example 3.9 For n G N, let Fn = {z : Rez > ± }. Then each Fn is closed, but | J n Fn = {z : Re z > 0 } which is not closed. The next theorem sheds some light onto questions such as "what do you end up with if you take a set and keep cutting it in half (and throwing away the half that has been cut off)?" It will make a crucial appearance in the middle of arguably the most important theorem in the subject, namely, Cauchy's Theorem. Definition 3.8 A sequence ( A „ ) „ 6 N of sets is said to be nested if it is true that An+i C An, for all n G N.

Fig. 3.2

A nested sequence of sets in C.

Definition 3.9 A non-empty set A C C is said to be bounded if there is some M > 0 such that \z\ < M for all z G A. If A is bounded, its diameter is defined to be the number diam A = sup{ \z — £| : 2 , ( e i } . So the set A is bounded if and only if it is contained inside some disc, A C D(0, M). Notice also that if \z\ < M for all z £ A, then it is true that \z — CI < \z\ + ICI < 2M for all z,C, £ A and so diam A is well-defined (and is not greater than 2M). Theorem 3.3 (Cantor's Theorem) Suppose that (Fn) is a nested sequence of (non-empty) bounded, closed sets such that diamF„ —> 0 as n —> co. Then fl^Li Fn consists of exactly one point.


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Proof. First we shall show that f]n Fn can contain at most one point. To see this, suppose that z\,z% G f]n Fn. Then zi,z2 € Fn, for every n G N, and \z\ — z2\ < diamF„. Now, diamF n —> 0 and so for any given e > 0 there is N G N such that diam Fn < e, whenever n > N. In particular, for any n > N, \zi — 221 < diamF„ < e. That is, 0 < |zi — 22I < £ for any given e > 0, which is only possible if \zi — zi\ = 0. We must have z\ = z2 which means that |~|n F n can contain at most one point. The proof is complete if we can show that f]n Fn is not empty, for then it must contain precisely one point. For each n G N, let zn be some point in Fn (any point at all). We claim that the sequence (zn) thus obtained is a Cauchy sequence. For this, let e > 0 be given. Since diamF„ —> 0, there is N € N such that diamF n < e whenever n > N. Let m,n > N. Then zm,zn G FJV+I, since both Fm C FN+I and Fn C FAT + I (because the -F„s are nested), and therefore \zm — zn\ < diamF^v+i < £. Hence (zn) is a Cauchy sequence. Now, we know that any Cauchy sequence converges, that is, there exists £ in C such that zn —» £ as n —> 00. We shall complete the proof by showing that £ G f) n F„. Let k G N be given. For m = 1,2,... define £ m = Zfc+m. Then clearly £ m —> £ as m —> 00. (For given e > 0 there is iV such that |C — 2„| < e whenever n > N. In particular, for any m > N, we have that m + k > N and so |£ — £ m | = |C — zfc+m| < £•) Furthermore, Cm = Zk+m G Fk+m Q Fk for each m and, by hypothesis, F/c is closed. Hence ( e F j . This holds for any k £ N and so we finally deduce that ( G f|„ Fn, as required. • Example 3.10 For each n € N, let Fn be the "half-plane" defined by Fn — { z : Rez < —n}. Each F„ is a closed set and evidently F n + i C Fn. However, f]n Fn = 0 . This does not contradict Cantor's Theorem because Fn is not bounded.

3.7

Compact Sets

A further very important concept is that of compactness. Definition 3.10 A set i f C C i s compact if every sequence in K has a convergent subsequence with limit in K. Thus, K is compact if and only

42


Analysis

if whenever (zn) is a sequence in K, there is a subsequence (znfc)fc€N> say, and some z € K such that zni! —> z as fc —> oo. Evidently, a set A fails to be compact if it possesses a sequence which has no convergent subsequence with limit in A. It follows that the empty set, 0 , is compact because it cannot fail to be compact. Example 3.11 The set A = {z : 0 < \z\ < 1} is not compact; for example, the sequence (zn = ^)n€N converges to 0 and so every subsequence will also converge to 0. However, 0 ^ A. Compactness of sets in C is characterized by the following very useful criterion. Theorem 3.4 bounded.

A subset K C C is compact if and if K is both closed and

Proof. Suppose first that K is closed and bounded. Let (zn) be any given sequence in K. Since K is bounded, there is M > 0 such that \z\ < M for all z e K. In particular, \zn\ < M, for all n. Writing zn = xn + iyn, it follows that \xn\ < M and also \yn\ < M, for all n, i.e., (xn) and (yn) are bounded sequences of real numbers. From "real analysis", we know that every bounded sequence of real numbers has a convergent subsequence (Bolzano-Weierstrass theorem). Hence, there is some subsequence (xnk) of (xn) and i e R such that xnk —> x as k —> oo. We are unable to say anything about the convergence of the subsequence {Vnk) of (yn)- However, we do know that \ynk | < M, for all k. For notational convenience, write Uk = xnk and Vk = ynk- Then (v,k) converges to x, as k —> oo. Again by the Bolzano-Weierstrass theorem, (vk) has a convergent subsequence; say, {vkj)j£N converges to y, as j —> oo. But then (ufc^) is a subsequence of (uk) and so also converges to x, as j —+ oo. It follows that Ukj + ivkj —* x + iy as j —> oo. Furthermore, by hypothesis, K is closed and Ukj + ivkj £ K for each j . Hence x + iy G K. Since (u^. + wÔjeN is a subsequence of (zn), we conclude that K is compact. For the converse, suppose that K is compact. First we shall show that K is closed. Let (zn) be any sequence in K such that zn —> z. We must show that z € K. By compactness, there is a convergent subsequence; znk —> Q, with C G -ft', as fc —> oo. But z„fc is a subsequence of the convergent sequence (z n ) and so must have the same limit, that is, Q = z and so z € K. It follows that K is closed. To show that K is bounded, we suppose the contrary, i.e., suppose that K is not bounded. Then for any M > 0 there is some z £ K satisfying


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\z\ > M. In particular, for each n € N, there is some zn, say, in K such that \zn\ > n. Since K is compact, (zn) has a convergent subsequence, (znk), say, with znk —> £, as fc —> oo, for some ( S if. We shall show that this is incompatible with the inequalities \znk\ > nk- Indeed, convergence to C, implies that certainly |£ — znk\ is eventually smaller than 1. Hence, for all sufficiently large k, k 00. By Cantor's theorem, theorem 3.3, it follows that D^Li -^n = {C}> for some £. In particular, C, & F\ = K. The family {Ga}aei is a cover for K and so there is at least one a, say a0, such that ( € Gao. Now, Gao is open and so there is some r > 0 such that D((,r) C Gao. But £ G Fn, for every n and diamF n —» 0 and so, for all sufficiently large n (in fact, for n with d/2n~1 < r), we have that Fn C D(C,r) C Gao- This contradicts the impossibility of covering any Fn with a finite collection of Gas—we can do it with just one!


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We conclude that there is some finite collection, say a\,... ,am, such that K C Gai U • • • U Gam, as required. For the converse, suppose that any open cover of K possesses a finite subcover. For each z e K, let Gz be the disc Gz = D(z,l). Evidently, {GZ}Z£K is an open cover of K and therefore, by hypothesis, has a finite subcover. That is, there is z\,... zp such that K C D(z!,l)

Li • • • U

D(zp,l).

It follows that K is bounded (in fact, K C D(0, r), where the radius r is given by r = max{ \zi\ : 1 < i < p } + 1).

Fig. 3.4

K is closed.

To prove that K is closed, we will show that C \ K, the complement of the set K, is open. For this, let £ e C \ K (note that C \ K is not empty because K is bounded). For each n € N, let Gn = { z : \z — Q > ^ }. Then Gn is open, and Gn C Gn+\. Furthermore, every z ^ ( is contained in some Gn (in fact, z is a member of every Gn for which n \z — £| > 1) and so certainly {G n } n e N is an open cover of K (because C, ^ K). By hypothesis, there is a finite subcover, K C Gni U • • • U Gnm for some m , . . . , n m € N. Let k = m a x { m , . . . , n m } . Then Gni C Gfe for each i = 1 , . . . , m and we have K C Gfc. It follows that .D(C, \) Q C\K and we conclude that C \ K is open and so the proof is complete. •

46


Analysis

Remark 3.11 We therefore have three equivalent formulations, each one expressing the compactness of a given set K. • Any sequence in K has a subsequence convergent to some point in the set K. • K is both closed and bounded. • Every open cover of K has a finite subcover. In the more general context of topological spaces, the third version is taken as the definition of compactness. It should be noted that in this more general situation (i.e., in a general topological space) these three statements about the set K are no longer necessarily equivalent. (In fact, boundedness may not even be denned.)

Proposition 3.6 Let G be a non-empty proper open subset of C. Then there is a nested sequence Kn C Kn+i of compact sets in G such that G = U^Li Kn- (The sequence (Kn) is a compact exhaustion of G.) Proof. Let Fn = { z : z £ G and \z — w\ > \ for all w £ Gc}. It is clear that Fn C Fn+\. We claim that Fn is closed. To show this, suppose that (zk) is a sequence in Fn and Zk —» C> a s k ~^ °°- Then for each w £ Gc, \zk — w\ —> |C — w\. But \zk — w\ > ^ for all k and so |£ — w\ > i . In particular, £ ^ Gc and so £ € G. But then this means that £ £ Fn and so Fn is closed. Let Kn = Fn n D(0,n). Certainly, Kn C G and Kn is bounded. Kn is also closed because it is the intersection of closed sets. Hence Kn is compact. It is also clear that Kn C Kn+i. Finally, suppose that z £ G. Since G is open, there is r > 0 such that D(z,r) C G. In particular, for any w E Gc, \z — w\ > r. Choose n £ N such that n > 1/r and n > \z\. Then z £ Fn and also z £ D(0,n), that is, z £ Kn. It follows that G — U^Li Kn, as required. • By considering the discs D(0,1) and D(2i, 1) or even D(0,1) and the closed disc D(2i, 1), for example, we see that it is quite possible for two sets A and B to be disjoint but be "almost touching" in the sense that there are points a £ A and b £ B as close together as we wish. This cannot happen for closed sets if one of them is compact, as we now show. (We shall give two proofs, illustrating different aspects of compactness.)


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Proposition 3.7 Suppose that K and F are disjoint subsets of C with K compact and F closed. Then there is some r > 0 such that \z — w\ > r for every z £ K and w £ F. In other words, there is some r > 0 such that the distance between any point in K and any point in F is not less than r. Proof. Suppose the contrary, that is, for any r > 0 there is some z £ K and w £ F with \z — w\ < r. In particular, for any n £ N, there is zn £ K and wn £ F such that \zn — wn\ < ^ (taking r = £). By compactness of K, the sequence (zn) has a convergent subsequence (znjJfceNi sav> znk —* C, as k —> oo, with ( £ K. We claim that the (sub)sequence (wnk)k£N also converges to £. This is to be expected. After all, for large n, the wns are close to the zns, so the wnks get dragged along with the znks towards C,. To show this, let e > 0 be given. There is fco £ N such that

K~znk\

< \e

whenever k > ko. Choose TV £ N such that N > ko + -• We have |C - Wnk | < |C - *nk | + \znk - Wnk |

whenever k > N (since nk > k and ^ < | e , in this case). It follows that wnk —> £, as k —> oo, as claimed. However, each wnk £ F and, by hypothesis, F is closed. It follows that C, £ F, that is, £ is a common point of both K and F. This contradicts the hypothesis that K and F are disjoint. We conclude that there must be r > 0 such that \z — w\ > r for any z £ K and any w £ F. Alternative Proof. By hypothesis, K and F are disjoint. This means that K C C \ F. Since F is a closed set, G = C \ F is open. Hence, for every z £ K, there is some rz > 0 such that D(z,rz) C G. The family { D(z, ^ r 2 ) : 2 £ K } of open discs is clearly an open cover of the compact set K (each point of K is the centre of one of the discs). By compactness, this family contains a finite subcover, so it follows that there is a finite set of points zi,...,zm in K such that the discs D(z\, ^ rZl),..., D(zm, \ rZjn) cover K.

48


Analysis

Set r = m i n { r 2 l , . . . , rZjn} so that r > 0 and consider the collection {D(z, \ r) : z G K} of open discs, labelled by the points of K. We claim that for any z G K, the disc D(z, \ r) belongs to one of the m discs D(zj,rZj), j = 1 , . . . , m. To see this, first note that for given z G K there is 1 < j < m such that z G D(ZJ, \ rZj) (since these m discs cover K). Let w be any point in the disc D(z, \ r). Then

\W-Zj\

< \W-Z\

i r, for any z € K. •

Example 3.12 Let Fi = {z = x + iy : x > 1 and xy > 1 } and let F2 = {z = x — iy : x > 1 and xy > 1 } . (F2 is the set of complex conjugates of points in F\, that is, its mirror image in the real-axis. We can also think of F\ as all those points (x, y) in the plane with x > 1 and lying above the hyperbola y — l/x.) Evidently, F\ D F2 = 0 . We claim that Fi and F2 are closed. To see this, suppose that (zn) is a sequence in Fi such that zn —»iy. Let a;n = a;„ + iy n and w = u + iv. Then we know that xn —* u and that y n —> v and so x n y n —> uu. But x„y„ > 1, for all n, and therefore uv > 1. Furthermore, x n > 1 and so M > 1. It follows that w = u + iv £ F\ and we conclude that F\ is closed. Similarly, we see that Fi is closed. Now, Cn = n + - i G F\ and Cn £ ^2 and the distance between these two points is \(n — (n\ = ^, which can be made as small as we wish by choosing n sufficiently large. It follows that inf{ \z - w\ : z G Fi, w G F2 }, the distance between the two closed, disjoint sets Fi and F 2 , is zero.


3.8

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49

Polygons and Paths in C

Definition 3.11

The line segment from ZQ to z\ is the set

[z0, z\\ = { z : z =

ZQ

+ t(zi - z0), 0 < t < 1}

It consists of all those points in the complex plane lying on the straight line between ZQ and z\. V

z = z0+ t(zi - z0)

Fig. 3.5

The line segment joining ZQ to z\.

Remark 3.12 By looking at the real and imaginary parts, we can express this in terms of (familiar) cartesian coordinates. Let z = x+iy £ [ZQ, Z\\ and write ZQ = xo + iyo and z\ = x\+iyi, z can be written as z = zo + t(zi — ZQ) for some 0 < t < 1, which gives the pair of equations x=

XQ

+ t(xi -

y = yo+t(yi

XQ)

-t/o).

Eliminating t shows that (x, y) lies on the straight line y = yo

(2/i - Z/o) (x — x0). (xi - x0)

Moreover, the restriction 0 < t < 1 means that x varies between XQ (when t = 0) and x\ (when t = 1). Definition 3.12

A polygon in C is a set of the form [z0, zi] U [zi,z2] U • • • U [zn-i,zn]

for some n £ N and points ZQ, Z\, ...,

zn.

50


Analysis

In other words, a polygon is a finite collection of line segments placed end to end. We say that the polygon joins the points ZQ and zn. Notice that zn need not be the same as ZQ.

Fig. 3.6

A polygon joining ZQ to zn.

Definition 3.13 A path in C is a continuous function ip : [a,b] —• C, where [a, b] is some closed interval in R. We say that if joins its initial point, 0 such that D( 0 such that 0. Any point w, say, in D(z',r') is stepwise connected to z'. liw were stepwise connected to zo, then z' would be as well (via w). We conclude that no point of D(z', r') can be stepwise connected to zo, i.e., D(z', r') C B, showing that B is open. But G = A\J B and G is connected, by hypothesis. It follows that B = 0 . Hence G =• A and the proof is complete. • What's going on? The equivalence of these various notions of a set being all in one piece is only claimed to hold for open sets in C. They are not the same in general. However, our main concern will be with open sets, so we can appeal to whichever version we find appropriate at any particular time, to suit our own convenience.

56


Analysis

We can now easily answer the question "which sets are both open and closed"? Theorem 3.8 The only subsets o / C which are simultaneously open and closed are 0 and C itself. Proof. By way of contradiction, suppose that A is a non-empty proper subset of C and that A is both open and closed. It follows, in particular, that C \ A, the complement of A, is open. But then we can write C as the disjoint union of two open sets A and B = C\A, C = A U B . Since A is a proper subset of C it follows that B is non-empty. This means that C is disconnected. However, C is open and evidently polygonally connected (indeed, [ZQ, Z\] joins the points ZQ and Z\). But this means that C is also connected, by theorem 3.7. This gives a contradiction and we deduce that no such nonempty proper subset A exists. D

3.10

Domains

Definition 3.18 An open connected set is called a domain (also sometimes known as a region).

Fig. 3.11 Examples of a star-domain and a convex domain.

Definition 3.19 A domain D is (star-like or) a star-domain if there is some ZQ € D such that for each z £ D the line segment [z0,z] lies in D. Any such point ZQ is called a star-centre.


Plane

57

A domain D is convex if for any pair of points z, C G D, the line segment [z, C] lies in D. Evidently, if D is convex, then it is star-like and each of its points is a star-centre. The converse is false, in general. Example 3.13 The set D = { z : z + \z\ ^ 0 } is star-like, but not convex. In fact, D is the whole complex plane with the negative real-axis (including the point 0) removed. (The complement of D is the set { z : z = — \z\ }, which is the set of complex numbers which are real and negative (or 0).) D is not convex because the line segment joining, say, the point (—1 — i) to the point (—1 4- i) crosses the negative real-axis at z = — 1 which does not belong to D. On the other hand, we see that, for example, ZQ = 1 is a star-centre. (For any z G D, the line segment [1,2:] lies entirely in D.) Example 3.14 Let L\ and Li be the semi-infinite line segments given by L\ = {z : z = r, r > 1} and Li = { z : z = ir, r > 1} and let D = C \ (Li U L2). Evidently D is a star-domain with star-centre ZQ = 0. Moreover, the point ZQ = 0 is the only star-centre for D. Example 3.15 For 0 < r < R, the ring (annulus) A = { z : r < \z\ < R} is pathwise connected. To see this, let z\ and ZQ. be any pair of points in the annulus A. Let pi = |zi| and P2 = \z2\- Then z\ can be connected to IUI = pi (on the positive real axis) in A by the path given by

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