Foundations of Complex Analysis [2 ed.] 1842652230, 9781842652237

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S. Ponnusamy

Foundations of Complex Analysis With 114 Illustrations

Dedi ated to

Professor O.P. Juneja and Professor M.S. Ranga hari

Prefa e

The main aim of this book is to present the on epts and te hniques of

omplex fun tion theory in a way that will give the reader maximum assistan e in mastering the fundamentals of the theory. The book is designed to serve as a text for a rst ourse on the lassi al theory omplex fun tions or as a supplement to other standard texts. The sele tion and the sequen ing of the ontents are the result of the experien es I had during the ourse of my studies and tea hing. As a prerequisite the reader is expe ted to have adequate knowledge of the elements real analysis. However, in an eort to make the book a

essible to a wider audien e, I have tried my best to minimize the prerequisites and to keep the exposition at an elementary level. Thus I have in luded a number of examples motivating the ideas involved in most of the theorems and de nitions. Most of the exer ises have been provided with hints for their solutions. Some theorems have been given more than one proof to help the reader a quire a deeper understanding of the theory. The present edition has a large number of illuminating new examples, observations, exer ises, and some additional materials overing some advan ed topi s. Within ea h hapter, all the numbered items (ex ept gures) eg. Corollary, equation, Example, Lemma, Proposition, Remark, Theorem are numbered onse utively as they appear. For the sake of onvenien e, the sign signals the end of the proofs of Theorem, Corollary, Lemma and Proposition whereas the sign  indi ates the end of Remark, Observation and Example. It is a great pleasure for me to express my gratitude to Prof. M.S. Ranga hari who has not only en ouraged me to write the rst edition of the book but also has read several drafts with are and patien e. During my do toral work at the Indian Institute of Te hnology Kanpur, I had opportunities to be a tutor to graduate and undergraduate students on this subje t. In this ontext, I would like to thank my tea her Prof. O.P. Juneja who introdu ed me to the theory and has been a sour e of inspiration sin e then. Many people have given me help, en ouragement, and inspiration. I should take this opportunity to mention just a few. I express my deep gratitude to Professors R. Balasubramanian, O. Martio, St. Rus heweyh, C.S. Seshadri, V. Singh, M. Vuorinen, and G.P. Youvaraj whose ontinued en ouragement has been invaluable for widening my resear h interest. Parti ularly, I am deeply indebted to Prof. M. Vuorinen for his ontinued support in bringing out this edition. In addition, a number of my resear h visits to work with him has helped me to learn a lot. It is my pleasant duty to thank Prof. G.P. Youvaraj who has been my

best friend and to this work. He has proofread the nal manus ript and his

onstru tive riti isms has improved the presentation lot. I take this opportunity to thank Prof. Roger W. Barnard for his support and friendship during my visit to Texas Te h University. I greatly appre iate his areful reading of the nal version of the manus ript and making suggestions. I wel ome and appre iate any suggestions for improvements and omments regarding errors and misprints. I must also re ord my appre iation due to my daughter Abirami and son Ashwin for their understanding and love during the long period that I have taken to omplete this se ond edition of the book. Above all, my deepest gratitude goes to my wife Booma (alias Geetha) for her in nite patien e,

ontinued support, and loving en ouragements in all walks of my life. Finally, I wish to express my thanks to the Center for Continuing Edu ation at the Indian Institute of Te hnology, Madras, India, for its support in the preparation of the manus ript. Also, I thank Mr. N.K. Mehra, publisher and Managing Dire tor of Narosa Publishing House Pvt. Ltd. who has been superb. S Ponnusamy IIT Madras, India

Contents

1

2

3

4

Prefa e . . . . . . . . . . . . . . . . . . . . . . . . . . . Complex Numbers . . . . . . . . . . . . . . . . . . . . . 1.1 De nition of Complex Numbers . . . . . . . . . . 1.2 Geometri Interpretation . . . . . . . . . . . . . 1.3 Square roots . . . . . . . . . . . . . . . . . . . . 1.4 Rational Powers of a Complex Number . . . . . 1.5 Topology of the Complex Plane . . . . . . . . . . 1.6 Sequen es and Series . . . . . . . . . . . . . . . . 1.7 Exer ises . . . . . . . . . . . . . . . . . . . . . . Fun tions, Limit and Continuity . . . . . . . . . . . . . 2.1 One-to-one and Onto Fun tions . . . . . . . . . . 2.2 Con epts of Limit and Continuity . . . . . . . . 2.3 Stereographi Proje tion . . . . . . . . . . . . . . 2.4 Sequen es and Series of Fun tions . . . . . . . . 2.5 Exer ises . . . . . . . . . . . . . . . . . . . . . . Analyti Fun tions and Power Series . . . . . . . . . . . 3.1 Dierentiability and Cau hy-Riemann Equations 3.2 Harmoni Fun tions . . . . . . . . . . . . . . . . 3.3 Power Series as an Analyti Fun tion . . . . . . 3.4 Exponential and Trigonometri Fun tions . . . . 3.5 Logarithmi Fun tions . . . . . . . . . . . . . . . 3.6 Inverse Fun tions . . . . . . . . . . . . . . . . . . 3.7 Exer ises . . . . . . . . . . . . . . . . . . . . . . Complex Integration . . . . . . . . . . . . . . . . . . . . 4.1 Curves in the Complex Plane . . . . . . . . . . . 4.2 Properties of Complex Line Integrals . . . . . . . 4.3 Cau hy-Goursat Theorem . . . . . . . . . . . . . 4.4 Consequen e of Simply Conne tivity . . . . . . . 4.5 Winding Number or Index of a Curve . . . . . . 4.6 Homotopy Version of Cau hy's Theorem . . . . . 4.7 Cau hy Integral Formula . . . . . . . . . . . . . 4.8 Morera's Theorem . . . . . . . . . . . . . . . . . 4.9 Existen e of Harmoni Conjugate . . . . . . . . . v

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

iii 1 1 6 14 16 20 27 36 39 39 44 54 65 70 73 73 89 101 112 118 129 131 137 137 142 157 166 167 170 176 190 192

vi

CONTENTS

5

6

7

8

9

4.10 Taylor's Theorem . . . . . . . . . . . . . . . . . . . . 4.11 Zeros of Analyti Fun tions . . . . . . . . . . . . . . 4.12 Laurent Series . . . . . . . . . . . . . . . . . . . . . 4.13 Exer ises . . . . . . . . . . . . . . . . . . . . . . . . Conformal Mappings and Mobius Transformations . . . . . 5.1 Prin iple of Conformal Mapping . . . . . . . . . . . 5.2 Basi Properties of Mobius Maps . . . . . . . . . . . 5.3 Fixed Points and Mobius Maps . . . . . . . . . . . . 5.4 Triples to Triples under Mobius Maps . . . . . . . . 5.5 The Cross-Ratio and its Invarian e Property . . . . 5.6 Conformal Self-maps of Disks and Half-planes . . . . 5.7 Prin iple of Symmetry and Mobius Maps . . . . . . 5.8 Exer ises . . . . . . . . . . . . . . . . . . . . . . . . Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Maximum Modulus Prin iple . . . . . . . . . . . . . 6.2 Hadamard's Three Cir les/Lines Theorems . . . . . 6.3 S hwarz' Lemma and its Consequen es . . . . . . . . 6.4 Liouville's Theorem . . . . . . . . . . . . . . . . . . 6.5 Doubly Periodi Entire Fun tions . . . . . . . . . . . 6.6 Fundamental Theorem of Algebra . . . . . . . . . . 6.7 Zeros of ertain Polynomials . . . . . . . . . . . . . 6.8 Exer ises . . . . . . . . . . . . . . . . . . . . . . . . Classi ation of Singularities . . . . . . . . . . . . . . . . . 7.1 Isolated and Non-isolated Singularities . . . . . . . . 7.2 Removable Singularities . . . . . . . . . . . . . . . . 7.3 Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Further Illustrations through Laurent's Series . . . . 7.5 Isolated Singularities at In nity . . . . . . . . . . . . 7.6 Meromorphi Fun tions . . . . . . . . . . . . . . . . 7.7 Essential Singularities and Pi ard's Theorem . . . . 7.8 Exer ises . . . . . . . . . . . . . . . . . . . . . . . . Cal ulus of Residues . . . . . . . . . . . . . . . . . . . . . . 8.1 Residue at a Finite Point . . . . . . . . . . . . . . . 8.2 Residue at the Point at In nity . . . . . . . . . . . . 8.3 Residue Theorem . . . . . . . . . . . . . . . . . . . . 8.4 Number of Zeros and Poles . . . . . . . . . . . . . . 8.5 Rou he's Theorem . . . . . . . . . . . . . . . . . . . 8.6 Exer ises . . . . . . . . . . . . . . . . . . . . . . . . Evaluation of ertain Integrals . .. . .. .. .. . .. .. .. R 2+ R( os ; sin ) d . . . . . . . 9.1 Integrals of Type R 9.2 Integrals of Type R 11 f (x) dx . . . . . . . . . . . . . 9.3 Integrals of Type 11 g(x) os mx dx . . . . . . . . . 9.4 Singularities on the Real Axis . . . . . . . . . . . . .

194 199 207 217 223 224 234 241 245 247 250 261 266 269 269 275 279 291 297 299 301 304 307 307 311 314 317 320 324 327 333 337 338 348 350 355 361 364 369 369 375 385 388

vii

CONTENTS

10

11

12

9.5 Integrals Involving Bran h Points . . . . . . . . . 9.6 Estimation of Sums . . . . . . . . . . . . . . . . 9.7 Exer ises . . . . . . . . . . . . . . . . . . . . . . Analyti Continuation . . . . . . . . . . . . . . . . . . . 10.1 Dire t Analyti Continuation . . . . . . . . . . . 10.2 Monodromy Theorem . . . . . . . . . . . . . . . 10.3 Poisson Integral Formula . . . . . . . . . . . . . 10.4 Analyti Continuation via Re e tion . . . . . . . 10.5 Exer ises . . . . . . . . . . . . . . . . . . . . . . Representations for Meromorphi and Entire Fun tions . 11.1 In nite Sums and Meromorphi Fun tions . . . . 11.2 In nite Produ t of Complex Numbers . . . . . . 11.3 In nite Produ ts of Analyti Fun tions . . . . . 11.4 Fa torization of Entire Fun tions . . . . . . . . . 11.5 The Gamma Fun tion . . . . . . . . . . . . . . . 11.6 The Zeta Fun tion . . . . . . . . . . . . . . . . . 11.7 Jensen's Formula . . . . . . . . . . . . . . . . . . 11.8 The Order and the Genus of Entire Fun tions . . 11.9 Exer ises . . . . . . . . . . . . . . . . . . . . . . Mapping Theorems . . . . . . . . . . . . . . . . . . . . . 12.1 Open Mapping Theorem and Hurwitz' Theorem 12.2 Basi Results on Univalent Fun tions . . . . . . 12.3 Normal Families . . . . . . . . . . . . . . . . . . 12.4 The Riemann Mapping Theorem . . . . . . . . . 12.5 Bieberba h Conje ture . . . . . . . . . . . . . . . 12.6 The Blo h-Landau Theorems . . . . . . . . . . . 12.7 Pi ard's Theorem . . . . . . . . . . . . . . . . . . 12.8 Exer ises . . . . . . . . . . . . . . . . . . . . . . Index of Spe ial Notations . . . . . . . . . . . . . . . . . Hints and Solutions for Sele ted Exer ises . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

396 399 405 407 407 417 422 432 435 437 438 446 453 457 469 473 480 487 498 503 503 506 512 518 527 530 538 542 547 551

viii

CONTENTS

Chapter 1

Complex Numbers

In this hapter we review basi results su h as fundamental algebrai and topologi al properties of omplex numbers. We assume that the reader is a quainted with the familiar properties of the real number system. In Se tion 1.1, we introdu e omplex numbers. Se tion 1.3 provides a way of obtaining solutions of a quadrati equation in omplex variable. Se tion 1.2 dis usses polar representation of omplex numbers whereas Se tion 1.4 gives an easy method of nding solutions of a rational power of omplex numbers. In Se tion 1.5, we introdu e topologi al properties of the omplex plane. In Se tion 1.6, we present several fundamental theorems on erning

onvergen e of sequen es and series of omplex numbers.

1.1 De nition of Complex Numbers Consider ordered pairs of real numbers (x; y). The word `ordered' means that (x; y), (y; x) are distin t unless x = y. We denote the set of all ordered pairs of real numbers by C . We shall all C as the set of all omplex numbers. In C , we de ne addition (+) and multipli ation ( or juxtaposition) between two su h ordered pairs (x1 ; y1), (x2 ; y2 ) by (1.1)

(x1 ; y1 ) + (x2 ; y2 ) := (x1 + x2 ; y1 + y2 )

and (1.2) (x1 ; y1 )(x2 ; y2 ) := (x1 x2 y1 y2 ; x1 y2 + y1 x2 ): If z1 = (x1 ; y1) and z2 = (x2 ; y2 ), then we say that

z1 = z2 () x1 = x2 and y1 = y2 :

In parti ular, z = (x; y) = (0; 0) () x = 0 and y = 0: We an easily he k the following simple properties for equality of ordered pairs making it an equivalen e relation: For any z1; z2 and z3 in C,

2

Complex Numbers

(i) z1 = z1 (ii) z1 = z2 =) z2 = z1 (iii) z1 = z2 and z2 = z3 =) z1 = z3 . The asso iative and ommutative laws for addition and the multipli ation and distributive laws et ., follow easily from the properties of the eld of real numbers R. Further, it is lear from (1.1) and (1.2) that (0; 0) is the additive identity, (1; 0) is the multipli ative identity, ( x; y) is the additive inverse of z = (x; y) and 

1 x y := 2 2 ; 2 2 z x +y x +y



is the multipli ative inverse of z = (x; y) = 6 (0; 0).1 Given two omplex numbers z1 and z2 ,

 there is a unique omplex number, say z3 , su h that z1 + z3 = z2: If 

zj = (xj ; yj ) (j = 1; 2; 3), then z3 = (x2 x1 ; y2 y1 ) and is denoted by z2 z1 . [Subtra tion℄ for z2 6= (0; 0), there is a unique z3 su h that z1 = z2 z3: In fa t, z3 = z1:(1=z2 ) sin e z2 z3 = z2 :z1 :(1=z2) = (z2 :(1=z2)):z1 = 1:z1 = z1 : The

omplex number z3 is otherwise written as z3 = z1 =z2. [Division℄

The symbol ommonly used for a omplex number is not (x; y) but x + iy, x; y real. Following Euler, we de ne i := (0; 1) in the omplex number system C of ordered pairs. We write a real number x as (x; 0). Then a

ording to (1.2)

i2 = (0; 1)(0; 1) = ( 1; 0); i3 = i2
i = ( 1; 0)(0; 1) = (0; 1); and i4 = i2
i2 = (1; 0). Also, x + iy = (x; 0) + (0; 1)(y; 0) = (x; y): The above dis ussion shows that C is also a eld. Further, writing a real number x as (x; 0) and noting that (x1 ; 0) + (x2 ; 0) = (x1 + x2 ; 0) and (x1 ; 0)(x2 ; 0) = (x1 x2 ; 0);

R turns out to be a sub eld of C . The asso iation x 7! (x; 0) shows that we

an always treat R as a subset of C . Complex numbers of the form (x; 0)

are said to be purely real or just real. Those of the form (0; y) are said to be purely imaginary whenever y 6= 0. In parti ular, we have, with the above identi ation of R, i2 = 1: Every ( omplex number) z = (x; y) 2 C , denoted now by x + iy, admits a unique representation2 (x; y) = (x; 0) + (0; 1)(y; 0) = x + iy; with x; y 2 R: 1 We use `:=' to abbreviate 2 From now on if we write

stated.

\de ned by" or \written as". z = x + iy 2 C , x; y are real numbers unless otherwise

3

1.1 De nition of Complex Numbers

`Zero' viz. (0; 0) = 0 + i0 is the only omplex number both real and purely imaginary. The onjugate of a omplex number z = x + iy is the omplex number z := x iy. Note that z = z if and only if x + iy = x iy, i.e. y = 0; i.e. z is purely real. The inverse or re ipro al z 1 of a omplex number z = x + iy 6= 0 is 

1 z x iy x = = 2 2= 2 2 z zz x + y x +y







y i 2 2 ; x +y

whi h was de ned earlier as the multipli ative inverse of z . We all x and y the real part and imaginary part of z = x + iy, respe tively. We write

z+z z z and Im z = : 2 2i We know that ordered pairs of real numbers represent points in the geometri plane referred to a pair of re tangular axes. We then all the olle tion of ordered pairs as R2 and the two axes as the x-axis and y-axis. Be ause (x; 0) 2 R2 orresponds to real numbers, the x-axis is alled the real axis and sin e iy = (0; y) 2 R2 is purely imaginary, the y-axis is alled the imaginary axis. Now, we an visualize C as a plane with x + iy as points in R2 and we simply refer to it as the nite omplex plane or simply omplex plane. Depending on the problems on hand, we use x + iy or (x; y), to represent a omplex number. Re z := x and Im z := y; Re z =

1.3. Theorem. The eld C annot be totally ordered in onsisten e with the usual order on R. (Total ordering means that if a 6= b, then either a < b or a > b): Proof. Suppose that su h a total ordering exists on C . Then for i 2 C we would have either i > 0 or i < 0 sin e i 6= 0. This means that in either

ase 1 = i
i = ( i)( i) > 0

whi h is not true in R. This observation shows that su h an ordering is impossible in C .

Theorem 1.3 means that the expressions z1 > z2 or z1 < z2 have no meaning unless z1 and z2 are real.

1.4. Con epts of modulus/absolute value. The modulus or absolute value of x 2 R is de ned by

jxj :=



x if x  0 x if x < 0:

4

Complex Numbers

As it stands, there is no natural generalization of j
j to C , be ause, as we have seen in Theorem 1.3, there is no total ordering on C . However, we interpret jxj geometri ally as the distan e from x to the origin (zero) of the real line. It is this fa t whi h leads p us topde ne the modulus of a omplex number z = x + iy 2 C by jz j := zz = x2 + y2 .It is easy to derive the following simple fa ts: (i) jz j = 0 i3 z = 0. (ii) jRe z j  jz j; jIm z j  jz j; jz j = jzj, z1 + z2 = z1 + z2 ; z1 z2 = z1 z 2 . Equality holds in the rst two inequalities i Im z = 0 and Re z = 0, respe tively. z jz j (iii) jz1z2 j = jz1 j jz2 j; 1 = 1 (z2 6= 0). z2 jz2 j (iv) Re (z1 + z2 ) = Re z1 + Re z2 and Im (z1 + z2 ) = Im z1 + Im z2 :

1.5. Example. It is easy to see that if z1 z2 and z1 z2 are both real, then either z1 and z2 are both real or one is the onjugate of the other. Indeed, there is nothing to prove if one of z1 and z2 is zero. Therefore, we may assume that both z1 and z2 are non-zero. By hypothesis z1 z2(z1

z2 ) = z1 (z1 z2

jz2 j2 )

whi h is real. Therefore, we must have either z1 is real or z1 z2 jz2 j2 = 0 whi h is equivalent to saying that either z1 is real or z1 = z2 . This proves the required on lusion. 

1.6. Example. We wish to show that (i) jz1 + z2 j  jz1j + jz2 j, (ii) j jz1j jz2 j j  jz1 z2 j for all z1 ; z2 2 C . It is also easy to see that (i) and (ii) are equivalent. To prove these, we rst observe that if z1 + z2 = 0, then there is nothing to prove. If z1 + z2 6= 0, then jz1 + z2 j 6= 0. Sin e Re z  jz j, we have (1.7)

z















z1 z2 +  Re z z+1 z + Re z z+2 z = 1 1 + z2 z1 + z2 1 2 1 2

from whi h (i) follows. To prove the se ond inequality, we write z1 = z2 + (z1 z2 ) so that, by (i), (1.8)

jz1 j  jz2 j + jz1 z2 j; i.e. jz1 j jz2 j  jz1 z2 j:

Similarly, we obtain (1.9) 3 We

jz2 j jz1 j  jz2 z1 j = jz1 z2 j: use Paul Halmos's notation `i' to abbreviate the words `if and only if'.

5

1.1 De nition of Complex Numbers

Combining (1.8) and (1.9) we obtain (ii). A tually, (ii) implies (i) an be seen similarly. Thus, (i) and (ii) are equivalent. Finally, we dis uss the equalities in (i) and (ii). If one of z1 ; z2 is zero, then equality in (i) holds obviously. If z1 + z2 = 0, then equality in (i) is possible only when z1 = z2 = 0. So we assume that z1 6= 0, z2 6= 0, z1 + z2 6= 0 and note that jz j = jRe z j () Im z = 0. From (1.7), we see that the equality in (i) holds i for ea h k = 1; 2, Im







zk zk = 0 and Re z1 + z2 z1 + z2



 0:

That is, i

jz1 z2 j = Re (z1 z2 ) ; i.e. jz z j = Re (z z ): 1 2 1 2 jz1 + z2 j2 jz1 + z2 j2 Equivalently, this holds when z1 z 2 is a non-negative real number. This is, in fa t, equivalent to the relation z1 = tz2 with t  0, sin e z1 z 2 = z1 jz2 j2 =z2. One an easily he k that the same ondition holds for the equality in (ii). This ompletes the problem.  Here is another proof of the two inequalities of Example 1.6:

jz1 + z2 j2 = (z1 + z2 )(z 1 + z2 ) = jz1 j2 + 2Re (z1 z 2 ) + jz2 j2  jz1 j2 + 2jz1z 2 j + jz2 j2 ; sin e Re z  jRe z j  jz j; = jz1 j2 + 2jz1j jz2 j + jz2 j2  (jz1 j + jz2 j)2 ; that is, jz1 + z2j  jz1 j + jz2 j whi h is known as the triangle inequality and it is lear that the equality holds i jz1 z2 j = Re (z1 z 2 ); i.e. z1 = tz2 , where t  0. Moreover, by nite indu tion, it is easy to see that n X zj j =1



n X j =1

jzj j:

The equality holds when zj (j = 1; : : : ; n) lie on the same ray emanating from the origin.

1.10. Remark. If z1 = tz2 where t  0, z1 = x1 + iy1 and z2 = x2 + iy2 6= 0, then it follows that x1 = tx2 and y1 = ty2 so that (0; 0), (x1 ; y1 ) and (x2 ; y2 ) lie on a ray from (0; 0). In other words, the points in R2 orresponding to z1 and z2 are on the ray from the origin. Note that this means that z1 ; z2 are linearly dependent. 

6

Complex Numbers y z π/3

π/3 π/3 x

O

p

p

Figure 1.1: Geometri representation of 3=2 + i= 2.

1.11. Remark. For z = x + iy, we have jz j2  2jxj jyj so that 2jz j2 = jz j2 + jz j2  x2 + y2 + 2jxj jyj = (jxj + jyj)2 :

p

Hen e, for any z = x +piy 2 C , we have jz j  jxj + jyj  2jz j and it is

lear that the onstant 2 annot be repla ed by a smaller number unless x = 0 or y = 0. The latter inequality be omes equality if x = y. 

1.2 Geometri Interpretation As we have pointed out in Se tion 1.1, for every point of the z -plane there is one and only one omplex number z and onversely. A omplex number z = x + iy may be thought of as a ve tor or as a dire ted line segment, in the omplex plane and pi tured as an arrow from the origin to the point (x; y) in C ; or as any ve tor obtained by translating this ve tor parallel to itself, i.e. one whose starting point is irrelevant. For instan e the point

p

3 i + 2 2 determines the ve tor p from the origin to this point and so the dire ted line segment fromp(a 3=2; b 1=2) to (a; b) also represents the same omplex number z = 23 + 2i for any (a; b) 2 R2 . Both ve tors have the same length and point in the same dire tion. Thus we have an in nite number of line segments that we an draw in order to represent a omplex number, be ause all su h segments have the same length and point in the same dire tion (see Figures 1.1 and 1.2). Here jz j is the length of the ve tor z . Some of the simple geometri relations between z , z and z are outlined in Figure 1.3. Geometri ally, addition of two omplex numbers orresponds to the ve tor addition of ve tors representing them. Let z1 = x1 + iy1 and z2 = x2 + iy2 be two omplex numbers su h that 0; z1 and z2 are not ollinear. Then

z=

z1 + z2 = (x1 + x2 ) + i(y1 + y2 ) = (x1 + x2 ; y1 + y2 ): We plot these ve tors in Figure 1.4. As indi ated in Figure 1.4, the ve tor z1 + z2 is the diagonal of the parallelogram with z1 and z2 as adja ent sides.

7

1.2 Geometri Interpretation y

y2 − y1

z2

z1 x2 − x1

O

Figure 1.2: z = z2 z

y

z1 = (x1 , y1 ) z2 = (x2 , y2 )

x

z1 = (x2 x1 ) + i(y2

y1 ).

z = the mirror image of z reflected across the real axis

O

x

−z = the mirror image of z reflected through origin

−z

z

Figure 1.3: Relations between z , z and z . y

z1 + z2

z2

z2 z1 O

−z2 z1 − z2

−z2 z1

x

Figure 1.4: Law of parallelogram.

+

z2 |

y

z1

|z 1

z2 |z2 |

O

|z 1|

x

Figure 1.5: Triangle inequality jz1 + z2 j  jz1 j + jz2 j.

8

Complex Numbers y

y

z = reiθ

O

θ = Arg z x

O

x θ = Arg z z = reiθ

Figure 1.6: Argument of z 6= 0.

Now the triangle inequalities (see Example 1.6 and Figure 1.5) are obvious. If r is the length of the ve tor represented by z , i.e. the distan e from the origin to z , and  is the angle measured from the positive x-axis to the radius ve tor joining z in the anti- lo kwise sense and z = x + iy, then the

oordinate systems, viz. the artesian and polar systems, are related by (see Figure 1.6) (1.12) x = r os  and y = r sin ; and hen e (1.13)

p

r = x2 + y2 and tan  = y=x:

Thus, we de ne z = r( os  + i sin ) = rei : Any value of  for whi h (1.12) (or (1.13)) holds is alled an argument of z 6= 0 written as  = arg z: Clearly z has an in nite number of distin t arguments. Any two distin t arguments of z dier by an integral multiple of 2. (Sin e z = 0 () jz j = 0, arg 0 is indeterminate). In order to spe ify a unique value of arg z , we may restri t its value to some interval of length 2. To make this statement pre ise, we introdu e the on ept of \prin ipal value" of arg z as follows: For an arbitrary z 6= 0, the prin ipal value of arg z is de ned to be the unique value that satis es  < arg z   and it will be denoted by Arg z . Thus, the relation between arg z and Arg z is given by arg z = Arg z + 2k; k 2 Z: For onvenien e the set of all the values of arg z is denoted by  arg z . For example,

  3 Arg i = ; Arg (1 i) = ; Arg ( 1) = ; Arg ( 1 i) = : 2 4 4 While inverting the se ond equation in (1.13) we should note the following

9

1.2 Geometri Interpretation y Arg z = π + Arctan (y/x)

Arg z = Arctan (y/x)

O Arg z = −π + Arctan (y/x)

x Arg z = Arctan (y/x)

Figure 1.7: Des ription of argument of a omplex number z = x + iy . y

θ = 7π/6 (wrong) √ − 3 − i = reiθ

x θ = −5π/6 (correct)

Figure 1.8: To spe ify the orre t value when  = Arg z = 5=6.

(see Figure 1.7): Arg z =

8 > > > > > < > > > > > :

Ar tan (y=x)  + Ar tan (y=x)  + Ar tan (y=x) =2 =2

if x > 0 if x < 0, y  0 if x < 0, y < 0 if x = 0, y > 0 if x = 0, y < 0

where Ar tan t is the prin ipal value of the ar tangent of a real number t satisfying the inequality =2 < ar tan t  =2. The following are then true: y x y sin(arg z ) = p 2 2 ; os(arg z ) = p 2 2 ; and tan(arg z ) = : x x +y x +y For instan e, we have the following: (a) arg z = arg z unless z is a negative real number, be ause otherwise arg z = arg z = . In fa t, we have  arg z = arg z + 2k; k 2 Z: p (b) If z = 3 i, then we have Arg z = 5=6 and so (see Figure 1.8)  arg z = 5=6 + 2k; k 2 Z:

p

( ) The omplex number z = 1 i may be written as z = 2ei ; where

 = Arg z + 2k = 3=4 + 2k; k 2 Z:

10

Complex Numbers

We get non-prin ipal values if and only if k 2 f1; 2; : : : g. p p p 10 + i 30 = 2 10ei with  = 2=3 + (d) Similarly we easily derive 2k, k 2 Z: Next we see that the polar representation of a omplex number simpli es the task of des ribing the produ t of two omplex numbers geometri ally. To do this we onsider two omplex numbers z1 and z2 with polar representations: z1 = r1 ei1 and z2 = r2 ei2 : Now,

ei1 ei2 = ( os 1 + i sin 1 )( os 2 + i sin 2 ) = ( os 1 os 2 sin 1 sin 2 ) + i(sin 1 os 2 + os 1 sin 2 ) = os(1 + 2 ) + i sin(1 + 2 ) = ei(1 +2 ) ; whi h gives (1.14) z1z2 = r1 r2 ei1 ei2 = r1 r2 ei(1 +2 ) : Note that jei j = 1 and ei = ei(+2k) , k 2 Z. Thus, we have

jz1 z2 j = jz1 j jz2 j

(1.15)

and (1.16) arg(z1 z2 ) = arg z1 + arg z2 (mod 2) in the sense that they are the same but for an integral multiple of 2. For z2 6= 0, we have z1 r1 ei1 r1 i(1 2 ) = = e : z2 r2 ei2 r2 Thus, for z2 6= 0, we have (1.17)

arg





z1 = arg z1 arg z2 (mod 2): z2

If we ombine (1.17) with the operation of subtra tion, learly the angle  de ned by   z2 z1  = arg z3 z1 represents the angle at the vertex z1 (see Figure 1.9). For instan e, let us

hoose z1 = 1 and z2 = i: Then z1 z2 = i and hen e Arg (z1 z2 ) = =2. Further Arg z1 + Arg z2 =  =2 = =2: Therefore, in this ase, we obtain Arg z1 + Arg z2 = Arg (z1 z2 ): As another example, suppose z1 = 1 and z2 = i. Then we see that Arg (z1 z2) = =2 and Arg z1 + Arg z2 = 3=2: Therefore, in this ase, we obtain the orre t answer by adding 2 to bring within the interval ( ; ℄. These two examples show that when the

11

1.2 Geometri Interpretation z1 z2

y

z3 z3 − z1

y

z2

O

θ1 +

z2

φ − z1 z1 z 2

θ2 θ1

x

θ2

z1 x

O

Figure 1.9: Angle at the vertex z1 . y

zeiφ

z φ

O

x

Figure 1.10: Rotation of z through an angle .

prin ipal arguments are added together in a multipli ation problem, the resulting argument need not be a prin ipal value. Conversely, we also see that when the non-prin ipal arguments are ombined, a prin ipal argument may result. From this observation, we also see that if we are given two

omplex numbers z1 and z2, then z1 and z2 are equal i jz1 j = jz2 j and arg z1 = arg z2 (mod 2): (i) Suppose that z1 = ei , where  is real and z2 = z , any omplex number. Then z1 z2 = ei z is obtained by rotating z through an angle  (see Figure 1.10).  (ii) The set of all points given by the equation < arg(z 2 3i)   6 is geometri ally represented in Figure 1.11. y

Arg z = π/6 (2, 3) Arg (z − 2 − 3i) = π O

x

Figure 1.11: The set of all points su h that 6 < Arg (z

2 3i)   .

12

Complex Numbers y z g(

θ=

a

Ar

−

a)

x

O

Figure 1.12: The straight line Arg (z

a) = , where a = m + in; m; n > 0.

y

θ+

π

θ a x

O

Figure 1.13: The straight line arg(z

a) =  +  .

(iii) The set of points des ribed by S = fz : j arg z =2j < =2g represents the upper half-plane, viz fz : Im z > 0g. Two more examples of this kind are given in Figures 1.12 and 1.13. (iv) Suppose z1 and z2 are non-zero omplex numbers. Then

z1z 2 + z 1 z2 = 0

() () ()





z1 z2 z2 z22 = ; i.e. 1 2 = z1 z2 jz1 j jz2 j2 2 arg z1 =  + 2 arg z2 + 2k arg z1 = arg z2 + (2k + 1)=2; k 2 Z: y z1

z2

O z3

x

z3

Figure 1.14: Des ription for perpendi ular and parallel ve tors.

13

1.2 Geometri Interpretation y ζ2 − ζ3

ζ3 ζ2

ζ1 − ζ2 ζ1

x

O

Figure 1.15: Des ription for parallel ve tors on the same straight line.

This means that the ve tors z1 and z2 are perpendi ular (see Figure 1.14) i z1 z2 + z1 z2 = 0, i.e. Re (z1 z 2 ) = 0 whi h is same as writing z1 = isz2; for some real number s. Note that perpendi ularity is equivalent to the Pythagorean equation jz1 z2 j2 = jz1 j2 + jz2 j2 : (v) The line through points 1 and 2 is perpendi ular to the line through points 3 and 4 i (the ve tors) 1 2 and 3 4 are perpendi ular. Thus, we on lude that the two lines are perpendi ular if and only if there exists a real number s su h that   1 2   1 2 = is; i.e. arg =  2k or  2k 3 4 3 4 2 2

for k 2 N 0 := N [ f0g. (vi) Similarly, for nonzero z1 ; z2 2 C it follows that

z1 z2

z1 z2 = 0

() 2Im(z1 z2 ) = 0 () Im zz1 = 0; i.e. zz1 is real 2 2 () arg z1 = arg z2 + k; k 2 Z:

This means ve tor z1 is parallel to ve tor z2 i z1 = tz2 , where t is real. (vii) Thus, three omplex numbers j (j = 1; 2; 3) lie on the same straight line i 2 3 = 1 (3 1 ) for some real 1 (see Figure 1.15). Similarly we also have

3

1 = 2 (1

2 ) and 1 2 = 3 (2

3 )

for the real numbers 2 and 3 . The points 1 ; 2 ; 3 are ollinear i

1 (3

1 ) + 2 (1

2 ) + 3 (2

3 ) = 0

for some 1 ; 2 ; 3 2 R. Equivalently, (2

1 )1 + (3

2 )2 + (1

3 )3 = 0:

14

Complex Numbers

Writing 2 1 = 1P , 3 2 = 2 and 3 = 3 , we an P 1 state the ondition as 3j=1 j j = 0 and 3j=1 j = 0 for some real j (j = 1; 2; 3) not all zero.

1.3 Square roots

Sin e the square of a real number is nonnegative, x2 = a has a real solution only if a  0. Now we onsider the question of square roots in C , for instan e w = i; i are the solutions of w2 = 1. Moreover, our next theorem shows that if z 2 C , then there is a solution w 2 C for w2 = z: Theorem 1.18 below gives a purely algebrai proof of this fa t. Later we obtain this result, using polar

oordinates, as a spe ial ase of a more general result.

1.18. Theorem. For a given z = x + iy 2 C , the solutions of w2 = z

are given by

w=

"r

r

#

jz j + x + i sgn (y) jz j x ; sgn (y) =  +1 if y  0 1 if y < 0: 2 2

Proof. We have to solve the equation (u + iv)2 = x + iy for u and v. This equation gives (1.19) u2 v2 = x and 2uv = y and therefore, u2 + v2 = jz j: Using this and the rst equation in (1.19), we have jz j + x and v2 = jz j x : u2 = 2 2 (Note that jz j  x  0). From the se ond equation in (1.19) we observe that y > 0 () uv > 0; y < 0 () uv < 0: Therefore, sele ting u and v so that their produ t has the same sign as that of y, we obtain the required on lusion. 1.20. Corollary. For every omplex number z with jz j = 1 and Re z  0, there exists p a omplex number w with jwj = 1 su h that w2 = z and jIm wj  (1= 2)jIm z j: Proof. Let z = x + iy with x  0 and jz j = 1. Then x2 + y2 = 1 and x  x2 . Using Theorem 1.18, we have (sin e jz j2 = 1 = jz j)

jIm wj2

=

r

2

jz j2 x = x2 + y2 x  x + y2 x = jIm z j2 2 2 2 2

whi h ompletes the proof.

15

1.3 Square roots

1.21. Remark. For a given omplex number z , the following fa ts are easy to obtain: (i) There are two values of w su h that w2 = z and these two values are

alled the square roots of z . (ii) Ea h of the two values of w is real i z is real and positive. (iii) Ea h of the two values of w is purely imaginary i z is real and negative. (iv) The two values of w oin ide with zero i z = 0. (v) For ; 2 C , the equation (1.22) z 2 + z + = 0 has solutions +w (1.23) ; z= 2 where w is su h that w2 = 2 4 as obtained in Theorem 1.18. Here the pro edure adopted to obtain the solution is the same as for equations with real oeÆ ients, viz. that of ompleting the square. (vi) Suppose that the equation (1.22) has real roots, say z = x. Then we have x2 + x + = 0, and x2 + x + = 0: Thus, eliminating x, the equation (1.22) has real roots if ( )2 = ( )(  ); i.e. (Im )2 + (Im )(Im ( )) = 0: Similarly we see that the ondition under whi h (1.22) has purely imaginary roots is that (Re )(Re ( )) = (Im )2 : (vii) On the other hand suppose that the equation (1.22) has a omplex root, say z = z1 su h that jz1 j = 1. Then we have (1.24) z12 + z1 + = 0: Sin e jz1 j = 1, z1 z1 = 1 so that z 1 = 1=z1. So (1.24) yields z1 +  + z 1 = 0 and z 1 +  + z1 = 0: Now, eliminating z 1 from the above two equations we get (1 j j2 )z1 = ( ): Thus, (1.22) has roots on jz j = 1 only if j j = j1 j j2 j: 

p

1.25. Remark. Let  a + ib denote the two square rootsp of a + ib given by Theorem 1.18. More generally we use the notation  n a + ib to denote the n n-th roots of a + ib. It is easy to see that p p p  3 + 4i = (2 + i);  3 + 4i = (1 + 2i);  2i = (1 + i):  In Se tion 1.4, De Moivre's Theorem will be used to solve the equation wn = z for w when z is given. In fa t, there are exa tly n n-th roots of any non-zero omplex number z .

16

Complex Numbers

1.4 Rational Powers of a Complex Number Repeated appli ation of (1.14) yields (rei )m = rm eim

(1.26)

or equivalently, jz mj = jz jm and arg(z m ) = m arg z (mod 2), where m is a natural number. If z = rei 6= 0, repeated appli ation of (1.17) shows that for a natural number m, z m = (rei ) m = r m e im : Of ourse, we de ne z 0 = 1 for z 6= 0. In parti ular, if we let jz j = 1, i.e. z = ei , then we obtain the following:

1.27. Theorem. (De Moivre's Theorem) If m is an integer, then (ei )m = eim , i.e. ( os  + i sin )m = os m + i sin m: Further, the identity (1.26) is espe ially useful for nding n-th roots of a omplex number z0 6= 0, when n is a natural number. For, if we have z n = z0 with z = rei and z0 = r0 ei0 , then

p

rn ein = r0 ei0 =) r = n r0 and n = 0 + 2k;

p

where r = n r0 is the unique positive n-th root of r0 (> 0). Hen e, all the roots of z = z01=n are given by (1.28)

p n

jz0 jei(0 +2k)=n

where k is any integer. Ea h value of k = 0; 1; : : : ; n 1 gives a dierent value of z . Any other value of k merely repeats one of the values of z

orresponding to k = 0; 1; : : : ; n 1, sin e e2ik = 1. Thus, there are exa tly n n-th roots of z0 6= 0. Also (1.28) shows that the n n-th roots of z0 a tually lie on a ir le entered at the origin and having radius equal p to n jz0 j. Ea h of the roots obtained from (1.28) has the same modulus and the arguments are equally spa ed. Geometri ally the n n-th roots of z0 6= 0 are lo ated atpthe verti es of a regular n-sided polygon ins ribed in the ir le of radius n jz0 j. Thus, we have proved

1.29. Theorem. Given a nonzero omplex number z = rei , the equation wn = z has pre isely n distin t solutions given by

p

wk = n rei(+2k)=n ; where k = 0; 1; : : : ; n 1; and and  = Arg z .

pn r denotes the positive n-th root of r = jz j

For instan e, the n-th roots of unity are given by (1.30)

!k = ei2k=n ; k = 0; 1; : : : ; n 1:

17

1.4 Rational Powers of a Complex Number 2π3

ω

=

i

y

e

1

ω1

ω2

ω0 x

ω2

ω3

Figure 1.16: The n-th roots of 1 when n = 3; 4. y ω2 ω3

ω1 ω0

ω4 ω5

x

ω7 ω6

Figure 1.17: The n-th roots of 1 when n = 8.

The n-th roots of 1 when n equals 4 and 8 are pi tured in Figures 1.16 and 1.17. =12 , k = 0; 1; : : : ; 5: Further, Similarly, the 6-th roots of i are ei(4k+1)p we easily derive that the values of [(1 i)=( 3 + i)℄1=6 are 2 1=12ei ;  = (4k 5)=12; k = 0; 1; : : : ; 5:

1.31. Remark. It is easily veri ed that (1.28) is valid when n is a negative integer, sin e z 1=n = (1=z ) 1=n. Further, we an easily show that if m and npare integers having no ommon fa tors then all the values of z0m=n are n jz0 jm ei[(m=n)0 +(2mk=n)℄ ; k = 0; 1; : : : ; n 1:  If we set ! = e2i=n , then all the n-th roots of unity are expressed by 1; !; !2; !3 ; : : : ; !n 1 : From (1.28) it is lear that if ! 6= 1 is an n-th root of unity, then the others are !2 ; !3 ; : : : ; !n 1 ; 1. Hen e if is any one of n-th roots of z0 , then all the n-th roots of z0 are given by

; !; !2; : : : ; !n 1 : Now, we see some of the immediate onsequen es of the above observations. Sin e !n 1 = (! 1)(1 + ! + !2 +


+ !n 1 ) = 0: If ! 6= 1, then

18

Complex Numbers

for n > 1, we have 1 + ! + !2 +




+ !n 1 = 0 (! = ei2=n ):

Thus, in parti ular, by equating real and imaginary parts on both sides 







nX1 2k 2k = 1 and sin = 0: n n k=1 k=1 Further, it is also evident that the sum of the produ ts of all n-th roots of unity, taken 2; 3; : : : ; n 1 respe tively at a time, is zero, sin e they are the roots of the equation z n 1 = 0: Now, for ! 6= 1, the ondition !n = 1 gives that nX1

os

(1 !)(1 + 2! +


+ n!n 1 ) 1 ! 1 + ! + !2 +


+ !n 1 n!n = 1 ! n = : 1 ! Let h be any integer whi h is not a multiple of n, i.e. h = np + q (0 < q < n). Then !h = !q ; q < n whi h implies that the sets 1 + 2! +




+ n!n

1

f1; !h; !2h ; : : : ; !(n

=

1)h g

and f1; !; !2; : : : ; !n

1g

are the same in some order. Hen e if h is not a multiple of n or h is an integer su h that 0 < jhj < n, then 1 + !h + !2h +




+ !(n

1)h

= 0:

If h is a multiple of n, then !h = !np = (!n )p = 1 and so we have 1 + !h + !2h +


+ !(n

1)h

= n:

Similarly, it is evident that




+ ( 1)n 1!(n

1 !h + !2h

1)h

=

1 ( 1)n !nh 1 ( 1)n = : 1 + !h 1 + !h

That is, if ! is an n-th root of unity, n X k=1

(

1)k 1 !(k 1)h

(

=

0 if n is even 2 if n is odd. 1 + !h

1.32. Example. We wish to prove that all ir les that passes through  and () 1 interse t the ir le jz j = 1 at right angles if jj 6= 1,  6= 0.

1.4 Rational Powers of a Complex Number

19

To do this, we onsider the equation of the ir le C with enter at z0 and radius r: C = fz 2 C : jz z0 j = rg: If C passes through the points  (6= 0) and 1=, then we have

j z0 j2 = r2 ; i.e. jj2 + jz0 j2 2 Re (z0 ) = r2 and

j1 z0j2 = jj2 r2 ; i.e. 1 + jj2 jz0 j2 2 Re (z0 ) = jj2 r2 : Subtra t the latter from the former to get (1 jj2 )(1 + r2 jz0 j2 ) = 0: This implies either jj = 1 or jz0 j2 = 1 + r2 holds. The se ond ondition yields the required on lusion if jj = 6 1.  1.33. Example. Dis uss the nature of the set (1.34)

S = fz : jz aj + jz + aj = 2 g:

For two omplex numbers z1 and z2 we know that

jz1 z2 j; jz1 + z2j  jz1 j + jz2 j so that j2aj = jz + a (z a)j  jz + aj + jz aj = 2 : Thus, there are

omplex numbers satisfying (1.34) only if jaj  . Suppose that z 2 S . Then 2jz j = jz + a + z aj  jz + aj + jz aj = 2 ; i.e. max jz j = : z2S Here the maximum is attained at z = e i where  = arg( =a). If z 2 S , then we see that the sum of the distan es from the point z to the given points a and a, is equal to the onstant 2 . This means that S represents an equation of an ellipse with fo i at a. For instan e, if a is real then the equation of the ellipse using re tangular oordinates is given by

y2 x2 + = 1:

2 2 a2 Its enterpis at the origin and its semi-major and semi-minor axes are equal to and 2 a2 , respe tively. If a is real, then we an dis uss the set des ribed in (1.34) in the following way: Let z = x + iy. Then, jz aj + jz + aj = 2 is equivalent to jx + iy aj + jx + iy + aj = 2

() jx + iy aj = 2 jx + iy + aj () jx + iy aj2 = (2 jx + iy + aj)2 () (x a)2 + y2 = 4 2 + (x + a)2 + y2 4 jx + a + iyj

20

Complex Numbers |z − p|

z

|z −

q|

p

q

Figure 1.18: Equidistant points from p and q : jz

pj = jz q j.

() () () ()

( jx + a + iyj)2 = ( 2 + ax)2

2 [(x + a)2 + y2 ℄ = ( 2 + ax)2 ( 2 a2 )x2 + 2 y2 = 2 ( 2 a2 ) x2 y2 + = 1:

2 2 a2 Using the te hnique of this example one an show that the set

S = fz : jz aj

jz + aj = 2 g

des ribes a hyperbola with fo i at z = a and z = a.



1.35. Example. If p and q are two distin t points in C , then it is easy to see that the set des ribed by fz : jz pj = jz qjg gives the straight line that is the perpendi ular bise tor of the line segment joining p and q (see Figure 1.18). Next we see that the set of points des ribed by

jz pj = jz qj (p 6= q; 0 <  < 1)

(1.36)

is a ir le. Upon squaring (1.36) we get

jz

pj2

= 2 jz

qj2

()

jz j2

() z

() z

 



p 2 q 2 jq2 j jpj2 2Re z = 1 2 1 2 2 p 2 q jp 2 qj2 2 jqj2 jpj2 = + 1 2 (1 2 )2 1 2 2 jp qj2 = ; (1 2 )2 p 2 q jp qj = : 1 2 1 2

Thus, the set of the points des ribed by (1.36) is a ir le.



1.5 Topology of the Complex Plane In the earlier se tions we dis ussed some of the algebrai and geometri properties of the omplex eld or plane as the ase may be. In this se tion

21

1.5 Topology of the Complex Plane

we study some of the topologi al properties of the omplex plane. This is required for our study of Complex Analysis. If X is any set, then the fun tion d : X  X ! R is alled a metri or a distan e fun tion if it satis es the following onditions for all a; b and in X : (i) (ii) (iii) (iv)

d(a; b)  0 d(a; b) = 0 () a = b d(a; b) = d(b; a) d(a; )  d(a; b) + d(b; ).

The set X together with a metri , i.e. (X; d) or, in short X , is alled a metri spa e. As we have seen earlier, the fun tion d : C  C ! R; (z; z 0) 7! jz z 0 j; has the following properties: (a) jz z 0 j  0 (b) jz z 0 j = 0 () z = z 0 ( ) jz z 0 j = jz 0 z j (d) jz wj  jz z 0j + jz 0 wj, where z; z 0; w 2 C , where d(z; z 0) = jz z 0 j is alled the Eu lidean metri . Thus, C is a metri spa e with the Eu lidean metri d. For instan e, we have (a) If X = R and d(x; x0 ) = jx x0 j, then (R; d) is a metri spa e.

(b) If Y  X and (X; d) is a metri spa e, then so is the restri tion (Y; d). ( ) Besides the Eu lidean metri we have another natural metri known as the maximum metri on C . This is de ned as d(x + iy; x0 + iy0) = maxfjx x0 j; jy y0 jg:

For a detailed dis ussion on metri spa es, we refer to the book by Ponnusamy [8℄. In the Eu lidean metri spa e (C ; d), an open ball
(z0 ; ) = fz 2 C : jz

z0j < g

is alled an open disk of radius  > 0 entered at z0 2 C or an  neighborhood or simply a neighborhood of z0 . Geometri ally,
(z0 ; ) is just the disk

entered at z0 onsisting of all points at a distan e less than  from z0 . Evidently,
(z0 ; 1) = C for any z0 2 C . We use the term deleted neighbourhood of z0 to denote a set of the form

fz 2 C : 0 < jz z0 j < g; i.e.
(z0 ; ) nfz0 g:

22

Complex Numbers y

y

Re z > 0

Im z > 0 x

O

O

x

Figure 1.19: Half planes. y

∂1

1

x

Figure 1.20: Des ription for a non-open set.

We de ne 
(z0 ; ) = fz 2 C : jz z0 j = g, the ir le of radius  > 0

entered at z0 . Throughout this book, we use the notation
R :=
(0; R) = fz 2 C : jz j < Rg and
:=
1 : The unit disk
, as we shall see in later hapters, plays a ru ial role in the theory of fun tions of a omplex variable. A subset S  C is alled open (in C ) if for every z0 2 S there is a Æ > 0 su h that
(z0 ; Æ)  S . This means that some disk around z0 lies entirely in S . For instan e, the interior of a ir le, the entire plane C , and the half-planes given by Re z < ; Im z >  and Im z < , are all examples of open sets (see Figure 1.19). Here  is an arbitrary xed real number. On the other hand \the interior of a ir le union ir umferen e" does not form an open set, sin e no neighborhood of a point on the ir umferen e lies entirely within the set; for instan e, S = fz : jz j  1g is not open (see Figure 1.20). Observe that R when onsidered as a subset of C is not open. To show that the disk
(z0 ; R) is open, let  2
(z0 ; R). If we hoose Æ su h that 0 < Æ < R j z0 j, then
( ; Æ) 
(z0 ; R). As  2
(z0 ; R) being arbitrary, this proves that
(z0 ; R) is open. On the other hand, this fa t is geometri ally lear. The omplement of a set S  C is C n S := fz 2 C : z 62 S g and is usually denoted by S . A set S  C is said to be losed if its omplement C n S is open. For ea h  > 0, the set
(z0 ; ) := fz 2 C : jz z0 j  g

23

1.5 Topology of the Complex Plane y

O

x

Figure 1.21: Des ription for onne ted set.

is losed and onsequently we all it a losed disk. We write
(0; ) as
 and
1 simply as
. Here are some examples of losed sets. (a)
= fz 2 C : jz j  1g;
= fz 2 C : jz j  1g; 
= fz 2 C : jz j = 1g; (b) fz 2 C : Re z  0g; fz 2 C : Im z = 1g; fz 2 C : jz 4j  jz jg. ( ) The entire plane C and the empty set ;. There is another way of hara terizing a losed set S using the notion of limit point of S . A point z0 is a limit point of a set S if every
(z0 ; )

ontains a point of S other than z0. The point z0 itself may or may not belong to the set S . For example, z0 = 0 is a limit point of   1 1 1 S = 1; ; ; : : : ; ; : : : 2 3 n but 0 = z0 62 S . Similarly if
= fz : jz j < 1g, then ea h point on jz j = 1 is a limit point of
but again does not belong to the disk
. Ea h z 2
is also a limit point of S .

1.37. Example. We easily have the following: (a) S = fz : z = x + iy with x and y rationalg is neither open nor losed. (b) S = fz : z = 2g [ fz : jz j < 2g is neither open nor losed. ( ) C , ; are both open and losed. (d) S = fz : 0 < jz j  1g is neither open nor losed..  As an alternative hara terization of losed sets in C , we have \a set S is losed () S ontains all its limit points." We also note that, not every point of a losed set S need be a limit point of S ; for instan e, if

S = fz : z = 0 or z = 1=n; for positive integers ng; then z = 0 is the only limit point of S (whi h is in S ) and therefore, S is

losed. Note that no other point of S is a limit point of S , sin e, for 1 1 z0 = and  = ; n n(n + 1)

24

Complex Numbers y

(−a, 0)

(a, 0)

x

Figure 1.22: The set des ribed by S1 . y

(−a, 0)

(a, 0)

x

Figure 1.23: Set des ribed by S2 and S3 .

the disk
(z0 ; ) ontains no point of S other than z0 itself. Points in a set S whi h are not limit points are alled isolated points of the set S . Further, it is lear from the de nition that ea h element of an open set is a limit point of the set. A boundary point of a set S is a point for whi h every neighborhood

ontains at least one point of S and at least one point not in S . The boundary of S , denoted by S , is de ned as the set of all the boundary points. For instan e, onsider

S = fz 2 C : jz 1j  1g: The point z0 = 1+ i is a boundary point of S sin e every Æ-neighborhood of it has a non-empty interse tion with both S and S . Although in this ase y

(−a, 0)

(a, 0)

x

Figure 1.24: The set des ribed by S3 .

1.5 Topology of the Complex Plane

25

z0 2 S , this need not always be so. For example, z0 = 1 + i is a boundary point but not in S = fz 2 C : jz 1j < 1g. The boundary S is always losed in C , and S , the losure of S , is de ned by S = S [ S . A point z0 is alled an interior point of S if there exists a Æ > 0 su h that
(z0 ; Æ)  S . The interior of S , denoted by Int S , is the set of all interior points of S . Thus, it is lear from the de nition that \A set is open () ea h of its points is an interior point." A set S  C is said to be separated (or dis onne ted) if there exist two nonempty disjoint open sets A and B su h that S  A [ B , S \ A 6= ;, and S \ B 6= ;. If S is not dis onne ted, it is alled onne ted. In any parti ular situation we generally an tell at a glan e if a given set S in the omplex plane is onne ted, as illustrated in Figure 1.21. If a > 0, S1 S2 S3 S4

= = = =

fz : jz fz : jz fz : jz fz : jz 2

aj  a or jz + aj  ag aj  a or jz + aj < ag; aj < a or jz + aj < ag; 1j < 1g;

then S1 and S2 are onne ted (see Figure 1.23) whereas S3 and S4 are not

onne ted (see Figure 1.24). The fun tion : [0; 1℄ ! C ; de ned by (t) = (1 t)z0 + tz1 is alled the line segment with end points z0 and z1 and is designated by [z0 ; z1 ℄. If (t) 2 S for ea h t 2 [0; 1℄, then the line segment [z0 ; z1 ℄ is said to be

ontained in S . A polygonal line from z0 to zn is a nite union of segments of the form [z0 ; z1 ℄ [ [z1 ; z2 ℄ [


[ [zn 1 ; zn ℄: The points z0 and zn are then said to be polygonally onne ted. If the segment [zk ; zk+1 ℄ is ontained in S , k = 0; 1; : : : ; n 1, then the polygonal line from z0 to zn is said to be ontained in S . A set S is said to be polygonally onne ted if any two points of S an be onne ted by a polygonal line ontained in S . In other words, S is onne ted i ea h pair of points z;  of S an be onne ted by an ar lying in S . For instan e, any open disk
(z0 ; Æ) is polygonally onne ted. For, if z1 ; z2 2
(z0 ; Æ) and (t) = (1 t)z1 + tz2 ,

j (t) z0 j = j(1 t)(z1 z0 ) + t(z2 z0 )j  (1 t)Æ + tÆ = Æ and so for ea h t 2 [0; 1℄, (t) 
(z0 ; Æ). A domain is a nonempty open onne ted set in C . A domain together

with some, none, or all of its boundary points is referred to as a region. For instan e, if

S5 = fz 2 C : Re z < a; a-realg and S6 = fz 2 C : Re z  a; a-realg;

26

Complex Numbers y

(−a, 0) O

x

(a, 0)

Figure 1.25: The set des ribed by S7 . y

R e z = Im −

O

x

z

ia

a

Figure 1.26: The set fz = x + iy : y < xg.

then S5 des ribes a domain whereas S6 is a region but is not a domain, sin e the set de ned by S6 is not open but onne ted. On the other hand, the set S7 = fz 2 C : jRe z j > a for some a > 0g does not onstitute a domain. Note that S7 is open but not onne ted (see Figure 1.25). Further, the set

S8 = fz 2 C : a < Re z  b for some a < bg;

alled an in nite strip, is not a domain but is a region. Also, S9 = fz 2 C : jIm z j < jRe z jg is dis onne ted while S10 = fz 2 C : jIm z j  jRe z jg is onne ted. Similarly, the set

S11 = fz 2 C : jz + iaj < jz aj for some a > 0g is onne ted and open. Note that (see Figure 1.26)

jz + iaj < jz aj () () () ()

jz + iaj2 < jz aj2 jz j2 + jaj2 + 2Re ( iza) < jz j2 + jaj2 2Re (za)

2Re (iza) < 2Re (za) y < x; sin e a > 0:

A set S is bounded if there is an R > 0 su h that S 
R : Geometri ally S is ontained in a losed disk entered at 0 and radius R. A set that annot

1.6 Sequen es and Series

27

be en losed by any
R for R > 0 is alled unbounded. A simple example is an in nite strip, S8 above. Sets whi h are losed as well as bounded in C are alled ompa t sets in C . Of ourse, the omplex plane C is not ompa t as it is not bounded in C . Note that C is losed be ause C n C = ; is open. Thus, C (with usual metri ) is not a ompa t metri spa e. On the other hand,
R is ompa t. A set S is ountable if the elements of S an be pla ed in a one-to-one

orresponden e with the set of positive integers. For instan e, ea h of the sets de ned by,

A = f2n : n = 1; 2; : : : g; B = f2n + 1 : n = 1; 2; : : : g; C = fr : r rationalg; is ountable whereas the set of irrationals and the set of reals are both not

ountable, i.e. un ountable.

1.6 Sequen es and Series An in nite sequen e of omplex numbers is a list of points z1 ; z2 ; : : : ; zn ; : : : of C listed in some order. More expli itly a mapping N ! C , n 7! zn ; is

alled a sequen e. This is brie y denoted by fzn gn1 (or simply by fzng when there is no onfusion) with the understanding that zn is the n-th term of the sequen e. A sequen e is thus merely an assignment of a spe i point zn to ea h n 2 N . Suppose fzn g is a sequen e of points ( omplex numbers) in C and that fnk g is a stri tly in reasing sequen e of natural numbers. Then the sequen e fznk g (think of znk as ak ) is alled a subsequen e of the sequen e fzng. For instan e,

fzk+1 g; fz2k g; fz2k+5 g; and fz2k g are some subsequen es of fzng. Roughly speaking subsequen es are ob-

tained by deleting some of the terms from the sequen e under onsideration. Of ourse, fzng is trivially a subsequen e of itself. For instan e, fz2k+1 g may be obtained by removing the terms z2 ; z4 ; z6 ; : : : . We remark that the notion of sequen es is not on ned to sequen es of omplex numbers. In later hapters, we shall also onsider sequen es of sets and sequen es of fun tions. A sequen e fzng is said to onverge to a point z0 , and write zn ! z0 , if for every  > 0 there exists an N () 2 N su h that

jzn z0j < ; for all n > N (): That is, zn ! z0 if jzn z0j ! 0, as a real sequen e. If a sequen e fails to

onverge, it is said to diverge.

28

Complex Numbers

z0 z4 ·z ·z3 2 ·z1

·z5

·z7

Figure 1.27: Des ription for a limit of a sequen e.

Geometri ally, zn ! z0 if every neighborhood of z0 ontains all but nitely many terms of the sequen e fzn g; su h a point z0 is alled a limit of the sequen e (see Figure 1.27). Sometimes we write lim z n!1 n

= z0

instead of zn ! z0 . A point z0 is a limit of the sequen e fzn g if there exists a subsequen e that onverges to z0 . To have a better understanding of series that will be introdu ed later, we must de ne the notion of limit superior (resp. limit inferior) of a sequen e frn g of real numbers. If frn g is bounded above (resp. bounded below) and has at least one onvergent subsequen e, then limit superior (resp. limit inferior), written

L = lim sup rn or limn!1 rn n!1





resp. l = lim inf r or limn!1 rn ; n!1 n

is the least upper bound (resp. the greatest lower bound) of the limits of all onvergent subsequen es of frn g. If no su h L (resp. l) exists, we set

L = lim sup rn = +1 n!1



resp. l = lim inf r = n!1 n



1 :

Note that this is possible only if frn g is an unbounded sequen e. Of ourse, if frn g happens to be a onvergent sequen e then we note that lim sup rn = nlim inf r : !1 rn = lim n!1 n n!1 For instan e, let

an = ( 1)3n ; bn = 1 + ( 1)n ; n = n( 1)n ; dn = 1 4 n : Then we have (i) lim supn!1 an = 1 and lim inf n!1 an = 1 (ii) lim supn!1 bn = 2 and lim inf n!1 bn = 0

29

1.6 Sequen es and Series

(iii) lim supn!1 n = +1 and lim inf n!1 n = (iv) lim supn!1 dn = 1 = lim inf n!1 dn .

1

How about the sequen e feng, where en = sin(n=2) + n os(n=2)? For any omplex number z , we have

jRe z j; jIm z j  jz j  jRe z j + jIm z j: Now, suppose zn = xn + iyn ! z0 = x0 + iy0: Then for a given  > 0, there exists an N () su h that

jzn z0 j = j(xn x0 ) + i(yn y0 )j <  for all n > N (): This implies

jxn x0 j <  and jyn y0 j <  for all n > N (); i.e. xn ! x0 , yn ! y0 . Conversely, let xn ! x0 and yn ! y0 : Then for a

given  > 0, there exist N1 () and N2 () su h that

jxn x0 j < =2 for all n > N1 () and

jyn y0 j < =2 for all n > N2 (): Hen e, for all n > N () = maxfN1(); N2 ()g, we have jzn z0 j  jxn x0 j + jyn y0 j < :

The above dis ussion shows that (1.38)

zn ! z0 () Re zn ! Re z0 and Im zn ! Im z0:

A sequen e an have at most one limit if it exists; for, let zn ! z0 and zn ! z0. Then for a given  > 0, we have jz0 z0j = j(zn z0) (zn z0 )j  jzn z0j + jzn z0 j < =2 + =2 = ; for suÆ iently large n: Re all that if x; y 2 R and x < y +  for any  > 0, then x  y. For, suppose that x y > 0. Then for  = x y, the hypothesis yields  < . This ontradi tion shows that x  y. Note also that z0 ; z0 are independent of n and jz0 z0j <  = 0 +  for every . So jz0 z0 j  0, and hen e we must have jz0 z0j = 0, i.e. z0 z0 = 0 or z0 = z0. This shows that a limit of a sequen e is unique.

30

Complex Numbers

1.39. Remark. If zn ! `, then the limit ` is a limit point of the sequen e fzng. However, the onverse is not true. For example, let n ; n  1: zn = ( 1)n + i n+1

Then 1 + i and 1 + i are the limit points of the set fz1 ; z2 ; : : :g, but zn 6! 1 + i, or 1 + i. Similarly, for zn = 21 n + n + ( 1)n n (n  1); we have zn 6! 0, but 0 is the limit point of the set S = fz1; z2 ; : : :g. 

1.40. Example. To illustrate the on ept of the limit of a sequen e, we shall present some more examples. p (a) For zn = n1 + (n n1)i , we have zn ! i sin e jzn ij = n1 j1 ij = n2 ! 0: If we write zn = xn + iyn with xn = n1 and yn = nn 1 as real sequen es, we have xn ! 0 = x0 and yn ! 1 = y0 (z0 = x0 + iy0 = i): Compare with the observation in (1.38). (b) If zn = in =n, then, for a given  > 0, there is an N () su h that4   jzn 0j = n1 <  for all n > N () = 1 + 1;

i.e. zn ! 0.

( ) For zn = 1+ i=n and zn0 = 1 i=n, we have zn ! 1 and zn0 ! 1 whereas Arg zn !  and Arg zn0 ! : (d) The sequen e fArg [( 1)n =n℄g is divergent, be ause the sequen e has the form , 0, , 0, : : : and hen e has no limit.  A sequen e fzn g is alled a Cau hy sequen e if for ea h  > 0 there exists an N () su h that jzn zmj <  for all n; m > N (): Using this de nition, we an easily on lude the Cau hy riterion for onvergen e in C: fzng onverges () fzng is a Cau hy sequen e: For, let zn = xn + iyn ! z0 = x0 + iy0 . Then, Re zn ! Re z0 ; Im zn ! Im z0 : As jzn zmj  jRe (zn zm )j + jIm (zn zm )j; fzng is learly a Cau hy sequen e. Suppose fzng is a Cau hy sequen e. As

jRe (zn zm )j; jIm (zn zm )jg  jzn zmj; we see that fRe zng and fIm zn g are Cau hy sequen es in R. Be ause of the ompleteness property (see [11℄) of R, fRe zn g and fIm zng onverge and hen e, fzng onverges. A onvergent sequen e with limit zero is alled a null sequen e. If the sequen e fzng onverges to z0 , then the sequen e fzn z0 g is a null sequen e. 4 Here

[x℄ denotes the greatest integer less than or equal to x

31

1.6 Sequen es and Series

Sin e fzn g = z0 + fzn z0 g; any onvergent sequen e may be written as a sum of a xed number and a null sequen e. Conversely, if fzn g = z0 + fzn0 g; where fzn0 g is a null sequen e, then jzn z0 j = jz 0 j <  for n > N () n

and so the sequen e fzn g is onvergent.

1.41. Example. It is easy to see that fz ng is a null sequen e for jz j < 1. To prove this, let  > 0 be given. We must nd a value of N su h that, for any n > N , jz n j = jz jn < : This is ertainly true if z = 0. So, for a xed z 6= 0, we simply need to nd an N su h that, for any n > N , n ln jz j < ln ; i.e. n > ln = ln jz j (sin e ln jz j is negative, for jz j < 1). This proves that if jz j < 1 then

jz jn <  for n > N = [ln = ln jz j℄ + 1: Thus, z n ! 0 as n ! 1 if jz j < 1. For jz j > 1, the sequen e fz ng does not onverge to any point of C . How about for points z on the unit ir le jz j = 1?  On many o

asions we will need to work with an in nite series. Given a sequen e fzn g of omplex numbers the sequen e fsng de ned by

sn =

n X k=1

zk

is alled the sequen e of partial sums of the (in nite) series

onvenien e we shall use the equivalent forms: 1 X X zk or zk k=1 k 1 P

P1

k=1 zk :

For

(whi h will be sometimes abbreviated as zk ). Sometimes it is onvenient to start the series 0 (or perhaps even with some other integer p). P with k =P 1 z whi hever the ase may be. We then writeP 1 z or k=1 k k=p k The series zk is said to be onvergent or summable or said to onverge to s if sn ! P s. The number s is then alled the sum of the series and we write s = zk : Otherwise, the series is said to be divergent or to diverge. If the series is onvergent, the sequen e of its partial sums is bounded.

1.42. Remark. As in the ase of sequen es, if we let zk = xk + iyk and s = u + iv; xk ; yk ; u; v real

32

Complex Numbers

then the series verges to v.

P

zk onverges to s i

P

xk onverges to u and

P

yk on-



From the Cau hy riterion, i.e. \a sequen e is onvergent i it is a Cau hy sequen e", we see that X

P

zk onverges

() fsn g is a Cau hy sequen e.

That is, \ zk onverges i for every  > 0 there exists an N su h that

jsn sm j = Writing zn = sn inequality

n X

k=m+1

zk

<  for n > m > N ."

sn 1 , we have, by hoosing m = n 1 in the above

X

zk onverges =) nlim !1 sn = s =) nlim !1 zn = 0: P That is, \a ne essary ondition for the series zn to onverge is that zn ! 0 as n ! is not suÆ ient for the P 1." However, this ondition P

onvergen e of zn as the harmoni series n1 1=n shows. 1.44. Example. P (i) As the n-th term ( 1)n of the series n1 ( 1)n does not onverge to zero, the series diverges. P (ii) Consider the series 1 n=1 an , where an = 1=[(n + 1)(n + 2)℄: Then (1.43)

n X

n  X



1 1 1 sn = ak = = 2 k=1 k=1 k + 1 k + 2 P1 whi h onverges to 1=2. Thus, n=1 an = 1=2: P 1=2 diverges to 1, be ause (iii) The series 1 n=1 k

sn =

n X

k=1

1 n+2

k 1=2 > number of terms times the last term

= n(n 1=2 ) = n1=2 ; so limn!1 sn  limPn!1 n1=2 = +1. A similar argument may be  diverges to 1 when  2 [0; 1). How used to show that 1 n=1 k about when  = 1? If  = 1, then we easily have     1 1 1 1 1 1 1 + + + + + s2n > 1 + + 2 4 4 8 8 8 8   1 1 +


+ n +


+ n 2 2 1 1 1 1 n = 1+ + + +


+ = 1+ : 2 2 2 2 2

33

1.6 Sequen es and Series

This shows that the partial sum of the series is unbounded and hen e

fPsng is divergent. Note also that, sin e k p Pk 1 for all p  1 and k1 k

  1.

1

diverges, we dedu e that the series

k1 k



diverges for



The following result is easily established from the de nition of onvergen e of a series. So we omit its proof. P

P

1.45. Theorem. If zk = s and wk = t then P and zk = s; where is any omplex onstant.

P

(zk  wk ) = s  t

P

As with real series, a series Pzk is said to onverge absolutely or be absolutely onvergent if the series jzk j is onvergent. Further, using the fa t that n n X X zk  jzk j; jsn sm j =

k=m+1



k=m+1

we on lude that \every absolutely onvergent series is onvergent." Note that the onverse of this result is generally false. For instan e, the series P k 1 k1 zk , where zk = ( 1) =k , is onvergent, but not absolutely. This is be ause

jsn sm j =

n X

k=m+1

zk

1 1 1 +


n1 m+1 m+2 m+3   1  m + 1 <  for n > m > N = 1  1 1 and so fsn g is a Cau hy sequen e. P Further P one1 an show that the sum of this series is, in fa t, ln 2. But jzP k j = k1 k diverges as shown earlier. On the other hand, the series zk , where zk = ( 1)k 1 =k2, is absolutely onvergent (and hen e onvergent).P Analogous to Remark 1.42 we have \ zk is absolutely onvergent i P P ea h of xk and yk is absolutely onvergent." The following results are often useful. =

P

1.46. Theorem. If Pk1 jwk j is onvergent and if jzk j  jwk j ex ept for nitely many k 's then k1 zk is absolutely onvergent. The above theorem, popularly known as the omparison test for onvergen e, an be easily dedu ed from the Cau hy riterion for onvergen e. For, let  > 0 be given. Then there is an N  1 su h that for all n > m  N ,

jsn sm j 

n X

km+1

jzk j 

n X

km+1

jwk j < :

34

Complex Numbers

This Pobservation proves the theorem. The series of jzk j.

P

jwk j is alled a majorant

in 2 3=2 1.47. P Example. If zn = e os(n )=n , then jzn j hen e zn is onvergent.

1.48. Theorem. Let

P

n

3=2

and



zk be a series with nonzero terms su h that

lim sup Ln = L and lim inf L = l where Ln = jzn+1 =znj: n!1 n n!1 Then we have the following:

(a) If L < 1, the series onverges absolutely. (b) If l > 1, the series diverges. ( ) If l  1  L no on lusion an be made on erning the onvergen e of the series.

Proof. Clearly L; l  0. Suppose that L < 1. Then, for  with L <  < 1, there exists an integer N su h that zn+1 z

n

so that

jzn j = jzN j


zN +1 z




N jzN jp for

<  for all n  N zN +2 z

N +1






zn z

n 1

<

jzN jn N :

P

Therefore, jzN +pj

1. Then, by hypothesis, there exists an integer N su h that l > k > 1 and zn+1 z > k for all n  N: n Therefore, for all n > N ,

jzn j = jzN j


zN +1 z

N




zN +2 z

N +1

P






zn z

n 1



> jzN jkn N

! 1:

Hen e, zn 6! 0 as n ! 1 and so zn diverges. This proves (b). The last ase follows by onsidering the series with zn = 1=n and zn = 1=n2.

1.49. Corollary.

su h that

Let

P

zk be a series of non-zero omplex terms

 = nlim !1 Ln ; where Ln = jzn+1 =znj:

35

1.6 Sequen es and Series

(a) If  < 1, the series onverges absolutely. (b) If  > 1, the series diverges. ( ) If  = 1, the series may onverge or diverge. P

1.50. Example. Consider a series zk with Ln = jzn+1 =znj: p p p p (i) If zn = (1 + i)n =n, then Ln = nn+12 = 2 n+12 ! 2 as n ! 1 and hen e the series diverges. p (ii) If zn = (1 + i)n =n!, then Ln = n+12 ! 0 as n ! 1 and so the series

onverges absolutely. p +2) (iii) If zn = (n + 1)(1 + i)n =n!, then Ln = 2 ((nn+1) 2 ! 0 as n ! 1 and therefore the series onverges absolutely. (iv) If a; b; and d are real su h that ja2 + b2 j < j 2 + d2 j, then we see that the series, with zn = (a + ib)n =( + id)n , is absolutely onvergent.  1.51. Theorem. Let

P

zk be a series of omplex terms su h that

lim sup jzn j1=n = L: n!1 (a) If L < 1, the series onverges absolutely. (b) If L > 1, the series diverges. ( ) If L = 1, the series may onverge or diverge.

Proof. (a) If L < 1, hoose  > 0 so that L <  < 1 Then for all suÆ iently large values of n, we have

jzn j1=n < ; i.e jzn j < n P and so the onvergen e of jzn j follows from the omparison test with the P

geometri series n . (b) If P L > 1, jzn j > 1 for in nitely many n so that zn 6! 0 as n ! 1 and thus, zn diverges if L > 1. ( ) Examples relating to the proof of Theorem 1.48 an be used to justify the last assertion. P

The Cau hy in nite series of omplex terms Pprodu t of two onvergent P a and b is the series

n n n n0 n0 n0 , where

n =

n X k=0

ak bn k ; n = 0; 1; 2; : : : :

Next we state the following theorem. Proof of (a) follows dire tly from the de nition and the hypothesis.

1.52. Theorem.

36

Complex Numbers

(a) If fzng has a limit point at z0 , then there exist a subsequen e fznk g of fzn g su h that fznk g ! z0 as k ! 1. P (b) If the sequen e f nk=1 ak g is bounded P and fbn g is a de reasing null sequen e of positive numbers, then n1 an bn onverges. Proof. To prove (b), by hypothesis, we note that there exists an M su h that n X an  M for all n:

k=1



Sin e fbng is a null sequen e, given any  > 0 there exists an N su h that bn < =(2M ) for all n  N . Now, for all n  N m X

k=n+1

ak bk

M

"

m X

k=n+1

#

(bk

bk+1 ) + (bn+1 + bm+1 )

 2Mbn+1 < :

P

Thus, f m k=1 ak bk g is a Cau hy sequen e and the result follows by the Cau hy riterion.

1.7 Exer ises 1.53. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) The set of points z su h that jz + bij < jz + bj is the half-plane fz = x + iy : y < xg. (b) The inequality jz1 z2 j < jz 1 + z2 j holds provided either Re z1 > 0 and Re z2 > 0, or Re z1 < 0 and Re z2 < 0. ( ) Two omplex numbers z1 and z2 whose sum and dieren e are real and purely imaginary, respe tively, must satisfy z2 = z1 . (d) The equation az + bz + = 0 has exa tly one solution, i jaj 6= jbj. (e) If the sum and produ t of two omplex numbers are both real then either both the omplex numbers are real or one is the onjugate of the other. (f) The inequalities Re z > 0 and jz 1j < jz + 1j are equivalent.   1+z (g) The inequalities jz j < 1 and Re > 0 are equivalent. 1 z 1 1 1 (h) The inequalities Re z  Æ > 0 and  are equivalent. z 2Æ 2Æ (i) The roots of the ubi equation (z +  )3 = 3 ( 6= 0) form p the verti es of a triangle ea h side of whi h is of length equal to 3jj. (j) For Re zj > 0 (j = 1; 2), Arg (z1 z2 ) = Arg z1 + Arg z2 .

37

1.7 Exer ises

(k) For any positive integer n, jIm z nj  njIm z j jz jn 1. (l) If jz j < 1, then both Re (1 + z ) > 0 and Re (1 + z )2 > 0 do not hold simultaneously. (m) If z 2 = (z)2 , then z is either real or purely imaginary. 1 + os  + i sin  (n) The equation = ei( =2) ot(=2) holds for ea h 1 os  + i sin   6= 2n, n 2 Z. (o) The non-real roots of (1 + z )4 = 16z 4 are ( 1  2i)=5. (p) The produ t of the distin t n-th roots of unity is ( 1)n 1. p (q) All the solutions of z 4 + 81 = 0 are 3[1  i℄= 2. (r) The omplex roots of a quadrati equation have the property that one is the square of the other. (s) If z = a + ib, where a and b both are integers, then (i) j1 + z + z 2 + (ii) j1 + z + z 2 +




+ z nj  jz jn if a > 0


+ z nj  jz jn if a < 0. (t) The set of omplex numbers z su h that p arg(z i) = =3 represents the equation of the straight line y = 3x + 1. p (u) If Arg (z + 3) = =3, then the least value of jz j is 3 3=2. (v) If jz (4 3i)j = 2, then the greatest and least value of jz j are 7 and

3, respe tively. (w) Convergen e of fzng implies the onvergen e of fZng, where Zn = 1 Pn z . n k=1 k n z an jz jn jajn  jz j jaj : (x) For z 6= a and n 2 N , we have z a (y) If ! is an n-th root of unity, then

nX1 j =0

jz1 + !j z2j2 = n[jz1 j2 + jz2 j2 ℄:

Note: For n = 2, this redu es to the Parallelogram identity. (z) If zn 6= 0 and zn ! ` 6= 0, then Arg zn ! Arg `. 1.54. For any two non-zero omplex numbers z1 and z2 , prove the following (a) Arg (z1 z2 ) = Arg z1 + Arg z2 + 2k1 (b) Arg (z1 =z2) = Arg z1 Arg z2 + 2k2 ; where kn (n = 1; 2), whi h depends on z1 and z2 , is given by

kn =

8 < :

1 if 2 < Arg z1 + ( 1)n 1 Arg z2   0 if  < Arg z1 + ( 1)n 1 Arg z2   1 if  < Arg z1 + ( 1)n 1 Arg z2  2:

38

Complex Numbers

1.55. Prove Lagrange's identity (using the method of indu tion): 2 m X zk wk

k=1

=

m X k=1

jzk

j2

!

m X k=1

jwk

j2

!

X

k j

jzk wj zj wk j2 :

1.56. If z1 ; z2; z3 are the verti es of an equilateral triangle, then show that z12 + z22 + z32 = z1 z2 + z2z3 + z3z1 : If z1 = 1+ i and z2 = 1 i, determine the two possible values of z3 so that z1 ; z2; z3 form an equilateral triangle. 1.57. Compute the limit of the sequen e fzng when zn equals p n1=n ; 3 n+i n ; 2n =n; n sin(1=n); n2 e n ; in!: P 1.58. Test the onvergen e of 1 z when z equals n(3 + i)n ; n3 3n(2 + i) n ; n

n=1 n 1 in ; 5 n ;

n

(n!)2 =(3n!); n 1=3 i3n :

1.59. Classify the following sets a

ording to the properties open,

losed, bounded, unbounded, ompa t: (a) (b) ( ) (d) (e)

S1 = fz : S2 = fz : S4 = fz : S5 = fz : S6 = fz :

z = e2ki=5 ; k = 0; 1; 2; 3; : : : g jz j > 2jz 1jg Re (iz ) < 1g jz + 2j  jz jg z 2 + z2 = 1g.

1.60. Supply the geometri des ription of the following subsets of C :

(a) S1 = fz : 0 < Re (iz )  1g     z z1 (b) S2 = z : Im = 0; for z1 ; z2 2 C z z2     z z1 ( ) S3 = z : Re = 0; for z1 ; z2 2 C z z2 (d) S4 = fz : jz 1j = 3jz + 1jg (e) S5 = fz : jz 1j = Re (z )jg (f) S6 = fz : jz + 1j > 2; 0 < Arg (z 2) < =4g (g) S7 = fz : =4 < arg(z k) < 3=2; k = 1; 2; : : : ; 5g (h) S8 = fz : =6 < arg(z 1=k) < 13=6; k = 1; 2; : : : ; 5g :

1.61. Find all ir les that are orthogonal to both jz

jz j = 1.

1j = 4 and

Chapter 2

Fun tions, Limit and Continuity

In this hapter we introdu e fundamental results on erning limits, ontinuity and uniform onvergen e of sequen es and series of fun tions. We also introdu e sets in the extended omplex plane. This hapter lays the ground work for a areful treatment of analyti fun tions of a omplex variable. In Se tion 2.1, we de ne fun tions in C and their elementary properties su h as one-to-one and onto. In Se tion 2.2, we brie y present basi fa ts about limits and ontinuous fun tions of a omplex variable. In Se tion 2.3, we formalize the notion of `the point at in nity'. This helps to extend the notion of limit and ontinuity for fun tions de ned on unbounded sets. This se tion also provides a onvenient way for dis ussing the behavior of fun tions as jz j gets large and for ertain generalization of domains in the

omplex plane.

2.1 One-to-one and Onto Fun tions Let A and B be two non-empty subsets of C . A fun tion from A to B is a rule, f , whi h assigns ea h z0 = x0 + iy0 2 A a unique element w0 = u0 + iv0 2 B . The number w0 is alled the value of f at z0 and we write w0 = f (z0 ): If z varies in A then w = f (z ) varies in B . We say that f is a

omplex fun tion of a omplex variable in A. We say that f is a fun tion de ned on A. We also write (2.1)

f : A ! B; z 7! w = f (z ):

Here z is alled the independent variable; w the dependent variable and A the domain of f . If S  A, we an have f : S ! B and we all this new fun tion the restri tion of f in (2.1) to S and denote it by f jS . Let us examine the graphi al representation of (2.1), i.e. w = f (z ). By de nition, for ea h z = x + iy,

w = f (z ) = Re f (z ) + i Im f (z );

40

Fun tions, Limit and Continuity

w being a omplex number. Identifying z = x + iy with (x; y) have the fun tions

2 R2 , we

(x; y) 7! Re f (z ); (x; y) 7! Im f (z ); or equivalently Re f : z 7! Re f (z ); Im f : z 7! Im f (z ); de ned on the domain A, now onsidered a subset of R2 . We observe that these are real-valued fun tions de ned on A  R2 and denote them by Re f; Im f , respe tively. Conversely, if A  R2 and we have two real-valued fun tions u : A ! R; v : A ! R; then, by de ning

f (z ) = u(x; y) + iv(x; y); (x; y) 2 A; we obtain f : A ! C ; where A is now onsidered as a subset of C . Thus, a omplex fun tion of a omplex variable is ompletely determined by the fun tions Re f; Im f , known as the real and imaginary parts. If we use the polar form for z , then f an be written as

f (z ) = u(r; ) + iv(r; ): To illustrate these ideas let f (z ) = 2z 2

z + 1, z 2 C . Then

f (z ) = [2(x2 y2 ) x + 1℄ + i[4xy y℄: Thus, (Re f )(z ) = 2(x2 y2) x + 1 and (Im f )(z ) = 4xy y: On the other hand if we write z = rei , then

f (z ) = [2r2 os 2 r os  + 1℄ + i[2r2 sin 2 r sin ℄: Sin e ea h of the variables z and w requires two dimensions for representation, a graphi al representation annot be given for a omplex fun tion. Be ause of this, we need to employ two dierent opies of the omplex plane to des ribe the nature of a omplex fun tion of a omplex variable. Thus, to every z = x + iy of its domain in the z -plane, we determine the resulting values of u = Re f and v = Im f (w = u + iv) and plot them in the w-plane (see Figure 2.1). If f is de ned on A and S  A, then f (S ) = ff (z ) : z 2 S g is alled the image of the set S under f . The set R = f (A) of all image points of A is alled the range of f . If R  B , the fun tion f on A to B is alled a mapping of A into B and if R = B , we say that the fun tion f maps A onto the range R sin e every element w 2 R is an image of at least one point in A. Clearly, a mapping

41

2.1 One-to-one and Onto Fun tions y

v

w = f (z)

w0

z0

O

x

O

u

Figure 2.1: Des ription for mapping.

whi h is \onto" is \into" but the onverse is not ne essarily true. Point w 2 R is alled an image of a point z 2 A and the point z is alled the pre-image of w, under the mapping5 w = f (z ): A mapping is said to be an open mapping if it maps open sets onto open sets. We know that every omplex number z 2 C n f0g an be uniquely written in polar form as

z (r; ) = rei (r = jz j;  2 ( ; ℄): If we in rease  by 2,

z (r;  + 2) = rei(+2) = rei = z (r; ) returning to its original value. With this observation, the de nition of single-valuedness of a fun tion takes the following simple form in polar

oordinates. A fun tion f is said to be single-valued if f satis es

f (z ) = f (z (r; )) = f (z (r;  + 2)):

(2.2)

Otherwise, f is said to be multiple-valued. Let us look at the fun tion f (z ) = z n: Then, when n is an integer, we know that for every w 6= 0 there are n values of z satisfying w = f (z ) = z n : We have

f (z (r; )) = rn ein ; f (z (r;  + 2)) = rn ein e2ni : So, the ondition (2.2) is satis ed i n is an integer. This shows that the fun tion f (z ) = z n is single-valued i n is an integer. If n is not an integer, then e2ni 6= 1 and so f in this ase is multiple-valued. If the elements of A are omplex numbers and those of B are real numbers, then we say that f is a real-valued fun tion of a omplex variable. Similarly, if the elements of A are real and those of B are omplex numbers, then f is a omplex-valued fun tion of a real variable. Whenever we speak of a fun tion we shall, unless otherwise stated expli itly, onsider a single-valued fun tion. 5 `Mapping'

is another word for fun tion, transformation.

42

Fun tions, Limit and Continuity

Suppose we have a mapping of a set A onto a set R. It may happen that some points of R have more than one pre-image. If ea h w 2 R is the image of pre isely one point in A; that is

f (z1) = f (z2 ) =) z1 = z2 ;

or z1 6= z2 =) f (z1) 6= f (z2 );

then the mapping w = f (z ) is said to be one-to-one. If the mapping w = f (z ) is one-to-one then the fun tion f is said to be univalent, a fan y term for one-to-one fun tions. (Note that the fun tion f is said to be univalent at z0 if it is univalent in a neighborhood of z0 . This will be dis ussed later in this book.) In this ase, we have a mapping from R into the z -plane with A as the range and R as the domain of de nition. Denoting the latter mapping, alled the inverse of f , by f 1 we write

z = f 1(w) if w = f (z ): Thus, if f maps A in a one-to-one fashion onto R, then there is an inverse mapping z = f 1 (w) on R onto A. Note that f 1 (w) = f 1(f (z )) = z: Observe that if f : A ! R is univalent on A, then f 1 is de ned on R and is univalent therein. A fun tion f with domain A and range B is alled a onstant fun tion if B ontains only one element, say . In this ase, we write f (z )  , z 2 A. A fun tion f (z ) is said to be bounded on a subset S  C if there exists an M > 0 su h that jf (z )j  M for all z 2 S . Suppose we have a fun tion f with domain D1 and another fun tion g with domain D2 . Suppose further that, for ea h z 2 D1 , f (z ) is in D2 . Then for every z 2 D1 the asso iation g Æ f de ned by (g Æ f )(z ) = g(f (z )); is a fun tion alled the omposition of f and g. We indi ate this by

D1 f! D2 g! C : gÆf For instan e, let f (z ) = 2z + 1 and g(z ) = z 2 + 2 on C : Then, g(f (z )) = (2z + 1)2 + 2 and f (g(z )) = 2(z 2 + 2) + 1; z 2 C : If f (z ) = u(z ) + iv(z )  u(x; y) + iv(x; y); where u; v are real-valued fun tions, then

f (z ) = u(x; y) iv(x; y) and f (z) = u(x; y) + iv(x; y): Thus, we note that f (z ) and f (z) are in general dierent fun tions. For instan e, if f (z ) = (1 i)z , f may be rewritten as f (z ) = (x + y) + i(y x) so that f (z ) = (1 + i)z and f (z ) = (1 i)z:

2.3. Examples. Consider w = f (z ) = z 2 ; jz j  1: Here the domain of de nition of f is
and the range is also
: An angular region with

43

2.1 One-to-one and Onto Fun tions y

v i

2i α

e iα

e

2α

α x

−1

u

Figure 2.2: Mapping under w = z 2 ; jz j  1.

vertex at 0 of  radians in the z -plane is mapped into an angular region of 2 radians in the w-plane, under this mapping (see Figure 2.2). The fa t that f (z ) = z 2 is not univalent on
an be seen as follows: let z = rei (0 < r  1), be the polar representation of the points on
. Then

w = r2 e2i = ei

with  = r2 and  = 2, 0    4.

So, z1 = rei , z2 = rei(+) = z1 are su h that

z12 = r2 e2i = r2 e2i+2i = r2 e2i(+) = z22: This shows that the fun tion is not univalent. In fa t, it maps
onto
twi e. Next, we onsider w = f (z ) = z 2 , z 2 D = fz : jz j  1; 0  arg z < g: Then the range is
: Thus, if z = rei then w = r2 e2i = ei with

p = r (0 <  < 1) and  = 2; 0   < 2:

If the two pre-images (why not more?) for w = ei , where  = r2 and  = 2 by f (z ) = z 2, are z1 = rei and z2 = rei(+) , then only one of z1 or z2 an lie in D sin e any two elements of D have their arguments diering by less than . This shows that the fun tion is univalent in D. Again we onsider the same fun tion f (z ) = z 2, but this time without indi ating the domain of de nition. Then, we have

f (z1 ) = f (z2) =) z12 z22 = 0 =) either z1 = z2 or z1 = z2: The points z1 and z2 su h that z1 = z2 are symmetri with respe t to the origin, i.e. lie at the same distan e from the origin on the same straight line through the origin. This shows that f (z ) = z 2 is univalent in a domain D i this domain does not have even a single pair of points symmetri with respe t to the origin. For instan e, f (z ) = z 2 is univalent in the upper half-plane U = fz : Im z > 0g, the lower half-plane L = fz : Im z < 0g, respe tively. Note also that f maps U and L, into the w-plane with a ut along the negative real semi-axis omitting the origin. 

44

Fun tions, Limit and Continuity

2.4. Examples. Consider w = f (z ) = z: Observe that the ee t of this mapping on the points of the plane is a re e tion on the real axis. This fun tion is one-to-one and the inverse is z = w. It maps the entire z -plane onto the entire w-plane. On the other hand, the fun tion w = g(z ) = jz j maps the entire z -plane onto the non-negative real axis of the w-plane. In fa t, it maps every ir le

entered at the origin to a point. It is not one-to-one and hen e no inverse exists.  2.5. Examples. It is easy to see that the fun tion f (z ) = 3z + z 2 is one-to-one in
. Indeed, for z1 ; z2 in
, f (z1 ) = f (z2) =) (z1

z2 )(z1 + z2 + 3) = 0 =) z1 z2 = 0

(z1 + z2 + 3 6= 0 in
, sin e Re (z1 + z2 + 3) > 1 1 + 3 = 1). Similarly, it is easy to see that the fun tion f (z ) = (1 + z )2 is univalent in
. More generally, f (z ) = z + z 2 is univalent for jz j < 1=(2jj) whi h is in fa t the largest disk entered at the origin on whi h f is one-to-one. This is easily seen using the argument relating to f (z ) = 3z + z 2 .  A domain D is said to be symmetri with respe t to the origin, if for every z 2 D, the point z 2 D. A fun tion f de ned on the domain D whi h is symmetri with respe t to origin is said to be even if f (z ) = f ( z ) is valid for every z 2 D. For example, f (z ) = z 2n (n 2 N ) is even in C . A fun tion f is said to be odd on D if f (z ) = f ( z ) for every z 2 D. For example, f (z ) = z 2n 1 (n 2 N ) is an odd fun tion in C . A domain D is said to be starlike with respe t to a point z0 2 D if for ea h z 2 D the line segment, [z0 ; z ℄, from z0 to z lies entirely in D. The point z0 is alled a star enter of D. A starlike domain is a domain whi h is starlike with respe t to the origin. A domain D is said to be onvex if for ea h pair of points ; z 2 D, the line segment, [; z ℄, joining z and  lies entirely in D. Obviously, a onvex domain is starlike with respe t to any of its points. An open disk, half-planes su h as Re z > 0, an open ellipse and an open re tangle are examples of onvex domain. An example of a starlike domain but non- onvex is the domain C n fz = x : x  1g, i.e. the plane minus the negative real axis from 1. How about the set C n fx + iy : x = 0; jyj  1g?

2.2 Con epts of Limit and Continuity The de nitions of limit, ontinuity and uniform ontinuity are analogous to those in Real Analysis. Suppose that a omplex-valued fun tion f is de ned on D  C and z0 2 D. Then f is said to have a limit ` as z ! z0 and we write lim f (z ) = ` or f (z ) ! ` as z ! z0 z!z0

45

2.2 Con epts of Limit and Continuity z D

z0

z z

z

z

Figure 2.3: z ! z0 in C .

i for any given  > 0, there exists a Æ = Æ(; z0 ) > 0 su h that

jf (z ) `j <  whenever z 2 D and 0 < jz z0 j < Æ; i.e. i for ea h  > 0 there exists a Æ > 0 su h that (see 1.6)

f (z ) 2
(`; ) whenever z 2 [
(z0 ; Æ) nfz0 g℄ \ D: It is straightforward to state lim f (z ) = l

z!z0

() zlim !z0 jf (z ) `j = 0:

Less pre isely stated, this means that if z is near z0 , then f (z ) is lose to `. First, it should be noted that the fun tion need not be de ned at z0 in order to have a limit at z0 . Se ondly, it is the pun tured disk
(z0 ; Æ) nfz0g whi h is involved in D, i.e. z0 need not be in D. Thirdly, even if the

ondition that z0 2 D holds, we may have f (z0) 6= `. In real variable theory, we do not have the freedom whi h a omplex variable produ es for, if z0 = x0 2 R, a neighboring point z = x ! x0 has only two possible ways either from the left or from the right. In the omplex ase, z an approa h z0 in any manner in the omplex plane (see Figure 2.3). As in Real Analysis, if the limit exists then it must be unique: Suppose that 0 0 lim f (z ) = ` and zlim z!z0 !z0 f (z ) = ` with ` 6= ` : Then for a given  > 0 there exist Æ1 ; Æ2 > 0 su h that

jf (z ) `j <  whenever z 2 D and 0 < jz z0 j < Æ1 and

jf (z ) `0 j <  whenever z 2 D and 0 < jz z0 j < Æ2 : Therefore, whenever z 2 D and 0 < jz z0 j < Æ = minfÆ1; Æ2 g, j` `0 j = j(f (z ) `0) (f (z ) `)j  jf (z ) `0 j + jf (z ) `j <  +  = 2:

Both the left and the right side of the above inequality are independent of Æ. As  is arbitrary, the inequality an hold i ` = `0 .

46

Fun tions, Limit and Continuity

Suppose that f; g are fun tions de ned on D fun tion de ned on D by

 C . Then f + g is the

(f + g)(z ) = f (z ) + g(z ); z 2 D: Similarly, we de ne (fg)(z ) = f (z )g(z ) for z 2 D, and  

f f (z ) (z ) = ; z 2 D when g(z ) 6= 0 in D: g g(z )

The following theorem gives the \ omplex" theory of limits from the \real" theory, and onversely.

2.6. Theorem. Let f (z ) = u(z ) + iv(z ), where u(z ) = u(x; y) and v(z ) = v(x; y) are real-valued fun tions, be de ned on D ex ept possibly at z0 . Then for ` and `0 2 R, lim f (z ) = `1 + i`2

(2.7) i

(2.8)

z!z0

lim

(x;y)!(x0 ;y0 )

u(x; y) = `1 and

lim

(x;y)!(x0 ;y0 )

v(x; y) = `2 :

Proof. The proof of this theorem follows immediately from Remark 1.42. However, we in lude the details here. Suppose that the limit (2.7) exists. Using the triangle inequality it follows that the inequalities

ju(x; y) `1 j; jv(x; y) `2 j  jf (z ) (`1 + i`2 )j and jx x0 j; jy y0 j  jz z0 j are satis ed. Now if we allow z ! z0 , i.e. (x; y) ! (x0 ; y0 ), it is evident that the

two limits (2.8) exist. Conversely, suppose that the two limits (2.8) exist. Then, for a given  > 0, there exist Æ1 and Æ2 su h that

ju(x; y) `1 j < =2 whenever 0 < j(x x0 ) + i(y y0 )j < Æ1 and

jv(x; y) `2 j < =2 whenever 0 < j(x x0 ) + i(y y0 )j < Æ2 : By the triangle inequality

jf (z ) `j = j(u(x; y) `1 ) + i(v(x; y) `2 )j  ju(x; y) `1j + jv(x; y) `2 j < (=2) + (=2) = 

whenever 0 < jz

z0j < Æ = minfÆ1 ; Æ2 g and the proof is omplete.

47

2.2 Con epts of Limit and Continuity

2.9. Example. For z 6= 0, onsider f (z ) =

z (z )2 x2 y2 2ixy = = : z jz j2 x2 + y2

If we set f (x + iy) = u(x; y) + iv(x; y), u(x; y); v(x; y) being real, then

u(x; y) =

x2 y2 2xy and v(x; y) = 2 2 : x2 + y2 x +y

For ea h point z0 6= 0, it is lear that limz!z0 f (z ) exists and equals f (z0 ). We shall see whether limz!0 f (z ) exists or not, by examining the limits as z ! 0 in many ways. Let m be any real number and allow z ! 0 along the line y = mx. Then

f (x + mxi) =



1 m2 1 + m2



i



2m 1 + m2



whi h learly shows that limz!0 f (z ) does not exist.



2.10. Theorem. Let f and g be de ned in a neighborhood of z0 ex ept possibly at z0 . Given lim f (z ) = ` and lim g (z ) = `0 ; we have z!z0 z!z0 0 (a) zlim !z0 [f (z ) + g(z )℄ = ` + ` 0 (b) zlim !z [f (z )g(z )℄ = `` 0

f (z ) ` 0 ( ) zlim !z0 g(z ) = `0 if ` 6= 0.

In parti ular, For a, b omplex onstants, lim (az + b) = az0 + b: z!z0

Proof. By hypothesis, given 1 > 0, there exists Æf = Æf (z0 ; 1 ) > 0 su h that (2.11) 0 < jz z0 j < Æf =) jf (z ) `j < 1 : Similarly, given 2 > 0, there exists Æg = Æg (z0 ; 2 ) > 0 su h that (2.12) 0 < jz z0 j < Æg =) jg(z ) `0 j < 2 : (a) Then for all z with 0 < jz z0 j < Æ = minfÆf ; Æg g it follows from the triangle inequality that jf (z ) + g(z ) (` + `0 )j  jf (z ) `j + jg(z ) `0 j < 1 + 2 : Given  > 0, hoosing 1 and 2 so that 1 + 2 = , we see that (a) follows from the last inequality.

48

Fun tions, Limit and Continuity

(b) Now for all z su h that 0 < jz z0 j < Æ, it follows from the triangle inequality that jf (z )g(z ) ``0j = jg(z )(f (z ) `) + `(g(z ) `0 )j  jg(z )j jf (z ) `j + j`j jg(z ) `0j  [jg(z ) `0j + j`0 j℄jf (z ) `j + j`j jg(z ) `0 j  [2 + j`0 j℄1 + j`j2 and the result follows as 1 and 2 are arbitrary. ( ) If we hoose 2 = j`0 j=2, as jg(z )j + `0  jg(z ) `0 j, (2.12) shows that (2.13) 0 < jz z0 j < Æg =) jg(z )j > j`0 j=2: So g(z ) 6= 0 in the deleted neighborhood
(z0 ; Æg ) nfz0g. Now for all z su h that 0 < jz z0 j < Æ, it follows that for 2  j`0 j=2 f (z )`0 g (z )` f (z ) ` = g (z ) `0 g(z )`0 0 (f (z ) `)` `(g(z ) `0 )  g(z )`0    [1 j`0 j + 2 j`j℄ j`20 j
j`10 j ; by (2:13); 2(1 j`0 j + 2 j`j) = : j`0 j2 Again the result follows as 1 and 2 are arbitrary. A fun tion f : D ! C is ontinuous at z0 2 D i limz!z0 f (z ) exists and equals the fun tion value f (z0 ). We say that f is ontinuous on D or f : D ! C is ontinuous when f is ontinuous at all points of D. Note that f is ontinuous at z0 i the following three onditions hold:

f (z0) is de ned, zlim !z f (z ) exists, and zlim !z f (z ) = f (z0 ). 0

0

In terms of our earlier notation, the de nition of ontinuity is that for a given  > 0, there exists a Æ > 0 su h that

jf (z ) f (z0 )j <  whenever z 2 D and jz z0 j < Æ; or equivalently, (2.14)

f (z ) 2
(f (z0 ); ) whenever z 2
(z0 ; Æ) \ D:

A fun tion f : D ! C is dis ontinuous (or has a dis ontinuity) at a point z0 if z0 2 D, yet f is not ontinuous at z0 .

2.2 Con epts of Limit and Continuity

49

2.15. Remark. If the domain D of f is su h that z0 2 D and
(z0 ; Æ)  D for some Æ > 0, then all the points in
(z0 ; Æ) for this Æ or a smaller Æ0 are to satisfy (2.14). It might happen that z0 2 D is su h that there exists a Æ > 0 with only z0 in
(z0 ; Æ) \ D; i.e. z0 is an isolated point of D. In this ase (2.14) is trivially satis ed and f is ontinuous at z0 . Avoiding this trivial ase we shall, hen eforth assume that when we

onsider z0 2 D for ontinuity of f : D ! C , z0 is an interior point of D. If we are on erned with ontinuity of f on D we shall assume D to be open (sin e D is open, there exists an r su h that z0 + h 2 D with jhj < r).  In the de nition of f at z0, the number Æ depends on z0 and . When, given  > 0, there exists a Æ = Æ(), independent of z0, satisfying (2.14) for all z0 2 D then we say that f is uniformly ontinuous on D. Clearly every uniformly ontinuous fun tion on D is ontinuous on D, but the onverse is not true in general (see Examples 2.28).

2.16. Example. Consider f : f (z ) = As z 2

3z

8 < :

10i = [z

f (z ) = so that

lim

C ! C de ned by

0 if z = (1 + 2i) 3z 10i if z 6= (1 + 2i). z + (1 + 2i)

z2

(4 + 2i)℄[z + (1 + 2i)℄; we may rewrite 

z! (1+2i)

z

0 if z = (1 + 2i) (4 + 2i) if z 6= (1 + 2i);

f (z ) = (5 + 4i) 6= f ( 1 2i):

To verify this using ` Æ' notation, let  > 0 be given. For z 6= 1 we obtain that jf (z ) + 5 + 4ij = jz + (1 + 2i)j: Therefore,

2i,

jf (z ) ( 5 4i)j <  whenever 0 < jz ( 1 2i)j < Æ =  so that limz!

(1+2i) f (z ) =

(5 + 4i):



In the de nition of limit, if ` 6= f (z0), then f is said to have a removable dis ontinuity at z0 2 D (i.e. if the value of f is \ orre ted" at the point z0 , it be omes ontinuous there). This an also be done when f (z0 ) is not de ned there. For instan e in the above example if we de ne f ( 1 2i) = 5 4i, then f be omes ontinuous at z = 1 2i and at all other points in C . As an immediate onsequen e of Theorem 2.6 we have

2.17. Theorem. The fun tion f (z ) = u(x; y) + iv(x; y) is ontinuous at z0 = x0 + iy0 i u(x; y ) and v (x; y ) are ontinuous at (x0 ; y0 ). In other words, Re f , Im f are both ontinuous i f is ontinuous.

50

Fun tions, Limit and Continuity

Consider f (z ) = jz j2 . As the real and imaginary parts of f are ontinuous fun tions of x and y for all (x; y) 2 R2 , f (z ) is ontinuous on C . As a result of the properties of limits (see Theorem 2.10), we obtain the following theorem.

2.18. Theorem. If f; g : D ! C are ontinuous at z0 2 D, then their sum f + g , produ t fg , quotient f=g where g (z0 ) 6= 0, and jf j are also

ontinuous at z0 . In parti ular, every polynomial a0 + a1 z +


+ an z n is

ontinuous for every z in C . Here are some examples of fun tions whose ontinuity follow from Theorem 2.18. The fun tion f de ned by

z3 + 3 z2 + 4 is ontinuous on C n f2i; 2ig. More generally, Theorem 2.18 shows that any rational fun tion p(z )=q(z ), where p and q are polynomials, is ontinuous on C n fz : q(z ) = 0g. For example, a fun tion of the form a n a (n 1) a + n 1 +


+ 1 + a0 + a1 z +


+ an z n zn z z is ontinuous on the pun tured plane C n f0g. f (z ) =

2.19. Theorem. If limz!z0 f (z ) = w0 and g is a fun tion whi h is

ontinuous at the point w0 , then limz!z0 (g Æ f )(z ) = g (w0 ): Proof. Continuity of g at w0 implies that for a given  > 0, there exists a Æg > 0 su h that (2.20)

jg(w) g(w0 )j <  whenever jw w0 j < Æg :

Further, by the ondition on f , for this Æg > 0, there exists a Æ > 0 su h that jf (z ) w0 j < Æg whenever z 2
(z0 ; Æ) nfz0g: Now if we let w = f (z ) in (2.20) we see that for all z 2
(z0 ; Æ) nfz0g,

j(g Æ f )(z ) g(w0 )j = jg(f (z )) g(w0 )j <  from whi h we obtain the required on lusion.

2.21. Corollary. The omposition of two ontinuous fun tions is

ontinuous; i.e. if f : D1 ! D2 is ontinuous at z0 2 D1 and if g : D2 ! C is ontinuous at w0 = f (z0 ), then g Æ f de ned by (g Æ f )(z ) = g (f (z )) is

ontinuous at z0 . Proof. The proof is a onsequen e of Theorem 2.19, see Figure 2.4.

51

2.2 Con epts of Limit and Continuity w = f (z) D1 z0

ζ = g(w)

f w0 =

(z 0) D2

g )=

(f (z

)

0)

g (w 0

Figure 2.4: Des ription for a omposite map.

An alternative and useful hara terization of a ontinuous fun tion in terms of sequen es is the following.

2.22. Theorem. A fun tion f is ontinuous at a point z0 2 D i f (z0 ) = limn!1 f (zn ) for every sequen e fzng su h that zn 2 D for n = 1; 2; : : : and zn ! z0 as n ! 1. Proof. =) : Consider a sequen e fzng in D su h that zn ! z0 as n ! 1. Continuity of f at z0 implies that, for a given  > 0 there exists a Æ with jf (z ) f (z0 )j <  whenever jz z0 j < Æ: For this Æ, sin e zn ! z0 , there exists an N su h that for all n  N , jzn z0 j < Æ. Thus,

jf (zn ) f (z0)j <  for all n  N from whi h it follows that f (zn ) ! f (z0 ), as desired. (= : Suppose that the onverse part is not true. Then, for some  > 0,

for every Æ > 0 there orresponds a point  su h that

j z0j < Æ and jf ( ) f (z0 )j  : Fix su h an . Then for ea h n 2 N there exists  2 D \
(z0 ; 1=n), denoted by n ,

jn z0 j < n1 and jf (n ) f (z0)j  : So, n ! z0 but f (n ) 6! f (z0 ) as n ! 1. This is a ontradi tion to our

assumption, and so the onverse part is proved.

Consider the fun tion f de ned by z f (z ) = ; z 2 C n f0g: z Note that the fun tion itself is not de ned at 0 and so, it is not ontinuous at 0. If zn = 1=n and zn0 = i=n, then f (zn ) = 1 and f (zn0 ) = 1. Note

52

Fun tions, Limit and Continuity

δ 

z0

ǫ f (z0 ) f ()

Figure 2.5: Des ription for ontinuous mapping.

that zn ! 0 and zn0 ! 0. Can this fun tion be made ontinuous at 0, by de ning its value suitably at the origin?

2.23. Theorem. Let f : D ! C be a fun tion. Then f is ontinuous on D i for every open set O  C , f 1 (O) = fz 2 D : f (z ) 2 Og is open in D. Proof. (= : Suppose that for ea h open set O  C ; f 1 (O) is open. Let z0 2 D and  > 0 be given. Then A =
(f (z0 ); ) is an open set in C and so by hypothesis f 1 (A) is open in D. Sin e f 1 (A) is open in D and z0 2 f 1 (A), there exists an open disk
(z0 ; Æ) su h that
(z0 ; Æ) \ D  f 1 (A): Hen e (see Figure 2.5), f (
(z0 ; Æ) \ D)  A =
(f (z0 ); ): Continuity of f at z0 is thus established. =) : Suppose that f is ontinuous on D and let O be any open set in C . If f 1 (O) = ;, then it is open. Otherwise, let z0 be any point of f 1 (O). Then z0 2 D and f (z0 ) 2 O. As O is open, there exists an open disk
(f (z0 ); )  O. Continuity of f at z0 then implies that

f (z ) 2
(f (z0 ); )  O whenever z 2
(z0 ; Æ) \ D: Thus,
(z0 ; Æ) \ D  f 1 (O) and this shows that f 1 (O) is open in D. One of the important results on erning ontinuous fun tions is that they preserve onne tedness.

2.24. Theorem. Let f : D ! C be a ontinuous fun tion and let D be a onne ted set. Then f (D) is a onne ted set. Proof. Suppose that the theorem is not true. Then there exist two non-empty disjoint open subsets V1 ; V2 of f (D) su h that f (D) = V1 [ V2 . By Theorem 2.23, U1 = f 1(V1 ) and U2 = f 1 (V2 ) are both non-empty open subsets of D su h that U1 \ U2 = ; and D = U1 [ U2 , ontradi ting the onne tedness of D.

2.2 Con epts of Limit and Continuity

53

Our next theorem is helpful parti ularly in onstru ting uniformly ontinuous fun tions.

2.25. Theorem. A ontinuous fun tion on a ompa t set D is uniformly ontinuous therein. Proof. Suppose that f is not uniformly ontinuous on D. Then, pro eeding as in the onverse part of Theorem 2.22, there exists an  > 0, and two sequen es fn g and fn g in D su h that for every n 2 N , (2.26) jn n j < n1 and jf (n ) f (n )j  : Sin e D is ompa t, fn g ontains a subsequen e fnk g onverging to a point z0 , say, where z0 is a point of D; i.e. nk ! z0 . Let fnk g be the

orresponding subsequen e of fn g. Then nk ! z0. For, the triangle inequality gives jnk z0 j  jnk nk j + jnk z0 j and so we have nk ! z0 . Therefore, for the subsequen es fnk g and fnk g, (2.26) gives (2.27) jnk nk j < n1 and jf (nk ) f (nk )j   k for every k. However, as f is ontinuous at z0, we have f (nk ) ! f (z0 ) and f (nk ) ! f (z0) as k ! 1;

ontradi ting (2.27). Hen e, f is uniformly ontinuous. 2.28. Example. There are uniformly ontinuous fun tions on a set whi h is not ompa t. For instan e, the identity fun tion z is obviously uniformly ontinuous on C . But f (z ) = z 2 is not uniformly ontinuous on C whereas it is uniformly ontinuous on
R = fz : jz j < Rg (also
R nf0g), where R is a xed positive number. To verify this, hoose two points z0 and z on
R . Then jz 2 z02 j = jz z0 j jz + z0j  jz z0 j(jz j + jz0 j)  2Rjz z0 j: Therefore given any  > 0 there exists a Æ = =(2R) su h that jz z0 j < Æ implies jz 2 z02j < ; thus proving the uniform ontinuity on
R . Suppose that z is not restri ted to
R and allow z; z0 to be C . We

hoose two points of C to be z = R + 1=R; z0 = 1=R so that jz z0 j = R, where R > 0. Then,   2 2 2 jz z0 j = R + R R = R2 + 2 > 2 and so, f (z ) = z 2 is not uniformly ontinuous throughout C . 

54

Fun tions, Limit and Continuity

2.3 Stereographi Proje tion When dealing with the real line, we frequently use the on ept of in nity and speak of +1 and 1. For instan e, the sequen e f2ng diverges to +1 whereas f ng diverges to 1 and fxn g = f( 1)nng is unbounded above and unbounded below, so lim sup xn = +1; lim inf x = 1: n!1 n n!1 Therefore, in the ase of real-valued fun tions, limits su h as 0 lim f (x) = 1; xlim x!a !1 g(x) = l; x!lim1 h(x) = l ; xlim !1 p(x) = 1 provide valuable information about the behavior of fun tions near these points, namely a; 1; 1. In dealing with the omplex plane C , we too speak of in nity and denote it by the usual symbol `1'. In C , we do not give a sign to the omplex in nity. One of the main reasons for this is that C has no natural ordering as R does. In Se tion 1.5, we have dis ussed the topologi al properties of the omplex plane C with the usual Eu lidean metri . It turned out that the

ompa t sets were the bounded and losed sets. However, the in nite set, for instan e S = fz 2 C : z = m + in; m; n 2 Ng has no limit in C and is losed but is non- ompa t. Nevertheless, it will be very mu h useful to develop the notion of limit points for su h unbounded sets. Therefore, we shall extend the omplex plane C by adjoining one extra point alled the \point at in nity" whi h we shall denote by 1, i.e. we onsider the extended omplex plane C 1 := C [f1g as a losed surfa e having a single point at in nity. We shall then introdu e a new metri to analyze the behavior of a omplex fun tion at in nity and to map the points in C into the surfa e of a sphere. This pro ess will be referred to as stereographi proje tion. In fa t, su h an extra point `1' is de ned so as to satisfy the following

omputational properties (see Figure 2.6): Whatever be z 2 C , z z 1 = 0; z + 1 = 1 (z 6= 1); 0 = 1 (z 6= 0);

z
1 = 1 (z 6= 0);

1 = 1 (z 6= 1):

z We do not de ne (or have no sense to de ne) 0 1 ; 1 + 1; 1 1; 0
1; and : 0 1 We de ne z 0 = 1 for z 2 C 1 , and for ea h n 2 N , we set 0 n = 1n = 1; 0n = 1 n = 0:

55

2.3 Stereographi Proje tion ∞

y ∞

x ∞

∞

Figure 2.6: Complex `1'.

We use the following onstru tion due to Riemann. There are two ommonly used methods. In one method, a orresponden e is set up between the points of C and those of a sphere of radius 1=2 with enter at (0; 0; 1=2) tangent to this plane. There is another method of orresponden e in whi h the sphere of radius 1 has enter at (0; 0; 0) and the plane passes through (0; 0; 0). We shall use the rst one. Let C be the omplex plane. Through the origin onstru t a line perpendi ular to C . Let this be  -axis of a 3-dimensional Eu lidean spa e in whi h a point has oordinates (; ;  ). Consider the sphere S of radius 1=2 with enter at (0; 0; 1=2). That is,

S = f(; ;  ) 2 R3 :  2 + 2 + (

1=2)2 = 1=4g:

It is a ommon pra ti e to all the points N and O with oordinates (0; 0; 1) and (0; 0; 0) the north pole and south pole of the sphere S , respe tively. The great ir le in the plane  = 1=2 is alled the equator. The plane  = 0 oin ides with the omplex plane C and the  and  axes are the x and y axes, respe tively. Let Q(x; y; 0) be any point in the plane C . Through the points N and Q we draw a straight line NQ interse ting the sphere S at a point, say, P (; ;  ). Then (; ;  ) is alled the stereographi proje tion, or image of (x; y; 0) on the sphere and is onsidered as the spheri al representation of z = x + iy. This pro edure assigns a unique point on S to every given omplex number z so that we are free to think of C as sitting inside R3 . Conversely, to ea h point (; ;  ) on the sphere other than N we an asso iate the omplex number z = (x; y; 0) where the line from (0,0,1) through (; ;  ) interse ts C . Now we immediately see that there is a one-to-one orresponden e between C and the points of S with one ex eption, namely, the north pole (0,0,1) itself. By assigning to the north pole N of the sphere to orrespond to the point at in nity, we obtain a one-to-one orresponden e between the points of the sphere S on one hand and the points of the extended omplex plane C 1 on the other. The sphere is often alled Riemann sphere or omplex sphere It is easy to obtain expli it equations expressing ;  and  in terms of

56

Fun tions, Limit and Continuity ζ N (0, 0, 1)

C(0, 0, 1/2) O

y, η

x, ζ

Figure 2.7: Riemann's sphere.

x and y. The line in R3 passing through (0; 0; 1) and (x; y; 0) is given by (2.29)

ft(0; 0; 1)+(1 t)(x; y; 0) : t 2 Rg  f((1 t)x; (1 t)y; t) : t 2 Rg:

Sin e this line interse ts the sphere S , we must have 1 4 so that (1 t)2 jz j2 = t(1 t): If (; ;  ) 6= (0; 0; 1), then we arrive at (1 t)2 x2 + (1 t)2 y2 + (t 1=2)2 =

t=

jz j2 ; i.e. 1 t = 1 : 1 + jz j2 1 + jz j2

Using the fa t that the points (0; 0; 1), (; ;  ) and (x; y; 0) are ollinear, (2.29) now yields 8 x z+z > >  = = ; > 2 2 > 1+x +y 2(1 + jz j2 ) > > > < y i(z z)  = = ; (2.30) 2 2 1+x +y 2(1 + jz j2 ) > > > > > x2 + y2 jz j2 ; > > : = = 2 2 1+x +y 1 + jz j2 that is z = x + iy 2 C orresponds to 

x y x2 + y2 ; ; 2 2 2 2 1 + x + y 1 + x + y 1 + x2 + y2 or equivalently to 

z+z i(z z) jz j2 ; ; 2 2(1 + jz j ) 2(1 + jz j2) 1 + jz j2





2S

2 S:

57

2.3 Stereographi Proje tion

p

For instan e, the images of 1; i; (1 i)= 2 on the sphere are respe tively given by Z1 ; Z2 ; Z3 , where

p

p

Z1 = (1=2; 0; 1=2); Z2 = (0; 1=2; 1=2); Z3 = ( 2=4;

2=4; 1=2):

Using the three equations in (2.30) it is easy to see that 1 =

(2.31) so that (2.32)

1  ; i.e. jz j2 = 1 + jz j2 1 

   + i ; y= ; or z = : 1  1  1  That is (; ;  ) orresponds to x=





1 



+i







1 

2 C:

The map z ! (; ;  ) is alled stereographi proje tion of C on S nf(0; 0; 1)g or vi e versa. In fa t if  : C ! S nf(0; 0; 1)g is the stereographi proje tion of C on S nf(0; 0; 1)g, then 



z+z i(z z) jz j2 (z ) = ; ; : 2(1 + jz j2 ) 2(1 + jz j2 ) 1 + jz j2 The inverse of , 

1

: S nf(0; 0; 1)g ! C , is given by

 1 (; ;  ) =



1 

+i



1 

:

2.33. Remark. If z is a omplex number orresponding to the point Q(; ;  ) on the pun tured sphere S nf(0; 0; 1)g, then (2.31) shows that (2.34)

2

jz j ? R ()  ? 1 +R R2 ; R > 0:

In parti ular, the images of fz : jz j < 1g and fz : jz j > 1g are the southern hemisphere and the northern hemisphere, respe tively.  Now suppose that (n ; n ; n ) is a sequen e of points of S whi h onverges to (0; 0; 1) and let fzng be the orresponding sequen e of points in C . It follows (as is obvious geometri ally) from (2.31) that jznj be omes very large; i.e. as (n ; n ; n ) ! (0; 0; 1), jzn j ! 1. Conversely, in view of (2.30), if jzn j ! 1 then (n ; n ; n ) ! (0; 0; 1). Thus, it is reasonable to introdu e the symbol \1" to orrespond to (0; 0; 1) 2 S . We an now think of C 1 either as `a plane + an ideal point' or as a sphere. Correspondingly, as was pointed out earlier, we may think of C as a plane, or as a sphere

58

Fun tions, Limit and Continuity ζ N

N P C O Q

C P y, η

→ → → → → ≡

(0, 0, 1) (ξ, η, ζ ) (0, 0, 1/2) (0, 0, 0) (x, y, 0) (ξ, η, 0)

Q x, ξ

Figure 2.8: Proje tion of a portion of the Riemann sphere.

without the north pole. Both points of view are useful, depending on the problem on hand. The notions su h as limit point of a sequen e and neighborhoods of a point an now be de ned for the extended omplex plane C 1 through this identi ation with the Riemann sphere S .

2.35. De nition. A sequen e fzn g is said to diverge to the limit 1, written zn ! 1 or limn!1 zn = 1, if for any R > 0, there exists a number N su h that jzn j > R for n > N: Sin e fz : jz j > Rg orresponds to a ` ap' (see also Figure 2.7) it makes sense to think of fz : jz j > Rg as a neighborhood of 1. Su h a

neighborhood be omes smaller as R gets larger and larger. This geometri intuition allows us to de ne the on ept of ontinuity on C 1 . We next give a justi ation for this notion. We de ne (z; z 0) = d(Z; Z 0 )) to be the Eu lidean distan e between Z = (; ;  ) and Z 0 = ( 0 ; 0 ;  0 ) in the three dimensional spa e whi h are respe tively the pre-images of z = x + iy and z 0 = x0 + iy0 (under the stereographi proje tion of the sphere S onto the omplex plane C ). Sin e (; ;  ) and ( 0 ; 0 ;  0 ) are on the sphere S , (2.36)  2 + 2 +  2 =  and  02 + 02 +  02 =  0 : The length of the segment joining Z and Z 0 , known as hordal distan e of z from z 0 , is de ned as (z; z 0) = d(Z; Z 0 ): Therefore, we have p (z; z 0) = (  0 )2 + ( 0 )2 + (  0 )2 p =  +  0 2( 0 + 0 +  0 ); by (2:36); jz pz 0 j = p ; by (2.30) and (2.36). 1 + jz j2 1 + jz 0j2

59

2.3 Stereographi Proje tion

In parti ular, if z 0 is the point at in nity then (z; 1), the spatial distan e between the images of z and (0; 0; 1), is given by (2.37)

8 > (z; > > >
> > > :

p

1 ; by (2:31); 1 + jz j2 = 0lim (z; z 0): z !1 =

p

Further, we note that (z; z 0) = 1

() jz z 0 j2 = (1 + jz j2 )(1 + jz 0 j2 ) () j1 + zz0 j = 0 () zz0 = 1: In other words, the points z and z 0 in C represent diametri ally opposite

(antipodal) points of the Riemann sphere S i zz0 = 1. By (2.34) and (2.37), we see that the set fz : jz j > Rg; R > 0, orresponds to the setpf(; ;  ) 2 S n(0; 0; 1) : (; ;  ) lies within the spheri al

ap of radius R= 1 + R2 g. Conversely, if (; ;  ) 6= (0; 0; 1), as noted above, we have p

 2 +  2 + (

1)2 =

p

1 =

p

1 : 1 + jz j2

p

Note that

1 1 2 p <  () j z j > ( > 0):  1 + jz j2 Therefore, the -neighborhood of (0; 0; 1) on S, namely, p

f(; ;  ) 2 S :  2 + 2 + ( 1)2 < g is nothing but N (1; ) = fz 2 C : jz j > -neighborhood of 1. Clearly, (a) (b) ( ) (d) (e) (f)

p1 2 g [ f1g; whi h we all an 

(z1 ; z2)  0 (z1 ; z2) = 0 () z1 = z2 (z1 ; z2) = (z2 ; z1 ) (z1 ; z3)  (z1 ; z2 ) + (z2 ; z3 ) (0; z1)  (0; z2 ) provided jz1 j  jz2 j  1 (z1 ; z2)  jz1 z2 j = d(z1 ; z2).

In parti ular, we all  de ned on C 1 , the hordal metri on C 1 . This allows us to treat the point 1 like any other point. Thus, we have

60

Fun tions, Limit and Continuity

2.38. Theorem. The hordal metri (: ; :) de ned on C 1 satis es the properties of a metri . By a ir le on the sphere S we mean the interse tion of S with some plane a + b +  + d = 0: Let us now nd the equation of the ir le on the sphere whi h is the stereographi image of the ir le fz : jz aj = Rg. By (2.32) and jz aj2 = R2 , the required equation is given by

 2 + 2 (1  )2

a



 i 1 



a





 + i = R2 1 

jaj2 ;

whi h upon simpli ation be omes (a + a) + i(a a) + (jaj2

2.39. Theorem. Suppose T of T on the Riemann sphere S is

R2 1) = 1 + jaj2 R2 :

 C 1.

Then the orresponding image

(a) a ir le in S not ontaining (0; 0; 1) if T is a ir le; (b) a ir le in S minus (0; 0; 1) if T is a line.

Proof. First we onsider the general equation of a ir le in the plane: (2.40)

T = f(x; y) : A(x2 + y2 ) + Bx + Cy + D = 0g:

Using stereographi proje tion, i.e. by (2.32) and (2.31), we have (A D) + B + C + D = 0 whi h is the equation of a plane in spa e when a general point has oordinates (; ;  ). Note that a plane and a sphere interse t in a ir le. Suppose A = 0, then (2.40) is a straight line in C . Thus, the orresponding set in the sphere S is given by the interse tion of

 2 + 2 +  2 = ; with B + C D + D = 0 whi h is a ir le minus (0,0,1). If we onsider T  C 1 , then the orresponding image set governed by the above equations is a ir le passing through the north pole (0,0,1). This proves (a). Suppose A 6= 0 and D = 0. Then T redu es to

A(x2 + y2 ) + Bx + Cy = 0; whi h is a ir le passing through (0; 0) and so the orresponding image is given by the interse tion of

 2 + 2 +  2 = 

61

2.3 Stereographi Proje tion

with B + C + A = 0: This is in fa t a ir le passing through the south pole (0,0,0). The general statement (b) is obvious. The onverse of Theorem 2.39 takes the following form.

2.41. Theorem. If TS is a ir le on the Riemann sphere S and TI is its stereographi proje tion on C 1 , then (a) TI is a ir le if (0; 0; 1) 62 TS (b) TI is a line if (0; 0; 1) 2 TS .

Proof. If TS is a ir le on the sphere S , then TS = f +  +  + Æ = 0g \ f 2 + 2 +  2 =  g: Thus, TS passes through (0; 0; 1) provided + Æ = 0. It follows from (2.30) that the orresponding set of points of the plane in C satis es (2.42)

( + Æ)(x2 + y2 ) + x + y + Æ = 0; (x; y) 2 TI :

Clearly, this equation represents the equation of a ir le if + Æ 6= 0. If

+ Æ = 0, then (2.42) is the equation of a line. The on lusion now follows from the fa t that + Æ = 0 () (0; 0; 1) 2 TS :

2.43. Example. Suppose that a ube has its verti es on the Riemann sphere and its edges parallel to the o-ordinate axes. Let us now nd the stereographi proje tions of the verti es. By hypothesis, the verti es are Z1 Z3 Z5 Z7

= = = =

(; ;  ); (; ; 1  ); ( ; ; 1  ); (; ;  );

Z2 Z4 Z6 Z8

= = = =

(; ; 1  ); ( ; ; 1  ); ( ; ;  ); ( ; ;  ):

Further the length of the sides of the ube gives the relation

j j = jj = j 1=2j :

(2.44)

For onvenien e, we let  > 0,  > 0 and  > 12 . As ; ;  lie on the Riemann sphere  2 + 2 + ( 1=2)2 = 1=4; by (2.44), we have  2 +  2 +  2 = 1=4; i.e. 3 2 = 1=4: Thus, we get

==

1



p = 

2 3



p

1 3+1 ; i.e.  = p 2 2 3

p

!

3 1 1 = p : 2 3

62

Fun tions, Limit and Continuity

Therefore, by (2.32), we have





 + i (1 + i) 1 z= =p = p +i 1  3 1 3 1 whi h is the stereographi proje tion of 1

1

p





p1

3 1

!

p ; p ; 3p+ 1 : 2 3 2 3 2 3 Similarly the stereographi proje tions of the other verti es are obtained as follows:

Z2 = Z4 = Z6 = Z8 =

1

p

!

p ; p ; 3p 1 ; 2 3 2 3 2 3 1

1

1

1

1

Z3 =

p

!

p

!

p ; p ; 3p 1 ; 2 3 2 3 2 3 p ; p ; 3p+ 1 ; 2 3 2 3 2 3 1

1

p

!

1

1

!

p

!

p

!

p ; p ; 3p 1 ; 2 3 2 3 2 3

Z5 = Z7 =

p

p ; p ; 3p 1 ; 2 3 2 3 2 3 1

1

1 1

p ; p ; 3p+ 1 ; 2 3 2 3 2 3

p ; p ; 3p+ 1 ; 2 3 2 3 2 3

and



1 Z2 ! z2 = p

Z3

!

Z4

!

Z5

!

Z6

!

Z7

!

Z8

!





1 +i p



3 + 1  3 + 1 1 1 i p z3 = p 3 + 1 3 +1     p 1 +i p 1 z4 = 3 + 1  3 + 1  1 p1 z5 = i p 3 + 1 3 + 1    p 1 +i p 1 z6 = 3 1  3 1  1 1 z7 = p i p 3 1 3 1     1 p1 z8 = i p : 3 1 3 1 



2.45. Example. We wish to prove that the points z and z 0 in the

omplex plane will orrespond to symmetri points with respe t to the equatorial plane (viz., the plane  = 1=2) i zz0 = 1.

63

2.3 Stereographi Proje tion

To prove this, we rst note that z and z 0 orrespond to symmetri points with respe t to equatorial plane i z and z 0 orrespond to (; ;  ) and (; ; 1  ), respe tively. This holds if and only if

z=

 + i  + i  2 + 2 and z 0 = ; i.e. zz0 = = 1: 1  1 (1  ) (1  )



We shall now brie y indi ate how to extend the limit on epts dis ussed in the previous se tion to the extended omplex plane.

2.46. De nition. (Limit at in nity) Let f be de ned on an unbounded set E . (Then for any R > 0, there exists z 2 E su h that jz j > R.) We say that f (z ) ! ` as z ! 1 if for every  > 0, there exists an R > 0 su h that jf (z ) `j <  whenever z 2 E and jz j > R: In this ase, we write lim f (z ) = ` or

z!1

lim f (z ) = `: jzj!1

Re all (see De nition 2.35) that a set of the form fz : jz j > Rg [ f1g is

alled a neighborhood of 1. For onvenien e, we may introdu e
(1; 1=R) := fz : jz j > Rg [ f1g = C 1 n
(0; R): The losed disk
(1; 1=R) and the pun tured disk
(1; 1=R) nf1g may be de ned similarly. For instan e,
(1; 1=R) nf1g := C n fz : jz j > Rg. Let us show that 1 1 (a) zlim (b) zlim !1 z = 0; !1 z 2 = 0. To do this, we rst note that 1=z and 1=z 2 are de ned everywhere in C n f0g. Then for every  > 0 there exists a R = 1= su h that 1 z

<  whenever jz j >



1 = R: 

For the se ond ase, wepnote that 1=z 2 < 

ase, we hoose R = 1= .

() jz j > 1=p so, in this

2.47. De nition. (In nite limit) Let f be de ned on D ex ept possibly at z0 of D. We say that f (z ) ! 1 as z ! z0 if for every R > 0, there exists a Æ > 0 su h that

jf (z )j > R whenever z 2 D \
(z0 ; Æ) nfz0g:

64

Fun tions, Limit and Continuity

In this ase, we write, lim f (z ) = 1 or zlim !z0 jf (z )j = 1:

z!z0

Note. Considering the real-valued fun tion jf j on D or D nfz0 g de ned by jf j(z ) = jf (z )j, z 2 D, we observe that lim f (z ) = 1 i zlim !z0 jf j(z ) := zlim !z0 jf (z )j = 1:

z!z0

1 1 2.48. Example. We have (a) zlim = 1 , (b) lim = 1. 2 !1 jz 1j z !0 z 2 Note that the fun tion f de ned by 1 f (z ) = 2 z 1 is de ned for all z 2 C n f 1; 1g. Let R > 0 be given. Then we must show that we an nd a Æ > 0 su h that jf (z )j = z 2 1 1 > R whenever 0 < jz 1j < Æ: Note that jf (z )j > R () 0 < jz 2 that

1j < 1=R: Now 0 < jz

1j < Æ implies

jz 2 1j = jz 1j jz 1 + 2j  jz 1j[jz 1j + 2℄ < Æ(Æ + 2) = (Æ + 1)2 1 p and therefore jz 2 1j < 1=R if Æ = 1 + R 1 1: Hen e p

jf (z )j > R whenever z 2 [C n f 1; 1g℄ \ [
(1; 1 + R

1

1) nf1g℄

and the on lusion follows. The se ond ase is lear, be ause 1 = 1 > R whenever jz j = jz 0j < Æ = p1 : z2 jz j2 R



2.49. De nition. Let f be de ned on an unbounded set E . If for every R > 0 there exists K > 0 su h that jf (z )j > R for jz j > K and z 2 E; then we say that f (z ) ! 1 as z ! 1 and write zlim !1 f (z ) = 1: For instan e, if f (z ) ! 1 as z ! 1, then jf (z )j ! 1 as z ! 1: In parti ular, zlim !1 z = 1.

2.50. Remark. Using the hordal metri  on C to see that the following are true: (a) zn ! z0 () (zn ; z0 ) ! 0

[ f1g, it is easy

(b) zn ! 1 () (zn ; 1) ! 0:

2.4 Sequen es and Series of fun tions

65

Part (b), in parti ular, helps us to give an alternate meaning to a statement su h as jf (z )j > R for jz z0 j < Æ in the form 1 (f (z ); 1) < p for (z; z0 ) < jz z0 j < Æ:  1 + R2 Let E be an unbounded set. Then f : E ! C 1 is ontinuous at z0 2 E i for every  > 0, there exists a Æ > 0 su h that

jf (z ) f (z0)j <  whenever z 2
(z0 ; Æ) \ E; or equivalently,

f (
(z0 ; Æ) \ E ) 
(f (z0 ); ): In this de nition, with the help of the above notation, we now admit the

ases where z0 = 1 and f (z0) = 1. We say f is ontinuous on E i f is ontinuous at all points of E . For instan e, a rational fun tion in z is a ontinuous fun tion from C 1 into C 1 . Many other basi notions may also extended to the extended omplex plane. We postpone our dis ussion on ertain aspe ts of fun tions on the extended omplex plane to a later

hapter.

2.4 Sequen es and Series of Fun tions Next we dis uss the uniform onvergen e of sequen es and series of fun tions. Consider a sequen e of omplex-valued fun tions ffn(z )g, fn : D  C ! C and n 2 N . For a xed z0 2 D, ffn (z0 )g is an ordinary sequen e of omplex numbers and so the onvergen e of the sequen e of these fun tions for ea h z0 2 D is as in the de nition of onvergen e of a sequen e of

omplex numbers in Se tion 1.6.

2.51. De nition. A sequen e ffng of fun tions is said to be onvergent at z0 2 D if the sequen e ffn (z0 )g onverges. We shall all this limit f (z0 ). The sequen e ffn(z )g of fun tions is said to onverge `pointwise' to f (z ) in D if ffn(z0 )g onverges to f (z0 ) at ea h point z0 2 D. Sin e the limit of a sequen e, when it exists, is unique, in the ase of `pointwise' onvergen e, we have a uniquely de ned fun tion f from D into C and we all f , the pointwise limit, or simply the limit fun tion of the sequen e ffn(z )g and write f (z ) = limn!1 fn (z ), z 2 D: An equivalent formulation of the above dis ussion is given by the following:

2.52. De nition. Let fn and f be fun tions from D into C . We say that fn(z ) ! f (z ) on D i for every  > 0 and every z 2 D, there exists N = N (; z ) su h that jfn (z ) f (z )j <  for all n  N: This onvergen e is said to be uniform if it is possible that N (; z ) an be hosen independent of z 2 D. That is, one N () works for all z 2 D.

66

Fun tions, Limit and Continuity

We often write fn ! f uniformly on D, or limn!1 fn (z ) = f (z ) uniformly on D to denote the uniform onvergen e of fn to f on D. Also note that the uniform onvergen e on D implies pointwise onvergen e. The onverse is false, for example onsider fn (z ) = z n for z 2
(see also Example 2.55). Then the sequen e fz ng onverges uniformly on jz j  r (0 < r < 1); for if  > 0 is given, then jz nj  rn <  whenever n > (ln )= ln r: Note that

f (z ) = nlim !1

zn



j jz j  1=n : Then jz n j   and so the hoi e of N depends on z . For instan e, given any positive integer n, one an exhibit points z with jz jn  1=3 (note that the hoi e z = exp( (1=n) ln 3) will do when  = 1=3). Alternatively, it suÆ es to observe that (sin e jz j < 1),

jz jn <  whenever n ln jz j < ln  or n > (ln )= ln jz j: Sin e ln jz j approa hes zero as jz j ! 1, the maximum value of n is in nite

and so there annot exist an N independent of z satisfying the ondition that is required for uniform onvergen e. P We are frequently on erned with series of the form n1 fn (z ) for z in some subset of C . In situations like this, as in the ase of series of omplex terms, we have P

2.53. De nition. The series k1 fk onverges `pointwise' in D to a P fun tion f in D if the orresponding sequen e of partial sums sn = nk=1 fk

onverges to f `pointwise' in D. Then we write 1 X f = fk P

k=1

and say that the series n1 fn is pointwise onvergent in D with sum f . This series is said to onverge to f uniformly in D i sn onverges to f uniformly in D. The series is said toP onverge absolutely or said to be absolutely onvergent in D if the series k1 jfk j is onvergent in D. P

2.54. Example. Consider the geometri series k1 z k 1 : If z 6= 1, then the n-th partial sum is 1 zn sn = 1 + z +


+ z n 1 = : 1 z

67

2.4 Sequen es and Series of fun tions

Sin e fz ng is a null sequen e for jz j < 1, it follows that X

k1

In fa t,

sn

zk

1

= nlim !1 sn = nlim !1

1 zn 1 = ; jz j < 1: 1 z 1 z

= z

jz jn <  whenever n ln jz j < ln(j1 z j): 1 j1 z j If 0 < jz j < 1, then ln jz j < 0 and so we have 1 ln(j1 z j) sn <  whenever n > N  : 1 z n ln jz j 1

Thus, the geometri series is pointwise onvergent in
with sum f (z ) = 1=(1 z ). If jz j  1, jz jn  1 and sin e jz jn 6! 0, the series diverges in this ase. These fa ts an also be veri ed with the help of Corollary 1.49 and Theorem 1.51. Next we wish to show that the geometri series is not uniformly onvergent in jz j < 1. To do this, we need to show that sn does not onverge uniformly in jz j < 1. Clearly, from the expression for sn , it suÆ es to show that fSng, where Sn (z ) = z n=(1 z ), is not uniformly onvergent in jz j < 1. Let  > 0 be given. If fSn g were uniformly onvergent for jz j < 1, then there would exist an N su h that n z 1 z <  for all n > N and all z 2
:  in
. Then for this point, Now, for a given  > 0 we hoose z0 = NN+++1 we note that

z0N +1 [1 1=(N +  + 1)℄N +1 1 (N + 1)=(N +  + 1) = > = 1 z0 1=(N +  + 1) 1=(N +  + 1) (sin e (1 x)n > 1 nx if 0 < x < 1 and n is a positive integer  1); so the onvergen e is not uniform for jz j < 1. 

2.55. Example. Consider the series S (z ) = z + z (1 z ) + z 2 (1 z ) +




+ z n 1(1 z ) +




Then the n-th partial sum is

sn (z ) = z + (1 z )

nX1 k=1

zk = zn

68

Fun tions, Limit and Continuity

so that the sequen e fsng (and, so the series) onverges to zero for jz j < 1. We have already shown that the sequen e fz ng does not onverge uniformly for jz j < 1. Consequently, the given series is not uniformly onvergent for jz j < 1.  Many results about the onvergen e of numeri al sequen es in C an be

arried over easily to the onvergen e of a sequen e of fun tions at a point and so, for pointwise onvergen e. We list a few of them.

2.56. Theorem. Let ffng and fgng be sequen es of fun tions de ned

on a set D  C .

! f pointwise in D () for ea h  > 0 and ea h z 2 D there exists an N (; z ) su h that jfn (z ) fm (z )j <  for all n; m > N (; z ): (b) If fn ! f and gn ! g both pointwise in D, then fn  gn ! f  g and fngn ! fg pointwise on D. (a) fn

f f ( ) If gn(z ) 6= 0 and g(z ) 6= 0 for ea h z in D and n 2 N , then n ! gn g pointwise in D. (d) ffn(z )g is uniformly onvergent in D i for every  > 0 and all z 2 D there exists an N = N () su h that jfn (z ) fm (z )j <  for all n; m > N (): (e) If fn ! f and gn ! g both uniformly in a ommon region D then fn  gn ! f  g uniformly in D and for a omplex onstant , fn ! f uniformly in D.

Theorem 2.56(d) is alled the Cau hy riterion for uniform onvergen e. Uniform onvergen e does not arry over to the produ t of fun tions, in general. To see this, onsider 1 1 1 fn (z ) = gn (z ) = + and f (z ) = g(z ) = in
nf0g: z n z Then, fn ! f and gn ! g both uniformly in
nf0g. Now 1 2 1 fn (z )gn (z ) f (z )g(z ) = 2 +
: n n z 1 2 In parti ular, for z = n , or n in
, we nd that (fn gn fg)(z )  2 for all n  1: Hen e, fn gn 6! fg uniformly in
.

2.57. Theorem. The limit fun tion of a uniformly onvergent sequen e of ontinuous fun tions is itself ontinuous. Proof. Let ea h fn be ontinuous in D and suppose that fn ! f uniformly in D. Let z0 2 D be given. Now, for all z 2 D and for all indi es n, the triangle inequality gives jf (z ) f (z0 )j  jf (z ) fn (z )j + jfn (z ) fn (z0 )j + jfn (z0 ) f (z0)j:

69

2.4 Sequen es and Series of fun tions

The laim now follows from this inequality upon using the hypothesis. P

2.58. Corollary. If n1 fn is uniformly onvergent in D and if fn P is ontinuous in D for ea h n, then so is the sum f = n1 fn . Corollary 2.58 makes no assertion about the sum of a series of ontinuous fun tions if the onvergen e is not uniform. The Cau hy onvergen e riterion for series of omplex P numbers dis ussed in Chapter 1 takes the following form: \The series n1 fn(z ) onverges uniformly in D i for every  > 0 there exists an N = N () su h that

n X

k=m+1

fk (z ) =

jsn sm j < 

for all z 2 D whenever n > m  N ":

The next theorem gives a suÆ ient ondition for the uniform onvergen e P of the series n1 fn (z ). This is one of the frequently used results in

omplex analysis. P

2.59. Theorem. (Weierstrass' M-test) Let n1 Mn be a onvergent series of positive real numbers su h P that jfn (z )j  Mn for all n 2 N and for all z in a set D. Then the series n1 fn (z ) onverges absolutely and uniformly in D. Proof. To prove this, we note that, for all n > N and z 2 D, (2.60)

n X

k=N +1

fk (z )



n X k=N +1

jfk (z )j 

n X k=N +1

Mk :

The Cau hy P riterion (see Theorem 2.56(a) and De nition 2.53) applied to the series n1 Mn shows that for every  > 0; there exists an N () su h P that nN +1PMk <  for all n > N = N () This observation, by (2.60), implies that n1 fn (z ) satis es the Cau hy riterion (see Theorem 2.56(a)) for uniform onvergen e on D and so the assertion follows. We illustrate the appli ation of Weierstrass' M-test P with an example whi h is already familiar for us. The geometri series n1 z n 1 onverges uniformly for jz j  r, where 0 < r < 1. Finally, we end this se tion with two more examples.

2.61. Example. De ne 2 1 + zn fn (z ) = = 1; n 2 N : 1 zn 1 zn Note that if jz j < 1, then jz jn ! 0 as n ! 1; and if jz j > 1, then jz jn ! 1. Therefore, if jz j > 1 then for suÆ iently large values of n, we

70

Fun tions, Limit and Continuity

have j1 z nj  jz jn exists and

1 ! 1 as n ! 1: Hen e, for jz j 6= 1, limn!1 fn (z )

f (z ) = nlim !1 fn (z ) =



1 if jz j < 1 1 if jz j > 1:

From this, we also on lude that it is not possible to de ne the limit fun tion f (z ) on jz j = 1 so that f be omes a ontinuous fun tion in C . 

2.62. Example. Let us dis uss the onvergen e of 1 X zk 1 (jz j 6= 1): k (1 z )(1 z k+1 ) k=1

For jz j 6= 1, the partial sums takes the form n X



zk 1 zk Sn (z ) = k 1 z k+1 k=1 1 z 1 z   1 1 zn = : 1 z 1 z 1 z n+1 1



Note that Sn (0) = 1. For 0 < jz j < 1, we have 1



z n jz jn = ! 0 as n ! 1  n +1 z 1 jz jn+1

and for jz j > 1, we get zn 1 = n n +1 1 z z z It follows that, as n ! 1,

Sn (z ) ! f (z ) =

8 > >
> : + if jz j > 1: 1 z 1 z z 



2.5 Exer ises 2.63. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) The fun tion f (z ) = z + 1=(z ) is univalent in fz : jz j > jj 1 g,  6= 0. (b) The fun tion f (z ) = (az + b)=( z + d); ad b 6= 0, is univalent in C n f d= g.

2.5 Exer ises

71

( ) The fun tion f (z ) = (1 + z )3 is not univalent in
. (d) For every positive integer n, the fun tion f (z ) = nz + z n is univalent in
. (e) If n is a positive integer, then jz j < 1=n is the largest disk entered at 0 on whi h the fun tion f (z ) = z + z n is univalent. (f) The Koebe fun tion k(z ) = z=(1 z )2 is univalent in
. (g) The range of the Koebe fun tion k(z ) = z=(1 z )2 is ( 1; 1=4℄. P P (h) The series an onverges to a () an onverges to a. (i) The fun tion f (z ) = z2 is uniformly ontinuous on
R (also
R nf0g), where R is a xed positive number, but not in C . Re z 2 Im z 2 and lim does not exist. (j) Ea h of the limits zlim 2 z!0 jz j2 !0 jz j (k) For f (z ) = [Re (z ) + Im (z )℄=jz j2, limz!0 f (z ) does not exist. (l) The fun tion f (z ) = (Re z )=(Im z ) is ontinuous for all z with Im z = 6 0 and dis ontinuous for all z with Im z = 0. Also, no dis ontinuity of f (z ) is removable. (m) The fun tion f (z ) = (2 + z )Arg z does not have removable dis ontinuities. (n) The fun tion f (z ) = (Arg z )2 is ontinuous on the pun tured plane C n f0g. (o) The fun tion F (z; h) = [(z + h)n z n℄=h nz n 1 (0 6= h; z 2 C ) satis es the inequality jF (z; h)j  F (jz j; jhj). (p) The sequen e ffn(z )g, where fn(z ) = 1=(1 + nz ), does not onverge to f (z ) = 0 uniformly in any losed region ontaining the origin. 1g (q) The sequen e f nz n1 onverges for 0 < jz j < 1, but not uniformly. If r > 0 is xed, the onvergen e is uniform for r < jz j < 1. (r) The sequen e ffn(z )gn1, for jz j < 1, where fn (z ) = z 3 z=n, onverges uniformly to the limit fun tion f (z ) = z 3 for z 2
. (s) The sequen e fz n=ngn1 onverges uniformly to 0. P (t) The series n0 3 n z n onverges uniformly for jz j  r, 0 < r < 3. P 2 2 n (u) The series 1 n=0 z =(1 + z ) onverges for all z exterior to the lem2 nis ate jz + 1j = 1. (v) The transformation w = 2 1 [z + 2 z 1℄ ( 2 R), maps the ir le jz j = r (r 6= ) into an ellipse in the w-plane. P (w) The series n0 z 2n =(1 z 2n) onverges uniformly for jz j  r, 0 < r < 1. (x) The mapping  de ned for stereographi proje tion is a homeomorphism (i.e. bije tive and bi ontinuous).

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Fun tions, Limit and Continuity

2.64. Dis uss the ontinuity of the following omplex-valued fun tions at z = 0: 8 < (Re z ) (Im z ) if z 6= 0 jz j2 (a) f (z ) = : 0 if z = 0: 8 < Im z if z 6= 0 (b) f (z ) = jz j : 0 if z = 0: ( ) f (z ) = (Re z )=(1 + jz j) for z 2 C . (d) f (z ) = jRe (z ) Im (z )j for z 2 C . 8 < sin jz j if z 6= 0 jz j (e) f (z ) = : 1 if z = 0: 8 2 < 1 exp( jz j ) if z 6= 0 jz j2 (f) f (z ) = : 1 if z = 0: 2.65. Des ribe the position of the following points in relation to z in the omplex plane, as viewed on the Riemann sphere:

 iz; z 2 z ; zz ; jzz j : Also, for ea h of the following points in C , give the orresponding points z; z; 1=z; (z + z)=2;

on the Riemann sphere: 0; 1 + i; 1 i; 2 + 3i; 2 3i:

2.66. Let D  C and f : D statements are equivalent or not:

! C . Che k whether the following

(a) f is not uniformly ontinuous in D (b) there exists an  > 0 su h that for every Æ > 0 there are points Æ and Æ in D su h that jÆ Æ j < Æ and jf (Æ ) f (Æ )j   ( ) there exists an  > 0, and two sequen es fn g and fn g in D su h that for every n 2 N , jn n j < 1=n and jf (n ) f (n )j  :

2.67. Show that the eld of 2  2 matri es to the eld of omplex numbers x + iy.



x y y x



is isomorphi

Chapter 3

Analyti Fun tions and Power Series

In Se tion 3.1 we dis uss the fundamental dieren e between the derivative of a fun tion of a real and that of a omplex variable, by providing ne essary and suÆ ient onditions for dierentiability. As a onsequen e, a ontinuously dierentiable fun tion (brie y C 1 -fun tion) f de ned in an open set D is analyti if fz (z ) = 0 on D. There is an interesting relationship between fun tions that are analyti in a domain and real-valued fun tions that are harmoni in that domain. In Se tion 3.2 we dis uss this relationship at an introdu tory level, espe ially in nding harmoni onjugates. In addition, we dis uss the polar form of the Cau hy-Riemann equations and the Lapla e equation whi h have many impli ations in various bran hes of applied mathemati s. In Se tion 3.3, we investigate `in nite polynomials'-a generalization of `polynomials' and the in nite series of omplex numbers. The dis ussion allow us to hara terize the behavior of the power series in a very natural fashion. Therefore, the aim of this se tion is to study the onvergen e of the power series. In Se tion 3.4, we examine various known properties of the so- alled standard or elementary fun tions from a real variable to a omplex variable. Se tion 3.5 illustrates the relationship between w = log z (z 6= 0) and z = ew and onsiders its lo al properties. Several standard fun tions are asso iated with exponential fun tions or logarithmi fun tions. As a onsequen e, in Se tion 3.5, we de ne the inverse trigonometri and the inverse hyperboli fun tions.

3.1 Dierentiability and Cau hy-Riemann Equations Dierentiation in C is set against a ba kground of limits, ontinuity and

so on. To some extent the rules for dierentiation of a fun tion of a omplex variable are similar to those of dierentiation of a fun tion of a real variable. Sin e C is merely R2 with additional stru tures of addition and multipli ation of omplex numbers, we an immediately transfer most of the on epts of R2 into those for the omplex eld C . In fa t, we have

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Analyti Fun tions and Power Series

already done so when we de ned the on ept of distan e (modulus), p jz z 0 j = (x x0 )2 + (y y0 )2 (z = x + iy and z 0 = x0 + iy0) whi h is the same as the Eu lidean distan e between the points z = (x; y) and z 0 = (x0 ; y0 ) in R2 . We now start with

3.1. De nition. We say that a omplex fun tion f de ned in a nonempty open set D is dierentiable (or omplex dierentiable) at z0 2 D if the limit f (z ) f (z0 ) f (z0 + h) f (z0) (3.2) = lim lim z!z0 h!0 z z0 h exists. When this is the ase, the value of the limit, denoted by f 0 (z0 ), is

alled the derivative of f at z0 . The number f 0 (z0 ) is generally a omplex number. The fun tion f is said to be dierentiable in (on) D if it is differentiable at every point of D. A fun tion whi h is dierentiable in the entire omplex plane is alled an entire fun tion. We observe that \formally" the limit de nition of the derivative given by (3.2) is identi al in form to that of the derivative of a real (or omplex)valued fun tion of a real variable. In terms of ` Æ' notation, the limit in (3.2) exists i given any  > 0, there exists a Æ = Æ(; z0) > 0 su h that f (z ) z

Letting

 (z ) =

f (z0 ) z0



f 0 (z0 ) <  whenever 0 < jz z0 j < Æ:

8
0 su h that

f (z ) f (z0) = (z z0 )f (z ) for z 2
(z0 ; Æ1 )  D1 ; where f ( f 0 (z0 ) + (z ) in (3.3)) is ontinuous in
(z0 ; Æ1 ) with lim  (z ) = f 0 (z0 ): z!z0 f Further, sin e g is ontinuous at w0 , there exists a Æ2 > 0 su h that g(w) g(w0 ) = (w w0 )g (w) for w 2
(w0 ; Æ2 )  D2 ; where g is ontinuous in
(w0 ; Æ2 ) with limw!w0 g (w) = g0 (w0 ): Now

hoose Æ > 0 su h that Æ < Æ1 and

jz z0 j < Æ1 ) jf (z ) f (z0)j < Æ2 : Then for z 2
(z0 ; Æ), we have by substitution

g(f (z )) g(f (z0 )) = (f (z ) f (z0 ))g (f (z )) = (z z0 )f (z )
g (f (z )) = (z z0 )gÆf (z ): By Corollary 2.21, g Æ f is ontinuous at z0 and its value at z0 is g (w0 ) = g0 (w0 ). By Theorem 2.18, the produ t f (z )
g (f (z )), i.e. gÆf (z ), is

ontinuous at z0 and 0 0 0 0 lim  (z ) = zlim z!z0 gÆf !z0 [f (z )
g (f (z ))℄ = f (z0 )
g (w0 ) = f (z0 )g (f (z0 )): The assertion now follows from (3.3). Theorem 3.9 leads qui kly to the following useful properties of analyti fun tions.

3.13. Theorem. Linear ombinations and nite produ ts of analyti fun tions in an open set D are all analyti in D. If f and g are analyti in

80

Analyti Fun tions and Power Series

D, then the quotient f=g is analyti in D ex ept for those z in D at whi h g vanishes. 3.14. Corollary. If f and g are entire then so are f  g, fg; and f Æ g is entire when it is de ned. We know that if F (z ) = U (x; y) + iV (x; y) where U and V are realvalued fun tions de ned in a neighborhood of z0 , then the partial derivative Ux(x; y) of U with respe t to x at (x; y) (where z = x + iy), if it exists, is de ned to be U (x + h; y) U (x; y) (3.15) Ux(x; y) = lim : h!0 h Similarly, the partial derivative Uy (x; y) of U with respe t to y at (x; y), if it exists, is de ned to be

U (x; y + h) U (x; y) : Uy (x; y) = lim h!0 h Sometimes (3.15) and (3.16) are denoted by the Leibnitz notation, (3.16)

U U U U (z ) = (x; y) and (z ) = (x; y); x x y y respe tively. Note that h in (3.15) and (3.16) is a non-zero real number near 0. The partial derivatives Fx (x; y) and Fy (x; y) of the omplex-valued fun tion of the omplex variable F (z ) = U (x; y) + iV (x; y) at z = x + iy are de ned by

Fx (z ) = Ux (x; y) + iVx(x; y) and Fy (z ) = Uy (x; y) + iVy (x; y); respe tively, provided the partial derivatives on the right side of the orresponding equations exist. We simply write these expressions as

Fx = Ux + iVx ; and Fy = Uy + iVy : If Fy = iFx, then, by equating imaginary and real parts, we have a pair of famous partial dierential equations

Ux = Vy ; Uy = Vx: Conversely, the later two equations imply Fy = iFx . These two extremely important partial dierential equations are alled the Cau hy-Riemann6 (brie y we write the C-R) equations. The former is a tually referred to as 6 These two equations are named in honor of Fren h a mathemati ian, Augustin-Louis Cau hy (1789-1857), who dis overed them, and in honor of a German mathemati ian, Georg Friedri h Bernhard Riemann (1826-1866), who made them fundamental in the development of the theory of omplex analysis. Riemann is onsidered one of the three founders of omplex fun tion theory; the others being Cau hy and Weierstrass.

3.1 Dierentiability and Cau hy-Riemann Equations

81

the C-R equations in Cartesian form. We note that the existen e of the derivative for real-valued fun tions of single real variable is a mild smooth

ondition while the same for omplex-valued fun tions of a omplex variable leads to the above pair of partial dierential equations. We usually write

f (z ) = f (x + iy) = u(x; y) + iv(x; y) = Re f (z ) + iIm f (z ) or f (z ) = u(z )+ iv(z ) instead. However, we may abuse the notation slightly by writing f (z ) = f (x; y), but when we write it like this, we a tually identify u(x; y) + iv(x; y) 2 C with (u(x; y); v(x; y)) 2 R2 and use results from two variable al ulus for our investigation. If  R2 and u : ! R is a ontinuous fun tion, then u is alled a 1 C - (or ontinuously dierentiable) fun tion in if ux and uy exist and are

ontinuous in . More generally, if k 2 N then u is said to be in C k ( ) (or simply a C k fun tion or k-times ontinuously dierentiable fun tion) if all the partial derivatives of u up to and in luding order k exist and are

ontinuous in . We indi ate this by writing u 2 C k . Note that C 0 ( ) denotes the set of all ontinuous fun tions in . A fun tion f : ! C is said to belong to C k ( ), or simply all it a C k -fun tion, if both u and v belong to C k ( ). From the inspe tion of fun tions su h as jz j, jz j2 and z Re z , we on lude that it is not ne essarily an easy task to determine whether a given fun tion does or does not have a derivative. Let us start deriving a simple riterion whi h helps us to handle this problem.

3.17. Theorem. If f (z ) = u(x; y) + iv(x; y) is dierentiable at z0 , then the C-R equations hold at z0 = x0 + iy0 : ifx(z0 ) = fy (z0 ); or equivalently,

ux(x0 ; y0 ) = vy (x0 ; y0 ) and uy (x0 ; y0 ) = vx(x0 ; y0 ):

Proof. Let f : ! C where  C is a neighborhood of z0 . If f 0 (z0 ) exists for some point z0 = x0 + iy0, then the limit f (z0 + h) f (z0 ) lim h!0 h exists and is independent of the path along whi h h = h1 + ih2 ! 0. In parti ular, we have f (x0 + h1 ; y0 ) f (x0 ; y0) f f 0 (z0 ) = lim = (z0 ); h1 !0 h1 + i0 x and f (x0 ; y0 + h2 ) f (x0 ; y0 ) 1 f f 0 (z0 ) = lim = (z ): h2 !0 0 + ih2 i y 0

82

Analyti Fun tions and Power Series

A omparison of the two expressions for f 0(z0 ) shows that the omplex dierentiability of f at z0 implies that not only the partial derivatives of f (with respe t to x and y) exist at z0, but also that they satisfy the C-R equations{in omplex form

f 1 f (z ) = (z ); i.e. ifx(z0 ) = fy (z0 ): x 0 i y 0

(3.18)

Equating the real and imaginary parts yields the C-R equations{in Cartesian form ux (z0 ) = vy (z0 ) and uy (z0 ) = vx (z0 ): Another onvenient notation is to treat the pair of onjugate omplex variables z and z as two independent variables by writing 



z+z z z x= ; y= i : 2 2 Now we introdu e the following dierential operators: 

  x  y 1  := + = z x z y z 2 x and



 i y





 x  y 1    := + = +i : z x z y z 2 x y It follows that if f = u + iv, then

(3.19)



f 1 f fz = = z 2 x

and similarly, (3.20)

fz =





f 1 i = ([ux + vy ℄ + i[vx uy ℄) y 2 

f 1 f f 1 = +i = ([ux vy ℄ + i[uy + vx ℄) : z 2 x y 2

Therefore, the C-R equations (3.18) are exa tly equivalent to fz (z0 ) = 0 whi h is also referred to as the omplex form of the C-R equations. With this notation, we have the following alternate form of Theorem 3.17.

3.21. Theorem. A ne essary ondition for a omplex-valued fun tion f = u + iv to be dierentiable at z0 is that fz (z0 ) = 0: In parti ular, if f 2 H(D), then C-R equations hold at every point z 2 D. The C-R equations are more helpful in proving non-dierentiability. For example, onsider f (z ) = z and g(z ) = Re z . Then fz (z ) = 1 6= 0 and writing g as z+z g(z ) = ; 2

3.1 Dierentiability and Cau hy-Riemann Equations

83

it follows that gz (z ) = 1=2 6= 0. Thus, both f and g are nowhere dierentiable. On the other hand, the onverse of Theorem 3.17 (equivalently, Theorem 3.21) is not true. We an demonstrate this by a number of examples (for instan e, see Examples 3.23 and 3.24). That is, the ne essary ondition stated in Theorem 3.17 for dierentiability at a point is not generally suÆ ient for dierentiability at that point.

3.22. Remark. Ea h of the fun tions f1 (z ) = jz j; f2 (z ) = Re z and f3 (z ) = Im z; is a non- onstant real-valued fun tion de ned in C (see Theorem 3.6). Ea h of them is nowhere analyti . If we rewrite these fun tions as

f1 (x; y) =

p

x2 + y2 ; f2 (x; y) = x; f3 (x; y) = y;

then, ex ept f1 , ea h of these fun tions are real dierentiable in R2 .



3.23. Example. It is easy to see that the fun tion f de ned by f (z ) = jRe z Im z j1=2 satis es the C-R equations at the origin, but is not dierentiable at this point. To see this, we may rewrite the given fun tion as

f (z ) =

jz 2 z 2 j1=2 or f = u + iv with u(x; y) = jxyj1=2 and v(x; y) = 0. 2

Note that f is identi ally zero on the real and imaginary axes. Therefore, it is trivial to see that ux(0; 0) = uy (0; 0) = vx (0; 0) = vy (0; 0) = 0: For example, u(s; 0) u(0; 0) ux (0; 0) = slim = 0: !0 s Thus, the C-R equations hold at z = 0: However, taking h = rei 6= 0 with r ! 0, we nd that lim h!0

f (h) f (0) jr2 os  sin j1=2 = e i j sin 2j1=2 p = rlim !0 r( os  + i sin ) h 2

whi h is learly depending upon  (e.g. take  = 0 and  = =4). We

on lude that f is not dierentiable at z = 0 even though f satis es the C-R equations at the origin. Here, sin e v(x; y) = 0, v is a C 1 -fun tion in R2 : Are the partial derivatives ux and uy ontinuous at the origin? How about the fun tions

f (z ) = jRe z Im z j1=3 and f (z ) = jRe z Im z j1=4?



84

Analyti Fun tions and Power Series

3.24. Examples. Consider 

8 3

>
h 1 ih1 > : = 1 + i for h = h1 + i
0. h1 This observation shows that the partial derivatives exist and satisfy the C-R equations at the origin even though the fun tion is not dierentiable there. A similar on lusion ontinues to hold for the following two fun tions: 8
3, then f 2 H(D) and f 0 (z ) = 0 in D, yet f is non- onstant in D. Here is another example. De ne 

Re z > a f (z ) = 0i for for Re z < a; where a 2 R is xed. Then f 2 H(D), where D = C n fz : Re z = ag, yet f is not onstant in D. Note that, in ea h of these two examples, Re f (z ) = u(x; y) = 0 in D but Im f (z ) = v(x; y) is not onstant.

3.2 Harmoni Fun tions We begin with the formal de nition of a harmoni fun tion. A real-valued fun tion  = (x; y) of real variables x and y is said to be harmoni in an open subset of C if it has ontinuous partial derivatives of se ond order and satis es the Lapla e's equation7 in two variables 52 = 0 throughout

, where 52 is the se ond order dierential operator given by 2

2

52 = x2 + y2 = xx + yy : 7 This equation that bears the name of the Fren h mathemati ian, Pierre Simon de Lapla e (1749-1827), had been found by Leonhard Euler in 1752 in onne tion with Euler's studies on hydrodynami s. Lapla e ontributed signi antly to the elds of

elestial me hani s and probability theory and his parents were farmers.

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Analyti Fun tions and Power Series

The operator  7! 52 is alled Lapla e operator or simply a Lapla ian. A fun tion v is alled a onjugate harmoni fun tion (or, more brie y, a harmoni onjugate) for a harmoni fun tion u in whenever f = u + iv is analyti in . Note that the word onjugate here is not the same as onjugate of a omplex number z . Further, the harmoni onjugate v is unique, up to an addition of a real onstant. Indeed, if v1 is another harmoni onjugate of u, so that F = u + iv1 is also analyti in , then the dieren e F f = i(v1 v) be omes analyti in . But then by Theorem 3.31(ii), v1 v is a onstant. Sin e if = v + i( u), we also observe that u is a harmoni onjugate of v whenever v is a harmoni onjugate of u.

3.32. Remark. As an be observed from the C-R equations, we

annot hoose two arbitrary harmoni fun tions u and v and laim that the resulting fun tion f = u + iv is analyti . For example, u(x; y)  x and v(x; y)  y are harmoni fun tions in C , but f = u + iv = x + i( y) = z is nowhere analyti . On the other hand, v + iu = y + ix = i(x + iy) = iz whi h is analyti in C . Further, it would be appropriate to have the Lapla e equation to be satis ed not just for any set of points but for an open set or a domain or more importantly a simply onne ted domain. For example, if u = x3 y3 then

2u 2u + = 6(x y) = 0 only when y = x x2 y2 and the set fz = x + iy : y = xg is not open in C and so, u annot be treated as a harmoni fun tion in any open subset of C .  Let us now dis uss some fa ts about this topi and onstru t simple examples of harmoni fun tions. We rst introdu e a omplex Lapla ian. Suppose that we are given a harmoni fun tion   z+z z z u(x; y) = u ; : 2 2i Then, formally treating z and z as independent variables, as before u u x u y 1 = + = [u iuy ℄ z x z y z 2 x so that   2u 1  x  y = (ux iuy )
+ (ux iuy )
zz 2 x z y z 1 = [(uxx + uyy ) + i(uxy uyx)℄ ; 4

91

3.2 Harmoni Fun tions

where we have used the notation          u  u  u  u uxy = ; uyx = ; uxx = ; uyy = : y x x y x x y y Therefore any harmoni fun tion u satis es the dierential equation 2u = 0: 52 u = zz The operator 52 is sometimes alled the omplex Lapla ian. Thus if u and its rst and se ond partial derivatives are ontinuous in an open subset

 C , then 52 u = (1=4) 52 u: Hen e, the Lapla ian (operator) in omplex form is given by     52 = 4 z (3.33) : z For example, if f = u + iv is a omplex-valued fun tion in a domain then        f 2 5 (zf ) = 4 z z (zf ) = 4 z f + z = 4fz + 4 52 f whi h shows that if f and zf are harmoni in , then fz = 0 in and hen e f is analyti in , by Theorem 3.26. We say that a omplex-valued

fun tion is harmoni in an open set if both its real and imaginary parts are harmoni thereat.

Harmoni fun tions play an important role in both mathemati s and physi s. There are (at least) two important reasons why harmoni fun tions oupled with the C-R equations are dis ussed as an important part of omplex analysis as demonstrated in the next two theorems. First, we re all that if f = u + iv is analyti in an open set , then the C-R equations hold throughout : (3.34) ux = vy ; uy = vx: The C-R equations have some interesting onsequen es. For instan e, if the se ond partial derivatives exist and are ontinuous (in fa t, we shall later see that the derivative of an analyti fun tion in a domain is itself analyti there more generally in nitely dierentiable, and so u and v both have

ontinuous partial derivatives of all orders) then by dierentiating the rst equation with respe t to x and se ond with respe t to y, we get

uxx = vxy ; uyy = vyx : The ontinuity of these partial derivatives implies that the mixed derivatives are equal (whi h is an important fa t from two variable al ulus) and, in parti ular, vxy = vyx and therefore,

uxx + uyy = vxy

vyx = 0:

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Analyti Fun tions and Power Series

In a similar way, dierentiating the rst equation in (3.34) with respe t to y and the se ond in (3.34) with respe t to x, we nd that v satis es the Lapla e equation 52v = 0: In on lusion we have

3.35. Theorem. Let be an open subset of C . Then the real and imaginary parts of an analyti fun tion in are harmoni in . At this point it should be noted that the mixed se ond partial derivatives do not oin ide in general. For instan e, for the real-valued fun tion u(x; y) de ned by 8 2 y2 ) < xy (x if (x; y) 6= (0; 0) 2 x + y2 u(x; y) = : 0 if x = y = 0; it is easily seen that uxy (0; 0) 6= uyx(0; 0):

3.36. Example. If (x; y) is harmoni in a domain D, then, by Theorem 3.26 with u (= x ) and v (= y ), we nd that f = x iy 2 H(D). Similarly, if (x; y) is harmoni in D then g = x i y 2 H(D). In parti ular, we have f  g = (x  x ) i(y  y ) 2 H(D) f  ig = (x  y ) i(y  x ) 2 H(D) if  g = (y  x ) + i(x  y ) 2 H(D): This example gives a way of obtaining analyti fun tions from harmoni fun tions.  Theorem 3.35 provides numerous examples of harmoni fun tions as we shall soon see. However, the fun tion u de ned by u(x; y) = x2 + y2 annot be a real part of an analyti fun tion sin e 52u = 4. We raise the question: Given a real-valued harmoni fun tion u in an open set , an it be written as a real or imaginary part of an analyti fun tion f : ! C ? If so, under what ondition this is possible. We need some preparation to answer this question.

3.37. De nition. A domain D in C is simply onne ted if its omplement with respe t to C 1 (i.e. C 1 nD) is a onne ted subset of C 1 . Topologi ally, a simply onne ted domain D in C an be ontinuously shrunk to a point in D. Note that the pun tured unit disk A = fz : 0 < jz j < 1g an be shrunk to an arbitrarily small domain, but not to a point in A. Intuitively, a simply onne ted domain in C means that it does not

ontain any \holes". Consequently, a

ording to De nition 3.37, exterior of ir les (i.e. omplement of losed disks in C ) are not simply onne ted

3.2 Harmoni Fun tions

93

in C . Note that, in the extended plane, the exterior of a ir le is simply

onne ted be ause it an be \shrunk" to the point at 1. Some simple examples of simply onne ted domains in C are (i) disks
(a; r) (ii) half-planes fz : Re (ei z ) > g (iii) onvex domains su h as in (i), (ii) and in nite parallel strips, su h as D = fz :  < Re (ei z ) < g for some  and with  < (iv) domains that are starlike with respe t to the origin su h as C n[0; 1) and C n fiy : jyj > 1g (v) the whole omplex plane. The annular domain E = fz : 1 < jz j < 2g is not simply onne ted, whereas the domain E with fz : 1 < Re z < 2; Im z = 0g removed from it (i.e. E nf(1; 2)g), is simply onne ted. We note that there are many equivalent de nitions of simply onne tedness. One su h equivalent form of it is the following (see also De nition 4.50):

3.38. De nition. Let D be a domain in C and D1 be the set

orresponding to D on the Riemann sphere S . Then D is alled simply

= S nD1 is onne ted and ontains the north pole.

onne ted if D1 It would have been ni e if every harmoni fun tion is a real part of some analyti fun tion. But this is not true in general. However, it is indeed true lo ally, but globally provided is a simply onne ted domain. More pre isely, we have

3.39. Theorem. Let be a simply onne ted domain and let  be harmoni in . Then  has a harmoni onjugate in . It is onvenient to supply a proof of Theorem 3.39 in Chapter 4 sin e a motivation for an expli it onstru tion for the harmoni onjugate omes from ( omplex) integration. On the other hand, one an give a simple and a dire t proof of Theorem 3.39 espe ially when (i) (x; y) is a real-valued harmoni polynomial in C (ii) (x; y) is a real valued harmoni fun tion in where is either an open disk or open re tangle, see Corollary 3.57.

3.40. Example. Consider u(x; y) = 4xy x3 + 3xy2 . Then ux = 4y 3x2 + 3y2 and uy = 4x + 6xy

94

Analyti Fun tions and Power Series

from whi h it is easy to see that u is harmoni in R2 . To nd the harmoni

onjugate v(x; y) in C for u(x; y), we may pro eed as follows. If f = u + iv is the orresponding analyti fun tion, then f 0 (z ) = ux iuy = 4y 3x2 + 3y2 i(4x + 6xy) = 3[x2 y2 + 2ixy℄ 4i(x + iy) = 3z 2 4iz

whi h gives f (z ) = z 3 2iz 2 +ik, where k is a real onstant. The harmoni

onjugate v(x; y) may be obtained by taking the imaginary part of the last expression. Alternately, use the C-R equations to obtain

vy = ux = 4y 3x2 + 3y2:

(3.41)

Integrate (3.41) with respe t to y to obtain (3.42)

v=

Z

vy dy + (x) = 2y2 3x2 y + y3 + (x);

where  is a fun tion of x. If  in (3.42) is dierentiable with respe t to x the equation vx = 6xy + 0 (x) (= uy ) is obtained whi h, together with the C-R equation uy = vx, gives 0 (x) = 6xy uy = 6xy (4x + 6xy) = 4x so that (x) = 2x2 + k, where k is some real onstant. Hen e, by (3.42), the harmoni onjugate fun tion v(x; y) is given by

v = (2y2 3x2 y + y3) + ( 2x2 + k): Now we an use x = (z + z )=2 and y = (z of z .

z)=2i to write f as a fun tion



3.43. Theorem. If u and v are harmoni onjugates to ea h other in some domain then u and v must be onstant there. Proof. By the de nition and the hypotheses, f = u + iv and g = v + iu are analyti in D. But f

ig = 2u; and f + ig = 2iv:

By Theorem 3.9, these imply that the real-valued fun tions of a omplex variable, namely u and v are analyti in D: Therefore, by Theorem 3.6, f 0 (z ) = 0 in D. Sin e D is a domain, it follows that f and g are onstants and hen e, u and v are onstants.

95

3.2 Harmoni Fun tions

3.44. Polar form of the C-R equations and Lapla ian. There are ways to obtain the C-R equations in polar form. Here is a dire t method. Let f (z ) = u(r; ) + iv(r; ) be dierentiable at a point z . Then f 0(z ) exists and equals f (z + h) f (z ) : f 0(z ) = lim h!0 h Set z = rei . The nearby points of z along the radial dire tion may be given by z + h = (r +
r)ei . Therefore, for h =
rei 6= 0, f (z + h) f (z ) u(r +
r; ) + iv(r +
r; ) u(r; ) + iv(r; ) = h
rei
rei   1 u(r +
r; ) u(r; ) = i e
r   i v(r +
r; ) v(r; ) + i : e
r Allowing
r ! 0 shows that 1 (3.45) f 0 (z ) = i [ur + ivr ℄: e Next hoose nearby points of z along the ir ular path through the point z so that z + h = rei(+
) = rei ei
 ; i.e. h = rei (ei


1)

and for
 6= 0, f (z + h) f (z ) u(r;  +
) u(r; ) v(r;  +
) v(r; ) = +i h rei (ei
 1) rei (ei
 1)   u(r;  +
) u(r; ) 1 = irei
   v(r;  +
) v(r; ) i
 +i : i

 e  1 Now allowing
 ! 0, it follows that 1 1 (3.46) f 0 (z ) = i [u + iv ℄ = i [v iu ℄: ire re Comparing (3.45) and (3.46) produ es the C-R equations in polar form: v u (3.47) ur =  and vr =  : r r i Alternatively, let f (z ) = f (re ), r 6= 0, be dierentiable at a point z0 : Then both f r 0 f (z0 ) = = (ur + ivr )e i z0 r z z=z0

96

Analyti Fun tions and Power Series

and





f  e i = (u + iv )  z z=z0 ir z0 must be the same. Therefore, equating the right hand side of the last two equations yields (3.47). If we dierentiate the rst equation in (3.47) with respe t to r and the se ond in (3.47) with respe t to  we get f 0 (z0 ) =

  1 (v ) = ur + rurr ; (v ) = u r   r r and so, using the ontinuity of se ond partial derivatives, these two equations give 1 1 urr + ur + 2 u = 0: r r whi h is the polar form of the Lapla e equation. If u(r; ) depends on r alone, then the above Lapla e equation be omes 1 urr + ur = 0: r For example, we have

 rn os n and rn sin n are harmoni for any positive integer n  ln r = ln jz j is harmoni in the pun tured disk C n f0g. We note that ln jz j has no harmoni onjugate in C n f0g, though it does have in C n[0; 1), see Se tion 3.5. Next we attempt to develop methods for nding an analyti fun tion f whose real part is a given harmoni fun tion u(x; y) whi h is a rational fun tion in x and y. For this we start with

f (z ) = f (x + iy) = u(x; y) + iv(x; y) so that

f (z ) = f (x + iy) = u(x; y) iv(x; y); where x = (z + z)=2 and y = (z z)=(2i). Adding the above two equations we nd that f (x + iy) + f (x + iy) = 2u(x; y): Re all that analyti fun tions are ompletely hara terized by the ondition: f (3.48) fz = = 0: z For example, if we substitute z for f then we see that fz = 0 and fz = 1: Thus we remark that f (z ), the onjugate of an analyti fun tion f (z ), has

97

3.2 Harmoni Fun tions

the derivative zero with respe t to z and so f (z ) may be onsidered as a fun tion of z alone. Therefore, we an denote f (z ) simply by f (z). That is

f (x + iy) = f (x iy); and 2u(x; y) = f (x + iy) + f (x iy): In parti ular, the last equation yields z z  z 1 h z z z i 1 u ; = f +i +f i = [f (z ) + f (0)℄ 2 2i 2 2 2i 2 2i 2 whi h gives z z  f (z ) = 2u ; f (0): 2 2i Also we observe that

u(0; 0) = (1=2)[f (0) + f (0)℄ = Re f (0) and so

f (0) = Re f (0) i Im f (0) = u(0; 0) ik; where k = Im f (0) is a real number. Consequently, the fun tion f may be

omputed by the formula z z  f (z ) = 2u ; u(0; 0) + ik: 2 2i Similarly, it follows that in a neighborhood of z0 , the analyti fun tion f = u + iv asso iated with the harmoni fun tion u(x; y) is given by 



z + z0 z z0 ; u(x0 ; y0 ) + ik; 2 2i where k is a real onstant. We shall illustrate the appli ation of this method with two examples. f (z ) = 2u

3.49. Example. Consider u(x; y) = x3 3xy2 : Then u is de ned in and u satis es Lapla e's equation for all points in R2 . Using the above method, the orresponding analyti fun tion f is given by   z z   z   z 2  z 3 f (z ) = 2u ; u(0; 0) + ik = 2 3 + ik = z 3 + ik; 2 2i 2 2 2i where k is a real number. 

R2

3.50. Example. We wish to nd the most general ubi form u(x; y) = ax3 + bx2 y + xy2 + dy3 (a; b; ; d-real); whi h satis es Lapla e's equation, and to determine an analyti fun tion whi h has u as its real part. To do this, for (x; y) 2 R2 , we ompute

uxx + uyy = 2[(3a + )x + (b + 3d)y℄

98

Analyti Fun tions and Power Series

and so, 52u = 0 in R2 provided 3a + = 0 = b + 3d = 0: Thus, the most general harmoni polynomial of degree three takes the form

u = ax3 3dx2 y 3axy2 + dy3 : Further, using the above method of onstru tion of analyti fun tion, we have z z  + ik f (z ) = 2u ; 2 2i     z 2  z   z   z 2  z 3  z 3 = 2 a 3d 3a +d + ik 2 2 2i 2 2i 2i = (a + id)z 3 + ik whi h is the required analyti fun tion having u as its real part, where k is a real onstant. 

3.51. Dis ussion on nding harmoni onjugates. Let us arefully look at the problem of whether a real-valued harmoni fun tion u, a C 2 solution of Lapla e's equation in some domain , is a real or imaginary part of some analyti fun tion. For the time being we suppose that a harmoni fun tion u is given and we have found an analyti fun tion F (z ) in su h that F (z ) = u + iv, z 2 : Then, the C-R equations give vx = uy and vy = ux whi h show that, vx and vy are ompletely determined from the given fun tion u. Therefore, v may be found up to an additive onstant. Again, in view of the C-R equations, we see that     uxx + uyy = 0 () (ux ) = ( uy ) () (vy ) = (vx ): x y x y Set f = uy and g = ux . This amounts to rephrasing our problem as follow. 3.52. Problem. Given f; g 2 C 1 ( ) with fy = gx, an we nd a fun tion v 2 C 2 ( ) su h that vx = f and vy = g in ? If so under what

onditions on , is this possible? First, we aim at giving a partial solution to this problem by dis ussing a spe ial ase when is an open re tangle (and the proof is similar when

is an open disk). 3.53. Theorem. Let = f(x; y) 2 R2 : jx aj < Æ; jy bj < Æ0 g be an open re tangle in R2 . Suppose that f; g 2 C 1 ( ) su h that (3.54)

fy = gx in :

Then there exists a fun tion v 2 C 2 ( ) satisfying the onditions

(3.55)

vx = f and vy = g

99

3.2 Harmoni Fun tions in . (We may take v as real-valued whenever f and g are also).

Proof. Choose an arbitrary point (x; y) 2 and de ne v(x; y) =

Z x

a

f (s; b) ds +

Z y

b

g(x; t) dt:

Observe that any other v may dier from the above merely by a onstant. What motivates us to de ne v(x; y) in this way? (see Remark 3.56). We need to show that v has the desired properties. By the fundamental theorem of al ulus for C 1 -fun tions,

vy (x; y) = g(x; y): Next, on e again using the fundamental theorem of al ulus for the rst integral and the theorem on dierentiation under the integral sign, it follows that Z y   vx (x; y) = f (x; b) + g(x; t) dt x b Z y  = f (x; b) + g(x; t) dt x b Z y  f (x; t) dt (sin e gx = fy ) = f (x; b) + b y = f (x; b) + f (x; y) f (x; b) so that vx (x; y) = f (x; y): Thus, we have found a fun tion v with vy = g 2 C 1 ( ) and vx = f 2 C 1 ( ) whi h means that v 2 C 2 ( ):

3.56. Remark. The fun tion v(x; y) that we need to de ne must satisfy (3.55). Indeed, integrating vy (x; y) = g(x; y) with respe t to y, we get Z y v(x; y) = g(x; t) dt + (x) C 1 -fun tion

b

where  is some of x. Sin e we also need vx = f , by the rst equation in (3.55), we ompute that Z y

Z y   0 vx (x; y) = g(x; t) dt +  (x) = f (x; t) dt + 0 (x) x y b b whi h gives f (x; y) = f (x; y) f (x; b) + 0 (x); i.e. 0 (x) = f (x; b):

This has the solution

(x) =

Z x

a

f (s; b) ds + ;

100

Analyti Fun tions and Power Series

where is some onstant. Hen e, v(x; y) must be of the form

v(x; y) =

Z x

a

f (s; b) ds +

Z y

b

g(x; t) dt + :



3.57. Corollary. If is either an open re tangle (with sides parallel to the axes) or open disk and if u is a real-valued harmoni fun tion in , then there exists an analyti fun tion F in su h that u = Re F: Proof. Set f = uy and g = ux: Then f; g 2 C 1 ( ) and, sin e 52u = 0 in , we have fy = gx in : By Theorem 3.53, there exists a real-valued fun tion v 2 C 2 ( ) su h that vx = f = uy and vy = g = ux in : By the theorem on suÆ ient onditions for analyti fun tions (see Theorem 3.26), we on lude that F = u + iv is analyti in . Theorem 3.53 is alled an antiderivative theorem (for real-valued fun tions). Now it is natural to raise the following

3.58. Problem. Given an analyti fun tion F in , an we nd an analyti fun tion G su h that G0 (z ) = F (z )? If the answer to this problem is yes, then we all the fun tion G a primitive or anti-derivative for F . Any other anti-derivative of F would dier from G by a onstant. Indeed, if both G1 and G2 are primitives of a fun tion F , then (G1 (z ) G2 (z ))0 = G0 (z ) G0 (z ) = F (z ) F (z ) = 0 1

2

and so by Theorem 3.31, we see that G1 (z ) G2 (z ) is a onstant. Here is a simplePillustration. The primitive of the polynomial fun tion p P de ned by p(z ) = nk=0 ak z k is P (z ) = nk=0 (ak =(k + 1))z k+1 + K; where K is an arbitrary onstant. Is there any restri tion on for the existen e of primitives? First we give an aÆrmative answer when is an open re tangle or onvex or open disk in C . However, we shall later dis uss a more general theorem whi h provides an aÆrmative answer to this problem whenever is a simply

onne ted domain.

3.59. Theorem. (Antiderivative Theorem) Let be either an open re tangle (with sides parallel to the axes) or an open disk. Then every analyti fun tion F (z ) in possesses a primitive in . Proof. Set F (z ) = u(z ) + iv(z ). Sin e F is analyti in , uy = vx () fy = gx;

101

3.3 Power Series as an Analyti Fun tion

where u = f 2 C 1 ( ) and v = g 2 C 1 ( ): By Theorem 3.53, there exists U 2 C 2 ( ) su h that

Ux = f (= u) and Uy = g (= v) in : As F is analyti , we have vy = ux where v; u 2 C 1 ( ). Again, by Theorem 3.53, it follows that there exists V 2 C 2 ( ) su h that

Vx = v and Vy = u in : Finally, de ne G(z ) = U + iV . Then, U; V 2 C 2 ( ) and the C-R equations are satis ed in : Hen e, by Theorem 3.26, G is analyti in . Note that G0 (z ) = Ux + iVx = u + iv = F (z ):

3.3 Power Series as an Analyti Fun tion In our earlier se tions we have seen that the polynomial of degree n given by (3.60) p(z ) = a0 + a1 z +


+ an z n (an 6= 0)

an be dierentiated term-by-term to get (3.61) p0 (z ) = a1 + 2a2 z +


+ nanz n 1 :

1

P

Consider a series of fun tions of the form n0 an ( 0 )n ; where  is a

omplex variable and 0 , an , n = 0; 1; 2; : : : , are xed onstants. Su h a series will be alled a (formal) power series with enter 0 and oeÆ ients an . The substitution z =  0 transforms the above series into the power series X (3.62) an z n : n0 So, for our dis ussion it is enough to onsider power series with enter 0 as the results about general power series an be obtained by translation. Now the polynomial (3.60) may be thought of as a power series at 0 with

oeÆ ients ak = 0 for all k > n and so we say that the family of polynomials is ontained in the family of power series. Further, we also note that a power series de ned by (3.62) is a spe ial ase of the limit of in nite sequen es of P fun tions, namely, ffn(z )g, where fn (z ) = nk=01 ak z k . In any ase, (3.61) suggests that for the power series (3.62) with sum f (z ), we should have

f 0 (z ) =

X

n1

nan z n

1 P

and in that ase we all the R.H.S the derived series of n0 an z n. This fa t of ourse requires justi ation and before giving a proof, we dis uss some important fa ts about the onvergen e of (3.62) to appre iate fully the ideas involved.

102

Analyti Fun tions and Power Series

Now, we onsider the polynomials

z2 z4 z 2n +


+ ( 1)n ; 2! 4! (2n)! z3 z5 z 2n+1 +


+ ( 1)n ; qn (z ) = z 3! 5! (2n + 1)! z2 zn rn (z ) = 1 + z + +


+ : 2! n! We know that these are entire. As n ! 1, we obtain that pn (z ) = 1

pn (z ) ! os z; qn (z ) ! sin z; rn (z ) ! ez and in all three ases, the orresponding limit fun tions are also entire. How about the polynomial 1 + z + z 2 +


+ z n when n ! 1? How about 1 + rz + r2 z 2 +


+ rn z n (r > 1) as n ! 1? We start with some interesting fa ts due to Abel: Suppose 0 6= z1 2 C P is su h that the power series n0 an z1n onverges. The terms of the series are then bounded (see Se tion 1.6). Indeed, as jan z1n j ! 0 as n ! 1, there exists an M > 0 su h that jan z1n j < M for all n  0. We have then, for all z with jz j < jz1 j, n





n

jan z nj = jan z1n j zz < M zz for all n: 1

P

1

Sin e jz=z1j < 1, the P geometri series n0 jz=z1jn onverges, so that, by the omparison test,P n0 jan z n j onverges for all z with jz j < jz1 j. For example, P if f (z ) = n1 z n =n then the series onverges for jz j < 1, sin e f ( 1) = n1 (P1)n =n onverges. If the series n0 an z2n diverges, then, for all z with jz j > jz2 j, we have jan z nj  jan z2n j for ea h n  0: P

So, by the omparison test, the series n0 an z n does not onverge. For P P example, f (z ) = n1 z n=n diverges for jz j > 1, sin e f (1) = n1 1=n diverges. The above fa ts allow us to hara terize the behavior of the power series (3.62) in a very natural fashion. Indeed, translating the series about `0' into a series about a 2 C , the above dis ussion gives (see Figure 3.1) P n 3.63. Theorem. If the series 1 n=0 an (z a) onverges at some point z1 (6= a), then the series onverges (absolutely) at all points in the disk
(a; jz1 aj). If the series diverges at z2 (6= a), then it diverges for all z with jz aj > jz2 aj. Be ause of the interesting information about the onvergen e from Theorem 3.63, it is natural to ask: what is the largest disk about a on whi h

103

3.3 Power Series as an Analyti Fun tion divergent ring of doubt z1 z2

a

convergent

Figure 3.1: Des ription for Abel's test. diverges do not know R

converges a

uniformly converges

Figure 3.2: Illustration for the disk of onvergen e. P

the series n0 an (z a)n onverges? We now make this issue more pre ise in the form of a de nition for a series about the origin. The radius of

onvergen e R of the given power series (3.62) is de ned by

R = supf :

X

n0

an z n onverges for all z satisfying jz j  g.

P

Note that R =P0 if n0 an z n onverges only for z = 0. If R = 1, then the series a z n onverges for all z 2 C . Thus, by de nition, if P nn0 n 0 < R < 1, jan z j onverges for all z su h that jz j < R and diverges for all z su h that jz j > R. The series may onverge for some or all points on the ir le jz j = R. The ir le jz j = R is then alled the ir le ofP

onvergen e be ause this is the greatest ir le about a = 0 inside whi h n0 jan z n j

onverges at ea h point (see Figure 3.2). The onventions 0 1 = 1 and 1 1 = 0 are observed so that R is the unique number in [0; 1℄.

3.64. Theorem. (Root Test) Let L (i) if L = 1, the series

P

1

= lim supn!1

p n

jan j. Then

n n0 an z onverges absolutely for all nite

and uniformly in any bounded set

z

104

Analyti Fun tions and Power Series

(ii) if L = 0, the series onverges only at z = 0 and diverges at all other points other than 0 (iii) if 0 < L < 1, the series onverges absolutely for jz j < L, uniformly for jz j  r, r < L and diverges for jz j > L (iv) L = R.

Proof. As the series onverges for z = 0, we need to onsider only the

ase z 6= 0. So, for z 6= 0, we have lim sup n!1

p n

jan z nj = jz j lim sup n jan j = jLz j : p

n!1

P

By Theorem 1.51, the series n0 an z n onverges absolutely for jz j < L and diverges for jz j > L. The uniform onvergen e of the series for jz j  r (r < L) follows from the Weierstrass M-test (see Theorem 2.59). Thus, (i) to (iii) follows. We leave (iv) as a simple exer ise.

3.65. Remark. Note that if nlim !1 p n

p n

jan j exists, then we have p

jan j = lim sup n jan j:

lim n!1

n!1

In this ase, the radius of onvergen e R is simply determined from

R

1

= nlim !1

p n

jan j:



1=n 3.66. Example. For ea h xed z 6= 0, we have nlim !1 jz j = 1. This

an be proved as follows: Let Æn = n1=n 1. Then for ea h n > 1, Æn is a positive real number. Now, for n  2, we have

)n

n = (1 + Æn =

n   X n k=0

k

Ænk > 1 +

n(n 1) 2 n(n 1) 2 n2 2 Æn > Æn  Æn : 2 2 4

This implies that 0  Æn2 < n2 and so we have Æn ! 0 as n ! 1; i.e. lim n1=n = 1. Applying this result for jz j  1, we get jz j1=n ! 1, sin e n!1 1  jz j1=n  n1=n for suÆ iently large n: If jz j < 1, apply the pre eding argument for 1=z .



3.67. Remark. The above example shows that fzng, where zn = n1=n 1, is a null sequen e. Sin e n1=n ! 1, we also have i 1h 1 + 21=2 + 31=3 +


+ n1=n ! 1 as n ! 1:  n

105

3.3 Power Series as an Analyti Fun tion

The determination of the radius of onvergen e for the series (3.62) will not always require an appli ation of Theorem 3.64, i.e. the Root test. In most of the ases, the following theorem alled the Ratio test will serve the purpose instead.

3.68. Theorem. (Ratio Test) If an 6= 0 for all but nitely many values of n, then the radius of onvergen e R of (3:62) is related by



an+1  1 l := lim inf n!1 an R

In parti ular, if

an+1 lim n!1 a

n





 lim sup aan+1 =: L: n!1

n

exists either as nite or +1 then



an+1 1 = lim sup jan j1=n = nlim a : !1 R n!1 n Proof. Proof of this theorem is an immediate onsequen e of Theorem 1.48. Again it suÆ es to onsider the ase z 6= 0. In this event, an+1 z n+1 lim sup a zn n!1

n

= jz jL and P

an+1 z n+1 lim inf n!1 a z n

n

= jz jl:

By Theorem 1.48, the series n0 an z n onverges absolutely for jz j < 1=L and diverges for jz j > 1=l. Thus, R must be at least 1=L and at most 1=l; that is R is related by 1=L  R  1=l: Consequently, if limn!1 jan+1 =an j exists then L = l and hen e an+1 = lim sup jan j1=n = lim inf jan j1=n = lim jan j1=n = 1 : lim n!1 an n!1 n!1 R n!1 This ompletes the proof. Note that if the sequen e fjan+1 j=janjg os illates, the limit does not exist and therefore the Ratio test be omes of no use. For instan e, for series su h as X

n0

an z n = 3 + z + 3z 2 + z 3 + 3z 4 +




+ 3z 2k + z 2k+1 +




the ratio jan+1 j=jan j is alternately 1=3 and 3, so the limit does not exist. In this parti ular series, even though the Ratio test indi ates that the value of R is lying between 1=3 and 3, it does not yield the exa t value of R. However, as ja2k j = 3 and ja2k+1 j = 1, Example 3.66 gives R 1 = lim sup jan j1=n = 1: n!1 Thus in the above two examples the Ratio test gives no information whereas the Root test gives the radius of onvergen e. We look now at a string of examples.

106

Analyti Fun tions and Power Series P

(i) Sin e (see Example 3.66) n1=n ! 1 as n ! 1, the series n1 nz n

onverges for jz j < 1 and diverges for jz j > 1. For jz j = 1, jnz nj = n ! 1 and so the series diverges for jz j = 1. In this ase we note that the series diverges at all points on the ir le of onvergen e. Also, we observe that the Root test is learly appli able. P (ii) Consider the series n0 ( 1)n z pn = 1 z p + z 2p


; where p  is a xed positive integer. Then

anp =





1 if n = 1; 3; : : : ( 1)m=k if m = 0; k; 2k; : : : ; i.e. am = 1 if n = 0; 2; : : : 0 otherwise:

In this ase lim supn!1 jan j1=n = 1. Hen e, R = 1 and the series onverges absolutely for jz j < 1, uniformly for jz j   < 1, and diverges for jz j > 1. Sin e the sequen e of terms of the series does not approa h zero when jz j = 1, it follows that the series diverges for jz j = 1. Note that the Ratio test is not appli able dire tly, but we ould obtain the radius of onvergen e by translating the given series into a new series with a new variable  ,  = zp. (iii) By means of the Ratio test we at on e see that ea h of the series X

X

( 1)n nk z n;

n1

( 1)n

n1

n ; nk

1z

X

n1

nk z n (k = 1; 2; : : : )

has 1 as the radius of onvergen e. For k  2, the se ond of the above

onverges absolutely on jz j = 1 sin e, on jz j = 1, X (



1)n z n X 1 = nk n1 nk

n1

onverges for k  2. For k = 1, the series be omes X

n1

( 1)n

1z

n

n

=z

z2 z3 + 2 3




; jz j < 1:

This series is alled the Logarithmi series (see Se tion 3.5). On the other hand, the series X zn (3.69) n1 n at z = 1 be omes 1 + 12 13 + 41


Pand so onverges for z = 1, but non-absolutely. At z = 1, this series is n1 n1 whi h is the harmoni series and we know that this is divergent. What happens on the rest of 
? To see this, we let z = ei ( 6= 0; 2) and ak = z k = eik . Then n X k z

k=1

1

=



z n+1  2 1 z j1 ei j

107

3.3 Power Series as an Analyti Fun tion

showing that the partial sums are bounded provided  6= 0; 2. Hen e, by Theorem 1.52(b) with bn = n 1 , the series (3.69) onverges for jz j = 1 P ex ept at z = 1. Consequently, the logarithmi series n1 ( 1)n 1 z n =n

onverges for jz j = 1 ex ept P at z = 1. (iv) Consider a series n0 qn z kn ; where k  is a xed positive integer and q 6= 0, independent of n. Note that

an =

 n=k q

if n = 0; k; 2k; 3k; : : : 0 otherwise:

Then, a

ording to the Root test, one has lim supn!1 jan j1=n = jqj1=k and so, the power series has the radius of onvergen e R = jqj 1=k . The series

onverges absolutely for jz j < jqj 1=k , uniformly for jz j   < jqj 1=k , and diverges for jz j > jqj 1=k . ForPinstan e, if q = 1 then an = ( 1)n for n  0 and the orresponding series n0 ( 1)n z kn onverges absolutely for jz j < 1, uniformly for jz j   < 1, and diverges for jz j > 1. In this spe ial ase, the terms of the series does not approa h zero when jz j = 1, and therefore, the series diverges for jz j = 1. P If we set q = 1=2 then we get the series n0 2 n z kn whi h onverges absolutely for jz j < 21=k , uniformly for jz j   < 21=k , and diverges for jz j > 21=k . For jz j = 21=k , j2 n z nk j = 1, showing that the series diverges. In parti ular, for k = 2 the power series is X

n0

2

n z 2n ;



= 1; 3; 5; : : : ; with an = 2 n=02 ifif nn = 0; 2; 4; : : :

p

and so R = 2 is the radius of onvergen e for this series. If we set q = 5, then the orresponding series X

n0

5n z kn =

X

n0

(5z k )n

onverges absolutely for jz j < 5 1=k , uniformly for jz j   < 5 1=k , and diverges for jz j > 5 1=k . For jz j = 5 1=k , j5n z nk j = 1, showing that the series diverges. In parti ular, for k = 3 the above dis ussion gives the power series  n=3 X if n = 0; 3; 6; : : : n 3 n 5 z ; with an = 5 0 otherwise ; n0 and so R = 5 1=3 is the radius of onvergen e for this series. P

3.70. Theorem. A power series n0 an z n and the k -times derived P series de ned by nk n(n 1)


(n k +1)an z n k have the same radius of onvergen e.

108

Analyti Fun tions and Power Series

Proof. Let An = n(n 1)


(n k + 1)an = n!an =(n k)!, k  1. Then 1=n n! n! 1=n 1 =n jAn j = (n k)! an = (n k)!
jan j1=n : Using the parti ular ase of Theorem 3.68 (see also Example 3.66) we have 

n!

lim sup n!1 (n k )!

1=n



(n + 1)!=(n + 1 k)! = nlim !1 n!=(n k)! n+1 = nlim !1 n + 1 k = 1:



Therefore, lim sup jAn j1=n = lim sup jan j1=n whi h proves our theorem. n!1 n!1 Sin e a power series of the form (3.62) with non-zero radius of onvergen e R onverges (absolutely) for jz j < P R, we an study its behavior as a fun tion f de ned by the sum f (z ) = n0 an z n: The power series obtained by dierentiating this series term-by-term gives

f 0 (z ) =

X

n1

nan z n 1 ; jz j < R:

Next we have P

3.71. Theorem. If n0 an z n has radius of onvergen e R > 0, then P f (z ) = n0 an z n is analyti in jz j < R, f (k) (z ) exists for every k 2 N and (3.72)

f (k) (z ) = k!ak +

X

n!

an z n k (jz j < R): ( n k )! nk+1

where ak = f (k) (0)=k .

For example, the geometri series (1 z ) 1 = for jz j < 1, after k-times dierentiation yields

P

n0 z

n

whi h onverges

 

X n X (m + k )! 1 n k= = z z m for jz j < 1: (1 z )k+1 nk k k ! m ! m0 P

In parti ular, z (1 z ) 2 = n1 nz n for jz j < 1: P Proof. Let f (z ) = n0 an z n with the radius of onvergen e R. We have to prove the existen e of f 0 (z ) in
R . By Theorem 3.70 with k = 1, the series X nan z n 1 n1

109

3.3 Power Series as an Analyti Fun tion

onverges for jz j < R and de nes a fun tion, say g(z ), in jz j < R. We show that f (z + h) f (z ) = g(z ) for all z 2
R : f 0 (z ) = lim h!0 h Let z 2
R be xed and let h 2 C , 0 < jhj < (R jz j)=2. Then

jz + hj  jz j + jhj < jz j + R 2 jz j = jz j 2+ R < R +2 R = R:

Now we onsider  (z + h)n f (z + h) f (z ) X = an h h n1

zn



;

where z and z + h are su h that maxfjz j; jz + hjg  r < R and jhj is positive. So, we must show that as h ! 0, f (z + h) h

f (z )

g(z )

 X (z + h)n = an h n2

zn

 nz n 1

! 0:

First we note that, for  6= 1 and n  2, we have the identity 1 n = 1 +  + 2 +


+ n 1 1  and dierentiating with respe t to  shows that 1 n nn 1 = (1 )[1 + 2 + 32 +


+ (n 1)n 2 ℄: 1  Repla ing  by z=w (w 6= z ) gives wn z n nz n 1 = (w z )[wn 2 + 2wn 3 z +


+ (n 1)z n 2℄ w z so that for maxfjwj; jz jg  r, n w w

zn z



1

 jw z j [jwjn 2 + 2jwjn 3 jz j +


+ (n 1)jz jn 2℄  jw z jrn 2 [1 + 2 + 3 +


+ n 1℄ n(n 1) n 2 = jw z j r : 2 P By Theorem 3.70, the derived series n2 n(n 1)jan jrn 2 is onvergent for ea h r su h that jz j  r < R. Using this we see that as h ! 0, nz n



 X (z + h)n a n h n2

zn

 nz n 1

 jhj

X

n2

jan j n(n2 1) rn 2 ! 0:

110

Analyti Fun tions and Power Series

Consequently, f 0(z ) exists and equals g(z ). Sin e z is arbitrary, this holds at any interior point in the disk of onvergen e. A repeated appli ation of this argument shows that all the derivatives f 0 , f 00 , : : : , f (k) ; : : : exist in jz j < R and (3.72) holds. Putting z = 0 in (3.72) we have the formula for the oeÆ ient ak : f (k) (0)=k = ak and we are done. P

Suppose z0 6= 0 and f (z ) = n0 an (z z0 )n , jz z0 j < R: Then

onsider a simple transformation w = z z0 so that z = z0 + w. Then by Theorem 3.71 we see that the fun tion  de ned by

(w) = f (w + z0 ) = and the k-times derived series

(k) (w) = k!ak +

X

n0

an wn

n! an wn k ; ( n k )! nk+1 X

have the same radius of onvergen e and note that d(w + z0 ) (k) (w) = f (k) (w + z0 ) = f (k) (z ); jwj < R: dz From this P it follows that if R is the radius of onvergen e of the series f (z ) = n0 an (z z0 )n , then X n! f (k) (z ) = k!ak + an (z z0 )n k ; jz z0 j < R; ( n k )! nk+1 ( k ) so that ak = f (z0 )=k!: Note that a0 ; a1 ; : : : ; ak ; : : : depend on z0 . Thus, f be omes X f (n) (z0 ) f (z ) = (z z0)n ; jz z0 j < R: n ! n0 This is often alled the Taylor series expansion for f in jz z0j < R. In the spe ial ase when z0 = 0, it is alled the Ma laurin series expansion. One of the remarkable results in Complex Analysis is that the onverse of Theorem 3.71 holds. If f is analyti in a domain D, then f an be represented (lo ally about ea h point z0 2 D) as a Taylor series expansion about z0: X f (z ) = an (z z0 )n ; n0 where z0 is a enter of the largest disk
(z0 ; R)  D and z 2
(z0 ; R). For instan e if f (z ) = z 1 , then f is analyti in the pun tured plane C n f0g. If z0 6= 0, then we have the Taylor expansion 1 X (z z0)n 1 = ( 1)n for jz z0 j < jz0 j: f (z ) = z z0 + z0 z0 n0 z0n

111

3.3 Power Series as an Analyti Fun tion

However, we shall prove a more general result in Chapter 4 (see Corollary 4.95) whi h states that for a given analyti fun tion de ned in a domain D and for ea h z0 su h that
(z0 ; r)  D there is always a power series

onverging in
(z0 ; r) whose sum is f (z ).

3.73. Corollary. If f is entire and if z0 2 C , then f (n) (z0 ) exists for n = 0; 1; : : : and has the power series expansion f (z ) =

f (n) (z0 ) (z z0 )n for all z: n ! n0 X

3.74. Remark. Now it is lear that entire fun tions P are just those fun tions whi h are de ned by the sums of the series n0 an z n where lim supn!1 jan j1=n = 0.  In the next theorem ( alled the Uniqueness Theorem for power series), we show that a power series representing a given fun tion, when obtained by whatever method, is unique. P

P

3.75. Theorem. Suppose f (z ) = n0 an z n and g(z ) = n0 bn z n

onverge for jz j < R1 and jz j < R2 , respe tively. If f (zk ) = g (zk ) for a sequen e fzk g of nonzero omplex numbers in 0 < jz j < Æ su h that zk ! 0 as k ! 1, or if f (z ) = g(z ) for all z with jz j < Æ, where 0 < Æ < minfR1 ; R2 g, then an = bn for every n 2 N 0 . Proof. Sin e f; g are ontinuous in jz j  Æ, being analyti thereon, zk ! 0 and f (zk ) = g(zk ) =) lim f (zk ) = lim g(zk ) k!1 k!1 =) f (0) = g(0): Thus, a0 = b0 . To omplete the proof, we use the method of indu tion. Now suppose that aj = bj for j = 0; 1; : : : ; m 1: Then

f (zk ) = g(zk ) =) =) =) where

F (z ) =

X

nm X

an zkn = an zkn m

X

nm

=

bn zkn

X

nm nm F (zk ) = G(zk );

X

nm

bn zkn m; sin e zk 6= 0;

an z n m and G(z ) =

X

nm

bn z n m :

112

Analyti Fun tions and Power Series P

P

Sin e the radius of onvergen e of n0 an z n and nm an z n m are the same, F is ontinuous for jz j  Æ. Similarly, G is ontinuous for jz j  Æ. Thus (see Theorem 2.22), F (zk ) = G(zk ) =) lim F (zk ) = lim G(zk ) =) am = bm k!1 k!1 and we on lude that an = bn for every n  0. Note that if f (z ) = g(z ) for jz j < Æ, we an nd a sequen e fzk gk1 2 fz : jz j < Æg su h that zk ! 0 as k ! 1 (see Theorem 2.22). The proof is omplete. P

3.76. Corollary. Suppose f (z ) = n0 an z n, with the power series having radius of onvergen e R > 0. If 0 is a limit point of the set of zeros of f , then f  0 for jz j < R. It will be shown, more generally later (Theorem 4.103) that if f; g are two analyti fun tions whi h agree at a sequen e of points fzk g having a limit point in their ommon domain of analyti ity, then f  g therein.

3.4 Exponential and Trigonometri Fun tions In this se tion we des ribe the omplex analogues of exponential and trigonometri fun tions of elementary al ulus. There are many approa hes whi h lead to de nitions of these fun tions although our approa h is intuitive and dire t. Let us rst re all the following fa ts from al ulus: (a) Trigonometri fun tions are de ned by means of the ratios of the sides of a right triangle; for instan e, sin2 x + os2 x = 1, x 2 R. d d d x (b) (sin x) = os x; ( os x) = sin x; (e ) = ex; dx dx dx ( ) Further, x2 x5 sin x = x +


; 3! 5! x2 x4

os x = 1 +


; 2! 4! x2 x3 ex = 1 + x + + +


: 2! 3! The properties (b) and ( ) show that sin x and os x are the unique solution of the se ond order ordinary dierential equation f 00 (x) + f (x) = 0 subje t to the onditions f (0) = f 0 (0) 1 = 0 and f (0) = f 0(0) 1 = 1, respe tively. On the other hand ex is de ned to be the solution of f 0 (x) = f (x); f (0) = 1: P In the same way we begin with a fun tion f (z ) = n0 an z n satisfying

113

3.4 Exponential and Trigonometri Fun tions

(i) f 0(z ) = f (z ) for z 2 C , i.e. f is analyti in C (ii) f (x) = ex; x 2 R. Using (i) we nd that an 1 = nan : Sin e f (0) = e0 = 1 = a0 , indu tion on n yields an = 1=n!: Thus we de ne the omplex exponential fun tion as

f (z ) = exp(z ) = ez =

(3.77)

X zn

n0

n!

:

Conversely, if the omplex exponential fun tion is given by (3.77) then (i)(ii) follow easily. By the Ratio test, the series (3.77) onverges absolutely for all z 2 C and hen e, ez is an entire fun tion. In the same way one an obtain the series representations of sin z and os z . As (exp z )0 = exp z , we have the following: if g 2 H(D), then so does F (z ) = exp(g(z )) and F 0 (z ) = g0 (z ) exp(g(z )) for z 2 D. Next, for a xed  2 C , we de ne

g (z ) = ez e z (= f (z )f ( z )): Then for every z 2 C , we have g0 (z ) = 0. Therefore, by Theorem 3.31, g is onstant. Sin e g (0) = e0 e = e , we must have ez e z = e : Taking z = z1 and  = z1 + z2 , we see that for every z1 ; z2 2 C ,

ez1 +z2 = ez1 ez2 :

(3.78)

The result (3.78) is alled the \Addition theorem" for the exponential fun tion whi h is also a onsequen e of the Uniqueness theorem (Se tion 4.11). This property an also be obtained dire tly using the Cau hy produ t of series:

ez1 ez2

" k # X X z2m z1p z2k p = = n0 n! m0 m! k0 p=0 p! (k p)! " # k   X 1 X k p k p = z1 z2 k0 k ! p=0 p X (z1 + z2 )k = = ez1 +z2 : k ! k 0 X zn X

1

For any positive integer m, by indu tion, (3.78) with z1 = z2 = z shows that (3.79) emz = (ez )m :

114

Analyti Fun tions and Power Series

For z = iy, y 2 R, (3.77) gives

eiy =

X in y n

n0

n!

=

X

k0

( 1)k

X ( 1)k y 2k+1 y2k +i : (2k)! k0 (2k + 1)!

Thus, we have the Euler formula

eiy = os y + i sin y; y 2 R:

(3.80)

Therefore, for z1 = x and z2 = iy, x; y 2 R, (3.78) shows that (3.81)

ex+iy = ex ( os y + i sin y); ez = exeiy :

Sin e ez e z = ez z = e0 = 1; ez 6= 0 for ea h z 2 C . Further,

jez j2 = ez (ez ) = ez ez = ez+z = e2Re z so that jez j2 = (eRe z )2 ; by (3.79). Thus, from (3.81) and the above, we write (3.82) jez j = eRe z ; arg(ez ) = Im z (mod 2): In parti ular, we have jez j = 1 () z 2 iR: For m; n integers and k = 0; 1; : : : ; n 1, we have (ez )1=n = (ex eiy )1=n = ex=n [ei(y+2k)=n ℄ = e(z+2ki)=n and so we write (3.83) (ez )m=n = em(z+2ki)=n : From (3.80) we see in parti ular that, if z is real then

eiz + e iz eiz e iz and sin z = : 2 2i Thus the extension of the exponential to the omplex plane suggests to make (3.84) as de nitions of os z and sin z for any z 2 C although these de nitions an be realized as a onsequen e of the Uniqueness theorem. Hen e we make the de nition{the famous Euler formula: (3.84)

os z =

eiz = os z + i sin z; z 2 C : Now onsider f (z ) = sin z for z 2 C : Then, by the de nition of sin z and the fa t that (exp z )0 = exp z , (3.84) gives

ieiz + ie iz eiz + e iz = = os z 2i 2 and so using the de nition of os z , f 0 (z ) =

f 00 (z ) =

ieiz

2

ie iz

=

eiz

e iz = sin z: 2i

115

3.4 Exponential and Trigonometri Fun tions

Consequently, we have



0 if n = 2k ( 1)k if n = 2k + 1 ; k 2 N 0 : Similarly, if f (z ) = os z , we obtain f 0(z ) = sin z and f 00 (z ) = that  2k + 1 ; k 2 N : ( n ) f (0) = ( 1)0k ifif nn = 0 = 2k

f (n) (0) =

os z so

Now we use the Ma laurin series expansion (see just above Corollary 3.73), (n) P namely f (z ) = n0 f n!(0) z n , to get the series expansions for os z and sin z . Therefore, for any z 2 C , the osine and sine series are now

os z =

(3.85) and (3.86)

sin z =

X

( 1)n

n0

X

( 1)n

n0

z 2n (2n)!

z 2n+1 ; (2n + 1)!

respe tively. Also, note that (3.85) and (3.86) follow from the de nition of eiz and (3.84). Ea h of the series onverges absolutely for every z 2 C , be ause X jz j2n X jz j2n+1 and n0 (2n)! n0 (2n + 1)! are (i.e. terms hosen from those of) of the onvergent series P the subseries n =n!℄ for z 2 C . The remaining ir ular fun tions are then de ned [ j z j n0 by  sin z 1   ; tan z = ; se z = z 6= (2k+1) 2

os z

os z

os z 1

ot z = ; s z = (z 6= k); sin z sin z where k 2 Z. These fun tions are analyti ex ept at points where the denominators vanish. The hyperboli fun tions are de ned by

os(iz ) = osh z and sin(iz ) = i sinh z so that

ez + e z ez e z ; sinh z = ; 2 2   sinh z 1 (2k + 1)i ; se h z = z 6= ; tanh z =

osh z

osh z 2

osh z 1

oth z = ;

s h z = (z 6= ki); sinh z sinh z

osh z =

116

Analyti Fun tions and Power Series

where k 2 Z. Again these fun tions are analyti ex ept at points where the denominators vanish. For z1 ; z2 2 C , the identity (3.87) ei(z1 z2 ) = eiz1 eiz2 yields the addition formulae (whi h will be proved later also by using the Uniqueness theorem) (3.88)

os(z1  z2 ) = os z1 os z2  sin z1 sin z2

and (3.89) sin(z1  z2 ) = sin z1 os z2  os z1 sin z2 : Putting z1 = z2 = z in (3.88) and (3.89), we get

os2 z + sin2 z = 1; os 2z = os2 z

sin2 z ; sin 2z = 2 sin z os z:

Re all the known fa ts about the sine and osine fun tions of a real variable indi ated by the following hart: Interval

os sin 1& 1 [0; =2℄ 0 0% 0& 1& [=2; ℄ 1 0 0 0 [; 3=2℄ & 1 1% 1 0 [3=2; 2℄ 0% 1% as also the properties os 2k = 1; sin 2k = 0, k 2 N 0 : Therefore, we have from (3.88) and (3.89) (3.90)

os(z + 2k) = os z and sin(z + 2k) = sin z

where k 2 Z. The above fa ts together with the equation in (3.84) immediately yield the following ne essary and suÆ ient onditions for ertain identities to be ome true: (a) os z = 1 () z = 2k, (b) sin z = 1 () z = (4k + 1)=2, ( ) tan z = 1 () z = (4k + 1)=4, (d) os z1 = os z2 () either z1 + z2 = 2k or z1 z2 = 2k, (e) sin z1 = sin z2 () either z1 + z2 = (2k + 1) or z1 z2 = 2k, where k 2 Z. Note that (d) () (e). This fa t an be easily seen by taking z1 = 1 =2 and z2 = 2 =2.

3.91. De nition. A fun tion f : D ! C is said to be periodi if there exists an ! 6= 0 su h that f (z + !) = f (z ) for all z 2 D. The

3.4 Exponential and Trigonometri Fun tions

117

omplex number ! is then alled a period of f . The fun tion f is alled doubly periodi in D if there are omplex numbers !1 ; !2 whi h are linearly independent over R su h that f (z + !1 ) = f (z + !2 ) = f (z ) for all z 2 D: Sin e e2ki = 1, we nd that ez+2ki = ez e2ki = ez : Suppose ! = u + iv is a period for ez . Then by de nition ez+! = ez ; i.e. e! = 1, and je! j = eu = 1 and so u = 0. Hen e

eiv = 1; i.e. os v = 1 and sin v = 0. By (3.90), this o

urs only for v = 2k, where k is some integer. This shows that ! is a period of ez i ! = 2ki. In parti ular, ez is a periodi fun tion with period 2i. Similarly, we have sin(z + !) = sin z; or os(z + !) = os z () ! = 2k; k 2 Z:

3.92. Example. Consider the fun tion f (z ) = ez : Then we have f (z1 ) = f (z2 ) =) ez1 z2 = 1 =) z1 z2 = 2ki for some k 2 Z: This shows that f is one-to-one in a domain D i D does not have even a single pair of distin t points z1 and z2 satisfying

z1 = z2 + 2ki; k = 1; 2; : : : : For instan e, f is univalent in jz j < ; but not in C . Similarly, we see that f is univalent within any horizontal strip of the form:

b < Im z < a; 0 < a b  2: In fa t we see that f is univalent in ea h strip

Dk = fz : 2(k 1) < Im z < 2kg; k 2 Z; and maps ea h strip onto the w-plane with a ut along (i.e. omitting) the positive real axis: 0  x < 1. 

3.93. Example. Note that for z = x + iy, we have 

x ib b, a onstant, ez = eea eeiy ifif xy = = a, a onstant.

As a further illustration of the exponential fun tion, we have (i) The exponential fun tion ez maps ea h horizontal line y = b onto a ray from the origin to in nity, namely, freib : r 2 (0; 1)g.

118

Analyti Fun tions and Power Series

(ii) The exponential fun tion ez maps ea h verti al line x = a onto a

ir le entered at the origin and radius ea. Observe that ea h point on this ir le is the image of in nitely many points on the verti al lines as eiy is periodi . Note that these two families of urves (and also their images) are orthogonal to ea h other. (iii) The line y = b and the line x = a interse t at (a; b). How about their images in the w-plane? (iv) The re tangle D = fz : a1 < Re z < a2 ; b1 < Im z < b2 g is mapped into D0 = fw : ea1 < jwj < ea2 ; b1 < arg w < b2g. In parti ular, exp z maps the in nite strip fz : 0 < Im z < =2g onto the open rst quadrant fw : Re w > 0; Im w > 0g. 

3.5 Logarithmi Fun tions It is natural to think of a logarithmi fun tion as the inverse of the exponential fun tion. Re all that, when x is real, (a) ex > 0 for all x 2 R (b) ex ! 1 as x ! 1 ( ) e x = 1=ex whi h by (a) gives ex ! 0 as x ! 1 d (d) (ex ) = ex and so ex is monotoni ally in reasing for all x. dx This shows that ex de nes a stri tly in reasing dierentiable fun tion from R onto R+ , the set of positive reals. Hen e it has a ontinuous stri tly in reasing inverse fun tion alled the natural logarithm (with base e) ln :

R+ ! R

with the property that ln x = y is the solution of ey = x. In parti ular, for ea h x > 0 there is exa tly one y su h that ey = x. We would like to mimi the real variable ase and say that a omplex number w in C is alled a logarithm of a omplex number z in C , denoted by w = log z , if ew = z holds. Sin e ew 6= 0 for any w 2 C , the number 0 has no logarithm; hen e, the equation ew = z has no solution in C when z = 0 and z = 1. Consider an arbitrary xed z 6= 0 in polar form

z = jz jeiArg z = rei (r = jz j > 0;

 <   ):

Let us now solve the equation w = log z . If we let w = u + iv, u; v real, then ew = z be omes eu+iv = rei whi h yields

eu = r; e(v )i = 1; i.e. u = ln r; v =  + 2k; k 2 Z: A

ordingly, we therefore have for z 6= 0, (3.94)

w = log z = ln jz j + i(Arg z + 2k); k 2 Z:

119

3.5 Logarithmi Fun tions

The prin ipal value of a logarithm, sometimes denoted by Log , orresponds to the value of the prin ipal value of the argument; that is for z 6= 0, (3.95)

Log z = ln jz j + iArg z;

 < Arg z  :

This agrees with the meaning of ` Log z ' for positive real numbers z = x, for whi h Arg z = 0, with whi h the reader is already familiar. We use the notation  log z to denote the in nite set of values

fln jz j + i(Arg z + 2k) : k 2 Zg: It is often onvenient to single out a parti ular member in this set. Thus, we formulate the following

3.96. De nition. If w z 6= 0, then w de ned by (3.97)

w =  log z =

2 C satis es ew = z for an arbitrary xed 8 > > ln > > < > > > > :

jz j + i(Arg z + 2k) (or) Log z + 2ki (or) ln jz j + i arg z;

is alled the logarithm of z , where k 2 Z. Note that z 7! Log z is a fun tion de ned in C n f0g and every non-zero

omplex number has in nitely many logarithms whi h dier from ea h other by integral multiples of 2i. This shows that w = log z is not a fun tion in general and is in fa t an in nitely-valued asso iation or relation, with in nitely many values for ea h z 6= 0. For ea h k, the strip of the type

Dk = fu + iv : (2k 1) < v  (2k + 1); k 2 Zg is alled a fundamental region for log z (Note that a region in the z -plane whose image just overs the w-plane on e is alled a fundamental region for the fun tion w = log z ). Sin e the union is the w-plane, the set fDk g1 k= 1 is a omplete set of fundamental regions for the logarithm. In parti ular, this observation proves that for ea h k 2 Z, exp(Dk ) = C n f0g: It is also easy to see that for ea h k 2 Z, exp : Dk ! C n f0g is one-to-one and onto. We have (a) (b) ( ) (d) (e)

p

Log (i) = i=2, Log (1 + i) = ln 2 + i=4 p Log ((1  i)= 2) = i=4, ln( 1) = i p Log (2 3i) = ln 13 iAr tan (3=2) p Log ( 2 + 3i) = ln 13 + i( Ar tan (3=2)) Log i1=4 = i=8, Log (z ) = ln  + Log z ( > 0).

120

Analyti Fun tions and Power Series

From these examples, we an see that it is not even true in general that Log (z1 z2 ) = Log z1 + Log z2 when we limit our attention to prin ipal values. However, for z1 = 6 0 and z2 6= 0, the statements log(z1 z2 ) = log z1 + log z2 (mod 2); log(z1 =z2) = log z1 log z2 (mod 2) hold. Suppose that z1 = jz1 jeiArg z1 and z2 = jz2 jeiArg z2 : Then, for the

omplex numbers z1 ; z2 , we write

z1 z2 = jz1 z2 jeiArg (z1 z2 ) : There is a k1 2 f 1; 0; 1g (see Exer ise 1.54(a)) su h that Log (z1 z2 ) = ln jz1 z2 j + iArg (z1 z2) = ln jz1 j + ln jz2 j + i(Arg z1 + Arg z2 + 2k1 ) = Log z1 + Log z2 + 2ik1 : Thus, Log (z1 z2 ) = Log z1 + Log z2 parti ular, this statement leads to

() Arg z1 + Arg z2 2 ( ; ℄: In

Log (z1 z2 ) = Log z1 + Log z2 for all z1 ; z2 2 C with Re z1 > 0 and Re z2 > 0. A serious trouble ensues if the onditions Re z1 > 0 and Re z2 > 0 are dropped in the last equation. This point an also be he ked on taking z1 = i and z2 = i. Similarly, we see that the ondition Arg z1 + Arg z2 2 ( ; ℄ is met whenever Im z1 < 0 and Im z2 > 0. This means that Log (z1 z2 ) = Log z1 + Log z2 for all z1 ; z2 with Im z1 < 0 < Im z2. In fa t, by Exer ise 1.54(b), we similarly have Log (z1 =z2 ) = Log z1

Log z2 + 2ik2 ;

where k2 = k2 (z1 ; z2 ) is de ned in Exer ise 1.54(b). For instan e, let z1 = i and z2 = 1: Then Log z1 = i=2; Log z2 = i and Log (z1 z2 ) = i=2: Therefore, Log (z1 z2 ) = Log z1 + Log z2 larly, we have

2i sin e k1 (i; 1) = 1. Simi-

(a) Log 1 = Log ( 1) + Log ( 1) + 2i; sin e k1 ( 1; 1) = 1.

121

3.5 Logarithmi Fun tions y z = x0 + iy

Lim Arg z = π

z→x0

z0

O

Lim Arg z = −π

x

z→x0

z′ = x0 + iy ′

Figure 3.3: Des ription for ontinuity of Arg z at z0 = x0 < 0.

(b) Log ( 1 i) = Log i + Log ( 1 + i) 2i, sin e k1 (i; 1 + i) = 1. ( ) Log (r1 r2 ei3=2 ) = Log (r1 ei3=4 ) + Log (r2 ei3=4 ) 2i. It is often onvenient to single out a parti ular set in the set  log z . One immediate possibility is to onsider the prin ipal value of the logarithm Log z and to study analyti ity of the fun tion Log z whose domain of de nition is C n f0g. For z 6= 0, let

z = rei and u + iv = Log z; where jz j = r and  <  = Arg z  . Re all that Log z = ln jz j + iArg z:

If Log z = u(x; y) + iv(x; y), u; v real, then (3.98)

u(r; ) = ln r and v(r; ) = :

The fun tion Log z is not ontinuous at points on the negative real axis, sin e Arg z fails to possess a limit at the points along this axis; for, let z0 = x0 < 0. Then, for z = x0 + iy with y > 0, we have lim Arg z = ylim !0 Arg (x0 + iy ) = 

z!z0

y>0

and for z = x0 + iy with y < 0, we have lim Arg z = lim Arg (x0 + iy) = : y!0

z!z0

y 0;  2 < arg < g:

Thus u and v satisfy the C-R equations (3.47). Moreover, the partial derivatives ur ; vr et ., are all ontinuous in D. This implies the existen e of the derivative of Log z in D and so Log z is analyti therein. Sin e

w dw = e i ; dz r we readily have

d  e i 1 1 ( Log z ) = e i (ln r + i) = = i = ; z 2 D: dz r r re z In the same way, we see that arg z with z = rei (r > 0,  2 <  < ), is ontinuous in the `slit plane' D = C n fRei : R  0g where  2 R is xed (see Figure 3.4). Thus, in D , the fun tion f := log de ned by (3.99)

f (z ) := log z = ln jz j + i arg z;

where arg z denotes the unique hoi e of arg z in the interval ( 2; ), is ontinuous with derivative d 1 (f (z )) = ; z 2 D : dz  z Clearly, f is not ontinuous at points on the ray frei : 0  r < 1g. Evidently, exp(f (z )) = z . For the analyti property of multiple-valued fun tions, we regard a multiple-valued fun tion as a olle tion of singlevalued fun tions. More expli itly we have

3.100. De nition. Suppose F is a multiple-valued fun tion de ned in S . A bran h of F is a single-valued analyti fun tion f in some domain D  S obtained from F in su h a way that at ea h point of D, f assumes exa tly one of the possible values of F .

123

3.5 Logarithmi Fun tions

For example, from our dis ussion above, for ea h xed  2 R, the fun tion f de ned by (3.99) represents a bran h of log z in the slit plane D . We have reated a ut along  = , so that the restri tion on  = arg z makes the fun tion f single-valued and analyti . This ut orresponding to  = , i.e. the semi in nite line fRei : R  0g,  2 R xed, is regarded as a bran h ut for the bran h f of log z . For  = , f be omes Log z in D and is often alled the prin ipal bran h of log z . The bran h ut for Log z is then the negative-real axis from the origin, i.e. f R : R  0g.

3.101. De nition. A multiple-valued fun tion F de ned in S is said to have a bran h point at z0 2 C if, when z des ribes an arbitrarily small

ir le about z0 , then for every bran h f of F , f (z ) does not return to its original value. For example every z 6= 0 is not a bran h point of log z , sin e on a suÆ iently small ir le en losing the point z , every bran h of log z returns to its original value.

3.102. De nition. A fun tion f 2 H(D), where D is a domain D, is alled an analyti /holomorphi bran h of log z in D if f is analyti /holomorphi in D and exp(f (z )) = z for ea h z 2 D. Let  2 H(D) su h that 0 2= (D). We say f 2 H(D) is an analyti logarithm of  if exp(f (z )) = (z ) for ea h z 2 D. Clearly, if (z ) = z in D then the on ept of analyti -logarithm oin ides with that of the (analyti ) bran h of log z . Analyti /holomorphi bran hes of log (z ) are

alled logarithmi fun tions. For instan e, in the slit plane D , the fun tion f de ned by (3.99) is an example of a logarithmi fun tion.

3.103. Theorem. Let D be a domain in C . Then, any two logarithmi fun tions f , fe : D ! C are related by (3.104)

fe(z ) = f (z ) + 2ki

for some k 2 Z. Conversely, if a logarithmi fun tion f : D ! C is related by (3.104) with fe : D ! C , then fe is also a logarithmi fun tion.

Proof. Let f and fe be two logarithmi fun tions in D. Then by de nition expffe(z )g = expff (z )g; i.e. expffe(z ) f (z )g = 1 for all z 2 D; whi h implies that

F (z ) =

fe(z ) f (z ) 2i

2 Z:

124

Analyti Fun tions and Power Series

Note that F is an analyti fun tion in D and, in parti ular, F is ontinuous in the open and onne ted set D. But F is an integer-valued and so F maps a onne ted set onto a onne ted subset of Z, viz. a single point, k, say. Thus, f and fe are related by (3.104). Conversely, if f is a logarithmi fun tion in D and fe : D ! C is related by (3.104) for some k 2 Z then for all z 2 D, we have expffe(z )g = expff (z )g expf2ikg = expff (z )g = z and so fe be omes a logarithmi fun tion in D, by de nition. Theorem 3.103 shows that, given a bran h of logarithm in a domain, one an obtain all the bran hes of logarithm on that domain. For example, we let  = 3=2 and onsider (3.99). Then f3=2 is an analyti bran h of log z for z 2 D3=2 . Thus to nd a numeri al value of f3=2 (z0 ), where z0 = 1 i, we write

p

p

5i : 4 From Theorem 3.103, we also note that the value of f3=2 ( 1 i) for any other bran h of log z whi h is analyti in the slit plane D3=2 is given by

f3=2( 1 i) = ln 2 + i arg3=2 ( 1 i) = ln 2 +

p

5i + 2ki; for some k 2 Z: 4 As another hara terization of logarithmi fun tions, we have ln 2 +

3.105. Theorem. Let f : D ! C be analyti in a domain D not

ontaining 0. Then f is a bran h of log z in D i f 0 (z ) = 1=z for all z 2 D and expff (a)g = a for at least one a 2 D. Proof. Suppose that f is a bran h of log z in D. Then, expff (z )g = z for all z 2 D so that f 0 (z ) expff (z )g = 1; i.e. f 0 (z ) = 1=z for all z 2 D: To prove the onverse, we need to show that (3.106)

expff (z )g = z for all z 2 D:

To do this, we onsider g(z ) = z expf f (z )g: Then g is analyti in D and satis es g0(z ) = [1 zf 0(z )℄ expf f (z )g = 0 for all z 2 D: By Theorem 3.31, g is a onstant, say k. Thus, z expf f (z )g = k for all z 2 D: Sin e expff (a)g = a, we must have k = 1.

125

3.5 Logarithmi Fun tions

3.107. Example. We now demonstrate that there an be no bran h of logarithm in the domain = C n f0g. This fa t an be proved by a number of ways. Suppose on the ontrary that f (z ) is a bran h of log z in . The restri tion of f to the slit plane D = C n ( 1; 0℄ is then a bran h of logarithm in D . As Log z is the prin ipal bran h in D , Theorem 3.103 shows that for z 2 D , f must be of the form f (z ) = Log z + 2ki for some k 2 Z. But then, f being analyti in , f is ontinuous at x0 < 0 so that

f (x0 ) = lim f (x0 + iy) = ylim !0 Log (x0 + iy ) + 2ki = (2k + 1)i y !0 y>0

y>0

and

f (x0 + iy) = ylim f (x0 ) = lim !0 Log (x0 + iy ) + 2ki = (2k 1)i y !0 y 0,

1 : 1 (t) = rt (0  t  1) and 2 : 2 (t) = reit (0  t  3=2); then, with the notation = 1 + 2 (see Figures 4.2 and 4.1), we write

1 + 2 : (t) =



rei(t

rt for 0  t  1 1) for 1  t  3(=2) + 1:

A urve : [a; b℄ ! C an be written, sin e we an hoose the parametri interval onveniently, equivalently in the form

e : [0; 1℄ ! C ; i.e. e(t) = (a + (b a)t); t 2 [0; 1℄:

140

Complex Integration γ2 (t) γ2

r = γ1 (1) = γ2 (0) γ2

0

t

3π 2

3π 2




0

1

Figure 4.2: Sum of two urves 1 , 2 .

A polygon with verti es z0 ; z1 ; z3 ; : : : ; zn is parameterized by

k (t) = (k + 1 t)zk + (t k)zk+1 ; t 2 [k; k + 1℄; where 0  k  n 1 and = 0 + 1 +


+ n 1 . A urve de ned on [a; b℄ is alled simple if it does not interse t itself, that is, if (t1 ) 6= (t2 ) for t1 6= t2 , where the possible ex eption (a) = (b) is allowed. In the latter ase the urve is said to be a simple losed urve. For example, ir les, ellipses, re tangles, and triangles are simple losed

urves. For instan e, urve de ned by (t) = os t (   t  ) is the segment [ 1; 1℄ traversed twi e from 1 to 1 and then from 1 to 1, and therefore onsidered as a losed urve. But ea h point of [ 1; 1℄ is a self interse tion point for the urve so that it is not simple. Note that a simple

losed urve is a Jordan urve. Now we shall deal with ontinuously dierentiable urves.

4.1. De nition. A urve : [a; b℄ ! C is said to be ontinuously dierentiable on [a; b℄ or a urve of lass C 1 on [a; b℄ or simply a C 1 - urve on [a; b℄ if the fun tion (t) = x(t) + iy(t) is ontinuously dierentiable on [a; b℄, i.e. x0 (t) and y0 (t) exist on [a; b℄ and are ontinuous fun tions on [a; b℄ (Note that (t) is dierentiable on [a; b℄ means that 0 (t) exists on (a; b), and

(a + h) (a)

(b + h) (b) lim ; lim h h h!0+ h!0 0 both exist. We denote these limits by (a+) (or +0 (a)) and 0 (b ) (or

0 (b)), respe tively. We all 0 (a+) and 0 (b ) as the right-hand derivative at a and the left-hand derivative at b, respe tively). A ontinuously dierentiable urve is referred to as a smooth urve.9 For example, (t) = t3 , t 2 [ 1; 1℄, is a Jordan ar of lass C 1 , sin e 1  s < t  1 implies (s) < (t) 9 While

de ning "smooth urve" some authors insist an additional ondition that

0 (t) = 6 0 on [a; b℄. So the reader is advised to be aware of this in onsisten y while referring other texts in omplex variables.

141

4.1 Curves in the Complex Plane γ (b) γ (t3 ) γ (t) γ (t2 ) γ (t1 ) a

t1

t2

t3

γ (a)

b

Figure 4.3: Des ription for pie ewise C 1 urve.

and (t) is a dierentiable fun tion of t on [ 1; 1℄. Similarly, (t) = t3 + it2, t 2 [ 1; 1℄, is also a Jordan ar of lass C 1 , sin e (t) is a ontinuously dierentiable fun tion of t on [ 1; 1℄ and 1  s < t  1 implies j (s)j =

p

p

s6 + s4 < t6 + t4 = j (t)j:

Further, the line segment [z1 ; z2 ℄ from z1 to z2 parameterized by

(t) = (1 t)z1 + tz2; t 2 [0; 1℄; is ontinuously dierentiable. Similarly, the ir ular ar parameterized by

(t) = z0 + reit ; t 2 [a; b℄  [0; 2℄; is ontinuously dierentiable. The ase a = 0 and b = 2 yields a ir le with enter z0 and radius r. A urve (t); a  t  b, is alled pie ewise C 1 (or pie ewise smooth

urve) if there is a subdivision a = t0 < t1 <


< tj


< tn = b of the interval [a; b℄ su h that the restri tion of to ea h subinterval [tj ; tj+1 ℄, 0  j  n 1 is a smooth urve (see Figures 4.3). A ontour is just a

ontinuous urve that is pie ewise smooth. Given a domain D in C and two points z1 and z2 in D (need not be distin t), there exists a ontour in D with initial point z1 and terminal point z2 . This fa t is lear be ause any two points in D an be onne ted by a polygonal path in D. Consider (t) = t + ijtj; t 2 [ 1; 1℄: Then (see Figure 4.4)

(t) =



t it if t 2 [ 1; 0℄ t + it if t 2 [0; 1℄:

It is easy to see that the restri tions of to [ 1; 0℄ and to [0; 1℄ are smooth, even though is not smooth be ause 0 (t) fails to exist at t = 0. Note that

0 (t) is dis ontinuous at 0 but is pie ewise ontinuously dierentiable, sin e 0 (t) = 1 i on [ 1; 0) and 0 (t) = 1 + i on (0; 1℄. A

ordingly, is a pie ewise smooth urve.

142

Complex Integration y(t) z(1) = 1 + i

z(−1) = −1 + i

1

−1

−1

O

1

x(t)

Figure 4.4: The urve z (t) = t + ijtj; t 2 [ 1; 1℄.

4.2 Properties of Complex Line Integrals We assume the following fa ts from real analysis: if a Rreal-valued fun tion F is ontinuous on [a; b℄, then the Riemann integral ab F (t) dt exists. It is a trivial exer ise to extend this de nition for a ontinuous fun tion F : [a; b℄ ! C , where F = U + iV . Indeed, Z b

a

F (t) dt =

Z b

a

U (t) dt + i

Z b

a

V (t) dt:

A omplex-valued fun tion f is said to be ontinuous on a ontinuously dierentiable urve : [a; b℄ ! C (or more generally on a ontour) if (t) = f (z ) = f ( (t)) = u(t) + iv(t) is ontinuous for a  t  b. Suppose f is a omplex-valued fun tion that is ontinuous on an open set D  C and that : [a; b℄ ! C is a ontour with ([a; b℄)  D. We de ne the omplex line integral or ontour integral of f along the ontour R

, denoted by f (z ) dz , as follows: (4.2)

Z

f (z ) dz =

Z b

a

n 1 Z tj+1

X f ( (t)) 0 (t) dt =

j =0 tj

f ( (t)) 0 (t) dt;

where a = t0 < t1 <


< tn 1 < tn = b, and [tj ; tj+1 ℄, j = 0; 1; : : : ; n 1, being the intervals in whi h is dierentiable, and the integrals in the sum are Riemann integrals. The ontour is alled the path of integration of the ontour integral. Note that the produ t given by F (t) = f ( (t)) 0 (t) is pie ewise ontinuous on [a; b℄. So the se ond integral in (4.2) is well-de ned. For example, if (t) = a+reit is a ir le, then for an arbitrary ontinuous fun tion f de ned on : jz aj = r, Z

jz aj=r

f (z ) dz =

Z 2

0

f (a + reit )ireit dt:

In the most important spe ial ase, namely, f (z ) = (z have Z Z 2 ireit f (z ) dz = it dt = 2i: jz aj=r 0 re

a) 1 , we easily

143

4.2 Properties of Complex Line Integrals

Note that the expression under the integral sign on the right of (4.2) an be obtained by the formal substitution z = (t); dz = 0 (t) dt: If f = u + iv and z = x + iy, x; y; u; v being real-valued, i.e.

f ( (t)) = f (x(t) + iy(t)) = u(x(t); y(t)) + iv(x(t); y(t)); then (4.2) is really Z

f (z ) dz =

nX1 (Z tj+1 tj

j =0

+i

[u(x(t); y(t))x0 (t) v(x(t); y(t))y0 (t)℄ dt

Z tj+1

tj

)

[v(x(t); y(t))x0 (t) + u(x(t); y(t))y0 (t)℄ dt :

Using a hange of variable in the de nition of the Riemann integral, one has Z

f (z ) dz =

Z

Z

[u(x; y) dx v(x; y) dy℄ + i [u(x; y) dy + v(x; y) dx℄;

sin e 0 (t) = x0 (t) + iy0(t). In the above equality it is ta itly assumed that in the integrands (x(t); y(t)) being on the urve , that either y is a fun tion of x or x is a fun tion of y. Also note that the limits of integration will depend on the values of x(a); x(b) or y(a); y(b) as the ase may be. It should be observed that the expressions under the integral signs an be formally equated as

f (z ) dz = (u + iv)(dx + i dy) = (u dx v dy) + i(v dx + u dy): Note that if in (4.2) is real-valued, then the path of integration is part of

R. On the other hand the path of integration is in the z -plane.

4.3. De nition. A pie ewise smooth urve with parametri interval [a; b℄ is said to be a reparameterization of (t) (A  t  B ) i there is a C 1 -map  : [A; B ℄ ! [a; b℄ su h that 0 (t) > 0, (A) = a, (B ) = b and (t) = ((t)). Sometimes and are said to be equivalent. The onditions, 0 (t) > 0, (A) = a and (B ) = b are to ensure the dire tion of tra ing as does. Suppose that f is ontinuous in an open set D ontaining all the points of (t). Then, we have Z

f (z ) dz =

Z

f ( (t)) 0 (t) dt

144

Complex Integration

= = = =

Z B

A

f f ((t)g 0((t))0 (t) dt

Z (B )

(A) Z b Za

f ( (t)) 0 (t) dt

f ( (t)) 0 (t) dt

f (z ) dz:

Therefore, it is immaterial whi h parameterization is used. Often the omputation is simpli ed if we use some parti ular equivalent path in evaluating an integral. For ertain situations, the hoi e is by an important property whi h will be stated in later theorems (see Theorem 4.16). Suppose we want to evaluate Z (4.4) I = f (z ) dz; f (z ) = z n ; where

(t) = eit , I = rn+1

0  t  2, and n is any integer. It follows that

Z 2

that is, (4.5)

0

ie(n+1)it dt = Z

8  it(n+1)  2 > < n+1 e r n + 1 0 > :

2i

if n 6= 1 if n = 1;



0 if n 6= 1; 2i if n = 1:

We also note that (4.5) ontinues to hold if and f in (4.4) are repla ed by any ir le entered at z0 and (z z0)n , respe tively. This means that  Z 0 if n 6= 1 and integer n (z z0 ) dz = 2i if n = 1: jz z0j=r We now have the following useful appli ation. For n 2 N , Z 2

0

f (z ) dz =

os2n (t) dt

Z 2  it e

 + e it 2n = dt 2 0 Z 2n   1 2 X 2n i(2n 2k)t = 2n e dt 2 0 k=0 k 

 Z 2

2n 1 X 2n 22n k=0 k    2n = 2n 1 : 2 n

=

R

0

e2i(n k)t dt

4.6. Example. We evaluate Ij = j x dz , j = 1 to 7, where

145

4.2 Properties of Complex Line Integrals

(i) 1 is the straight line segment from 0 to a + ib (a; b 2 R) (ii) 2 is the ir le jz j = R (iii) 3 is the boundary of the square [0; 1℄  [0; 1℄ with C onsidered as R2 (iv) 4 is the ellipse x2 =a2 + y2 =b2 = 1 (v) 5 is the line from 0 to 2i and then from 2i to 4 + 2i (vi) 6 is the line segment from 0 to 1, 1 to 1 + i and then from 1 + i to 0 (vii) 7 is given by 7 (t) = t + it2 on [0; 1℄. (i) 1 may be parameterized by 1 (t) = (a + ib)t; 0  t  1: Clearly,

I1 =

Z 1

0

[Re 1 (t)℄ 10 (t) dt =

Z 1

0

ta(a + ib) dt =

a(a + ib) : 2

(ii) Parameterizing 2 by 2 (t) = Reit , 0  t  2, we have

I2 =

Z 2

0

= iR2

(R os t)(iReit ) dt

Z 2

0

[ os2 t + i sin t os t℄ dt

Z 2



Z

1 + os 2t i 2 dt + sin 2t dt 2 2 0 0 "     # sin 2t 2 i

os 2t 2 1 2 t+ + = iR 2 2 0 2 2 0 = iR2: =

iR2

(iii) In this ase, we have (see Figure 4.5)

I3 =

Z A

O

+

Z B

A

+

Z C

B

+

Z O!

C

x dz:

For the sake of onvenien e, we parameterize OA; AB; BC and OC , as follows:

C1 (t) C2 (t) C3 (t) C4 (t)

= = = =

(1 (1 (1 (1

t)
0 + t
1 = t t)
1 + t
(1 + i) = 1 + ti t)
(1 + i) + t
i = 1 t + i t)
i + t
0 = (1 t)i;

where t 2 [0; 1℄ and 3 = C1 + C2 + C3 + C4 (see Figure 4.5). Utilizing these, i.e. without reparameterizing, we have

I3 =

4 Z 1 X

k=1 0

(Re Ck (t))Ck0 (t) dt

146

Complex Integration y C3

C(0, 1)

B(1, 1)

C4

C2

O

C1 A(1, 0) x

Figure 4.5: Curve 3 = C1 + C2 + C3 + C4 .

=

Z 1

0

t dt +

Z 1

0

1
i dt +

Z 1

0

(1 t)( 1) dt +

Z 1

0

0
( i) dt

1 1 +i + 0 = i: 2 2 (iv) Write 4 as 4 (t) = a os t + ib sin t; 0  t  2: Then, =

Z 2

I4 =

0

(a os t)( a sin t + ib os t) dt Z

Z

2 a2 2 sin 2t dt + iab

os2 t dt = iab: 2 0 0 The ases (v), (vi) and (vii) may be evaluated similarly and are left as a simple exer ise. 

=

4.7. De nition. If : [a; b℄ ! C is a smooth and re ti able urve su h that (t) = x(t) + iy(t), then its (Eu lidean) length L( ) is de ned by (4.8)

L( ) =

Z

jd (t)j =

Z b

a

j 0 (t)j dt =

Z bp

a

x0 (t)2 + y0 (t)2 dt:

If is merely pie ewise smooth and re ti able, then its length is the sum of the integrals (4.8) over all the smooth paths of . We onsider a few simple examples to demonstrate the use of the formula (4.8). The length of a ir le of radius r (use the parametri equation (t) = a + reit , 0  t  2, so that 0 (t) = ireit ) is found to be Z 2

0

jireit j dt = 2r

as expe ted. The line segment [z1 ; z2℄ parameterized by (t) = (1 t)z1 + tz2 , 0  t  1; has its length

L( ) =

Z 1

0

j 0 (t)j dt =

Z 1

0

jz2 z1 j dt = jz1 z2 j:

147

4.2 Properties of Complex Line Integrals

Similarly, the perimeter of the re tangle may be obtained using this formula with a onvenient parameterization. Finally, if is an ellipse given by

x2 y2 + = 1; a2 b2 then its parametri form is (t) = a os t + ib sin t, t 2 [0; 2℄, so that L( ) =

Z 2 p

0

a2 os2 t + b2 sin2 t dt = a

Z 2 q

0

1 + [(b2 =a2 ) 1℄ sin2 t dt:

What is the value of this integral? Does there exist a simple formula to

ompute the ar length of an ellipse with semi-axes of length a and b? There is a se ond type of line integral that may be introdu ed. Let

: (t), a  t  b, be a smooth urve and s(t) denote the ar -length fun tion R for . Let f be a ontinuous fun tion on D with ([a; b℄)  D. Then, f (z ) jdz j is de ned to be the approximating sums of the form

S ( ; P ) =

n X j =1

f (zj )js(tj ) s(tj 1 )j;

where P : a = t0 < t1 <


< tn 1 < tn = b ranges over all possible partitions of the interval [a; b℄. Here zj lies on between (tj 1 ) = zj 1 and

(tj ) = zj . The standard pro edure then shows that (with s0 (t) = j 0(t)j) Z

f (z ) jdz j = R

Z

f ds =

Z b

a

f ( (t))j 0 (t)j dt:

If f isR real-valued, then f ds is a real number. Clearly, f (z ) = 1 gives that f ds = L( ). As we have seen earlier, the omplex integral may be put in the form Z

f (z ) dz =

Z

Z

(u dx v dy) + i (v dx + u dy):

Therefore, the usual rules of integration for real integrals must also apply to ontour integrals. The following theorem summarizes some useful properties of omplex line integrals. The on lusion of Theorem 4.9 ontinues to hold if ; 1 ; 2 are pie ewise smooth (= ontour) although we state and prove the theorem for smooth urves for the sake of simpli ity .

4.9. Theorem. Let be a smooth urve de ned on [a; b℄ and let f and g be ontinuous fun tions on an open set D ontaining ([a; b℄) and let  be a omplex onstant. Then, (i)

Z

f (z ) dz =

Z

f (z ) dz:

148

Complex Integration

(ii)

Z

[f (z ) + g(z )℄ dz = 

Z

Z

f (z ) dz +

g(z ) dz:

(iii) If L = L( ) is the length of the urve and M = max jf ( (t))j; then t2[a;b℄ R f (z ) dz  ML: This property is alled the standard estimate for (iv)

integrals, or M-LZinequality. Z

f (z ) dz =

Z

f (z ) dz + f (z ) dz whenever 1 and 2 are two

1 + 2

1

2 smooth urves su h that 1 (b) = 2 (a) and j ([a; b℄)  D for j = 1; 2.

Proof. Note that

Z

f (z ) dz = =

Z b

a

Z a

b

=

f ( (b + a t)) d( (b + a t)) f ( (s)) d( (s)); by a hange of variable s = b + a t,

Z b Za

=

f ( (s)) d( (s))

f (z ) dz:

The ase (i) now follows. The ase (ii) follows from the de nition and the linearity property of the Riemann integral. R R To prove (iii), we noti e that L = jdz j = ab j 0 (t)j dt and for a realvalued Riemann integrable fun tion  on [a; b℄, we know that Z b (t) dt a

(4.10) R



Z b

a

j(t)j dt: R

If f (z ) dz = 0, there is nothing to prove. Therefore, we let f (z ) dz 6= 0 and write Z Z b f (z ) dz = f ( (t)) 0 (t) dt = Rei ; say,

where R > 0 and  = Arg

R=

Z b

a

e

a

R



f (z ) dz : We have

i f ( (t)) 0 (t) dt =

Z b

a

Re [e i f ( t)) 0 (t)℄ dt:

We apply (4.10), with (t) = Re [e i f ( (t)) 0 (t)℄, to get (t)  e i f ( (t)) 0 (t)℄

R so that R  ab jf ( (t))j j 0 (t)j dt: Sin e jf ( (t))j  M for all t 2 [a; b℄ and sin e for positive integrands the Riemann integral is larger for a larger integrand, we have the assertion in (iii).

149

4.2 Properties of Complex Line Integrals

(iv) As 1 and 2 are smooth urves with 1 (b) = 2 (a), = 1 + 2 is then de ned by

(t) =



1 (2t a) if a  t  (b + a)=2

2 (2t b) if (b + a)=2  t  b:

The assertion now follows from the de nition after noting that for 1 and

2 a reparameterization has been made. hR

i

R

4.11. Remark. In general, Re f (z ) dz 6= Re [f (z )℄ dz as an be seen by hoosing (t) = it and f (z ) = 1. This is to aution the reader to be areful while taking real and imaginary parts of an integral.  4.12. Example. From Theorem 4.9(iv) we obtain the following: (i) If (t) = (1 + i)t; t 2 [0; 1℄, the p line segment from 0 to 1 + i, then for any point  on we have j j  2, j 3 + 2j  23=2 + 2 = M , say, and j 3 + 2j  23=2 2. As a result of this, we have the estimates Z

and



(z 3 + 2) dz  M

Z

Z

p p jdz j = (23=2 + 2) 2 = 4 + 2 2

p



23=2

(z 3 + 2) 1 dz 

2

2

=

p

2

sin e the line segment [0; 1 + i℄ has length 2. R

1

p; 2



(ii) Consider (t) = eit , 0  t  . Then, z 1ez dz  ML( ) = e be ause the length L( ) is  and



M = max e (t) = (t) = max e os t = e: t2[0;℄ t2[0;℄ (iii) If (t) = (1 t)(1 + i) + t (1 + 3i), 0  t  1, the dire ted line segment R from 1+ i to 1+3i, then z 2 dz  ML( ) = 1 be ause the length L( ) is j1 + 3i (1 + i)j = 2 and



1 1 1 M = max 2 = max = : 2 t2[0;1℄ (t) t2[0;1℄ 1 + (1 + 2t)2 R



(iv) If jf (z )j  M on , then we dire tly have f (z ) dz  ML( ): (v) The following inequalities may also be he ked in a similar fashion:

150

Complex Integration Z z e dz jzj=1

Z 1 =z e dz jzj=1

Z jzj=1



1 dz  2. (a)  2e,  2e, z Later, we shall a tually see that the value of rst integral is 0 the se ond while and the third ea h has the value 2i. Z p 2 (b) (z + 1) dz  9 5, where is the line segment parameter

by ( t) = 2 t(2 i); t 2 [0; 1℄. ized Z 2  (e 1) ( ) eiz dz  , where (t) = eit , 0  t  =4. 4 e



(d)



Z



[(Re z )2 + i(Im z )2 ℄ dz  2, where is the interval [ i; i℄ on

the imaginary axis. 

4.13. De nition. Let be a urve with parametri interval [a; b℄ and let ffn g be a sequen e of fun tions on an open set D ontaining ([a; b℄). If for a fun tion f de ned on D, jfn ( (t)) f ( (t))j ! 0 uniformly on [a; b℄, we say that fn ! f uniformly on . If for a fun tion f on D, P jSn ( P (t)) f ( (t))j ! 0 uniformly on [a; b℄, where Sn = nk=1 fk , we say that n1 fn ! f uniformly on . 4.14. Theorem. (Inter hange of limit and integration) Let ffng be a sequen e of ontinuous fun tions de ned on an open set ontaining a

ontour . Suppose that fn ! f uniformly on . Then, Z

lim f (z ) dz n!1 n

=

Z

f (z ) dz:

Proof. Let  > 0 and let fn onverge uniformly on with parametri interval [a; b℄. Then, there is an N su h that

jfn ( (t)) f ( (t))j <  for t 2 [a; b℄ and n  N: First we observe that f is ontinuous on (see Theorem 2.57). Using Theorem 4.9(iv), we have Z

fn (z ) dz

Z

f (z ) dz

=



Z b [fn ( (t)) a Z b

< 



jfn ( (t)) f ( (t))j j 0 (t)j dt

a

Z b

a

j 0 (t)j dt for n  N:

As  > 0 is arbitrary, the proof is omplete. A straightforward proof gives



f ( (t))℄ 0 (t) dt

151

4.2 Properties of Complex Line Integrals

4.15. Corollary. (Inter hange of summation and integration) P Let n1 fn be a series ofP ontinuous fun tions de ned on an open set

ontaining a ontour and n1 fn ! f uniformly on a ontour . Then, XZ

n1

fn (z ) dz =

Z

f (z ) dz:

There are two versions of the fundamental theorem of al ulus for realvalued fun tions: Z x  d (i) f (t) dt = f (x) for x 2 [a; b℄ where f is ontinuous on [a; b℄ dx a and one-sided derivatives are meant at a or b Z b (ii) f 0 (t) dt = f (b) f (a) whenever f 0 (t) is ontinuous on [a; b℄. a

The following weaker form of Cau hy's Theorem (see Theorem 4.33), whi h is a tually the analogue of the se ond statement of the fundamental theorem of al ulus, is helpful, and integration of familiar fun tions is fa ilitated by this result.

4.16. Theorem. (Weak form of Cau hy's theorem) If f = u + iv is analyti in an open set D ontaining a ontour with parametri interval [a; b℄, i.e. ([a; b℄)  D, then Z

f 0 (z ) dz = f ( (b)) f ( (a)):

That is, the R value of the integral is independent of the path. In parti ular, we have f 0 (z ) dz = 0 if is losed.

Proof. Let f be analyti on D and be, initially, a smooth urve with

([a; b℄)  D. Then, we must have (see Corollary 2.21 and (3.3)) f ( (t)) = f ( (t0 )) + ( (t) (t0 ))f 0 ( (t0 )) + ( (t) (t0 ))( (t)); for t near t0 2 [a; b℄, where  Æ is a ontinuous fun tion of t on [a; b℄ su h that limt!t0 ( (t)) = 0. Therefore, as t0 is arbitrary, the Chain rule for dierentiation gives

d [f ( (t))℄ = f 0( (t)) 0 (t) dt on [a; b℄ and f ( (t)) is ontinuously dierentiable for t 2 [a; b℄. The result now follows from the \se ond statement of the fundamental theorem of

al ulus for real variable". Indeed, we let f ( (t)) = f (x(t)+ iy(t)) = u(x(t); y(t))+ iv(x(t); y(t)) = U (t)+ iV (t); say;

152

Complex Integration

so that

d d d [f ( (t))℄ = f 0 ( (t)) 0 (t) = (U (t)) + i (V (t)): dt dt dt Sin e is smooth, we integrate both sides of the last expression to get Z

Z b

d [f ( (t))℄ d t a dt b = U (t) + iV (t)ja = f ( (b)) f ( (a)):

f 0 (z ) dz =

Suppose that is pie ewise smooth. Then, we an by de nition hoose a partition P : a = t0 < t1 <


< tn = b su h that the restri tion k of

to (tk ; tk+1 ); k = 0; 1; : : : ; n 1, is smooth. In view of Theorem 4.9(iii) and what has been just proved, Z

f 0 (z ) dz =

nX1 Z k=0 k

f 0 (z ) dz =

= f ( (b)) f ( (a));

nX1 k=0

f ( k (tk+1 )) f ( k (tk ))

again as asserted. Theorem 4.16 is also known as the fundamental theorem of line integrals (or ontour integration) in the omplex plane. Moreover, Theorem 4.16 shows that, if F (z ) = f 0 (z ) then one has Z z2

z1

F (z ) dz = f (z2 ) f (z1):

In R parti ular, for losed urves independent of paths, we on lude that

F (z ) dz = 0. One of the obje tives in Se tion 4.3 is to extend this for those fun tions F (z ) for whi h no f su h that F (z ) = f 0 (z ) is at hand. The examples of su h fun tions are os(z 2), sin(z 2 ), exp(z 2), et . On the other hand, it will be shown that Z



os(z 2 ) dz

=

Z

sin(z 2) dz

=

Z

exp(z 2 ) dz = 0

for ea h losed ontour . R

4.17. Example. Consider Ij = j z dz; j = 1; 2; 3; where (i) 1 is the dire ted line segment from 0 to 1 + i; (ii) 2 is the ar of the ir le 2 (t) = 1 + eit joining 0 and 1 + i;

153

4.2 Properties of Complex Line Integrals

(iii) 3 is the dire ted line segment from 0 to 1 and then from 1 to 1 + i. R

In this example we observe that f (z ) = z is nowhere analyti and so z dz need not be independent of the hoi e of the urve onne ting the points 0 and 1 + i. In fa t, it an be he ked easily that

I1 = 1; I2 = 1 + i(=2 1) and I3 = 1 + i: Thus I1 6= I2 , I2 6= I3 and I1 6= I3 , even though 1 , 2 and 3 have the same initial and the same terminal points.  Let z1 = 1, z2 = 1 and z3 = i. Consider 1 = [z1 ; z2 ℄ [ [z2 ; z3 ℄, and

2 = [z1 ; z3 ℄. Then both 1 and 2 are urves of lass C 1 whose initial and the terminal points are the same. But it is easy to see that Z

1

z dz = i and

Z

2

z dz = i:

Re all that f (z ) = z is nowhere analyti and hen e, has no primitive in a domain ontaining the points 1; 1 and i. The above examples show that the integral of a omplex fun tion depends on the path of integration. However, there are a few important Corollaries to Theorem 4.16. Keeping in mind the de nition of primitive, we have

4.18. Corollary. If RF is a primitive of f on D and : [a; b℄ ! C is a smooth urve in D, then f ( (t)) 0 (t) dt = F ( (b)) F ( (a)): For instan e, if n  0 is an integer then Z z2

z1

z n dz =

z2n+1 z1n+1 n+1

sin e `z n+1=(n + 1) + onstant' is a primitive of z n on C . If n < 1 is an integer, the above equality holds provided the path of integration omits the R origin. In parti ular, z n dz = 0 if is any smooth losed path omitting the origin and n 6= 1.

4.19. Example. Let : [0; 1℄ ! C be de ned by

(t) = 1 t sin t + i(t + os t) whi h is an ar onne ting 1 + i to 0 as in Figure 4.6. To evaluate

I=

Z

z 2 dz =

Z 1

0

2 (t) 0 (t) dt

we rst note that, sin e z 2 is analyti in C , lies in the domain of de nition of z 2 . Sin e the value of the integral is independent of the hoi e of the

154

Complex Integration y i

0

1

1+i 0

1

x

Figure 4.6: Curve (t) = 1 t sin t + i(t + os t).

urve onne ting 1 + i to 0, we have

I=

Z 1

0

12 (t) 10 (t) dt;

where 1 (t) = (1 t)(1 + i) + t
0; t 2 [0; 1℄: Thus, we easily nd that

I=

Z 1

0

(1 + i)2 (1 t)2 [ (1 + i)℄ dt =

(1 + i)3 : 3

Indeed, by Corollary 4.18, we dire tly obtain that I = F ( 1 (1)) F ( 1 (0)); where F (z ) = z 3=3 + K , and so I = F (0) F (1 + i) = (1 + i)3 =3:  P

4.20. Corollary. Let f (z ) = n0 an z n with its radius of onverR gen e R > 0. Then, for any losed ontour in
R , we have f 0 (z ) dz = 0: Corollary 4.20 follows P from the fa t that f possesses a primitive F in
R , namely, F (z ) = n0 (n + 1) 1 an z n+1.

4.21. Corollary. Theorem 3.31(i) follows from Corollary 4.18. Proof. By the assumption of Theorem 3.31(i), we note that f is a primitive of the zero fun tion g(z ) = 0 on D. Therefore, for any disk
(z0 ; r) entirely within D and ea h  2
(z0 ; r), 0=

Z z0



0
dz = f ( ) f (z0 )

so that f (z ) = f (z0 ) for z 2
(z0 ; r). R

4.22. Example. Let us evaluate (z 2 + z ) dz where = 1 + 2 + 3 as in Figure 4.7. First we note that f (z ) = z 2 + z is analyti in C and so it is analyti in any domain ontaining the points 1 and 1. Therefore, Theorem 4.16 allows us to hoose any path onne ting 1 and 1. The most

onvenient path in this ase is the line segment onne ting 1 and 1:

(t) = (1 t)( 1) + t(1) = 2t 1; t 2 [0; 1℄:

155

4.2 Properties of Complex Line Integrals γ2 γ1 γ3

γ2 γ1

γ3 γ

−1

1

Figure 4.7: Curves onne ting 1 and 1. 2i

1 + 2i

−1 + i 1

−1

Figure 4.8: Region for the analyti ity of Log (1 + z ).

Therefore, as f ( (t)) 0 (t) = [(2t 1)2 Z

f (z ) dz =

Z 1

0

f ( (t)) 0 (t) dt = 8

(2t 1)℄(2) = 8t2 Z 1

0

t2 dt 4 z3

Z 1

0

t dt =

4t, we have 8 3

4 2 = : 2 3

z2

Moreover, sin e f (z ) = FR0 (z ) with F (z ) = 3 + 2 + K; we an dire tly use Corollary 4.18 to obtain f (z ) dz = F ( (1)) F ( (0)) = 2=3:  R

dz , where is any on4.23. Example. We next evaluate I = 1+ z tour in D = fz : Im z > 0g, whi h joins 1 + i to 1 + 2i (see Figure 4.8). Suppose that we onsider the prin ipal logarithm. We observe that (1 + z ) 1 is the derivative of F (z ) = Log (1 + z ) and, sin e F is analyti in C n f( 1; 1℄g, F is analyti in D. Then the value of the integral is independent of the path joining 1 + i and 1 + 2i and hen e, Z Z dz = F 0 (z ) dz

1+z

= F (1 + 2i) F ( 1 + i) = Log [2(1 + i)℄ Log i p   p = [ln 8 + i ℄ i = ln 8 i=4:  4 2

156

Complex Integration R

4.24. Example. Suppose we wish to evaluate the integral dzz , where is an ar joining 1 i to 1 + i. Then, in this ase, it is ne essary to onsider a domain D ontaining the ar . Note that the integrand 1=z is analyti in C n f0g. As we have observed before if  2 R is xed, D = C n fRei : R > 0g and arg z is the hoi e of arg z in ( 2; ), then f (z ) = ln jz j + i arg z is the anti-derivative of 1=z in D . Suppose we hoose  = . Then,

j arg z j < ; D = D = C n f R : R > 0g and therefore, f (z ) be omes the prin ipal logarithm Log z . So if is any

urve in D whi h joins 1 i to 1 + i, then by Corollary 4.18 1+i

1+i

I = f (z )j1 i = Log z j1 i whi h gives

I = Log (1 + i)

Log (1 i) = i[Arg (1 + i) Arg (1 i)℄ = i



4

+

 4

so that I = i=2: Suppose we hoose  = 2. Then, 0 < arg z < 2; D2 = C n fR : R > 0g: Therefore if is any urve in D2 whi h joins 1 i to 1 + i, then h 1+i I = f2 (z )j1 i = i[arg2 (1 + i) arg2 (1 i)℄ = i 4



2

 i 4

so that I = 3i=2: If 1 and 2 are given by

1 (t) = eit ; t 2 [ =4; =4℄ and 2 (t) = eit ; t 2 [=4; 2 =4℄; then 1 2 D and 2 this ase, we have Z

2 D2 . Then = 1 + 2 is a losed urve and, in Z

dz dz = z jzj=1

1 z

Z

dz i = 2

2 z R





3i = 2i: 2

With this idea, it is also lear that jzj=r 0, there exists Æ > 0 su h that f (z ) = f (z0 ) + (z z0 )f 0 (z0 ) + (z z0 )(z ); with  ontinuous and j(z )j <  for jz Z

T n

z0 j < Æ: Therefore, we have

f (z ) dz = [f (z0 ) z0 f 0 (z0 )℄ +

Z

T n

Z

(z

T n

dz + f 0 (z0 )

z0 )(z ) dz:

Z

T n

z dz

160

Complex Integration R

R

We know that if is any losed ontour, then dz = (z As the rst two integrals vanish, Z

T n

f (z ) dz =

Z

T n

(z

z0 ) dz = 0:

z0)(z ) dz:

By (i)0 , we also note that, for suÆ iently large n, the triangle T n is ontained in fz : jz z0 j < Æg. Using the standard estimate (see Theorem 4.9(iv)) together with the above observations, we get

jI (n) j = =

Z f ( z ) dz n T Z ( z z )  ( z ) dz 0 T n

  zmax jz z0 j
L(T n) 2T n  
diam (T n)L(T n)  n  n 1  
2 diam (T ) 12 L(T ); by (iii)0 and (iv)0 ; = 4 ndiam (T )L(T ):

Taking (ii)0 into a

ount we have the inequalities 4 njI j  jI (n) j  4 ndiam (T )L(T ): Consequently, we have jI j  diam (T )L(T ). Sin e  > 0 in this inequality is an arbitrary positive number, we must have I = 0 and so the proof is

omplete. If is a quadrilateral, it an be divided into two triangles T and T 0 (see Figure 4.10) so that Z

f (z ) dz =

Z

T

f (z ) dz +

Z

T 0

f (z ) dz = 0;

sin e the integrals along AC and CA an el ea h other. In general, if is any simple Z polygon then we an de ompose su h a polygon into triangles so that f (z ) dz = 0:

An important point in the above proof is that it is not ne essary to assume the ontinuity of the derivative f 0 (z ) of f (z ). Next we present a simple extension of Theorem 4.28 with a relaxed ondition on the dierentiability of f .

4.30. Theorem. Let D be an open set and let f be analyti on D ex eptR possibly at a 2 D. Assume that f is ontinuous on D. Then, we have T f (z ) dz = 0 for every losed triangle T in D.

161

4.3 Cau hy-Goursat Theorem D(z4 )

C(z3 )

A(z1 )

B(z2 )

D(z4 ) C(z3 ) A(z1 )

B(z2 )

Figure 4.10: Triangles T = [z1 ; z2 ; z3 ℄ and T 0 = [z1 ; z3 ; z4 ℄.

Proof. For a losed triangle T in D, we may simply assume that a lies in T , for the result is a onsequen e of Theorem 4.28 otherwise. Given a positive integer, we an subdivide T into n2 ongruent triangles Tjk by adjoining the midpoints of opposite sides. Then, we have Z

T

f (z ) dz =

n Z n X X j =1 k=1 Tjk

f (z ) dz

sin e the dividing segments

an el in pairs. If a is not a point of Tjk , then, R by Theorem 4.28, Tjk f (z ) dz = 0. If a belongs to the triangle Tjk , then the M-L inequality shows that Z Tjk

f (z ) dz



Z

Tjk

jf (z )j jdz j  M L(Tjk ) = MLn(T ) ;

where M = maxz2T jf (z )j. Note that jf (z )j is a ontinuous fun tion on the ompa t set C . Note that the point a at the worst an belong to one of the four triangles Tjk . It follows that Z

T

f (z ) dz =

Z X a2Tjk Tjk

f (z ) dz

R



X Z a2Tjk Tjk

f (z ) dz

 4MLn (C ) :



Sin e n was arbitrary, T f (z ) dz = 0 and the proof is omplete.

4.31. Theorem. Let D be a domain that is starlike with respe t to a and f be analyti on D. Then, there existsR an analyti fun tion F on D su h that F 0 (z ) = f (z ) in D. In parti ular, C f (z ) dz = 0 for every losed

ontour C in D. Proof. Sin e D is starlike (see Figure 4.11) with respe t to a, [a; z ℄  D for every z 2 D. De ne F on D by F (z ) =

Z

[a;z℄

f ( ) d:

162

Complex Integration

z+

|h| z

h b a

a

b

Figure 4.11: Starlike domain with respe t to a.

Note that F (a) = 0. Fix z 2 D nfag. Then there exists h 2 C with jhj suÆ iently small su h that
(z ; jhj)  D and [z; z + h℄ 
(z ; jhj)  D. As D is starlike, the triangle T = [z; a; z + h℄ lies in D. By Theorem 4.28, Z

0= or

T

f ( ) d =

Z z+h

a

Z a

f ( ) d

z Z z

a

Therefore, we have

F (z + h) F (z ) h

f (z ) =

F (z )

f (z )

a

f ( ) d =

+

Z z !

z+h

Z z +h

z

=



f ( ) d;

f ( ) d:

Z

f (z ) z+h d h z

Z z+h 1 [f ( ) h z "

f (z )℄ d

1 z+h f ( ) d h z

and so F (z + h) h

+

Z z +h

Z

#

1 jf ( ) f (z )j jhj; jhj  2[sup z;z+h℄

using the standard integral estimate (see Theorem 4.9(iv)). Continuity of f immediately yields that F (z + h) h

F (z )



f (z ) ! 0 as h ! 0;

and so F 0 (z ) exists and is equal to f (z ). Sin e z is arbitrary, F 0 (z ) = f (z ) in D. That is, F is a primitive of f on D, as required. The se ond part follows from Theorem 4.16. Sin e every disk is a starlike domain, Theorem 4.31 yields the "lo al form" of Cau hy's theorem.

163

4.3 Cau hy-Goursat Theorem z0 = zn zn−1

zn−2

zn−3

z1 z5 D z2

z4 z3

Figure 4.12: Closed polygon with verti es on C .

4.32. Corollary. (Cau hy's Theorem for a Disk) Let f be analyti in a disk
(z0 ; R) (or more generally, f is ontinuousRin
(z0 ; R) and analyti in
(z0 ; R) nfag for some a 2
(z0 ; R)). Then, f (z ) dz = 0 for every losed ontour in
(z0 ; R). The fun tion f (z ) = 1=z de ned in the annulus A = fz : 1 < jz j < 2g shows that the on lusion of Theorem 4.31 fails if D is not a starlike domain with respe t to a point in D. Further, the ontour in Theorem 4.31 an have self-interse tion. Now we are in position to prove the long waited \Cau hy integral theorem".

4.33. Theorem. (Cau hy's Integral Theorem) If f is analyti in a simply onne ted domain D, Rthen there exists a fun tion F in D su h that F 0 (z ) = f (z ). In parti ular, f (z ) dz = 0 for ea h simple losed ontour

in D. Proof. Let C be a simple losed ontour in D, and D0 = C [ int C . So, D0  D is ompa t and hen e, for every z 2 D0 there exists Æz > 0 su h that
(z ; Æz )  D. Now D1 = [z2D0
(z ; Æ) is ompa t and so f is uniformly ontinuous on D1 . Let  > 0 be given. Then there exists a Æ > 0 su h that jf (z ) f ( )j < 2L(C ) for all z;  2 D1 and jz  j < Æ. Here L(C ) denotes the length of C . Now without loss of generality we may assume that Æz  Æ=2 for every z 2 D0 . Sin e C is ompa t, there exist points z0 ; z1; : : : ; zn = z0 in C su h that C  [nk=01
(zk ; Æzk ): Let  be a losed polygon in D1 with verti es z0 ; z1 ; : : : ; zn , see Figure 4.12. Therefore, (4.34) jf (z ) f (zk )j < 2L(C ) for k = 1; 2; : : : ; n,

164

Complex Integration C2 γ C2 C1

γ

C1

Figure 4.13: Region bounded by two urves C1 ; C2 .

whenever z 2 [zk 1 ; zk ℄, k = 0; 1; 2; : : : ; n 1. Further, 0=

Z



f (z ) dz = = =

n Z zk X k=1 zk 1 n Z zk X

(4.35)

n X k=1

[f (z ) f (zk ) + f (zk )℄ dz

k=1 zk 1 n Z zk X

= that is,

f (z ) dz

k=1 zk 1 n Z zk X k=1 zk 1

f (zk )(zk

[f (zk ) f (z )℄ dz + [f (zk ) f (z )℄ dz +

zk 1 ) =

n Z zk X k=1 zk 1

n X k=1 n X k=1

f (zk )

Z zk

zk 1

f (zk )(zk

dz zk 1 );

[f (zk ) f (z )℄ dz:

By (4.34) and (4.35), we have n X f (zk )(zk

k=1

Sin e lim

n!1

n X k=1

zk 1 )

 2L(C )

f (zk )(zk

n X

jzk zk 1 j < 2 : k=1

zk 1 ) =

Z

C

f (z ) dz

and sin e  is arbitrary, the theorem is proved. Suppose f is analyti in a ring shaped spa e bounded by two simple

losed ontours C1 and C2 as shown in Figure 4.13. Let be any ontour or a line from C1 to C2 , also shown in Figure 4.13. Then the region bounded by C = C1 + C2 is simply onne ted.

1

165

4.3 Cau hy-Goursat Theorem

Cn C3

C4

C2

C C1

Figure 4.14: Illustration for Cau hy's deformation of ontour

From Theorem 4.33, 0=

Z

C

f (z ) dz = = R

Z

C1

Z

C1

f (z ) dz + f (z ) dz R

Z Z

f (z ) dz

C2

Z

C2

Z

f (z ) dz

f (z ) dz

f (z ) dz; by Theorem 4.9(i);

whi h gives C1 f (z ) dz = C2 f (z ) dz: This result is referred to as the Cau hy deformation theorem. In Figure 4.14, we illustrate Cau hy's theorem for domain with (n 1) holes. In a manner similar to that used above, we get Z (4.36) f (z ) dz = 0: C1 +C2 +


+ n Equation (4.36) an be written in the form I

C1

I

f (z ) dz = =

I

C2 C2

f (z ) dz +

f (z ) dz +




+




+

I

I

Cn Cn

f (z ) dz



f (z ) dz:

In other words, by integrating along ea h inner ontour in the ounter lo kwise dire tion, so that the (n 1) inner ontours have negative orientation, it follows that the value of the integral along the outer ontour is equal to the sum of the values along the inner ontours. Annulus regions are lassi ed as follows: Let a 2 C and 0  r < R  1. Then the open subset

D = D(a; r; R) = fz 2 C : r < jz aj < Rg

of C is alled the annulus or ir ular ring around a with inner radius r and outer radius R. If r = 0 and R < 1, then D is a disk with enter removed, i.e. the pun tured disk: D = D(a; 0; R) =
(a; R) nfag. If r = 0 and R = 1, then D(a; 0; 1) = C n fag. Finally, D de ned above is the exterior ir le ex luding the point 1 if R = 1 and r > 0. In parti ular, the theorem of Cau hy's deformation of ontour gives the following:

166

Complex Integration

4.37. Theorem. Let 0  R1 < r < R2  1, and g(z ) be analyti in the annulus domain D = fz 2 C : R1 < jz aj < R2 g: If Cr = fz : R jz aj = rg, then Cr g(z ) dz is independent of r.

4.4 Consequen e of Simply Conne tivity How do we produ e a non-vanishing analyti fun tion f in a simply onne ted domain ? Take an arbitrary analyti fun tion h in . Then, the desired fun tion f is given by f (z ) = exp(h(z )): In the following theorem we a tually show that every non-vanishing analyti fun tion f arises in this way.

4.38. Theorem. Let be a simply onne ted domain and f 2 H( ) with f (z ) 6= 0 on . Then, there exists a h 2 H( ) su h that eh(z) = f (z ):

Proof. As f (z ) 6= 0 on , f 0 (z )=f (z ) is analyti on . By Theorem 4.33 (see also Corollary 4.61) there exists an h 2 H( ) su h that f 0 (z ) h0 (z ) = for z 2 : f (z ) We laim that f (z )e h(z) = 1 for z 2 . To do this, we de ne g(z ) = f (z )e h(z). Clearly, g 2 H( ) and g0 (z ) = (f 0 (z ) f (z )h0(z ))e h(z) = 0 for z 2 : Fix 2 . Then,

g(z ) g( ) =

Z z

g0( ) d = 0

so that f (z )e h(z) = g( ) or f (z ) = eh(z)g( ): As g( ) = 6 0, we an set g( ) = ek for some k. Then

f (z ) = exp(h(z ) + k) = exp(H (z )); where H (z ) = h(z ) + k. Another onsequen e of this result is the following square root property.

4.39. Theorem. Assume the hypotheses of Theorem 4:38. Then, f has an analyti square root- that is there exists a g 2 H( ) with g 2 (z ) = f (z ) for z 2 : Proof. The desired on lusion follows if we hoose h as in the previous theorem and set g(z ) = exp(h(z )=2). We have a dire t proof of Theorem 4.38 at least when = C .

167

4.5 Winding Number or Index of a Curve

4.40. Theorem. Let f 2 H(C ) with f (z ) = 6 0 on C . Then, there exists a h 2 H(C ) su h that f (z ) = eh(z) . Proof. By hypothesis, f 0 (z )=f (z ) belongs to H(C ) and therefore, it admits a Taylor series (about the origin) onverging in the whole of C : 1 f 0 (z ) X = a z k for z 2 C : f (z ) k=0 k De ne (4.41)

1 X

ak k+1 z = g(z ); k+1 k=0

Sin e

an 1=n lim sup n!1 n + 1

i.e.

f 0 (z ) = g0 (z ): f (z )

= lim sup(n + 1) 1=n jan j1=n = lim sup jan j1=n = 0; n!1 n!1

the radius of onvergen e of the series (4.41) that represents the fun tion g is in nity and therefore, g 2 H(C ). Set H (z ) = eg(z) : Then, f 0 (z ) H 0 (z ) = g0 (z ) = ; i.e. f (z )H 0 (z ) f 0 (z )H (z ) = 0 for z 2 C ; H (z ) f (z ) whi h gives





d H (z ) = 0 for z 2 C ; dz f (z ) so that H (z ) = kf (z ) for some onstant k. Note that H (0)=f (0) = k, k 6= 0 and so there exists a 2 C su h that ea = k. Thus, H (z ) = ea f (z ) so that eg(z) a = f (z ): Hen e, h(z ) = g(z ) a is the desired fun tion.

4.5 Winding Number or Index of a Curve Suppose that is a losed ontour in C . Let a be a given point in C n f g.

Then, there is a useful formula that measures how often winds around a. For example if : (t) = fz : z a = reit ; 0  t  2kg, then en ir les the point a k times ( ounter lo kwise). Further, Z

dz

z a

=

Z 2k

0

Z

ireit 1 dz dt = 2ki; i.e. = k: it re 2i z a

From this we also observe that if en ir les the point a k-times in the

lo kwise dire tion, then Z

1 dz = k: 2i z a

168

Complex Integration b a

c b

n(γ ; a) = 1 n(γ ; b) = 0

a

a n(γ ; a) = −1 n(γ ; b) = 0

b

n(γ ; a) = 1 n(γ ; b) = 0 n(γ ; c) = 2

Figure 4.15: Des ription for winding number. R

In either ase, 21i zdza is an integer. Here is the analyti de nition of the winding number of a, whi h aptures the intuitive notion of \the number of times wraps around a in the ounter lo kwise dire tion" (see Figure 4.15):

4.42. De nition. Let be a losed ontour in C that avoids a point a 2 C . The index (or winding number) of about a, denoted by n( ; a) or Ind ( ; a), is given by the integral Z

dz 1 : n( ; a) = 2i z a A tually, from our later dis ussion, Cau hy's theorem will imply that n( ; a) = n( 0 ; a) for all losed urves 0 that are homotopi to as losed

urves in C n fag. In the following we will olle t some properties of the index n( ; a).

4.43. Theorem. For every losed ontour in n( ; a) is an integer.

C

and a

2 C n ,

Proof. If the parametri interval of is [0; 1℄, then (0) = (1). Consider the fun tions g : [0; 1℄ ! C and h : [0; 1℄ ! C , de ned by Z t 0

(s) ds (4.44) g(t) = and h(t) = ( (t) a)e g(t) ; 0 (s) a respe tively. Then, g(0) = 0 and g is ontinuous on [0; 1℄. Likewise h is

ontinuous on [0; 1℄. Moreover, for t 2 [0; 1℄, g is pie ewise smooth and has the derivative

0 (t) g0 (t) =

(t) a at every point t where 0 (t) is ontinuous. Consequently, h has the derivative h0 (t) = [ 0 (t) g0 (t)( (t) a)℄e g(t) = 0

169

4.5 Winding Number or Index of a Curve

at every point t where 0 (t) is ontinuous. Be ause is pie ewise smooth, h0 (t) = 0 fails to hold only at a nite number points in the interval [0; 1℄. Thus, by the ontinuity of h, it follows that h must redu e to a onstant k on [0; 1℄. In parti ular,

h(0) = h(1); or (0) a = e g(1) ( (1) a): Sin e (1) = (0), the last equation means that e g(1) = 1. We on lude that g(1) = 2ik for some integer k. Hen e, by (4.44), we have n( ; a) = k 2 Z. To avoid the little te hni ality, one ould simply supply the proof of Theorem 4.43 just for smooth urves as it is easy. For onvenien e, we shall repla e a in Theorem 4.43 by  and obtain

4.45. Theorem. If is a losed ontour in C , then the mapping ! n( ;  ) is a ontinuous fun tion of  at any point  62 . Proof. Let D be an open set ontaining no point of , a 2 D, and Æ = dist (a; ). Then Æ > 0 and for z 2 , jz aj  Æ. Now, for all h with jhj < Æ=2, we have jz a hj  jz aj jhj  Æ Æ=2 7

and so, using the standard estimate for integrals (see Theorem 4.9(iv)), we have  Z  1 1 jn( ; a + h) n( ; a)j = 21 dz

z a h z a Z  j2hj (z a 1h)(z a) jdz j

j hj  L( )

2(Æ=2)Æ

whi h tends to zero as h approa hes 0, where L( ) denotes the length of . Hen e, as a is arbitrary, n( ;  ) is a ontinuous fun tion of  in C n f g. Integer-valuedness and the ontinuity of n( ;  ) yield

4.46. Corollary. The fun tion n( ;  );  the omponents of C n f g.

2 C n f g, is onstant in

4.47. Theorem. We have n( ;  ) = 0 in the unbounded omponent

of the losed ontour .

170

Complex Integration

Proof. Suppose that 
R . Then, for any  in the unbounded

omponent of C n
R , we have

jn( ;  )j =

Z 1 2

z

dz

< 

1 L( ) ; 2 j j R

sin e jz  j  j j jz j > j j R. So jn( ;  )j < 1 for j j suÆ iently large (for instan e for j j > L2( ) + R). Sin e jn( ;  )j is a non-negative integer and is onstant for every  , it follows that n( ;  ) = 0 for every  in the unbounded omponent of . R

Returning to the initial example where we had 21i zdza = k 2 Z whenever (t) = a+re2ikt , we an use this result to on lude that n( ; a) = k for jz aj < r and n( ; a) = 0 for jz aj > r. The on ept of the winding number is useful to hara terize what is meant by the inside (interior) and the outside (exterior) of a losed urve

, respe tively, in the following way: Int ( ) = fz 62 : n( ; z ) 6= 0g and Ext ( ) = fz 62 : n( ; z ) = 0g: Moreover, a losed urve : [a; b℄ ! C is said to be positively oriented if n( ; z ) > 0 for every z inside and is negatively oriented if n( ; z ) < 0 for every z outside . As we have seen, for the ir le 
(z0 ; r) about z0 , the positive orientation is ounter lo kwise. More pre isely, we have following results whi h we state without proof.

4.48. Theorem. If onsists of nitely many losed ontours 1 ; 2 ; : : : ; k in C , then for every a 62 k (i.e. not on any one of the j ), (i) n( ; a) = n( 1 ; a) + n( 2 ; a) + (ii) n( 1; a) = n( 1 ; a):




+ n( k ; a)

4.6 Homotopy Version of Cau hy's Theorem In this se tion we dis uss more generalR onditions under whi h we may vary the urve ontinuously and so that f (z ) dz un hanged when f 2 H(D) and is ontained in D. For this, we introdu e the important notion of homotopy of ontinuous maps. We onsider two urves whi h are parameterized by the same interval [0; 1℄ and formulate the following de nition. Let 0 ; 1 be two urves in a domain D  C with parametri interval [0; 1℄ having ommon initial and terminal points:

a = 0 (0) = 1 (0); b = 0 (1) = 1 (1): The set of all urves in D whi h onne t a and b is denoted by (D; a; b). If a = b, then we say that the urve is losed and a is alled the base point.

171

4.6 Homotopy Version of Cau hy's Theorem γ1 a = γ0 (0) = γ1 (0)

γ0

b = γ0 (1) = γ1 (1)

Figure 4.16: Des ription for homotopi urves.

(D; a; a) is then the family of all losed urves (loops in D with base point a).

4.49. De nition. Let 0 and 1 2 (D; a; b). We say that 0 and

1 are homotopi (or that 0 is homotopi to 1 ) with xed end points if there exists a ontinuous map F : [0; 1℄  [0; 1℄ ! C su h that (i) F (t; u) 2 D for all 0  t; u  1 (ii) F (t; 0) = 0 (t) and F (t; 1) = 1 (t) for all 0  t  1 (iii) F (0; u) = a and F (1; u) = b; for all 0  u  1. We write 0 ' 1 (or F : 0 ' 1 ) to indi ate that 0 is homotopi to 1 . To have a better understanding of the de nition, we write

u (t) = F (t; u) and say that F is a homotopy between 0 and 1 . Note that, for ea h u 2 [0; 1℄, u : [0; 1℄ ! D is a ontinuous map with u (0) = a and u (1) = b. At the start, u = 0 and u = 0 ; as u varies, the map u varies ontinuously so that at the end u = 1 we have u = 1 , i.e. 0 an be transformed

ontinuously into 1 in D (see Figure 4.16). We observe the following: (i) De nition 4.49(iii) means that the homotopy F xes both initial and terminal points. (ii) If 1 2 (D; a; b) is a losed urve (i.e. a = 1 (0) = 1 (1) = b),

0 (t)  a for all t 2 [0; 1℄ (i.e. 0 is a onstant urve) and 0 ' 1 , then we say that the urve 1 an be ontinuously deformed into the point a. The point a is alled the base point of 1 . In other words, if

is a losed urve with base point a then is said to be homotopi to 0 in D, or simply null-homotopi (written as ' 0) if ' 0 . The family of losed urves in D with base point a is denoted by (D; a). (iii) Homotopy is an equivalen e relation in (D; a; b). Clearly 0 ' 0 . For 0 ; 1 2 (D; a; b), we have

0 ' 1 =) 1 ' 0 :

172

Complex Integration

To he k this, let F : 0 ' 1 be a homotopy arrying 0 into 1 . Then, H de ned by H (t; u) = F (t; 1 u) is a homotopy arrying 1 into 0 , i.e. H : 1 ' 0 . For 0 ; 1 ; 2 2 (D; a; b); we have

0 ' 1 and 1 ' 2 =) 0 ' 2 : To verify this, let F : 0 ' 1 and G : 1 ' 2 . De ne H by 

F (t; 2u) if 0  u  1=2 H (t; u) = G (t; 2u 1) if 1=2  u  1: Then, H : 0 ' 2 . This proves our laim.

4.50. De nition. A domain D is said to be simply onne ted if every

losed urve in D is homotopi to a point in D. If D is a simply onne ted domain, then for any pair 0 ; 1 2 (D; a; b) we have 0 ' 1 by (iii) above, whereas any losed urve in D is homotopi to zero. Suppose D is starlike with respe t to a, and 1 be a losed urve in D. Putting F (t; u) = u 1(t) + (1 u)a; u 2 [0; 1℄; we see that F (t; u) is a ontinuous fun tion de ned on the re tangle f(t; u) : t; u 2 [0; 1℄g su h that F (t; u) 2 D. Also

F (t; 0) = 0 (t); F (t; 1) = 1 (t); for all 0  t  1 F (0; u) = u 1 (0) + (1 u)a = ua + (1 u)a = a and

F (1; u) = u 1 (1) + (1 u)a = ua + (1 u)a = a; sin e 1 2 (D; a). So, 1 ' 0 in D and thus D is simply onne ted. If D is a onvex domain then it is starlike and so it is simply onne ted. Therefore, if 0 and 1 are any two losed urves in (D; a) then 0 ' 1 in D. 4.51. De nition. Let D be an open set and 0 and 1 be two urves de ned on [0; 1℄. We say that 0 and 1 are lose together if there exists a partition P of [0; 1℄, P : 0 = t0 < t1 <


< tn = 1, and a sequen e of disks Dj ; j = 0; 1; : : : ; n 1, ontained in D su h that for ea h j = 0; 1; : : : ; n 1, Dj ontains the images 0 ([tj ; tj + 1℄) and 1 ([tj ; tj + 1℄). Let 0 and 1 be losed urves in D that are lose together. Let 0 (tj ) = zj , 1 (tj ) = j : Consider f 2 H(D) and let gj be a primitive of f on Dj ,

4.6 Homotopy Version of Cau hy's Theorem

173

zn = γ0 (tn ) = γ1 (tn ) = ζn zn zj +1

zj

ζj +1 ζj

z0

z0 = γ0 (t0 ) = γ1 (t0 ) = ζ0

Figure 4.17: Des ription for lose together urves.

whi h exists by Theorem 4.31. Further, if 0 and 1 are losed urves of

lass C 1 and are lose together, then

I0 = and

I1 =

Z

0 Z

1

f (z ) dz =

f (z ) dz =

nX1 j =0

[gj (zj+1 ) gj (zj )℄

nX1 j =0

[gj (j+1 ) gj (j )℄

so that I0 I1 = [gn 1(zn ) gn 1 (n )℄ [g0(z0 ) g0 (0 )℄: Sin e 0 and

1 are losed and sin e the primitives gn 1 and g0 dier by a onstant, I0 I1 = 0 and hen e I0 = I1 . The following theorem summarizes the above dis ussion.

4.52. Theorem. Let D be an open set in C and 0 and 1 be two

losed ontours in D. Suppose that 0 Rand 1 are lose together. Then, for R ea h f 2 H(D), we have 0 f (z ) dz = 1 f (z ) dz: Let : [0; 1℄

ompa t. De ne

! D be a urve in an open set D. Then, ([0; 1℄) is

(t) = min j (t)  j:  2C nD Then, (t) > 0 on [0; 1℄. Clearly  is ontinuous and  has a minimum on [0; 1℄. Further, is uniformly ontinuous on [0; 1℄. Let R = min0t1 (t). Therefore, for  = R=2 > 0 there exists a Æ > 0 su h that

j (t) (u)j < R=2 whenever jt uj < Æ (t; u 2 [0; 1℄): Choose points t0 = 0 < t1 <


< tn = 1 in [0; 1℄ su h that jtj tj+1 j < Æ. Then (see Figure 4.17),

([tj ; tj+1 ℄) 
( (tj ); R=2) = Dj  D:

174

Complex Integration

For all su h j , let j : [tj ; tj+1 ℄ ! D be the restri tion of to [tj ; tj+1 ℄. Sin ethe disk
( (tj); R=2) is onvex and lies in D, by Theorem 4.31, we R Pn 1 R have j =1 j f (z ) dz = 0 for ea h analyti fun tion f on D. That is, Z

(4.53)

f (z ) dz =

nX1 Z

j =1 j

f (z ) dz:

This also follows from the properties of the Riemann integral. Note also that, by Theorem 4.31, there exists an Fj on Dj su h that Fj0 (z ) = f (z ) on Dj . Thus, if j is of lass C 1 , (4.53) and Corollary 4.18 give Z

f (z ) dz = = = =

nX1 Z j =1 j nX1 Z j =1 j nX1 Z

f ( j (t)) j0 (t) dt Fj0 ( j (t)) j0 (t) dt

d[Fj ( j (t))℄ j =1 j nX1 [Fj (zj+1 ) Fj (zj )℄ j =1

where (tj ) = zj . Our next result is the famous Cau hy's theorem.

4.54. Theorem. (Homotopy Version of Cau hy's Theorem) Let D be domain in C and 0 and 1 be two losed R ontours in DR su h that

0 ' 1 in D. Then, for ea h f 2 H(D), we have 0 f (z ) dz = 1 f (z ) dz: Proof. Let F : 0 ' 1 be a homotopy in D. Sin e F is ontinuous on the square R = [0; 1℄  [0; 1℄ whi h is ompa t, the image F (R) is ompa t and F is uniformly ontinuous on R. Hen e, F (R) has a positive distan e from C n D. Choose partitions u0 = 0 < u1 <


< um = 1 t0 = 0 < t1 <


< tn = 1 and let Rjk = [tj ; tj+1 ℄  [uk ; uk+1 ℄ (j = 0; 1; : : : ; n 1; k = 0; 1; : : : ; m 1), a re tangle. Then, F (Rjk ) = Djk  D. De ne k by k (t) = F (t; uk ); k = 0; 1; : : : ; m: Then, k 's are ontinuous and the urves k , k+1 are lose together. By Theorem 4.52, Z

k

f (z ) dz =

Z

k+1

f (z ) dz; k = 0; 1; : : : ; m 1:

175

4.6 Homotopy Version of Cau hy's Theorem

As

0

= 0 and m = 1 , the desired equality follows.

4.55. Corollary. If D is a simply onne ted domain in C , then, for R 2 H(D) and any losed ontour in D, we have f (z ) dz = 0:

any f

R Proof. Let 0 be a onstant urve. Then, 00 (t) = 0 and so 0 f (z ) dz = 0: The result follows from Theorem 4.54 upon taking 1 = .

Another version (see also Corollary 4.61) of Corollary 4.55 is the following.

4.56. Corollary. If D is a simply onne ted domainR in C , then, 2 H(D), and 0 ; 1 2 (D; a; b) in D, we have 0 f (z ) dz = R f ( z ) dz:

1 for any f

R

Proof. Using Corollary 4.55, we obtain 0 1 f (z ) dz = 0 and the desired on lusion now follows. 4.57. Corollary. If 0 and 1 are two losed ontours in a domain D of C su h that 0 ' 1 then, for ea h a 2 C n D, we have n( 0 ; a) = n( 1 ; a): 4.58. Example. Let 0 and 1 be de ned by

0 (t) = e2it and 1 (t) = e 2it ; t 2 [0; 1℄: Then, 0 and 1 are the ir le jwj = 1. De ne F (t; u) = e2it 2iu sin(2t): Then, F : 0 ' 1 in D = C . Further, if D = C n f0g and f (z ) = 1=z then Z

0

f (z ) dz =

Z

1 dz = z

0

Z 1

0

1 2ie2it dt = 2i e2it

so that n( 0 ; 0) = 1. Similarly we see that n( 1 ; 0) = 1. Thus, 0 and 1 are not homotopi in D = C n f0g. 

4.59. Example. Consider two ir les

0 (t) = Re2it and 1 (t) = (R + 1)e2it (t 2 [0; 1℄); where R > 0 is xed. De ne F (t; u) = (R + u)e2it : Then, for t; u 2 [0; 1℄,

F (t; 0) = 0 (t); F (t; 1) = 1 (t) and F (0; u) = F (1; u) = R + u: Further, we also have jF (t; u)j  R + 1 for all u 2 [0; 1℄. Therefore, F is a  homotopy of 0 to 1 in any region D ontaining
R+1 .

176

Complex Integration

4.60. Example. Let 0 (t) = 2t and 1 (t) = 1 + ei(1 t) , t 2 [0; 1℄: De ne F (t; u) = (1 u) 0 (t) + u 1 (t): Then, jF (t; u)j  2 for all u 2 [0; 1℄.  Clearly, F : 0 ' 1 in any region D ontaining
2 . Next we give an alternate proof of Theorem 4.33 in the following form.

4.61. Corollary. If f is analyti in a simply onne ted domain D, then there is a fun tion F in D su h that F 0 (z ) = f (z ), and F is unique up to an additive onstant. Proof. Let z0 2 D. For ea h z 2 D, let z be a urve in D from z0 to z . De ne F by Z F (z ) = f ( ) d:

z

By Corollary 4.57, the value of F (z ) is independent of the hoi e of z . Therefore, for any ontour in D from z1 to z2 we have

F (z2 ) F (z1 ) =

(4.62)

Z

f ( ) d:

In parti ular, (4.62) holds for any line segment in D onne ting z1 and z2 . That is Z z2 F (z2 ) F (z1 ) = f ( ) d: z1

Here z2 is suÆ iently lose to z1 so that [z1 ; z2℄  D. It follows, from the method of proof of Theorem 4.31 (take z = z1 , z + h = z2 in Theorem 4.31), that F 0 (z ) = f (z ) in D. Now let G be any other fun tion in D su h that G0 (z ) = f (z ). Applying Corollary 4.18, it follows that for any points z1 ; z2 in D and any ontour in D from z1 to z2 , we have

F (z2 ) F (z1 ) = and therefore, F

Z

F 0 (z ) dz =

Z

f (z ) dz =

Z

G0 (z ) dz = G(z2 ) G(z1 )

G is onstant in D.

4.7 Cau hy Integral Formula The Cau hy integral formula expresses a remarkable fa t about an analyti fun tion. Its values everywhere inside a simple losed ontour are ompletely determined by its values on the boundary. The integral representation allows us to show that analyti fun tions are in nitely dierentiable. In fa t the values of ea h derivative of an analyti fun tion are determined just by the values of the fun tion on the boundary. Later in Se tion 4.10, we use the integral representation to obtain power series expansions for analyti fun tions.

177

4.7 Cau hy Integral Formula

4.63. Theorem. (Cau hy Integral Formula) If D is a simply onne ted domain and is a losed ontour in D, then for f 2 H(D) and a 2 D nf g, Z 1 f ( ) d: 2i  a Proof. Let f 2 H(D) and onsider

f (a)n( ; a) =

(4.64)

8
1. On the other hand, we know that 

Z

dz 0 if n 2 Z nf1g, J= = n 2i if n = 1: ( z a ) jz aj=r Here the non-zero result arises be ause of the fa t that the integrand in J is not analyti at z = a and this point lies inside the ir le jz aj = r whi h violates the ondition of Cau hy's theorem. A partial onverse of Cau hy's theorem is the following. The power of this theorem resides in the fa t that only the ontinuity of f is assumed. 4.86. Theorem. (Morera's Theorem) R Suppose that f is ontinuous in an open set D with the property that C f (z ) dz = 0 for ea h losed

ontour C in D. Then, f is analyti in D. Proof. Let a be an arbitrary xed point in D. Sin e the integral f C (z ) dz = 0 holds for ea h losed ontour C , for any two ontours C1 and C2 in D (see Figure 4.21) whi h onne t a with z1 , we have R

0=

Z

C1 C2

f (z ) dz =

Z

C1

f (z ) dz R

Z

C2

f (z ) dz:

This means that the value of the integral az1 f (z ) dz does not depend upon the path whi h onne ts a with z1 . De ne

F (z ) =

Z z

a

f ( ) d for z 2 D:

191

4.8 Morera's Theorem

Sin e D is open, there exists an r > 0 su h thatR
(z ; r)  D. Let jhj be suÆ iently small so that z + h 2
(z ; r). Sin e az f ( ) d is independent of the path, as usual, we may write

F (z + h) F (z ) = so that

Z z+h

z

f ( ) d

Z

F (z + h) F (z ) 1 z+h f (z ) = [f ( ) f (z )℄ d: h h z Sin e f is ontinuous, as in the proof of Theorem 4.31, it follows that F (z + h) h

F (z )



f (z ) ! 0 as jhj ! 0

and so F is dierentiable with F 0 (z ) = f (z ) in
(z ; r). Sin e F is analyti in
(z ; r), it follows from the Cau hy integral formula for derivatives that f is analyti in
(z ; r). Sin e z was an arbitrary point of D, we on lude that f is also analyti in D.

4.87. Remark. Suppose that f (z ) =

8 < :

(z

1

1)2

if z 2
(1; r) nf1g

1 if z = 1

8 <

os z if z 2
nf0g z2 ; g(z ) = : 0 if z = 0:

Then (see Example 4.4), for any simple losed ontour C R

in
(1; r), we have f ( z ) dz = 0 : However f is not analyti for z 2
(1; r), sin e f is not C even ontinuous at z = 1. Note that Morera's Theorem is not appli able sin e the ontinuity requirement is not satis ed. R Similarly for the g(z ) de ned above, we know that jzj=1 g(z ) dz = 0: Again g is not analyti in
, sin e g is not ontinuous at z = 0.  Most often we use Morera's Theorem but for a dierent situation other than that is stated.

4.88. Corollary. (Riemann's Removability Theorem) Suppose that f is ontinuous on a domain D and analyti on D nfz0 g for some z0 2 D. Then, f is analyti on D. Proof. It is enough to prove theR result for a disk jz z0 j < . By Cau hy's theorem for a disk, we have C f (z ) dz = 0 for all losed ontours C inside the disk jz z0j < . Hen e, f is analyti on jz z0 j <  by Morera's Theorem.

192

Complex Integration

z z0

R1 R2

Figure 4.22: Curve in a ring shaped region.

4.9 Existen e of Harmoni Conjugate Corollary 4.75 helps us in nding a harmoni onjugate of a harmoni fun tion. We shall now give the idea behind the formal solution to the statement of Theorem 3.39. Suppose  = (x; y) is the real part of an analyti fun tion f in a simply onne ted domain D and suppose we an nd = (x; y) su h that f =  + i . The C-R equations would then imply f 0 = x iy and so if is a path whi h onne ts z to z0 in D,

f (z ) f (z0) =

Z

f 0 ( ) d =

Z

[x

iy ℄ d

whi h is the only possible solution if we are given . The following example shows that we annot hoose D to be just a domain. p

4.89. Example. For u(x; y) = ln x2 + y2 in D = fz : R1 < jz j < R2 g where R2 > R1  0, there does not exist a v(x; y) su h that f = u + iv is analyti in D. To prove this assertion, we suppose that f = u + iv is analyti in D. Then, f 0 is also analyti in D. By hypothesis, x y ux(x; y) = 2 2 and uy (x; y) = 2 2 x +y x +y so that x iy z 1 f 0 (z ) = ux(x; y) + ivx(x; y) = ux (x; y) iuy (x; y) = 2 2 = = : x +y zz z Rz 0 So, by Theorem 4.16, the value of the integral z0 f (z ) dz is independent of the path joining z0 and z in D (see Figure 4.22). Thus, we have Z z Z z 1 f (z ) f (z0 ) = f 0 ( ) d = d: z0 z0  In parti ular, if we hoose the losed ontour j j = R with initial point z0 = R and the terminal point z = Re2i (R 2 (R1 ; R2 ) ), then the above gives that Z Z 2 1 iRei 2 i 0 = f (Re ) f (R) = d = i d = 2i j j=R  0 Re

193

4.9 Existen e of Harmoni Conjugate z = (x, y)

(x, y0 )

z0 = (x0 , y0 )

Figure 4.23: Determination of onjugate harmoni .

and this ontradi tion shows that no su h v(x; y) an exist in D su h that f = u + iv is analyti in D. 

Proof of Theorem 3.39. Sin e  is harmoni in D, Mx Ny = 0; where M = x and N = y . Now M and N are dierentiable fun tions of (x; y) in D and Ndx + Mdy is an exa t dierential for (x; y) 2 D, i.e. there exists a fun tion su h that d = Ndx + Mdy; for (x; y) 2 D: With respe t to a referen e point (x0 ; y0 ) 2 D, let be a path (see Figure 4.23) onsisting of the line segment onne ting (x0 ; y0 ) to (x; y0 ) and (x; y0 ) to (x; y). De ne (x; y) = (4.90)

=

Z

N dx + M dy + k

Z x

y (t; y0 ) dt +

x0

Z y

y0

x (x; s) ds + k

where k is some real onstant. The partial derivative of (4.90) with respe t to y is given by

 y (x; y ) = y

Z y

y0



x (x; s) ds = x (x; y);

sin e the rst integral in (4.90) is independent of y. Similarly, taking the partial derivative of (4.90) with respe t to x yields (4.91)

 x (x; y ) = y (x; y0 ) + x

Z y

y0



x (x; s) ds :

Using the dierentiation formula under the integral sign for the se ond term in (4.91), we have (as in the proof of Theorem 3.53) x (x; y ) =

y (x; y);

y (x; y ) = x (x; y ):x

194

Complex Integration

Thus,  and satisfy the C-R equations for F =  + i and is harmoni in D. By Theorem 3.26, F =  + i is analyti in D. Thus, a harmoni

onjugate always exists in D and the proof is ompleted.

4.92. Remark. We observe the following: (i) If the domain is multiply onne ted and  is harmoni there, then the

onjugate fun tion be omes multiple-valued. For instan e if p

(x; y) = ln x2 + y2 ; then the orresponding multiple-valued onjugate fun tion is y  (x; y) = Ar tan + 2k + onstant; k 2 f1; 2; : : : g: x (ii) From Theorem 4.33, we note that an arbitrary harmoni fun tion  in a simply onne ted domain an always be onsidered as the real (or imaginary) part of an analyti fun tion in D. 

4.10 Taylor's Theorem

P

In Theorem 3.71, we have shown that every power series n0 an (z a)n is in nitely dierentiable in its disk of onvergen e
(a; R). Moreover, an = f (n) (a)=n! and it does not depend on R. Now, we use omplex integration to show that every analyti fun tion in a domain an be expressed lo ally as a onvergent power series. This fa t opens the door to a systemati dis ussion of the lo al stru tural properties of analyti fun tions.

4.93. Theorem. analyti in
R , then f has a Ma laurin series P If f is n a z n0 n for all z 2
R , where

expansion f (z ) =

an =

Z

f (n) (0) 1 f ( ) = d; n = 0; 1; 2; : : : n! 2i  n+1

with = f : j j = rg and 0 < r < R.

Proof. For a given z 2
R , hoose r su h that r < R and let be

= f : j j = rg. By the Cau hy integral formula, f (z ) =

Z

1 f ( ) d: 2i  z

We know that, for all jz= j < 1,



1

z

=

X zn 1 1
=  1 z= n0  n+1

195

4.10 Taylor's Theorem

and the onvergen e is uniform on , and for a xed z . Now f ( ) is bounded on , sin e it is a ontinuous fun tion on the ompa t set . We form

f ( ) X f ( ) = f ( ) with fn ( ) = n+1 z n :  z n0 n 

(4.94)

Viewed as a series of fun tions of  , the series (4.94) onverges uniformly on . Indeed, in
 , f ( ) is bounded for ea h  < r, with bound M , say, and " #  n jfn ( )j  sup jfn ( )j  rnM+1 n = Mr r j j= P

and so n0 fn ( ) is uniformly onvergent on
 , by the Weierstrass M test. Therefore, the series (4.94) may be integrated term-by-term so that 

Z



Z

X X 1 f ( ) 1 f ( ) d = d z n = an z n: n +1 2i  z 2 i 

n0 n0 P

Sin e z 2
R is arbitrary, we have f (z ) = n0 an z n for all jz j < R and so the assertion now follows from Cau hy's integral formula for derivatives. Uniqueness of the oeÆ ients an follows from the Uniqueness theorem for Taylor series. By Theorem 4.16, the value of the integral is independent of the hoi e of the urve in jz j < R. Using the simple transformation z a = w, we obtain the following result whi h shows that every analyti fun tion an be expressed lo ally as a onvergent power series. However, we observe that several power series (sometimes perhaps in nitely many) may be required to represent f throughout the domain.

4.95. Corollary. (Taylor's Theorem) If f is analyti in a domain D then for z 2
(a; R)  D, f has the Taylor series expansion f (z ) =

X

n0

an (z

where C = f : j

a)n ;

Z

f (n)(a) 1 f ( ) an = = d; n! 2i C ( a)n+1

aj = rg and 0 < r < R.

If f is analyti in a domain D P with a 2 D, then f admits a Taylor n series expansion about a: f (z ) = 1 n=0 an (z a) : What is the radius of onvergen e of the series we have obtained? It may happen that the

ir le of onvergen e en loses points outside D. By Corollary 4.95 what we know is that if f is analyti in
(a; r), then the series about a onverges to f (z ) in that disk. The series onverges in at least the largest disk entered at a that is ontained in D. Clearly, r may be in reased until the ir le

196

Complex Integration

jz aj = r en ounters a singularity of f (z ). Thus, the radius of onvergen e R is the largest number R su h that f (z ) extends to be analyti on the disk jz aj < R and the extended fun tion is alled an analyti ontinuation of f . We shall have a preliminary dis ussion on this issue in Chapter 10. However, the example given below will give a little avor of this idea. When f (z ) is a single-valued analyti bran h of a multi-valued fun tion, then bran h points give obsta les just as mu h as a singularity. Thus, the radius of

onvergen e is the distan e from the enter of the expansion to the nearest singularity or bran h point. 4.96. Examples. We know that f (z ) = Log (1 z ) is analyti in the ut plane C n fx + i0 : x  1g. In parti ular, f is analyti in jz j < 1 and has a Taylor series expansion in jz j < 1 about 0: 1 X f (n) (0) f (z ) = an z n ; an = : n! n=0 Note that 1 is the largest number R su h that f extends to be analyti in the disk jz j < R. It follows that f (0) = Log 1 = 0 and for n  1,

f (n) (0) = So, for jz j < 1,



1
2


(n 1) = (n 1)!: (1 z )n z=0 Log (1 z ) =

1 zn X n=1

n

:

Note that an appli ation of the Ratio/Root test qui kly on rms that the series onverges absolutely for jz j   < 1. Next, we onsider the fun tion g de ned by Log (1 + z ) g(z ) = : z Then g 2 H(D), D = C n (fx + i0 : x  1g [ f0g). Although z = 0 is a singularity, g an be extended to be analyti at z = 0 (it is a removable singularity of g but we shall dis uss various singularities in Chapter 7) be ause 1 X zn Log (1 + z ) = ( 1)n 1 ; jz j < 1: n n=1 Moreover, if we let

1 Log (1 + z ) X = an (z 6)n z n=0 then the radius of onvergen e of this series is R = 7 (the distan e from the bran h point 1 to 6) not the distan e from 0 to 6. 

197

4.10 Taylor's Theorem

4.97. Example. Let f (z ) = Logp z for z 2 C n fx + i0 : x  0g and p a = 1 + i = 2e3i=4 . Then f (a) = ln 2 + 3i=4 and for n  1, f (n) (z ) =

( 1)n 1 (n 1)! zn

and

f (n) (a) ( 1)n 1 3in=4 = e n! n2n=2 so that the power series expansion of f about a is 1 ( 1)n 1 e 3in=4 X f (z ) = (z a)n : n=2 n 2 n=0 an =

It is easy to see (for example, by the Ratio p test) that the radius of onvergen e of the series on the right is R = 2. Again, this does not ontradi t the dis ontinuity of Log z at theppoint z = 1, be ause Log z extends to be analyti for jz ( 1 + i)j < 2 although the extension does not oin ide with Log z in the part of the disk that lies in the lower half-plane, see Chapter 10 for a dis ussion on analyti ontinuation. 

4.98. Example. Consider f : f (x) =

R ! R given by 1 : 1 + x4

This fun tion admits a real power series about any point a 2 R. Yet the power series about a = 0, given by 1 X f (x) = ( 1)n x4n n=0

has ( 1; 1) as its interval of onvergen e. On the other hand, its omplex analog 1 f (z ) = 1 + z4 has singularities at zk = ei(1+2k)=4 ; k = 0; 1; 2; 3. Clearly, the distan e from 0 to the nearest singularity is 1, whi h ne essarily is the radius of

onvergen e for the orresponding series about 0.  Consider the fun tion f (z ) = z 1: Then, f is analyti on C n f0g. Therefore, for all z 6= 0,

f (n) (z ) =

( 1)n n! ; n = 0; 1; 2; : : : ; z n+1

198

Complex Integration

and so by Taylor's Theorem, we have the Taylor series expansion of f about z0 6= 0, 1 X ( 1)n = (z z0 )n for z 2
(z0 ; jz0 j): z n0 z0n+1 After k-fold dierentiation this gives 1

z k+1

n 1 ( 1)n k (z k z0n+1 nk X

=

z0 )n k

for z 2
(z0 ; jz0 j). The substitution z0 = 1 with the transformation z = w 1 yields the Ma laurin series of (1 w) k 1 : (4.99)

X n 1 = wn k for w 2
; (1 w)k+1 nk k P

whi h is the k-fold dierentiation of the geometri series n0 wn . We remark that the oeÆ ient formulae (see Corollary 4.95) may not be dire tly useful in writing the Taylor expansion for analyti fun tions. The formula (4.99), the orresponding formulae for ez , sin z , os z et ., and formulae for the Cau hy produ t are more often useful in su h problems.

4.100. Example. Let us develop the fun tion f de ned by f (z ) =

1 1 z z2

into a Taylor series about 0. By partial fra tion de omposition we nd that

f (z ) = =



1 1



1

5 1 = ; = 2 Note that  < j j,  = 1 and  = of f about 0 is 1



 z  z      1 1 1 1 ;  1 z=  1 z=

p

where

1

2

X

zn n+1

p

5+1 : 2 p 5. Therefore, the Taylor series

X

3

zn 5 n+1

4 (jz j < );  n0 n0   1 X n+1 n+1 n = p z (jz j < ); 5 n0 ( )n+1

f (z ) =



199

4.11 Zeros of Analyti Fun tions

sin e the rst series onverges for jz j < j j while the se ond series onverges for jz j <  so that the ombined series onverges for jz j < minf; j jg = . As  = 1, we rewrite the above equation as 2

1 1 4 =p 1 z z2 5 n0 X

p

5+1 2

!n+1

p

1

2

5

!n+1 3

5 zn;

jz j < :

p

For jz j <  = ( 5 1)=2, we may write 1 = a z n ; i.e. 1 = (1 z 1 z z 2 n0 n X

2

z 2) 4

X

n0

3

an z n5 :

By uniqueness of Taylor's oeÆ ients of f about 0, on equating the oeÆ ients of z n on both sides, we see that

a0 = a1 = 1 and an+1 = an + an 1 : Note that the sequen e fan gn0 obtained here gives the Fibona

i sequen e 1; 1; 2; 3; 5; 8; 13; : : : . 

4.11 Zeros of Analyti Fun tions We now dis uss the zeros of analyti fun tions using the Taylor series expansion as a tool. Suppose that f is analyti , f (z ) 6 0 in an open set D and f vanishes at some point a 2 D. Then f admits a Taylor series about a: f (z ) = a1 (z a) + a2 (z a)2 +


; jz aj < R; where fz : jz aj < Rg  D. Sin e f (z ) 6 0, not all the oeÆ ients ak an vanish. This shows that there is a positive integer m  1 su h that

a1 =




= am 1 = 0; but am 6= 0:

Then we say that f has a zero ( nite) of order m at a. Also, the integer m is referred to as the multipli ity of the zero of f at a. Zeros of order 1 are often alled simple zeros. In ase f has a zero of order m at a, we may rewrite f (z ) as

f (z ) = am (z a)m + am+1 (z a)m+1 +




= (z a)m g(z );

where g(z ) is analyti at a and g(a) = am 6= 0 (Note that the radius of

onvergen e of am + am+1 (z a) +


is exa tly same as that of

am (z a)m + am+1 (z a)m+1 + The above dis ussion leads to




):

200

Complex Integration

4.101. Proposition. A fun tion f analyti at a has a zero of order m at a i f (z ) = (z a)m g(z ), where g is analyti at a and g(a) 6= 0. Further, one an use this Proposition to establish the L'H^ospital rule for omplex fun tions, see Exer ise 4.161. A zero of an analyti fun tion f is said to be isolated if it has a neighborhood in whi h there is no other zero of f . An important onsequen e of the following result is the Identity theorem for analyti fun tions.

4.102. Theorem. Every zero of an analyti fun tion f (6 0) is iso-

lated.

Proof. Suppose that f has a zero of order m at a. Then there exist an R > 0 su h that f (z ) = (z a)m g(z ); jz aj < R; where g is analyti at a and g(a) 6= 0. Let jg(a)j = 2 > 0. Then for this , sin e g is ontinuous at a, there exists a Æ > 0 su h that jg(z ) g(a)j <  whenever jz aj < Æ: Therefore when jz aj < Æ we have jg(z )j = jg(a) [g(a) g(z )℄j  jg(a)j jg(z ) g(a)j > 2  = : Thus, g(z ) 6= 0 in
(a; Æ) (We remind the reader that we have already noti ed this point while dis ussing the limit of a fun tion, see Theorem 2.10). But jz ajm 6= 0 in 0 < jz aj < Æ. Hen e, f (z ) = g(z )(z a)m 6= 0 in this neighborhood ex ept at a. This ompletes the proof. The following theorem plays an important role in omplex fun tion theory. In simplest terms this theorem, whi h is an extension of Theorem 3.75, ompletely hara terizes an analyti fun tion in a domain D just by its behavior in a small subset of D.

4.103. Theorem. (Identity/Uniqueness Theorem) Suppose that f is analyti in a domain D. If S , the set of zeros of f in D, has a limit point z  in D. Then f (z )  0 in D. The hypothesis that D is onne ted in Theorem 4.103 is ne essary. For example, if D = C n fz : 1  jz j  3g and if f : D ! C is de ned by  for jz j < 1 f (z ) = 20 for jz j > 3,

then f 2 H(D) and its zero set ( jz j < 1) has a limit point in D, yet f is not identi ally zero in D.

4.104. Remarks. The following observations are important:

201

4.11 Zeros of Analyti Fun tions

(i) Consider a real-valued fun tion f of a real variable de ned by 

1=x 0 f (x) = e 0 ifif xx  < 0:

For x 6= 0, it is lear that f has derivatives of all orders. In fa t, it is a simple exer ise to see that f is in nitely dierentiable at all points of R and, in parti ular,

f (n) (x) = 0 for x  0 and all n = 0; 1; 2; : : : :

However, f (x) does not vanish in R. Thus, in the ase of real-valued fun tions the behavior of in nitely dierentiable fun tions in one region of its domain of de nition has no ee t on its behavior on some other region. However, the Uniqueness theorem shows that this is not the ase with fun tions of a omplex variable as we shall soon see its remarkable role in a number of elementary onne tions as well. (ii) The hypothesis that the limit point a lies in D is not super uous. For example, onsider   1 f (z ) = sin : 1 z Then, f 2 H(
) and the zeros of f are given by z = 1 1=(n) (n 2 Z). But the zeros of f that lie inside
are 1 zn = 1 (n 2 N ) n and zn ! 1 as n ! 1 yet, f (z ) 6 0. Obviously, f is not analyti at 1 and 1 62
. (iii) Consider f (z ) = exp (z=(1 z )) 1 for z 2
. Then, f 2 H(
). The zeros of f are obtained from solving z=(1 z ) = 2ni: This gives 2ni zn = (n 2 Z) 1 + 2ni so that f has in nitely many zeros and, sin e jzn j < 1 for ea h n 2 Z, it follows that ea h of them lies inside
. In fa t, zn







1 1 1 + 2ni 1 = = 2 2 1 + 2ni 2

whi h implies that the zeros of f a tually lie in
\ 
(1=2; 1=2). Thus, f (z ) 6 0 and yet f has in nitely many zeros in
on whi h f is analyti . Obviously, f is not analyti at z = 1 and the point 1 is a limit of point of the above sequen e of zeros of f . In fa t, we may rewrite f (z ) as   1 f (z ) = e 1 exp 1: 1 z

202

Complex Integration

The reader with knowledge of isolated singular points an qui kly re ognize the point z = 1 as an essential singularity of f and also of the fun tion f onsidered in (ii). (iv) From the last two examples in (ii) and (iii), we also observe that there are non- onstant analyti fun tions in
having in nitely many zeros in
.  A dire t onsequen e of Theorem 4.103 is that if a set of zeros of an analyti fun tion in a domain D ontains an in nite sequen e of distin t points that has its limit point in D, then the fun tion is identi ally zero in D.

Sets su h as an interval (a; b) of R and an open disk in C always an ontain an in nite sequen e of points that onverges in the set itself. Consequently, the Uniqueness theorem is often helpful in he king the validity in the

omplex plane of ertain fun tional identities known to be true in R or in an open subset of R. For instan e, onsider

f1 (z ) = sin2 z + os2 z 1 and f2(z ) = osh2 z sinh2 z 1: We know from elementary mathemati s that the identities

f1 (z ) = 0 and f2 (z ) = 0

(4.105)

hold when z is real. Sin e trigonometri fun tions sin z; os z; osh z and sinh z are all entire, both f1 and f2 are analyti in C . Sin e the x-axis

ontains a sequen e of distin t points (for example 1=n, 1=n) onverging to an element in it, the Uniqueness theorem immediately shows that the identities in (4.105) hold for all z 2 C (Note that every point of R is a limit point). Similarly, we an easily derive the following identities as easy onsequen es of the Uniqueness theorem: 1 + tan2 z 1 + ot2 z 1 tanh2 z 1 + s h 2 z

= = = =

se 2 z; for all z 2 C n f(2k + 1)=2 : k 2 Zg ;

s 2 z; for all z 2 C n fk : k 2 Zg; se h 2 z; for all z 2 C n f(2k + 1)i=2 : k 2 Zg ;

oth2 z; for all z 2 C n fki : k 2 Zg;

(see also Se tion 3.4). Another interesting identity we derived earlier in (3.78) is ez1 +z2 = ez1
ez2 : Assume that this holds when z1 and z2 are real. Then to he k the validity of this for all z1 and z2 in C , using the Uniqueness theorem, we pro eed as follows: Let z2 = x2 , a xed real and z1 = z , a omplex variable. Then

ez+x2 = ez
ex2

4.11 Zeros of Analyti Fun tions

203

holds when z is real and so, by the Uniqueness theorem, this identity is true for all z 2 C . Fixing z as z1 , in parti ular, ez1+x2 = ez1
ex2 : This being true for all real x2 , it follows, by the same argument that ez1 +z = ez1
ez for all z 2 C : Taking z = z2 we get ez1 +z2 = ez1
ez2 : Similarly, it an be seen that sin(z1 + z2 ) = sin z1 os z2 + os z1 sin z2

os(z1 + z2 ) = os z1 os z2 sin z1 sin z2 ; whi h hold for all real values of z1 and z2 , ontinues to hold for omplex values of z1 and z2 . At this point we remind the reader that not all the familiar properties of the fun tions sin, os, tan et . as fun tions of a real variable remain true when these are viewed as fun tions of a omplex variable. We know that sin x and os x are bounded by 1 for all real x. On the other hand if y > 0 we have y2 ey e y ey + e y

os(iy) = >1+ and j i sin(iy)j = >y 2 2 2 whi h shows that there exists no onstant K for whi h j os z j < K and j sin z j < K in C . Our next example deals with the binomial expansion. Let X n 1 f (z ) = ; and g(z ) = zn k: (1 z )k+1 k nk Then, from real variable al ulus, we know that f (z ) g(z ) = 0 when z = x is real and jxj < 1. Clearly f is analyti for z 2 C n f1g. The radius of onvergen e of the series with sum g is 1. Therefore, in parti ular, f and g are analyti for z 2
. The Uniqueness theorem immediately yields f (z ) = g(z ) for all z 2
, i.e.   X n 1 = z n k for all z 2
: (1 z )k+1 nk k Using the same method, we an obtain a generalized version of the binomial expansion a(a 1) 2 a(a 1)(a 2) 3 z + z


; jz j < 1; (1 z )a = 1 az + 2! 3! where a is an arbitrary omplex number (see also Theorem 3.112). Proof of Theorem 4.103. Let fzng be a sequen e of zeros of f in D su h that zn ! z  , where z  is also a point in D. Then we note that, sin e zn ! z , f (zn ) = 0 for all n, and f is ontinuous at z  , f (z ) = nlim !1 f (zn) = 0:

204

Complex Integration

Then, by Theorem 4.102, either f (z )  0 in a neighborhood of z  or f (z ) 6= 0 in some pun tured neighborhood
(z  ; Æ) nfz g  D. The se ond part of the statement ontradi ts f (zn) = 0 sin e, for suÆ iently large n, zn lies in this pun tured neighborhood of the limit point z  . So we must have f (z )  0 in any disk
(z  ; R)  D. To omplete the proof we have to show that f (z )  0 in the whole of D, and for this we split D into two sets: A = f 2 D :  is a limit point of S g B = f 2 D :  62 Ag; where S is the set of zeros of f in D. Then D = A [ B and A \ B = ;: Let  2 A. Then, sin e  2 A is a limit point of S , f (z )  0 in a neighborhood of  . Thus ea h point in A is an interior point of A. Indeed, if z 0 2
( ; Æ) for some Æ > 0, then jz 0  j < Æ. Set  = Æ jz 0  j and z 2
(z 0 ; ), i.e. jz z 0 j < : Then jz  j  jz z 0j + jz 0  j <  + jz 0  j = Æ: It follows that
(z 0 ; ) 
( ; Æ); so A is open and is non-empty, by hypothesis. To show B is open, let  0 2 B . Sin e  0 is not a limit point of S , by

ontinuity of f at  0 , there exists Æ > 0 su h that f (z ) 6= 0 throughout
( 0 ; Æ)  D. Sin e D is onne ted and nonempty, it annot be written as the union of two non-empty disjoint open sets. Hen e we must have either A = ; or B = ;. But by hypothesis z  2 A. It now follows that A 6= ; and therefore B = ;. Thus A = D and every  2 D is a limit point of the zeros of f so that f  0 in D. The following result is also referred to as the Uniqueness theorem whi h is onsidered to be a fundamental result for an introdu tion to the on ept of analyti ontinuation (see Chapter 10).

4.106. Theorem. Suppose that f and g are analyti in a domain D. If S , the set of zeros of f g in D, has a limit point z  in D, then f (z )  g(z ) in D. Proof. Apply Theorem 4.103 to h = f g. 4.107. Example. The Uniqueness theorem provides another example of fundamental dieren es between omplex-valued fun tions and realvalued fun tions. For example, for k 2 N , de ne fk : R ! R by  2k x sink (2=x) for x 6= 0 fk (x) = 0 for x = 0:

4.11 Zeros of Analyti Fun tions

205

Then, for ea h k, fk is ontinuously dierentiable on R and fk (1=n) = 0 for all n 2 N , yet fk and fm are distin t fun tions for k 6= m. Next, we onsider a new fun tion gk : C ! R de ned by gk (z ) = fk (jz j); where fk is as above. Then, for ea h k, gk is ontinuous in C and gk (1=n) = 0 for all n 2 N , yet gk and gm are distin t fun tions in C for k 6= m. This example demonstrates that two dierent ontinuous fun tions in a domain D an assume the same values on an in nite set whi h has a limit point in D. It follows that the Uniqueness theorem does not hold for ontinuous fun tions. 

4.108. Corollary. Suppose that f and g are analyti in a domain D su h that f (z )g(z ) = 0 for ea h z 2 D. Then either f (z ) = 0 for z 2 D or g (z ) = 0 for ea h z 2 D.

Proof. Suppose that f (a) 6= 0 for some a 2 D. Then, by the ontinuity of f , there exists a disk
(a; Æ)  D su h that f (z ) 6= 0 in
(a; Æ). But then g(z ) = 0 for all z 2
(a; Æ). By the Uniqueness theorem, g(z )  0 in D. 4.109. Corollary. Suppose that f is analyti in a domain D and f 0 (z ) = 0 in some disk ontained in D. Then f (z ) is onstant in D. Proof. By Theorem 3.31, f is a onstant in a disk D1  D, say . Therefore, every point of D1 is a limit point of fz : f (z ) = 0g. By the Uniqueness theorem, f (z ) = 0 throughout D. 4.110. Example. Let S = f1=n : n = 1; 2; : : : g. Then S and S has a limit point 0.

 [0; 1℄

(i) Suppose that f 2 H(C ) and f (z ) = os z for z 2 S . Then by the Uniqueness theorem, sin e 0 2 C , f (z ) = os z in C . Note that f is determined throughout C just by its values at the points 1=n. (ii) Let f (z ) = e2i=z 1, z 6= 0. Then f is analyti for all z 2 C n f0g and f (zn) = 0 for zn = 1=n, n 2 Z. Even though zn ! 0 as n ! 1, we annot laim f (z )  0, sin e the ondition that the limit point 0 must lie in C n f0g is not satis ed. Noti e that zn is an isolated zero of f for ea h n = 1; 2; : : : whereas 0 is not in C n f0g. In this example, f annot even be de ned at z = 0 so as to make f to be ontinuous at 0. Similarly, the fun tion F de ned by F (z ) = sin(=z ) for z 6= 0, is analyti in C n f0g and F (1=n) = 0 for n 2 Z. Is F identi ally zero? Is it possible to extend F to make it ontinuous at 0? (iii) Let f (z ) = expf Log z g. Sin e Log z is seen to be analyti in D , where D = C n fz = x : x  0g, f is analyti in D . We know that expf Log z g = z for z = x > 0. Therefore, we have expf Log z g = z for z 2 D ;

206

Complex Integration

as we have obtained this fa t earlier but by a dierent approa h.  In general, using the Uniqueness theorem, we on lude the following:

4.111. Corollary. If f1 (z ), f2(z ), : : : are analyti fun tions de ned in a domain D and satisfy a ertain algebrai identity (or fun tional identity su h as in (4.105)) on a set with a limit point in its domain of analyti ity, then these fun tions satisfy the same identity throughout the domain of analyti ity. 4.112. Example. If f 2 H(
) and jf (z )j  1 jz j in
, then it is easy to see that f (z ) = 0 in
. Indeed, if a 2
r (0 < r < 1) is xed, then, by the Cau hy integral formula f (a) =

Z

1 f ( ) 1 d = 2i j j=r  a 2

Z 2

0

f (rei )rei d rei a

so that the standard estimate (see Theorem 4.9(iv)) gives

jf (a)j  (1r jra)jr ! 0 as r ! 1: Thus, f (z ) = 0 on
r as a is arbitrary. By the Uniqueness theorem, f (z ) = 0 on
.  By the method of proof of Theorem 4.103, we next show that an analyti fun tion in a domain annot vanish together with all its derivatives at some point inside the domain unless it is identi ally zero.

4.113. Theorem. (Uniqueness Theorem for Power Series) Let f be analyti in a domain D. Suppose that at some point a 2 D, f (n) (a) = 0 for all n = 0; 1; 2; : : : . Then, f (z )  0. Proof. Let a 2 D. There exists an r > 0 su h that
(a; r)  D. By Corollary 4.95, X f (n) (a) f (z ) = (z a)n n ! n0 holds for z 2
(a; r)  D so that, by hypothesis, f (z ) = 0 on
(a; r); whi h in turn implies that f (z )  0 on D by Theorem 4.103.

We now dis uss analyti fun tions in relation to the simple operation of multipli ation on their Taylor series.

207

4.12 Laurent Series P

4.114. Theorem. Suppose that the power series n0 an z n and n n0 bn z are onvergent for jz j < R1 and jz j < R2 with sums f (z ) and g(z ), respe tively. Then, we have

P

f (z )g(z ) =

X

n0

n z n ; n =

n X k=0

ak bn k ; for jz j < minfR1 ; R2 g = R:

Proof. By Theorem 3.71, f and g are analyti for jz j < R with an = f (n) (0)=n! and bn = g(n) (0)n!

and fg is analyti for jz j < R. Therefore, by the Cau hy produ t

f (z )g(z ) =

X

n0

n z n

onverges for jz j < R and hen e is analyti . Thus, by Taylor's theorem

n =

n n X 1 n! (fg)(n)(0) X = ak bn k = f (k) (0)g(n k) (0): n! n ! k !( n k )! k=0 k=0

4.12 Laurent Series Suppose that f is not de ned, or is not analyti , at a point a. Then, we annot express it in a neighborhood of a as a onvergent power series P expansion of the form f (z ) = n0 an (z a)n ; for, if we ould do so then, by Theorem 3.71, f would be analyti at a. For instan e, onsider f (z ) = sin(1=z ): The series representation for f is obtained by onsidering the series for sin z and repla ing z by 1=z , whi h gives a series involving negative powers of z : 1 1 1 f (z ) = +


; z 6= 0: z 3! z 3 It onverges for all z with z 6= 0. More generally, a series of the form (4.115)

X

n0

bn (z a) n

an be thought of as a power series in the variable 1=(z a). Letting  = 1=(z a), the above series be omes an ordinary power series in  : (4.116)

X

n0

bn  n :

The next theorem shows how the properties of power series in negative powers su h as (4.115) an be dedu ed from the orresponding properties of ordinary power series. In view of (4.116) and Theorem 3.64, we have

4.117. Theorem. Let r = r 1 = lim supn!1 jbn j1=n .

208

Complex Integration

(i) If r = 0, then the series (4.115) onverges absolutely for every z 2 C 1 nfag. (ii) If 0 < r < 1, then the series (4:115) onverges absolutely for all z with jz aj > r, the onvergen e being uniform on jz aj   > r and diverges for jz aj < r. (iii) If r = 1, then the series (4:115) diverges for all nite z . P1 We say that anP\in nite series" n= 1 An onverges to P of the form a limit L i both n0 An and n1 A n onvergePand the sum of their limit is L. Equivalently, we say that the double series 1 n= 1 An onverges to L i given  > 0 there exists an N su h that m X

k= n

Ak

<  whenever m; n  N:

It is important to observe thatPm and n are independent here. In fa t, the existen e of the limit limn!1P nk= n Ak does not in general imply that the

orresponding double series 1 n= 1 An onverges. 1 Let R = lim supn!1 jan j1=n and r = r 1 = lim supn!1 jbn j1=n and suppose that 0 < R; r < 1. Then, by Theorems 3.64 and 4.117, the series

f1 (z ) =

X

n0

an z n = a0 +

X

n1

an z n

onverges absolutely for jz j < R and diverges if jz j > R, while

f2 (z ) =

X

n0

bn z n = b0 +

X

n 1

b nzn

onverges absolutely for jz j > r and diverges if jz j < r. So there is a non-empty region of onvergen e for the series of the form (4.118)

f (z ) =

1 X n=

1

An z n ; An =

8
|b|

Laurent-1 Taylor

a

O |z| < |a|

b

|a| < |z| < |b|

Figure 4.24: Des ription for Laurent's series.

4.119. Remark. By Theorem 3.64, f1 onverges uniformly for jz j   < R and by Theorem 4.117, f2 onverges uniformly for jz j  0 > r. Sin e both series onverge uniformly for all z satisfying 0  jz j  , the series de ned by (4.118) onverges uniformly to f for 0  jz j  .  A Laurent series about a is a series of the form 1 X X X An (z a)n := An (z a)n + A n (z n= 1 n0 n1

a) n

whi h represents an analyti fun tion in the annulus r < jz aj < R. The numbers An are the orresponding oeÆ ients about a. The series of the form (4.118) is then a Laurent series about z = 0. As a motivation for a Laurent series we onsider the fun tion 1 (4.120) f (z ) = ; a 6= b: (z a)(z b) Then, f is analyti everywhere ex ept at z = a; b and therefore, we are unable to express it in the neighborhood of a as a onvergent series of positive powers of z a. To obtain a Laurent series for f about z = 0, we rewrite (4.120) as (see Figure 4.24)

f (z ) =

1



1

1

a b z a z b If 0 < jaj < jz j < jbj (so that jz=bj < 1; ja=z j < 1), 2

X  a n 41 f (z ) = a b z n0 z

1



: 3

1 X  z n 5 b n0 b

210

Complex Integration

= =

1

2 4

X an 1

a b n1 z n 1

2 4

X

a b n

1

This may be written as (4.121)

f (z ) =

1 X n=

1

n+1 n0 b

a n 1zn 8 > >
> :

zn

X

X

3 5

zn

3

n+1 n0 b

5:

1 if n  0 (a b)bn+1 1 if n  1: (a b)an+1

Note that the expression in (4.121) involves both positive and negative powers of z and (4.121) is the Laurent series of (4.120) valid for 0 < jaj < jz j < jbj. If jz j > jbj > jaj (so that jb=z j < 1, ja=z j < 1), then we have

f (z ) =

1

2 4

X

an

a b n0 z n+1

X

n0 z

bn

n+1

3 5

=

1

2 4

X an

a b n0 z n+1

For example, we also note that 1 1 1 1 ; ; + + (z z (z 1)2 (z 2)3 (z 2)2

3

bn 5

:

1)2 ;

are themselves the Laurent expansion around 0, 1 or 2 as the ase may be.

4.122. Example. Consider the series 1 zn 1 1 zn X X zn X = + 2 2 2: n= 2004 n n= 2004 n n=1 n We note that the rst series on the right (whi h ontains only a nite number of terms) onverges for all z 6= 0 whereas the se ond series onverges for jz j < 1, diverges for jz j > 1 and onverges for all z with jz j = 1. Thus, the given double series onverges for all 0 < jz j  1 and diverges for jz j > 1. On the other hand if we onsider 1 1 z n X 1 zn X zn X 2 = 2 + 2; n=1 n n=1 n n= 1; n= 6 0n then it follows that the rst series on the right onverges for jz j  1 whereas the se ond series onverges for jz j  1 so that (the ombined series) the double series onverges only for jz j = 1. 

211

4.12 Laurent Series

We next dire t our attention to obtain an analogue of the Cau hy-Taylor representation theorem whi h shows that a fun tion analyti in an annulus, say D = fz : R1 < jz aj < R2 g, an be expanded into a Laurent series whi h onverges to the fun tion for every z in the annulus D.

4.123. Theorem. (Laurent's Theorem) If f is analyti in the annulus: R1 < jz j < R2P(where 0  R1 < R2  1), then f has a unique representation f (z ) = n2Zan z n for any z in the annulus, where an =

(4.124)

Z

1 f ( ) d; n 2 Z; 2i C  n+1

with C = f : j j = rg and R1 < r < R2 .

4.125. Corollary. If f is analyti in the annulus: R1 < jz aj < R2 (where R1  0); then, for any z in the annulus, f has a unique representation

f (z ) =

1 X n=

1

an (z

a)n ;

Z

f ( ) 1 d (n 2 Z); an = 2i j aj=r ( a)n+1

with R1 < r < R2 .

Before proving Theorem 4.123, we note that the above orollary is a

onsequen e of applying this result to g(z ) = f (z + a), whi h is analyti in the annulus R1 < jz j < R2 . Laurent's theorem ontinues to hold if R1 = 0 or R2 = 1 or both. In

ase R1 = 0, the Laurent series represents an analyti fun tion in a deleted neighborhood:
(a; R2 ) nfag = fz : 0 < jz aj < R2 g. When R2 = 1 and R1 > 0, the series represents the fun tion in a deleted neighborhood of in nity: [
(a; R1 )℄ = fz : jz aj > R1 g. When R1 = 0 and R2 = 1, we say that the series represents the fun tion in the pun tured plane namely, fz : jz aj > 0g.

Proof of Theorem 4.123. For a given z in the annulus R1 < jz j < R2 ,

hoose r1 and r2 su h that R1 < r1 < jz j < r2 < R2 and Cm = f : j j = rm g for m = 1; 2; the positively oriented ir les. Note that z lies between C1 and C2 . By making two ross- uts from C1 to C2 avoiding the point z , we have (see Figure 4.25) (4.126)

Z

Z

Z

f ( ) f ( ) f ( ) d = d + d C2  z

 z C1  z

212

Complex Integration y r2

z r1 R2 x

R1

Figure 4.25: Annulus region.

where is a small ir le ontaining the point z inside . Here the ontributions to the integral of the urve along the ross- uts an el in pairs. But, by the Cau hy integral formula, Z

f ( ) d = 2if (z ):  z

Hen e, (4.126) be omes

Z

Z

f ( ) 1 f ( ) 1 (4.127) f (z ) = d d: 2i C2  z 2i C1  z Pro eeding exa tly as was done in proving Taylor's theorem, on C2 , we have (see Weierstrass' Theorem, namely Theorem 4.85) (4.128) with (4.129) Now, (4.130)



Z

Z



X X 1 f ( ) 1 f ( ) d = d z n = an z n n +1 2i C2  z 2 i  C 2 n0 n0

an =

Z

1 f ( ) d for n = 0; 1; 2; : : : : 2i C2  n+1

X n X zn 1 1
= = ; n +1  z z 1 =z n0 z  n+1 n 1 sin e j j < jz j on C1 , and the onvergen e is uniform in  on C1 for a xed z . Using (4.130), we easily have

(4.131)

1

=

Z

X 1 f ( ) d = an z n; 2i C1  z n 1

where an is the same as de ned in (4.129) for n = 1; 2; 3; : : : but over the ontour C1 . Sin e g( ) = f ( )= n+1 is analyti on the annulus domain

213

4.12 Laurent Series

R1 < j j < R2 , by Cau hy's deformation theorem (see Theorem 4.37), we may repla e C1 and C2 in the expression for an by C (as spe i ed in the statement of the theorem) in the al ulation of the oeÆ ients an . This observation P togethern with (4.128) and (4.131) shows that (4.127) be omes f (z ) = 1 n= 1 an z ; where an is de ned by (4.124). The integral formula for the oeÆ ients, namely (4.124) allows us to show that the Laurent series representation for a given fun tion f is unique. That is to say that if we have another Laurent series expansion 1 X (4.132)

n z n for R1 < jz j < R2 ; f (z ) = n= 1 then an = n for all n, where an is given by (4.124). Now for ea h n, by (4.132), (4.124) may be rewritten as ( ) Z Z X 1 1 X 1 1 1 k an =

 d =

 k n 1 d: 2i C  n+1 k= 1 k 2i C k= 1 k We again make use of the fa t that the inter hange of summation and integral signs is permissible (see Theorem 4.85). This is be ause the onP n vergen e of f (z ) = 1

n= 1 n z in the annulus R1 < jz j < R2 implies that it onverges uniformly along C . We also know that Z

C

 m d =



0 for any integer m 6= 1 2i for m = 1:

Thus, for ea h integer n, we have  Z Z 1 1 X 1 k n 1 an =

 =

 2i n= 1 k C 2i n C

1 d



= n :

4.133. Remark. Equation (4.127) expresses a tually the following:

\Every analyti fun tion f in the annulus R1 < jz j < R2 an be uniquely de omposed into a sum f (z ) = f (z ) + f+ (z ), where f+ (z ) is analyti for jz j < r2 (< R2 ), and f (z ) is analyti for jz j > r1 (> R1 )". The uniqueness

assertion in the proof implies that the de omposition is independent of r1 and r2 , so that both f+ (z ) and f (z ) are de ned and analyti in the annulus R1 < jz j < R2 . 

4.134. Remark. The formula, for R1 < 1 < R2 , gives the interesting relation between Laurent and Fourier series expansions. Let f be analyti in some neighborhood, say D = fz : 1  < jz j = 1 < 1 + g,  > 0, of the unit ir le jz j = 1. Then, for z in this neighborhood, we get Z Z 1 X 1 f ( ) 1 2 i in f (z ) = an z n ; an = d = f (e )e d: 2i j j=1  n+1 2 0 n= 1

214

Complex Integration

In parti ular if we let f (eit ) = F (t) and z = eit , we have Z 1 X 1 2 int (4.135) F (t) = F (t)e int dt: an e with an = 2 0 n= 1 The series in (4.135) is the Fourier series of F in the omplex form. P



4.136. Remark. Again, we note that the part fP+ (z ) = n0 an z n de nes an analyti fun tion for jz j < R2 , while f (z ) = n 1 an z n de nes an analyti fun tion for jz j > R1 . The Laurent series representation for f in two dierent domains will, in general, be dierent. If f is analyti at z = 0 then the orresponding f (z ) = 0 and the Laurent series be omes the Taylor series about 0. In this ase, we have an = f (n)(0)=n!. On the other hand, we annot set an = f (n) (0)=n! in Theorem 4.117 as we did with Taylor's series representation where we had assumed that f is analyti for jz j < R for some R. In fa t, f (n)(0) is not even de ned in Theorem 4.123 sin e 0 is not in the annulus. Similar omments apply for the Laurent series about z = a. The oeÆ ients an are not often obtained by using the integral formula de ned by (4.124). Be ause of the uniqueness property of the Laurent series expansion, it is enough to nd a valid expansion for the analyti fun tion in the same annulus by some other easy te hniques by omputation or otherwise. Therefore, whenever possible, we an make use of a simple known expansion to obtain the Laurent expansion of quite ompli ated fun tions.  4.137. Example. The fun tion f de ned by f (z ) = 1=z is itself a Laurent series at z = 0 in the annulus 0 < jz j < 1. To determine the Laurent expansion of f at z = a (6= 0) we pro eed as follows:   1 1 1 =
z z a 1 + a=(z a) 1 X an =
( 1)n ; jz aj > jaj; z a n0 (z a)n valid for jz aj > jaj. Similarly, for jz aj < jaj, we have   X 1 1 1 (z a)n = = ( 1)n n+1 z a 1 + (z a)=a a n0 whi h is the Taylor series expansion of f at a valid in jz aj < jaj.  4.138. Example. Let 2 C be xed su h that j j > 1. Let us dis uss the onvergen e of 1 X zn fn (z ); fn (z ) = jnj :

n= 1

215

4.12 Laurent Series

Note that X

n0

fn ( z ) =

X  z n

n0

=

1

= for jz j < j j: 1 z= z

Similarly, X

n 1

fn (z ) =

1 1 = for jz j > j j 1 : m (

z )

z 1 m1 X

P Therefore, the ombined series 1 n= 1 fn (z ) onverges for j j and diverges at various other values of z .

1

< jz j < j j



4.139. Theorem. If f is analyti in a neighborhood of in nity and f is bounded there, then the Laurent oeÆ ients an = 0 for n = 1; 2; 3; : : : , where ea h an is given by (4:129): Proof. By hypothesis there exists R > 0 su h that f is analyti for

jz j > R, and jf (z )j  M for some M > 0. Then, for r > R, f (z ) =

1 X

Z

f ( ) 1 d an z n; with an = 2 i  j j=r n+1 n= 1

and therefore, we have jan j  Mr n : But r an be hosen as large as we please; so r n ! 0 as r ! 1 for n > 0, whi h means an = 0 for n > 0.

4.140. Example. Let  2 R and f (z ) = exp( 12 (z f 2 H(C n f0g). Moreover, exp and exp

 z 

2

=

1  z n for all z n0 n! 2 X



  X ( 1)n z = 2z 2 n0 n!



z 1)). Then,

1 n

for all z 6= 0:

Therefore, the Laurent series expansion of f in C n f0g is given by Z 1 X 1 f ( ) (4.141) f (z ) = an z n ; an = n+1 d: 2 i  C n= 1 Choosing C = f : j j = 1g, i.e.  = ei , 0    2, we nd that

an =

1 2i

Z 2

0

e 12 (e e ) i
ie d = 21 ei(n+1) i

i

Z 2

0

ei( sin  n) d:

216

Complex Integration

By the hange of variable  = 2 , we note that Z 2

0

sin( sin  n) d = = =

R 2

Z 2

0

sin( sin(2

Z 2

) n(2 )) d

sin(2n + ( sin  n)) d

0

Z 2

sin( sin  n) d;

0

n) d = 0: This observation implies that Z 1 X 1 2 1 1 n exp (z z ) = an z with an =

os( sin  n) d: 2 2 0 n= 1 Now, for ea h z 2 C n f0g, we an write 1 1 (4.142) f (z ) = exp( z ) exp( z 1 ): 2 2 Sin e the Laurent expansion is unique we may ompare the oeÆ ients of z n in (4.141) and (4.142) to obtain (use the Cau hy produ t of two onvergent series) 1 ( 1)m   2m+n X an = : m=0 (m + n)!m! 2 Similarly, it is easy to show the following:   P n (i) If  2 R, then exp 12 (z + z 1) = 1 n= 1 an z for z 6= 0 with Z 1   2m+n X 1 2  os  1 an = e

os(n) d = = a n: 2 0 m=0 (m + n)!m! 2 so that 

0

sin( sin  

(ii) If  2 R, then sinh[(z + z 1)℄ = Z

P1

n=

n 1 An z for z 6= 0, with

1 2 An =

os(n) sinh(2 os ) d: 2 0 Note that the Laurent oeÆ ients in (i) and (ii) remain un hanged when n repla es n. Therefore, we an rewrite (i) and (ii) respe tively as follows:   1 X 1 1 exp (z + z ) = a0 + an (z n + z n ) 2 n1 and


sinh (z + z 1 ) = A0 +

X

n1

An (z n + z n ):



217

4.13 Exer ises

4.143. Example. Consider f (z ) = Log (z n=(z n 1)) for jz j > 1, where n is a xed positive integer. Write f as 



1 f (z ) = Log 1 (1=z )n = Log 1 Log (1 z n); sin e j1=z j < 1, 1 X 1 nm z ; jz j > 1; = m m= 1 whi h is the Laurent expansion for f .



4.13 Exer ises 4.144. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) (t) = t2 e2i=t ; t 2 (0; 1℄, with (0) = 0, is a Jordan ar of lass C 1 . (b) (t) = t2 ei=4 ; t 2 (0; 1℄, is a non-simple smooth ontour. ( ) If 1 : [a; b℄ ! C and 2 : [a; b℄ ! C are two ontours, then so is the sum 1 + 2 only if 1 (b) = 2 ( ). (d) The inequality jez 1j < jz j holds for all z 2 D = fw : Re w < 0g. (e) The inequality jea eb j  ja bj holds for a; b 2 D = fw : Re w  0g. Z

jz rj jdz j = 8r2 : jzj=r (g) If f is a omplex-valued ontinuous fun tion on C , then (f) I =

I=

Z

f (z ) f (1=z ) dz = 0: z jzj=1

(h) If p(z ) is a polynomial of degree n in z with omplex oeÆ ients, then

I=

Z

jzj=1

p(z) dz = 2ip0 (0):

(i) There is an entire fun tion f (z ) su h that ef (z) = 25(e2z +1)= os(iz ): (j) There is an entire fun tion f (z ) su h that ef (z) = 5(e2z 1)= sin(iz ). (k) If f is analyti and nowhere zero in = fz : Re z < 2003g, then ln jf j is harmoni in . (l) If g 2 H(
) and su h that jg(z ) z j  jz j on jz j = 1, then one has the estimate jg0 (a)j  1 + (1 jaj) 2 for ea h a 2
.

218

Complex Integration

(m) Let be the losed square given by fz : 5  Re z; Im z  5g. Then, there annot exist a fun tion f whi h is analyti on a domain that

ontains su h that maxz2 jf (z )j = 5 and f 00 (1) = 1. R (n) If f is an entire fun tion su h that 02 jf (rei )j d  r for some xed  > 0, and for all r > 0, then f (z )  0 in C . (o) Let f be analyti on C n f1g su h that f (
nf1g)  R, then f is a

onstant fun tion. P 1 n (p) The Taylor series 1 n=1 n (z 3) onverges to Log (4 z ) for jz 3j < 1.P P1 n (n) (q) If f (z ) = 1 n=0 an (z a) has the property that n=0 f (a) onverges, then f is ne essarily an entire fun tion. P n (r) If a power series 1 jz j < 1 and if bn 2 C is n=0 an z onverges for P 2 n su h that jbn j < n jan j for all n  0, then 1 n=0 bn z onverges for jz j < 1. (s) If fangn0 is a sequen e of real numbers su h that 1 X 1 = a (z + 2)n ; (1 z )2 n=0 n (t) (u) (v) (w) (x) (y) (z)

P n then the radius of onvergen e of the series 1 n=0 an z is 3. P1 The power series n=0 an (z a)n (a 6= 0) an onverge at z = 0 and diverge at z = b, whenever jb aj < jaj. An entire fun tion that takes real values on the real axis and purely imaginary values on the imaginary axis must be an odd fun tion: f (z ) = f ( z ) for all z 2 C . If f is entire and f (z ) = f ( z ) for all z , then there exists an entire fun tion g su h that f (z ) = g(z 2 ) for all z 2 C . If f is analyti in a neighborhood
Æ of 0 and f (z ) = f ( z ) for all z 2
Æ , then there exists an analyti fun tion g in
Æ su h that f (z ) = zg(z 2) for all z 2
Æ .  n  P z n3 The Laurent series 1 n=0 n! + zn onverges only at z = 0 and nowhere else. If f is an analyti fun tion on the losed disk jz j  R for some xed positive number R > 0, then f an never satisfy the inequality f (n) (0)  n!nn for all n 2 N : If f 2 H(D) and a 2 D, then the inequality jf (n)(a)j  n!nn annot hold for all n  1.

4.145. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample.

4.13 Exer ises

219

(a) sin z = 0 () z = k, k 2 Z. (b) For z = x + iy, osh x is never zero and osh z has in nitely many zeros when y 6= 0. ( ) The zeros of sin(1=z ) are z = 1=n (n 2 Z) and ea h zero is isolated. (d) The Uniqueness theorem does not ne essarily hold for harmoni fun tions. (e) If the zeros of an analyti fun tion are not isolated then f (z )  0 throughout the domain of analyti ity. (f) If D is a domain and f 2 H(D) vanishes throughout any neighborhood of a point in D, then f (z )  0 in D. (g) If D is a domain and f 2 H(D) su h that f (z ) = 0 at all points on an ar inside D, then f (z )  0 throughout D. P (h) If f (z ) = n0 an z n onverges in
R , R > 0, su h that f 2 (z ) = f (z ) for every z in the open interval (0; R), then f (z ) is either 1 or 0 for all z 2
R . (i) If f 2 H(
) su h that f (x) = f ( x) for every real number x in
, then f (z ) = f ( z ) for all z in
. (j) There exists no fun tion f that is analyti in the unit disk
su h that f (1=n) = f ( 1=n) = n 2k+1 for n = 2; 3; : : : ; where k 2 N is xed. (k) There exists a fun tion f that is analyti in a neighborhood of z = 0 su h that f (1=n) = f ( 1=n) = n 2k for all suÆ iently large n, where k 2 N is xed. (l) There exists an analyti fun tion in the unit disk
su h that     1 1 1 f =f = for n  2: 2n 2n + 1 n (m) There exists an analyti fun tion f in
su h that f (1=n)) = (n + 1)=(n 1) for n  2: (n) There exists an analyti fun tion f in
su h that f (in =n) = n 2 for n  2: (o) There exists no analyti fun tion f in
su h that f (z ) 6 0 and f (nin =(n + 1)) = 0 for n  1: (p) There
exists an analyti fun tion f in
su h that f ( 1=2) = 3, f n 2 = 5 for n  2: (q) Suppose that f; g 2 H(
), and neither f nor g has a zero in
. If (f 0 =f )(1=n) = (g0 =g)(1=n) for all n = 2; 3; : : : ; then f (z ) = g(z ) in
for some onstant . (r) If f 2 H(
), fzngn1 is a sequen e of non-zero omplex numbers su h that zn ! 0 as n ! 1 and f (zn) = f ( zn) for all n 2 N, then f is even.

220

Complex Integration

(s) If fzngn1 is a sequen e of distin t omplex numbers in
su h that zn ! 0 as n ! 1, there exists an entire fun tion f su h that f (zn ) = zn for all n 2 N and f (5) = 0. (t) If f and g are entire fun tions whi h agree on some interval [a; b℄  R, then f (z ) = g(z ) in C . R

4.146. Compute Ij = j jz j2 dz; j = 1; 2; 3; 4; where (a) 1 (t) = a os t + ib sin t; a; b 2 R; 0  t  2 (b) 2 (t) = t2 + it; 0  t  1 ( t if 0  t  1 ( ) 3 (t) = it= 2 ie if 1  t  2 (d) 4 (t) = it; 0  t  1 (e) 5 (t) = t + it3 ; 0  t  1. R

4.147. Compute j jz j2 dz , j = 1 to 7, over the same paths used in Example 4.6. 4.148. For = fz : jz j = 1g, evaluate the following integrals

Z

dz ; j z

jn

Z

jdz j ;

Z

(z)n dz;

Z

zn

where n 2 N is xed, and jj 6= 1.

(z)n jdz j;

Z

Re z jdz j; z 

Z

Im z jdz j z 

4.149. Let f 2 H(
) and have no zeros in
. Suppose that (i) jf (ei )j  M1 for =2    =2 and, (ii) jf (ei )j  M2 for =2    3=2. Find an upper bound for jf (0)j. 4.150. Let f be analyti inside and on the square Q entered at the origin. Suppose that jf (z )j  Mj for ea h z 2 Sj (1  jp 4), where Sj denotes the enumeration of its sides. Show that jf (0)j  4 M1M2 M3 M4 . 4.151. Let f 2 H(
). Using the Cau hy integral formula for derivatives, evaluate the following integrals: I =

Z 2

0

f (ei ) os2 (=2) d Z

and Is =

Z 2

0

f (ei ) sin2 (=2) d:

z 2 + 3z 7 dz for jaj 6= 4; determine f (a) and 4.152. If f (a) = jzj=4 (z a)2 also f 0(1 + i) and f 0 (1 i).

221

4.13 Exer ises

4.153. Use Cau hy's theorem and/or Cau hy integral formula to evaluate the following integrals: Z

Log (z + 1) (a) dz z 3 Zjz 2j=2 4 z (b) dz ( z i)3 jzj=4 Z z+5 ( ) 2 3z 4 dz z jzj=5

(e) (f) (g)

Z

z+3

z 4 + az 3 Z jzj=1 n z jzj=2 z

Z

jz

1

3

(jaj > 1)

dz

z 1=2 dz ij=5=4 z 1

4.154. If f 2 H(
) and jf (z )j  (1 jz j)  for all z 2
and for some  > 0, then show that there exists a positive real number M (independent of f ) su h that jf 0 (z )j  M (1 jz j)  1 for all z 2
: Does it also work for  < 0? 4.155. If f 2 H(
) and jf 0(z )j  (1 jz j)  1 for all z 2
and for some  > 0, then show that jf (z )j  M (1 jz j)  for some M > 0, independent of f . Does it also work if  < 0? 4.156. Find the radius of the series on the right hand P of onvergen e n ; where Log z denotes the prin ipal side of (1 z ) 1 Log z = 1 a ( z 3) n n=0 bran h of the logarithm. Will your answer hange, if we repla e Log z by another bran h of log z whose bran h ut is the ray rei=4 (r  0), rather than the negative real axis. 4.157. Using the Cau hy inequality, dis uss the following statements: (a) There does not exist a fun tion f su h that f is analyti on the losed disk jz + 1j  5, f 00 ( 1) = i and max jf (z )j = 5. jz+1j5 (b) Answer the same question when f 00 ( 1) = i with max jf (z )j = 25 jz+1j5 as well as when f 00 ( 1) = 1=3 with max jf (z )j = 5: jz+1j5

4.158. Let f (z ) = 1 + z 2 + z 4 + z 6 + P


, z 2
, and fan g be a n sequen e of real numbers su h that f (z ) P = 1 n=0 an (z 5) . Find the 1 n radius of onvergen e of the series f (z ) = n=0 an z . 4.159. Find the radius of onvergen e R of the Taylor series, about z = 1, of the fun tion f (z ) = (1 + z 3 + z 6 + z 9 + z 12 ) 1 : 4.160. f (z ) =

Find the Taylor expansions about 0 for

z 1 sin z 1 ; h(z ) = ; and g(z ) = z2 z 1 1 + z2 1 z + z2

222

Complex Integration

and determine the radius of onvergen e of the orresponding series.

4.161. (L'H^ospital rule) Suppose that f and g are analyti at a, g(k) (a) = 0 = f (k) (a) for k = 0; 1; 2; : : : ; n 1 but both are not identi ally equal zero. If g(n) (a) 6= 0, show that f (z ) f (n) (a) lim = (n) z!a g (z ) g (a) provided the limit on the right exists.

p

4.162. What are the zeros of z (if it has any)? 4.163. Determine all f 2 H(
) su h that f 00 (1=n) + e1=n = 0 for all n = 2; 3; : : : . Justify your answer. 4.164. Suppose that f is an entire fun tion su h that f 0 (0) = 0 and 00 f (1 + 1=n) = 7 3(1=n) for ea h n = 1; 2; 3; : : : . Find all f that satisfy these properties. 4.165. Does Theorem 4.123 provide a Laurent series expansion of any bran h of log z in an annulus r < jz j < R? 4.166. Find the Laurent series expansion of ea h of the following: 1 1 1 (d) f (z ) = + z (z 2 + z 2) 1 + z2 2 + z 1 1 1 1 (b) f (z ) = + + (e) f (z ) = 2 z z 2 (z + 1)2 (z 1)(z 2 9) 3z 5 z ( ) f (z ) = 2 (f) f (z ) = 2 : z + 5z 6 z + 7z 8 In ea h ase, how many su h expansions are there? In whi h region is ea h of them valid? Find the Laurent oeÆ ients expli itly for ea h of these expansions. (a) f (z ) =

4.167. Does tan(1=z ) have a Laurent series onvergent in a region 0


5.3. Proposition. Let f

a smooth urve



2 H( ), where is a domain ontaining

: (t); t 2 [0; 1℄; passing through a point z0 2 and f 0 (z0 ) 6= 0. Then the tangent to the

urve

at w0 = f (z0 ) is

(5.4)

: (t) = f (z )jz= (t) = (f Æ )(t); t 2 [0; 1℄; (f Æ )0 (t0 ) = f 0 (z0 ) 0 (t0 )

(Note that urves are regarded as mappings so that the transformed urve is simply the omposite map = f Æ ).

Proof. First we note that is a smooth urve that passes through z0 (when t = t0 ). If 0 (t0 ) 6= 0, then 0 (t0 ) determines a well de ned tangent

228

Conformal Mappings and Mobius Transformations

ve tor at z0 and, be ause 0 (t0 ) 6= 0, (t) 6= (t0 ) for t near t0 , t 6= t0 . So we may write

f ( (t)) f ( (t0 )) f ( (t)) f ( (t0 )) (t) (t0 ) = t t0

(t) (t0 ) t t0 and, if we allow t ! t0 , we have (f Æ )0 (t0 ) = f 0 (z0 ) 0 (t0 ): If 0 (t0 ) = 0, we obtain (f Æ )0 (t0 ) = 0 and the above formula ontinues to hold. If, in Proposition 5.3, f 0 (z0 ) 6= 0, then the formula (5.4) gives (f Æ )0 (t0 ) 6= 0 whi h determines a new tangent ve tor 0 (t0 ) at the point w0 = f (z0) = (t0 ) on the transformed urve, as indi ated in Figure 5.4. The relation (5.4) shows that the tangent to the image urve depends on the tangent to the original urve passing through z0 and the xed omplex number f 0 (z0 ). An analyti mapping w = f (z ) in a domain that preserves the angle (both in size and in sense) at z0 is alled onformal at z0 . More pre isely, f (z ) is said to be onformal at z0 2 if, whenever 1 and 2 are two parameterized urves interse ting at z0 = 1 (t0 ) = 2 (t0 ) with non-zero tangents, then the following holds: (i) the two transformed urves 1 = f Æ 1 and 2 = f Æ 2 have non-zero tangents at t0 (ii) the angle from 01 (t0 ) = (f Æ 1 )0 (t0 ) to 02 (t0 ) = (f Æ 2 )0 (t0 ) is the same as the angle from 10 (t0 ) to 20 (t0 ). If it is onformal at ea h point of , then we say that f is onformal in . A fun tion that preserves the size of the angle but not sense (i.e. orientation) is said to be isogonal. An example of the latter lass of fun tions is given by f (z ) = z . Indeed, if

1 = ft : t  0g and 2 = ftei=4 : t  0g are two urves in C , then the image urves under f (z ) = z are 1

= ft : t  0g and

2

= fte i=4 : t  0g;

see Figure 5.5. Although the two urves interse t at an angle =4 in ea h plane, the fun tion f (z ) = z reverses the angle of orientation. So the mapping in this ase is not onformal, but is isogonal.

229

5.1 Prin iple of Conformal Mapping y

v i

γ2

te

4 π/

f (z) = z

π/4

−π/4

x

u

te−iπ/4

Figure 5.5: Demonstration for isogonal mapping.

Let f 2 H( ), where is a domain ontaining a smooth urve passing through a point z0 2 where f 0 (z0 ) 6= 0. We wish to show that this

ondition is suÆ ient to show that f is onformal at z0. To do this, we

onsider two smooth urves

1 : 1 (t) and 2 : 2 (t); t 2 [0; 1℄; that pass through z0 = 1 (t0 ) = 2 (t0 ) with non-zero tangents at t0 . Then the transformed urves 1

= f Æ 1 and

2

= f Æ 2

pass through w0 = f (z0 ) in the w-plane when t = t0 , and, by Proposition 5.3, the tangents to these urves are given by 0 (t0 ) = (f Æ 1 )0 (t0 ) = f 0 (z0 ) 0 (t0 ) 1

and

1

0 (t0 ) = (f Æ 2 )0 (t0 ) = f 0 (z0 ) 0 (t0 ); 2

2

respe tively. Note that the tangents to the transformed urves 1 and 2 are obtained by multiplying the respe tive tangents to 1 and 2 by the non-zero fa tor f 0 (z0 ). Thus, the arguments of both tangents are in reased by the same angle, namely the argument of f 0 (z0 ). Consequently, Arg



 0   0  02 (t0 )  f (z0 ) 20 (t0 )

2 (t0 ) 01 (t0 ) = Arg f 0 (z0 ) 10 (t0 ) = Arg 10 (t0 )

so that the angle between 1 and 2 at z0 measured from 1 to 2 is equal to the angle between 1 and 2 at f (z0 ) measured from 1 to 2 . We have in fa t proved the following result.

5.5. Theorem. Let f f is onformal at z0.

2 H( ) and z0 2 su h that f 0 (z0 ) 6= 0. Then

230

Conformal Mappings and Mobius Transformations

Conformality is onsidered a lo al property of analyti fun tions. Further, sin e f (z ) f (z0) f 0 (z0 ) = zlim !z0 z z0 ; and w = f (z ) maps a urve : z (t) through z0 to another urve : w(t) = f (z (t)) through w0 = f (z0 ), we have f (z )







w w0 f (z0 ) 0 = zlim !z0 z z0 = jf (z0 )j: z z0 This equation shows that jf 0 (z0 )j is a lo al s aling fa tor of the fun tion at z0 , and is independent of the urve . Moreover, if jz z0 j is small, then jf (z ) f (z0 )j  jf 0 (z0 )j jz z0 j

lim z!z0

from whi h we see that \small" neighborhoods of z0 are mapped onto roughly the same on guration, magni ed by the fa tor jf 0(z0 )j. For example, "small triangle" ontaining z0 is mapped geometri ally onto a similar " urvilinear triangle" magni ed by the fa tor jf 0 (z0 )j. Thus, arg f 0 (z0 ) measures the rotation while jf 0 (z0 )j measures (for points nearby) the magni ation or distortion of the image.

5.6. Example. To see how onformality may fail at a point z0 where f 0 (z0 ) = 0, we onsider the fun tion f (z ) = z 2 . Then f 0 (0) = 0. Now, if

1 is the positive real axis from 0 to 1 and 2 is the (positive) imaginary axis in the upper half-plane, then f ( 1 ) = 1 is the positive real axis from 0 to 1, and f ( 2 ) = 2 is the negative real axis from 0 to 1. Note that the angle between 1 and 2 is =2, whereas the angle between their image

urves 1 and 2 is . Thus, f (z ) = z 2 is not onformal at 0 although it is

onformal at every other point of the omplex plane. To demonstrate this fa t, let a + ib 6= 0 and onsider two smooth urves 1 and 2 given by

1 = ft + ib : t  ag; and 2 = fa + it : t  bg:

They interse t at z0 = a + ib and the angle between them is =2. Further, as f 0 (z0 ) = 2z0 6= 0, f is onformal at z0 . Then, under the mapping f (z ) = z 2 , we have

f ( 1 (t)) = (t + ib)2 = t2 b2 +2itb and f ( 2 (t)) = (a + it)2 = a2 t2 +2iat: Letting f ( (t)) = u + iv, it follows that the images are the parabolas des ribed by 1

= fu + iv : v2 = 4b2(u + b2 )g and

2

1

and

2

of 1 and 2

= fu + iv : v2 = 4a2(a2

u)g;

respe tively. An inspe tion of Figure 5.6 indi ates that the angle between 1 and 2 at f (z0 ) is =2. Note that as a and b vary, we obtain two families of parabolas interse ting orthogonally at ea h point where a + ib 6= 0. 

231

5.1 Prin iple of Conformal Mapping Ŵ2

y

γ2 1 + i = z0

γ1 x

Ŵ1

v

f (z) = z2

2i = w0 −1

1

u

−2i

Figure 5.6: Conformality of f (z ) = z 2 at 1 + i.

If f (z ) = z n, then f magni es the angle at z = 0 by a fa tor of n and maps the disk jz j < r onto the disk jz j < rn in an n-to-one manner. Let us now onsider a general situation. If f is analyti at z0 su h that f 0 (z0 ) = 0, then the onformal hara ter fails. Su h a point z0 is alled a riti al point of f . Let us now examine the behavior of an analyti fun tion in a neighborhood of a riti al point. More generally, let w = f (z ) be analyti at z0 su h that f (k) (z0 ) = 0 for k = 1; 2; : : : ; n 1 and f (n) (z0 ) 6= 0. We wish to show that angles are not preserved at z0 but are multiplied by n. By hypotheses, we have

f (z ) f (z0 ) = (z z0 )n [an + an+1 (z z0 ) +


℄ = (z z0 )n g(z ) where g is analyti at z0 with g(z0 ) = an 6= 0. Thus, arg(w

w0 ) = arg(f (z ) f (z0 )) = n arg(z z0 ) + arg g(z ):

Suppose  is the angle that the tangent ve tor to a smooth urve at z0 makes with the positive x-axis, and is the angle that the tangent to the image urve under w = f (z ) at w0 = f (z0 ) makes with the positive u-axis. If z ! z0 along , then w = f (z ) ! w0 = f (z0 ) along so that the last equation gives

= n + arg g(z0 ) = n + arg an : This relation shows that the tangent to the image urve depends on the tangent to original urve as well as the order of the derivatives of f and the argument of the rst non-zero oeÆ ient in the series expansion of f 0 (z ) at the point in question. Let 1 and 2 be two smooth urves passing through z0 and let 1 and be their respe tive images under w = f (z ). Suppose that the tangent 2 to the urves k and k make an angle k and k with the real axis of the z -plane and of the w-plane, respe tively. Then, we have

1 = n1 + arg an and 2 = n2 + arg an ; i.e. = n

232

Conformal Mappings and Mobius Transformations

where  = 1 2 and = 1 2 are respe tively the angles between the

urves 1 , 2 and the respe tive image urves 1 , 2 . Now we have proved

5.7. Theorem. Suppose that f is analyti at z0 and f 0(z ) has a zero of order n 1 at z0 . If two smooth urves interse t at an angle  in the z -plane, then their images interse t at an angle n in the w-plane. From Theorem 5.7, we obtain that no analyti fun tion an be onformal at its riti al points.

5.8. Example. Consider f (z ) = sin z . Then f is entire and f 0 (z ) =

os z so that f 0 (zn ) = 0 for zn = (2n+1)=2, n 2 Z. Thus, f is onformal on

= C n f(2n + 1)=2 : n 2 Zg. Note that f 00 (z ) = sin z and f 00 (zn ) 6= 0 for ea h n 2 Z. A

ording to Theorem 5.7, the angle between any two smooth urves interse ting at zn (n 2 Z) is in reased by a fa tor of 2 by w = f (z ).  5.9. The transformation w = sin z. First we note that sin z is periodi with period 2, sin z = sin( z ) and sin(z + ) = sin z . In view of these observations, it suÆ es to understand the mapping behavior of sin z on a verti al strip of width . Further, sin z is learly onformal on D = fz : jRe z j < =2g and maps D one-to-one onto the domain C n f( 1; 1℄ [ [1; 1)g. With z = x + iy and w = u + iv, w = sin z gives (5.10)

u = sin x osh y and v = os x sinh y:

Let us dis uss the behavior of w = sin z on the horizontal and verti al line segments in = fz 2 C : jRe z j  =2g. First, we onsider the horizontal line segment Jb = fx + iy : y = b; jxj  =2g: By (5.10), the image of Jb is given by (5.11)

u = sin x osh b and v = os x sinh b (jxj  =2):

If b = 0, then the image of the line segment J0 = [ =2; =2℄ des ribed by (5.11) redu es to u = sin x and v = 0; sin e osh 0 = 1 and sinh 0 = 0. As we move x from =2 to =2 along the line segment J0 , the image in the w-plane advan es from 1 to 1 along the line v = 0 and thus, sin(J0 ) = [ 1; 1℄: Next we x b 6= 0 and onsider Jb . In this ase, the image set des ribed by (5.11) gives, by eliminating x, (5.12)

u2 v2 + =1 2

osh b sinh2 b

5.1 Prin iple of Conformal Mapping

233

whi h is the equation of an ellipse entered at the origin of the w-plane. Its major and minor axes have lengths 2 osh b and 2 sinh b, respe tively. Further, the major and the minor axes lie on the u- and the v-axes, respe tively. If b > 0, (5.11) indi ates that v > 0 showing that the image of Ib for b > 0 is the upper-half of the ellipse de ned by (5.12). Similarly, the image of Jb for b < 0 is the bottom half of the ellipse. The images of the segments

orresponding to b; b 6= 0, t together to form a omplete ellipse. Let us now dis uss the image of the verti al lines. Fix a with jaj  =2, and let Ia = fx + iy : x = a; y 2 Rg to represent a verti al line. By (5.10), the image of Ia under w = sin z is des ribed by the onditions (5.13)

u = sin a osh y; v = os a sinh y (y 2 R):

Elimination of the variable y yields

u2 v2 =1 2 sin a os2 a whenever sin a 6= 0 and os a 6= 0. The last onditions are satis ed when a 2= f0; =2; =2g. We dis uss separately the ase when a 2 f0; =2; =2g. Clearly, the lo ation of the image of I0 des ribed by (5.14)

u = 0 and v = sinh y (y 2 R); is a tually a parametri equation of the imaginary axis of the w-plane (as sinh(R) = R). For a = =2, the image of the line I=2 is the set of points given by u = osh y; v = 0 (y 2 R): As we move y from 0 to 1 along the line x = =2, these equations indi ate that u moves from 1 to 1 along the line v = 0. As osh y = osh( y), it follows that the image of the line I =2 is the set of points u  1 on the negative real axis. Next, we dis uss the ase when a 2= f0; =2; =2g. In this ase, the image of Ia is the hyperbola des ribed by (5.14) with verti es at ( sin a; 0) and slant asymptotes v = ( ot a)u. If 0 < a < =2, then the rst equation reveals that u > 0, and if =2 < a < 0 then u < 0. This observation shows that for ea h xed a; 0 < a < =2, the image of Ia is the right bran h of the hyperbola ontaining the point (sin a; 0) while the image of Ia , for =2 < a < 0, is the left bran h of the hyperbola ontaining the point ( sin a; 0). Then the image of the pair of verti al lines x = a and x = a with jaj < =2 onstitute the full hyperbola given by (5.14). Finally, the mapping w = os z an be analyzed using the equation

os z = sin(z + =2):

234

Conformal Mappings and Mobius Transformations

5.2 Basi Properties of Mobius Maps What is a Mobius transformation? They are simply a omposition of one, some or all of the following spe ial types of transformations.

 

 

Translation: It is a map of the form z 7! z + ,  2 C n f0g: If  = 0, then it is an identity map. Magni ation: It is a map of the form z 7! rz , r 2 R n f0g: Noti e that for r = 1, this is the identity map whereas for r = 0 it is a

onstant map. If r > 0, then w = rz multiplies the modulus of z by r and leaves its argument un hanged. Thus if r > 1, then this is a \magni ation" and if 0 < r < 1, it is a \shrinking/ ontra tion" rather than saying it a \magni ation". If r < 0, then w = rz gives the re e tion through the origin followed by su h a \magni ation" or \shrinking" depending on r < 1 or 1 < r < 0. Rotation: It is a map of the form z 7! ei z;  2 R: This map produ es a rotation through an angle about the origin with positive sense if  > 0. The rotation oupled with magni ation is referred to as dilation: z 7! az (a 6= 0). Inversion: It is a map of the form z 7! 1=z whi h produ es a geometri inversion (or re ipro al map or the inversion map).

Mobius transformations, named in honor of the geometer A.F. Mobius (1790-1868), are rational fun tions of the form (5.15)

T (z ) =

az + b (a; b; ; d 2 C ; ad b 6= 0)

z + d

where a; b; ; d are omplex numbers su h that ad b 6= 0: Note that (5.15) does not determine the oeÆ ients a; b; ; d uniquely. If we let T (z ) := Tab d(z ) and if  2 C n f0g, then a, b,  , d orrespond to the same Mobius transformation as

Tab d(z ) = T(a)(b)( )(d) (z ) so that if 2 = 1=(ad b ) then (a)(d) (b)( ) = 1: In other words, the behavior of T does not hange when a; b; ; d are multiplied by a non-zero

onstant and thus, we may assume that ad b = 1 whenever there is a need for this normalization to simplify our studies. Certainly, T is analyti on C n f d= g. Note that if = 0, then (5.15) redu es to

T (z ) =

a

d

z+

 

b := z + (ad 6= 0; i.e.  6= 0): d

A fun tion of this form is alled a linear map. Clearly, the identity transformation orresponds to a = d 6= 0 and b = = 0. If 6= 0, then we an

235

5.2 Basi Properties of Mobius Maps

de ompose (5.15) as

 





d ad 1 +b

(z + d= )   a ad b 1 = (5.16) ; 6= 0:

2 z + d= Thus, as mentioned in the beginning, a Mobius transformation is a omposition of magni ation, rotation and translation, and T (z ) redu es to a onstant whenever ad b = 0. So, throughout the dis ussion in this

hapter, the ases for whi h ad b = 0 should be ruled out in order to ex lude the trivial ase for whi h T (z ) redu es to a onstant. It an be also observed that the ondition ad b 6= 0 is required sin e otherwise T ( d= ) = 0=0, whi h is unde ned. Note also that T 0(z ) exists for all z where z + d 6= 0, and ad b (z 6= d= ); T 0(z ) = ( z + d)2 so that the ondition ad b 6= 0 simply guarantees that T (z ) is not a

onstant. Therefore, T de nes an analyti fun tion on C nf d= g. Clearly, T (z ) is obtained by su

essive appli ations of the following four mappings wi = Ti (z ) for i = 1; 2; 3; 4, where   d 1 ad b a w1 = z + ; w2 = ; w3 = w2 ; w4 = T4 (z ) = + w3

w1

2

unless = 0 where we write T (z ) = (a=d)z + (b=d): It follows that T (z ) is a tually given by the omposition of these simpler transformations. More pre isely, we have T = T4 Æ T3 Æ T2 Æ T1 and thus, these four types (translation, rotation, magni ation and an inversion) generate the group of Mobius transformations. Moreover, the fun tion w = T (z ) de ned by (5.15) an be written as T (z ) = a z +

wz

az + dw b = 0

and the expression on the left is linear in both variables z and w, and so, Mobius transformations are also alled Bilinear transformations. When

= 0, the transformation is learly linear.

5.17. Matrix interpretation and the group stru ture. There is a strong relationship between Mobius transformations and matri es. Indeed, ea h Mobius transformation of the form (5.15) an be asso iated with a 2  2 matrix via the map 

z 7! A = a db



! T (z ):

236

Conformal Mappings and Mobius Transformations

The Mobius transformation T0 given by T0 (z ) = z is the identity transformation whi h orresponds to the 2  2 identity matrix. If T and S are two Mobius transformations given by az + b a0 z + b0 T (z ) = and S (z ) = 0 ;

z + d

z + d0 then the omposition T Æ S is de ned by   0 a z + b0 +b a 0 (aa0 + b 0 )z + ab0 + bd0

z + d0  (T Æ S )(z ) = T (S (z )) =  0 = 0 : 0 az+b ( a + d 0 )z + b0 + dd0

0 +d

z + d0 Note that if A and B are the orresponding matri es asso iated with the transformations T and S , then  0 0 ab0 + bd0  aa + b AB = a0 + d 0 b0 + dd0 whi h implies that T Æ S orresponds to the matrix produ t AB . Moreover, if det A = ad b 6= 0 and det B = a0 d0 b0 0 6= 0 then det(AB ) = det A
det B = (ad b )(a0 d0 b0 0 ) 6= 0: In parti ular, we have the following simple result.

5.18. Proposition. Composition of two Mobius transformations is

a Mobius transformation.

The fa t that matrix multipli ation orresponds to omposition an be reformulated in the language of group theory. Now, let GL(2; C ) denote the general linear group onsisting of 2  2 invertible matri es A with omplex entries:     a b GL(2; C ) = A = d : a; b; ; d 2 C ; det A 6= 0 : Then, GL(2; C ) forms a subgroup of the group of all Mobius transformations under the operation of matrix multipli ation. If we assign ea h A 2 GL(2; C ) the Mobius transformation de ned by (5.15), then the map A 7! TA is an isomorphism; i.e. for all A; B 2 GL(2; C ), TAÆB = TA Æ TB : Clearly, the inverse of A is 1

ad b



d

b a



and TA ÆTA 1 = TAA 1 = TI . Thus, all the (non- onstant) Mobius transformations are invertible. In parti ular, they are one-to-one. Indeed if = 0,

237

5.2 Basi Properties of Mobius Maps

then T (z ) = (a=d)z +(b=d) whi h is trivially one-to-one in C . If 6= 0 then, by (5.16), for z1 ; z2 2 C n f d= g, we obtain

) z +1 d= = z +1 d= ) z1 = z2 : 1 2 Therefore, T is 1-1 on C n f d= g. Is a (non- onstant) Mobius transformation a surje tion onto C ? Is there a point in C missing from the image T (z1 ) = T (z2)

under a Mobius transformation? As T is one-to-one, its inverse always exists. The inverse of T is obtained by solving the equation

w = T (z ) =

az + b (z 6= d= )

z + d

for z . This gives

dw b (w 6= a= )

w + a where da ( b)( ) = ad b 6= 0: We have z = T 1(w) =

5.19. Proposition. The inverse of a Mobius transformation is also

a Mobius transformation.

Moreover, T : C n f d= g ! C analyti ) mapping with (5.20)

z = T 1 (w ) =

n fa= g is a bianalyti (i.e. T and T

dw b

w + a

and T 0(z ) =

1

ad b : ( z + d)2

This is an important property of Mobius transformations whi h is quite spe ial for omplex-valued fun tions. The later ondition shows that the Mobius transformation T is not only a bianalyti map of C n f d= g onto C n fa= g but is also onformal on C n f d= g. Is it possible to enlarge the domain of de nition of T in order to onsider it as a map de ned on the extended omplex plane? Observe that if = 0, then T maps C onto C and, T (z ) ! 1 if and only if z ! 1. Thus, it makes sense to de ne T (1) = 1 when = 0. If 6= 0, then T (z ) ! 1 as z ! d= and T (z ) ! a= if z = 1. In view of these observations, it is natural to introdu e the limiting values (in the usual sense)

a T (1) = lim T (z ) = and

jzj!1



T ( d= ) = 1 for 6= 0 T (1) = 1 for = 0:

Similarly, by the rst transformation in (5.20), we de ne

d T 1(1) = lim T 1 (w) = and

jwj!1



T 1(a= ) = 1 for 6= 0 T 1(1) = 1 for = 0;

238

Conformal Mappings and Mobius Transformations w=

y

1 z

= 1r e−iθ

v

eiθ 1 2

x

1

1

u

2 e−iθ

Figure 5.7: Images of disks under inversion. y w=

−1 −r O

r 1

x

v

1 z

− 1r −1

O

1

1 r

u

Figure 5.8: Images ertain sets under inversion.

so that we onveniently regard a Mobius map T as a one-to-one mapping of the extended omplex plane C 1 onto itself. Equivalently, we say that T maps the Riemann sphere onto itself; i.e. T de nes a onformal self-map of C 1 . In parti ular, we on lude that T (C 1 ) = C 1 , and for all z and w in C 1 , we have T 1(T (z )) = z and T (T 1(w)) = w:

5.21. Images of ir les and lines under Mobius maps. There are many properties of Mobius transformations that have onsiderable importan e in physi al appli ations. Let us rst study some basi properties of the inversion fun tion w de ned by w = z 1 (z 6= 0). Clearly, this fun tion establishes a one-to-one orresponden e between the non-zero points of the z , and the w-planes. To start with, we let z = rei (r > 0,  = Arg z ). Then 1 1 z w = = e i = 2 : z r jz j Under this transformation, we easily have the following (see Figures 5.7 and 5.8): (i) points in the upper half-plane (Im z > 0) are mapped onto points in the lower half-plane (Im w < 0) and vi e-versa

239

5.2 Basi Properties of Mobius Maps

(ii) the right half-plane (Re z > 0) is mapped onto itself (iii) the left half-plane (Re z < 0) is mapped onto itself (iv) points on the ir le jz j = R are mapped onto points on the ir le jwj = 1=R, the disk jz j < R is mapped onto the disk jwj > 1=R, and the points z su h that jz j > R are mapped onto points in the disk jwj < 1=R. So, it is natural to ask the following simple question: what happens to general ir les and straight lines under inversions, and more generally under Mobius transformations? Consider the equation in (x; y)- oordinates of the form

(x2 + y2 ) 2ax 2by = R2 (a2 + b2 ) (; a; b 2 R; R > 0) whi h is a ir le or a (open) straight line depending on whether  6= 0 or  = 0. In fa t  = 1 a tually orresponds to the ir le jz (a + ib)j = R, whereas  = 0 gives a line. As 2x = z

z and 2y = i(z z);

in terms of the omplex variable, we an onsider all possible ir les and straight lines in the form

jz j2

(a ib)z

(a + ib)z + a2 + b2

R2 = 0;

or equivalently in the form jz j2 + 2Re ( z ) + = 0, ; 2 R; 2 C :

5.22. Lemma. (Cir le-preserving Property) Every Mobius transformation maps ir les and straight lines in the z -plane into ir les or lines. Proof. It is quite obvious that among the four elementary transformations, the three of them, namely, translation, magni ation (s aling) and rotation do preserve ir les and straight lines as it is lear that ea h of these three transformations sends ir les to ir les and lines to lines. Therefore, from the equivalent expression for Mobius transformation given by (5.16), it only remains to verify the statement for the inversion given by w = 1=z: Every ir le or line in C an be des ribed in the form (5.23)

jz j2 + 2Re ( z ) + = 0; ; 2 R; 2 C :

Obviously, this is an equation of a ir le if  6= 0. To obtain the enter and the radius, omplete the square. If  = 0, it is a straight line in C , whi h is a ir le through 1. Now, if z 2 C nf0g and w = 1=z (w 6= 0) then z = 1=w so that (5.23) transforms to 



 + 2Re + = 0: 2 jwj w

240

Conformal Mappings and Mobius Transformations

Multiplying both sides by ww = jwj2 , this equation assumes the form

jwj2 + 2Re ( w) +  = 0:

(5.24)

Clearly, the desired on lusion follows from (5.23) and (5.24). We observe that Lemma 5.22 neither laims that every ir le in C is mapped to a ir le in C nor does it laim that every line in C is mapped to a line in C . Further, from Lemma 5.22, it follows that every Mobius transformation, being one of four spe ial transformations, arries the families of ir les in C 1 onto itself. Here we regarded straight line (as a limiting

ase of ir le) in the extended omplex plane as a ir le on the Riemann sphere{using the stereographi proje tion. That is to say that line on the extended omplex plane is a ir le of in nite radius (meaning that it is a

ir le through the point at in nity). So, we an think of ir les and lines as belonging to the same lass, and reformulate Lemma 5.22 as

5.25. Theorem. Every Mobius transformation maps ir les in C 1

onto ir les in C 1 .

It is worth it to have a bit more detail on the two equations (5.23) and (5.24). Case (i): Assume that  = 0 and = 0. In this ase, from (5.23) and (5.24), we see that the straight line that passes through the origin given by Re ( z ) = 0 is transformed into a straight line that passes through the origin given by Re ( w) = 0: Case (ii): Let  = 0 and 6= 0. In this ase, (5.23) is equivalent to 2Re ( z ) + = 0 whi h is an equation of a straight line that does not pass through the origin. By (5.24), this straight line is transformed into w +



j j i.e. = ;

whi h is a ir le passing through the origin. Note that = 0 is not possible in both the ases. Case (iii): Let  6= 0 and 6= 0. In this ase, (5.23) is equivalent to

jwj2 + 2Re ( w) = 0;

z +



r

j j2  =  2 where j j >  . This ir le does not pass through the origin and by (5.24), the image of this ir le is given by w +



=



s

j j2  ; j j2 >  :

2

5.3 Fixed Points and Mobius Maps

241

Note that this ir le does not pass through the origin. More pre isely, the above dis ussion leads to

5.26. Proposition. Under the fun tion w = 1=z , we have

   

the image of a line through the origin is a line through the origin the image of a line not through the origin is a ir le through the origin the image of a ir le through the origin is a line not through the origin the image of a ir le not through the origin is a ir le not through the origin.

In parti ular, under the inversion w = 1=z , the verti al line Re z =  ( 6= 0) maps onto the ir le jw 1=(2)j = 1=(2jj) while the horizontal line Im z =  ( 6= 0) maps onto the ir le jw + i=(2)j = 1=(2jj):

5.3 Fixed Points and Mobius Maps Let D be a subset of C 1 and f : D ! C 1 . A point z0 2 D is said to be a

xed point of f if f (z0 ) = z0 . The set of all xed points of f is denoted by Fix (f ). For example, we have

(i) the fun tion f (z ) = z 2 has exa tly three xed points, namely, 0; 1 and 1 whereas the fun tion f (z ) = z 1 has two xed points namely 1 and 1. (ii) the fun tion f (z ) = z 1 has no xed points in C whereas it has one xed point in C 1 , namely the point at 1. (iii) the re e tion z 7! z is not a Mobius transformation but f (z ) = z has all the points on R as its xed points. What are the xed points of f (z ) = iz? (iv) the fun tion f (z ) = iz=jz j, z 6= 0, has no xed points in C n f0g. (v) every non- onstant real-valued ontinuous fun tion f : ( 1; 1) ! ( 1; 1) has a xed point in ( 1; 1). However, a similar result does not hold for fun tions f :
!
. For example  :
!
, jj = 1, de ned by z   (z ) = 1 z has no xed points in
.

5.27. Proposition. Every Mobius transformation T : C 1 ! C 1 has at most two xed points in C 1 unless T (z )  z . Equivalently, if a Mobius transformation leaves three points in C 1 xed, then it is none other than the identity fun tion.

242

Conformal Mappings and Mobius Transformations

Proof. Suppose that T (z ) = is su h that T (z ) 6 z . Then

az + b

z + d

() z 2 + (d a)z b = 0: If 6= 0, then T ( d= ) = 1 and T (1) = a= , so neither d= nor 1 is a T (z ) = z

xed point of T . So T has at most two xed points in this ase and they are obtainable from the quadrati equation. If = 0, then

a b T (z ) = z + : d d In this ase, T has 1 and b=(d a) are the only xed points. 5.28. Corollary. If S and T are two Mobius transformations whi h agree at three distin t points of C 1 , then S = T . Proof. Suppose that S (zj ) = T (zj ) for z1 ; z2 ; z3 in C 1 . Then, (S

1 Æ T )(z

j ) = zj

and, by Proposition 5.27, S

T = (S Æ S

1) Æ T

= (T

1ÆT

1 Æ S )(z

j)

for j = 1; 2; 3

= I . Therefore,

= S Æ (S

1 Æ T) = S Æ I

= S:

The following result is a tually a reformulation of Proposition 5.27, but we will provide an alternate proof be ause of its independent interest.

5.29. Proposition. Every Mobius transformation whi h xes 0; 1; 1 is ne essarily the identity map. Proof. Suppose that T is given in the form (5.15), and T xes 0; 1; 1. It follows that (i) T (1) = 1 gives = 0 whereas T (0) = 0 gives b = 0 (ii) T (1) = 1 gives (a + b)=( + d) = 1 so that a = d, by (i). We on lude that T is the identity map. A Mobius transformation whi h has a unique xed point in C is alled paraboli . If it has exa tly two xed points, then it is alled loxodromi .

5.30. Chara terizations of Mobius maps in terms of their xed points. Let us hara terize a given Mobius transformation a

ording to its

243

5.3 Fixed Points and Mobius Maps

xed points (or point). Let S (z ) 6 z be a Mobius map. It might be

onvenient to use the following equivalent form: (5.31)

S (z ) =

8 > > < > > :

a

a b z + = z + if = 0 d d    ad b 1 1 = Æ if 6= 0;

2 z + d= z + d=

where (for onvenien e) we have used the notation

a b a ad b ( 6= 0):  = ; = ; = ( 6= 0); Æ = d d

2

Case (i): If 1 2 Fix (S ), then a= = 1 (i.e. = 0), and so (5.32)

S (z ) = z + ; 0 6=  2 C ; 2 C :

If  = 1 (i.e. a = d), then S (z ) = z + and 1 is the only xed point. If  6= 1 (i.e. a 6= d), then 1 and z0 = =(1 ) = b=(d a) are the only xed points. Observe that

S (z ) z0 = (z

z0 ) + ( 1)z0 + = (z z0 ):

Thus, S an be written in the form S (z ) z0 = (z z0), where  is a

omplex number whi h is neither 0 nor 1, and z0 is a xed point of S . Case (ii): If 1 2= Fix (S ) (i.e. 6= 0), then S has at most two xed points in C and are in fa t obtained from solving the xed point equation S (z ) = z , i.e. z 2 + (d a)z b = 0: This gives

p

(a d)  D ; D = (a d)2 + 4b = (a + d)2 4(ad b ): 2 Therefore, we need to deal with two ases, namely, D = 0 and D 6= 0. If D = 0, then S has only one xed point, say z0 , and it is given by

z0 =

z0 =

a+d a d ; i.e. z0 + d = : 2 2

So, we have ( z0 + d)2 = (a + d)2 =4 = ad b : Using this we an write 



1 1 z0 + d= z + d= (ad b )(z z0 ) = (sin e Æ = (ad b )= 2) ( z0 + d)( z + d) ( z0 + d)(z z0) = (sin e ( z0 + d)2 = ad b ):

z + d

S (z ) S (z0 ) = Æ

244

Conformal Mappings and Mobius Transformations

Therefore, if 1 62 Fix (S ) and z0 2 Fix (S ) with (a + d)2 = 4(ad b ) then, by the last equation, the Mobius transformation S redu es to the expression (note that S (z0) = z0 )

z + d + (z z0 ) 1

1 1 = 0 = + = + k; S (z ) z0 ( z0 + d)(z z0 ) z z0 z0 + d z z0 where k 6= 0: The dis ussion in Case (ii) gives

5.33. Theorem. If z0 is the oin ident xed point in C of w = S (z ),

then

1 1 = + k; k 6= 0: S (z ) z0 z z0

This result gives

z z0 (k 6= 0) 1 + k(z z0 ) whi h is the general form of a paraboli transformation that xes the nite point z0. Also, we note that, S (z ) takes 1 to z0 + 1=k and z0 1=k to 1, whi h means that z0, f (1) and f 1 (1) are ollinear points in C . Finally, suppose that D 6= 0, i.e. (a + d)2 6= 4(ad b ). Then S has two distin t xed points in C , say z1 and z2 . In this ase, we may sele t a Mobius transformation T (z ) su h that z1 7! 0 and z2 7! 1. This gives z z1 w = T (z ) = : z z2 Then, z = T 1(w) and f = T Æ S Æ T 1 has 0 and 1 as its only xed points. Hen e, by Case (i) with z0 = 0, f (w) = w; or equivalently,   z z1 T (S (T 1(w))) = w; i.e. T (S (z )) =  ; z z2 where  is neither 0 nor 1. As S (z1 ) = z1 and S (z2 ) = z2 , the last expression would then omplete the proof of the following S (z ) = z0 +

5.34. Theorem. Every Mobius transformation S (z ) whi h has exa tly two distin t xed points z1 and z2 in C 1 an be written as 8 >
: S (z ) z1 = (z z1 ) if z2 = 1;

where  2 C n f0; 1g.

This result may be obtained from another result on erning the invarian e property of the ross-ratio that will be dis ussed later, see Theorem 5.39.

245

5.4 Triples to Triples under Mobius Maps

5.4 Triples to Triples under Mobius Maps The Mobius transformation given by (5.15) has four oeÆ ients a; b; ; d. One of them an be adjusted without hanging the transformation. Indeed, as ad b 6= 0, both a and annot be zero simultaneously and so, we an rewrite the homogeneous expression of the oeÆ ients of T (z ) as 8 > >
(a= )z + (b= ) > : if 6= 0: z + (d= ) In either ase, T is a tually determined by only three onstants. Therefore, it is natural to expe t that we require only `three degrees of freedom' to determine T uniquely. As T maps a ir le onto another ir le and from elementary geometry we know that just three points determine a ir le. These observations will be made pre ise in the following two results whi h imply that three independent omplex parameters in C 1 are suÆ ient to des ribe Mobius maps uniquely, namely, the image of three pres ribed points.

5.35. Theorem. Given three distin t points z1 ; z2; z3 in C 1 , there exists a unique Mobius transformation T (z ) su h that T (z1) = 0; T (z2) = 1; and T (z3) = 1: Proof. To establish the existen e of su h a fun tion is easy. Regardless of the hoi e of the onstant k, T : C 1 ! C 1 de ned by T (z ) = k



z z1 z z3



sends z1 7! 0 and z3 7! 1. Adjust the onstant k su h that T (z2) = 1. This gives the desired map

T (z ) =

(5.36)

(z (z

z1 )(z2 z3 ) Az + B =: ; z3 )(z2 z1 ) Cz + D

whenever z1; z2 ; z3 2 C . Note that

AD BC = (z2 z1 )(z2

z3 )(z1

z3 ) 6= 0:

How do we de ne T if any one of the zj 's is not nite? If one of the zj 's is 1, then the orresponding T (z ) is de ned by the natural limiting value. For example if z1 = 1, then we need to write T (z ) as

T (z ) = z lim !1 1

(z (z

z1)(z2 z3 ) [(z=z1) 1℄(z2 z3 ) z2 z3 = lim = : z3)(z2 z1 ) z1 !1 (z z3 )[(z2 =z1) 1℄ z z3

246 Similarly, we de ne

Conformal Mappings and Mobius Transformations 8z 2 > > > z > < z

z3 if z1 = 1 z3 z1 if z2 = 1 (5.37) T (z ) = z z3 > > > > : z z1 if z3 = 1. z2 z1 Thus T (z ) de ned by (5.36) or (5.37) is the desired Mobius transformation whi h takes z1 to 0, z2 to 1 and z3 to 1. To omplete the proof, it remains to show that T is unique. Suppose that S (z ) is a Mobius transformation satisfying the ondition S (z1 ) = 0; S (z2 ) = 1; and S (z3 ) = 1: It follows that f = S Æ T 1 is again a Mobius transformation and that f xes 0; 1; 1. By Proposition 5.29, f (z ) = z whi h means that S = T . Note that T 1 Æ S xes three distin t points z1 ; z2 ; z3 and so, by Proposition 5.27, it is the identity transformation. Theorem 5.35 immediately implies an important mapping property whi h asserts that three distin t points uniquely determine a Mobius transformation.

5.38. Theorem. If fz1; z2 ; z3 g and fw1 ; w2 ; w3 g are two sets of triplets of distin t points in C 1 , then there exists a unique Mobius transformation taking zj to wj (j = 1; 2; 3) and that it is given by (w w1 )(w2 w3 ) (z z1 )(z2 z3 ) = : (w w3 )(w2 w1 ) (z z3 )(z2 z1 ) Proof. A

ording to Theorem 5.35, there exists two Mobius transformations S and R su h that S (z1) = R(w1 ) = 0; S (z2 ) = R(w2 ) = 1; and S (z3 ) = R(w3 ) = 1: They are given by (w w1 )(w2 w3 ) (z z1 )(z2 z3 ) S (z ) = and R(w) = ; (z z3 )(z2 z1 ) (w w3 )(w2 w1 ) respe tively. If any one of the zj 's and wj 's is 1, then the above are formally treated as a limiting ase. Now the existen e of an f sending zj to wj (j = 1; 2; 3) is lear, as the Mobius transformation f = R 1 Æ S does the job. As for the uniqueness, suppose that there are two Mobius transformations, say f and g, whi h enjoy the stated property; that is (see Figure 5.9), f (zj ) = g(zj ) = wj for j = 1; 2; 3: Then it follows that R Æ g Æ S 1 xes 0; 1; 1 and so it is the identity map. Thus, g = R 1 Æ S , whi h proves that f is unique.

247

5.5 The Cross-Ratio and its Invarian e Property S −1 0 1 ∞

R

◦g

g◦ ◦S −

1

0 1 ∞

z1 z2 z3

S −1

R

g w1 w2 w3

Figure 5.9: Uniqueness of f = R

1

Æ S.

5.5 The Cross-Ratio and its Invarian e Property For the set of three distin t points z1 ; z2 ; z3 of C 1 , the expression (z (z

z1 )(z2 z3 )(z2

z3 ) (z z1 )=(z z3 ) = z1 ) (z2 z1 )=(z2 z3)

is alled the ross-ratio of the four points z; z1; z2 ; z3 and is denoted by (z; z1; z2 ; z3 ), where if one of the four points is 1, the fa tors ontaining it should be omitted (see the limiting form in (5.37)). Considering z as a variable point and treating the three distin t points z1 ; z2 ; z3 as xed omplex numbers in C 1 , we obtain that the ross-ratio (z; z1; z2 ; z3 ) is pre isely the Mobius transformation whi h sends z1 ; z2 ; z3 into 0; 1; 1, respe tively. In view of the uniqueness of the map as des ribed in Theorem 5.35, the

ross-ratio is well de ned. Also, we observe that (z; z1 ; z2; z3 ) = (z1 ; z; z3; z2 ) = (z2 ; z3 ; z; z1) = (z3 ; z2; z1 ; z ): Note that there are 4! = 24 ross-ratios orresponding to the permutations performed on the four points z; z1; z2 ; z3 . But it an be easily seen that only six of these are dierent. In addition, we stress that a ross-ratio (z4 ; z1 ; z2 ; z3) asso iated with four distin t points z4; z1 ; z2 ; z3 in C 1 is a nite quantity dierent from 0 and 1. The invarian e of ross-ratios under Mobius transformations is the subje t of our next result whi h provides a way to represent a Mobius transformation that arries three distin t points to pres ribed image points w1 , w2 and w3 .

5.39. Theorem. The ross-ratio is invariant under Mobius transfor-

mations.

Proof. Let w = S (z ) be a Mobius transformation de ned by (5.31). Let fz1; z2 ; z3 g be a set of three distin t points in C 1 , and let fw1 ; w2 ; w3 g

248

Conformal Mappings and Mobius Transformations

be their images under this map, i.e. wj = S (zj ) for j = 1; 2; 3. Now, for ea h j = 1; 2; 3, we have 8 >
if 6= 0; : (z + d= )(zj + d= ) where  = a=d and Æ = (ad b )= 2 . Therefore, it follows easily that (w w1 )(w2 w3 ) (w w3 )(w2 w1 ) (z z1 )(z2 z3 ) = (z z3 )(z2 z1 ) = (z; z1 ; z2; z3 )

(w; w1 ; w2 ; w3 ) =

as asserted.

5.40. Example. Let us nd the Mobius transformation whi h sends 0 to 1, i to 0 and 1 to 1. To do this, for z 2= f0; i; 1g, we may appeal to (z; 0; i; 1) = (w; 1; 0; 1); i.e.

z 0 (w 1)(0 + 1) = i 0 (w + 1)(0 1)

and so, we arrive at the formula w = (1 + iz )=(1 iz ): There is a dire t approa h to problems of this type. Sin e i 7! 0, we an normalize a to be 1 and write the Mobius transformation in the form z i w= :

z + d The onditions 0 7! 1 and 1 7! 1 yield i 1 1= and 1 = ; i.e. d = i and = 1: d

The desired formula follows. 

5.41. Example. To demonstrate Theorem 5.34, we onsider w = T (z ) = 1=z . Then the xed points of T (z ) are z1 = 1 and z2 = 1. Note that w1 = T (z1) = 1, w2 = T (z2 ) = 1. Sele t the third point z3 distin t from z1 and z2 su h that T (z3) 6= 1. Be ause 1 7! 1, 1 7! 1 and z3 7! w3 , by the invarian e property of the ross-ratio, [w; 1; w3 ; 1℄ = [z; 1; z3; 1℄ so that   (w 1)(w3 + 1) (z 1)(z3 + 1) w 1 z 1 = ; i.e. = (w + 1)(w3 1) (z + 1)(z3 1) w+1 z+1

5.5 The Cross-Ratio and its Invarian e Property

249

for some  2 C . To nd , substitute z = i (so that w = i) into this equation. This gives  = 1. Note that the hoi e z = 0, and w = 1 qui kly gives  = 1. 

5.42. Remark. Suppose that z; z1; z2 ; z3 are four distin t points in

C 1 and one of them is 1, say z1 . Then, z z 1 (z; 1; z2; z3 ) is real () 2 3 = for some real t z z3 t () (1 t)z3 + tz2 = z showing that, (z; 1; z2; z3 ) is real i z; 1; z2 and z3 are on a line through 1.

Next, we re all that the equation arg(z b) =  (a onstant) de nes a half straight line issued from the point b 2 C . If a 2 C also lies on this line, then arg(a b) =  and therefore, z b = t; or z = ta + (1 t)b, for some real t: a b This observation implies that if z1; z2 ; z3 are distin t and lie on a line L in C , then we see that the quantity (z2 z3 )=(z2 z1 ) is a real number. Consequently, 1 z1 =z (z; z1 ; z2; z3 ) is real () is real 1 z3 =z () either z = 1 or zz zz1 = 1 1s for some s 2 R 3 () either z = 1 or z = sz1 + (1 s)z3 : The last impli ation learly shows that if we assume that z1 ; z2 ; z3 are distin t and lie on a line L in C , then the ross-ratio (z; z1; z2 ; z3) is a real number i either z = 1 or z is on the line L. 

5.43. Theorem. The four distin t points z; z1; z2 ; z3 in C 1 all lie on a ir le or on a line i their ross-ratio (z; z1 ; z2 ; z3 ) is a real number.

Proof. Suppose that z; z1; z2 , and z3 are four distin t points in C 1 and w = T (z ) = (z; z1; z2 ; z3 ): As (w; 0; 1; 1) = w, the last equation is equivalent to writing (w; 0; 1; 1) = (z; z1; z2 ; z3 ): But we know that T maps generalized ir les into generalized ir les and the points 0; 1; 1 are ollinear points so that the real axis in the w-plane is the image of the line or ir le through the points z1; z2 ; z3 , respe tively. Therefore, the point z is on this line or ir le i w = T (z ) is real. In other words, (z1 ; z2 ; z3 ; z4 ) 2 R i z1 ; z2 ; z3 , and z4 lie on a generalized ir le.

250

Conformal Mappings and Mobius Transformations

5.6 Conformal Self-maps of Disks and Half-planes Our results in this se tion mainly will be of the following type: Given two simply onne ted domains and 0 , onstru t an analyti onformal mapping from onto 0 . An analyti map f : ! 0 is said to be homeomorphi if it has an analyti inverse map g : 0 ! , i.e. f Æ g = I and g Æ f = I 0 . In addition, if = 0 then we say that f is an automorphism of . More pre isely, a onformal self-map of a domain

is an analyti fun tion from into that is one-to-one and onto. Every

onformal self-map of a domain is alled an automorphism of that domain. We denote by Aut ( ), the set of all automorphisms of . The set Aut ( ) forms what is alled a \group", with omposition as the group operation. The identity element is the identity map given by I (z ) = z .

5.44. Mappings of half-planes onto disks. There are a number of ways to hara terize all onformal mappings of the upper half-plane H + onto the (open) unit disk
and the extended real line R1 onto the unit

ir le 
. Suppose that  2 R and 2 C are xed numbers su h that Im > 0. Then, it is a simple exer ise to see that f , de ned by w = f (z ) = ei





z ; z

maps H + onto
and R1 onto 
. Indeed,

jwj < 1 () jz j2 jz j2 > 0 () 2Re (z ( )) = 4(Im )(Im z ) > 0: For example, for = i, we see that g(z ) = ei ((z i)=(z + i)) is a mapping of H + onto
. Consequently, as (z ) = ei=2 z maps the right half-plane fz : Re z > 0g onto H + , the omposition g Æ  given by (g Æ )(z ) = g(ei=2 z ) = ei



z 1 z+1



maps the right half-plane fz : Re z > 0g onto the unit disk
and the imaginary axis onto the unit ir le jwj = 1. We are a tually interested in the following:

5.45. Problem. Do all one-to-one, onto analyti maps of H + to
pre isely have the form of f for some 2 H + with some  2 R? The answer is yes. First, let us give a dire t proof. Consider

f (z ) =

az + b ; ad b 6= 0:

z + d

251

5.6 Conformal Self-maps of Disks and Half-planes

First, we re all that f (1) = 1 i = 0. Sin e we require a map whi h takes R1 into 
, we must have 6= 0 (and hen e a 6= 0). Consequently, we write   a z + b=a : f (z ) =

z + d= To obtain the expli it map, we translate our requirements into the equations a (5.46) jf (1)j = 1 = jf (0)j; i.e. = 1 = db : Note that both b and d annot be zero, sin e jb=dj = 1. Now, sin e ja= j = 1, we an write a= = ei for some  2 R so that 



z z  with  = d= and = b=a: Moreover, jf (1)j = 1 gives f (z ) = ei

j1 j = j1 j ; i.e. 1 + j j2 2Re = 1 + jj2 2Re : Sin e jj = j j, by (5.46), the last equation yields Re  = Re . Therefore,

either  = or  = : Consequently,  = as the alternate is not possible be ause otherwise ad b = 0. Thus, the desired map f turns out to be of the form   z f (z ) = ei z for some  2 R. Finally, we note that f ( ) = 0 2
. Therefore, if Im > 0, i.e. 2 H + , then we have proved the following result.

5.47. Theorem. All onformal mappings whi h map unit disk
su h that 2 H + maps onto 0 are given by   z (5.48) f (z ) = ei z

H+

onto the

for some  2 R. The inverse mapping f 1 :
! H + of f is given by

f 1 (w) =

ei ei

w : w

Using this result, we an nd the most general Mobius transformation sending the unit disk
onto itself (see also Theorem 5.59). We re all that every Mobius transformation is one-to-one on C 1 and the inverse fun tion exists and is also one-to-one on C 1 . Observe that f de ned by (5.48) 2 belongs to H(C n f g). On e again, as z jz j2 = 4(Im )(Im z ); it follows that \ jf (z )j < 1 () (Im ) (Im z ) > 0" and \ jf (z )j = 1 () (Im ) (Im z ) = 0":

252

Conformal Mappings and Mobius Transformations ζ = f (w) = eiθ

w = g(z) = zeiπ



w−β w−β¯



1

−1 φ(z) = (f ◦ g)(z)

Figure 5.10: Mapping of lower half-plane onto unit disk.

Thus, it is often interesting to study the mapping properties between lines or ir les.

5.49. Corollary. All onformal mappings whi h map the lower halfplane H = fz 2 C : Im z < 0g onto the unit disk
su h that b 2 H maps onto 0 are given by (see Figure 5.10) f (z ) = ei





z b ;  2 R: z b

5.50. Theorem. All onformal mappings whi h map the right halfplane fz 2 C : Re z > 0g onto the unit disk
su h that (Re > 0) maps onto 0 are given by f (z ) = ei





z ;  2 R: z+

Proof of this theorem is easy if one pro eeds exa tly as in Theorem 5.47. Alternatively, if (z ) = iz and f is as in Theorem 5.47, then the

omposition (f Æ )(z ) gives the desired map.

5.51. Example. Choosing  = 0 and = i in (5.48) we have (5.52)

w = f (z ) =

z i ; z+i

whi h is alled a Cayley mapping of H + onto
and the inverse is given by

f 1 (w) = i









1+w 1+w = ei=2 : 1 w 1 w

(i) By rotation, we see that the fun tion g given by 1+w g(w) = e i=2 f 1(w) = 1 w

253

5.6 Conformal Self-maps of Disks and Half-planes v

u≤v

i

f (z) =

C+

z−i z+i

u

C−

=

v

y

x

u

u

−1 D 1

= v −

−i u≤v

Figure 5.11: Image of D under Cayley map.

de nes an analyti map of the unit disk
onto the right half-plane f : Re  > 0g. Further, the image of jwj = r (r 2 (0; 1)) under  = g(w) may be omputed as the ir le 



1 + r2 2r = : 2 1 r 1 r2

Thus, jwj < r for r 2 (0; 1) is mapped onto the interior of this ir le. On the other hand, if r 2 (1; 1) then the image of jwj < r is seen to be the domain given by 



2r 1 + r2 > : 1 r2 j1 r2 j

What is the image of 1 < jwj < r under  = g(w)? (ii) Consider the two ir les de ned by

C+ = fz : jz aj = Rg and C = fz : jz + aj = Rg;

p

where a > 0 is xed and R = 1 + a2 . Clearly, these two ir les interse t at z = i and at z = i. Set

D+ = int C+ =
(a; R); D = int C =
( a; R) and let D = D+ \ D denote their ommon region of interse tion, see Figure 5.11. Let us nd the image of D under the Cayley map given by (5.52). Observe the following:

 f ( i) = 1 and so, ea h ir le that passes through the point 

z = i is mapped onto a straight line. In parti ular, the image of ea h of the ir les C+ and C is a straight line these two straight lines ne essarily pass through the origin, sin e both the ir les pass through z = i, where f (i) = 0

254

Conformal Mappings and Mobius Transformations

 f (i ) = (1 )=(1+ ) > 0 whenever 1 < < 1. In parti ular, the open verti al line segment onne ting i and i is mapped onto the positive real axis (0; 1).

Finally, by (5.52), it follows that

jz aj T R () () () () ()

  1+w i 1 w

a

TR

jw(a + i) (a i)j2 T R2 j1 wj2

Re [w(R2 (a + i)2 )℄ T 0 (sin e R2 = 1 + a2 ) Re [w(1 ia)℄ T 0 Re [we i ℄ T 0 (with Arg (1 + ia) = ):

Similarly, jz + aj T R () Re [w(1 + ia)℄ T 0 () Re [wei ℄ T 0: Now, it is lear that the domain D is mapped onto the in nite wedge bounded within the two lines Arg w =  and Arg w = . For instan e, if a = 1 then Arg (1 + ia) = =4 = Arg (1 ia). 

5.53. Example. Consider w = f (z ) = (i z )=(i + z ): Then we have (i) f maps the unit disk
onto the right half-plane fw : Re w > 0g (ii) f maps the upper half-plane fz : Im z > 0g onto the unit disk
(iii) f maps the open rst quadrant fz : Im z > 0; Re z > 0g onto the upper semi-disk fw : jwj < 1; Im w > 0g: Indeed, we see that w=

1 + iz 1 iz

2 () iz = 11 + ww = 1 jwj1j + w2j2iIm w

so that

 jz j < 1 () jw 1j < jw + 1j () Re w > 0  Im z > 0 () Re ( iz ) > 0 () jwj < 1  Im z > 0 and Re z > 0 () jwj < 1 and Im w > 0:



5.54. Example. Let us try to nd a map whi h takes the ir le

fz : jz 1j = 1g onto the line fw = u + iv : v = ug, see Figure 5.12. Our problem does not insist whi h points on the ir le should map whi h points on the line. Sin e a ir le is to map to a line, we must have 6= 0 and so, we an normalize to be 1. This suggests to onsider the Mobius transformation in the form az + b : f (z ) = z+d

255

5.6 Conformal Self-maps of Disks and Half-planes v

y 1+i

f (z) = e

−i3π/4

z z−2


O

2x

O

1+i u w0 = e−iπ/4

z0 = 1

Figure 5.12: Mapping from fz : jz

1j = 1g into the line fw = u + iv : v = ug.

It follows that our requirements translate into the equations (for example) by onsidering 0 7! 0, 2 7! 1, and 1 + i 7! 1 + i:

b 0 = f (0) = ; d

1 = f (2) = 22a++db ; f (1 + i) = 1 + i:

The rst two equations show that b = 0 and 2 + d = 0. So az f (z ) = : z 2 Finally, the last requirement f (1+ i) = 1+ i yields that a = 1+ i = e3i=4 : Hen e, the desired map is given by (5.55)

f (z ) = e3i=4



z

z 2



:

Note that f (1) = 1 i and therefore, be ause of the prin iple of onformal map (preservation of sense and magnitude), the map f given by (5.55) takes the disk fz : jz 1j < 1g onto the half-plane fw : Im w < Re wg: 

5.56. Automorphisms of the unit disk. Let  2 R; z0 2
be xed and  be given by   z z0 i (z ) = e (5.57) : 1 z0z Note that  is one-to-one in C n f1=z0 g. If jz j = 1, then z 1 = z so that j(z )j = 1: In addition, sin e every Mobius transformation maps a ir le onto a

ir le or a straight line,  must map the unit ir le jz j = 1 onto itself. Also, (z0 ) = 0 and (
) =
, be ause

j(z )j < 1 () jz z0 j2 < j1 zz0 j2 () (1 jz0 j2 )(1 jz j2 ) > 0 () jz j < 1 (sin e jz 0 j < 1)

256

Conformal Mappings and Mobius Transformations

so that
must be mapped onto itself by (z ). Now  has the inverse given by   w + ei z0 1 i :  (w) = e 1 + e i z0 w whi h has a similar form as  and so,  1 shares similar properties as that of . Organizing these observations together, we an assert that for ea h z0 with jz0 j < 1 and  2 R,  is a bije tive self mapping of the unit disk
. This result raises the following

5.58. Problem. Do all one-to-one, onto analyti maps of
to itself pre isely have the form of  for some jz0 j < 1 with some  2 R? The answer to this problem is yes (see also Example 5.81 and Theorem 6.45). The question of mapping the upper half-plane H + onto itself an also be treated easily (see Theorem 5.69).

5.59. Theorem. All onformal mappings whi h map the unit disk
onto itself and the point z0 , jz0 j < 1, onto 0 must be of the form (5:57) for some  2 R. Equivalently, Aut (
) =



ei



z z0 : z 2
; 0    2 : 1 z0z 0

Proof. We start with the Mobius transformation az + b T (z ) = (a; b; ; d 2 C ; ad b 6= 0):

z + d For ea h  2 ( ; ℄, the ondition jT (ei )j = 1 implies that jaei + bj2 = j ei + dj2 : That is (5.60)

jaj2 + jbj2 + 2Re (abei ) = j j2 + jdj2 + 2Re ( dei )

whi h must be true for ea h  in ( ; ℄. Choosing  = 0; , it follows that

jaj2 + jbj2 + 2Re (ab) = j j2 + jdj2 + 2Re ( d) and

jaj2 + jbj2 2Re (ab) = j j2 + jdj2 2Re ( d): Adding the last two equations, we nd that jaj2 + jbj2 = j j2 + jdj2 so that (5.61) jaj2 j j2 = jdj2 jbj2 and therefore, by (5.60), we get that Re (abei ) = Re ( dei ) for ea h  2 ( ; ℄:

257

5.6 Conformal Self-maps of Disks and Half-planes

Again, hoosing  = 0 or , and  = =2, we see that

ab = d:

(5.62)

Now using the two equalities (5.61) and (5.62), we nd that 2 2 jwj2 1 = jaz + jb zj + dj zj2 + dj (jaj2 j j2 )jz j2 (jdj2 jbj2 ) + 2Re (z (ab d)) = j z + dj2 2 2 2 (jz j 1)(jaj j j ) = j z + dj2 and therefore the requirement that \jz j < 1 ) jwj < 1" gives rise to the

inequality

0 < jaj2 j j2 ; i.e. j j < jaj (Note that a annot be zero, but ould be zero). Therefore, from (5.61), we also have d 6= 0 (but b ould be zero). Thus, by (5.62), we write  

b

= = k; say, a d

(5.63) with jkj < 1 and so 1

2

a

=1

2 b ; d

jaj2 j j2 = jdj2 jbj2 jaj2 jdj2

i.e.

so that by (5.61), we have jaj = jdj whi h, by (5.61), again gives jbj = j j. Thus, we an write a=d = ei for some  so that 

a z + b=a z z0 T (z ) =
= ei d 1 + ( =d)z 1 z0 z



where, by (5.61) and the fa t that jaj = jdj,

z0 =

b = a

ba jaj2 =

d jdj2 =

 

d

:

By (5.61), we also observe that if = 0, then b must be zero sin e a 6= 0. In that ase, we have S (z ) = ei z: On the other hand, if 6= 0 then b 6= 0 and therefore, b

b b z0 = =
= k ; i.e. jz0 j = jkj < 1: a a

There are several alternate proofs of Theorem 5.59, whi h gives a hara terization of all onformal self mappings of the unit disk
. For a better

258

Conformal Mappings and Mobius Transformations

understanding on these te hniques, we mention two more su h proofs. The se ond one follows from the prin iple of symmetry (see Example 5.81). The third proof is a onsequen e of S hwarz' lemma (see Theorem 6.45).

5.64. Example. Let us dis uss the question of nding the image of the open upper semi-unit disk = fz : jz j < 1; Im z > 0g under the Mobius map de ned by a z (5.65) : w = a (z ) = 1 az In our dis ussion, we restri t our attention to a spe ial situation a = i, 1 <  6= 0 < 1; and pro eed to dis uss the desired mapping properties. Note that a z a w (a w)(1 aw) a w + ajwj2 a2 w w= () z= = 1 az 1 aw j1 awj2 = j1 awj2 and jz j < 1 () jwj < 1. With a = i, Im z > 0

() Im [a w + ajwj2 a2 w℄ > 0 () j"wj2 (1 + 2 )Im w +  > 0 () () ()

 w "  w 8 > > > < w > > > : w



 2





1 + 2 i 2





1 + 2 2

2

!#

1

>0

 #

1 2 2 1 + 2 2 i >0 2 2   1 + 2 1 2 i > if 0 <  < 1 2 2 

1 + 2 i 2




0 under i=2 (z ) orresponds to the set of points given by

fw : jw (5=4)ij > 3=4g : Thus, the image of the open upper semi-disk fz : jz j < 1; Im z > 0g under

i=2 (z ) is the shaded region in Figure 5.13 and the image of the lower semidisk fz : jz j < 1; Im z < 0g under the same map is also indi ated in Figure 5.13. Clearly, the image of the open segment ( 1; 1) under i=2 (z ) is the open ar of the ir le jw (5=4)ij = 3=4 indi ated in the same gure. What happens when  = 1?  5.66. Example. Let us start onstru ting a bije tive analyti map  taking the upper semi-unit disk D = fz : jz j < 1; Im z > 0g onto the

259

5.6 Conformal Self-maps of Disks and Half-planes y

v 2i 5 i 4 3 i 4

−1

1

x

u

1

−1

Figure 5.13: Mapping from
under i=2 (z ) = (i 2z )=(2 + iz ).

unit disk
, leaving the points 1; 1; i xed, for example. Can there be a Mobius transformation arrying D onto
? As 1 + z (1 + z )(1 z ) 1 jz j2 f (z ) = = 1 z j1 z j2 = j1 z j2 + i





2Im z j1 z j2 ;

f maps jz j < 1 onto Re w > 0, and Im z > 0 onto Im w > 0. This observation implies that (be ause f : C 1 ! C 1 is a bije tive onformal map) the upper semi-unit disk maps onto the open rst quadrant ,

= fw : Im w > 0; Re w > 0g: This also follows from the fa t that 1+x iy z = x () f (x) = ; and jz j = 1 () f (z ) = 1 x j1 z j2 :

If we let g(z ) = z 2 , then g( ) = H + = fw : Im w > 0g. We know that z i h(z ) = z+i is an analyti bije tion of Im z > 0 onto the unit disk
. Thus, the omposition (1 + z )2 i(1 z )2 (z ) = (h Æ g Æ f )(z ) = (1 + z )2 + i(1 z )2 transforms the (open) upper semi-unit disk
\ H + onformally onto the unit disk
, leaving the points 1; 1; i xed, see Figure 5.14. This idea an be used to show that the transformation (1 + z n )2 i(1 z n)2 (5.67) (z ) = ; n 2 N; (1 + z n )2 + i(1 z n)2 maps the domain fz : jz j < 1; 0 < Arg z < =ng onformally onto the unit disk
. What are its xed points? Are all points z su h that z n = 1 or z n = 1 xed points, in parti ular? 

260

Conformal Mappings and Mobius Transformations i

f (z) =

1+z 1−z

α φ = h◦g◦f

−1

O

1

O

g(w) = w2

−i h(ζ ) =

ζ −i ζ +i

2α

O

Figure 5.14: Conformal map of upper semi-unit disk onto the unit disk.

5.68. Automorphisms of the upper half-plane H + . There are several dierent proofs of the following result and all these proofs follow if we pro eed with the idea of the proof of the results on erning automorphisms of disks. 5.69. Theorem. Every Mobius transformation of the form T (z ) =

az + b

z + d

is a onformal self-map of the upper half-plane H + i a; b; ; d are real numbers satisfying the ondition ad b > 0. Equivalently,

Aut (H + ) =





az + b : a; b; ; d 2 R; ad b > 0 :

z + d

Proof. The required map should take the real axis in the z -plane onto the real axis in the w-plane. As every Mobius map is one-to-one, the desired map should arry three distin t real numbers on the x-axis onto three distin t real numbers on the u-axis. Then there always exist three distin t real numbers x1 ; x2 ; x3 su h that 0; 1; 1 are their images. By the de nition of the ross-ratio and its invarian e property (see Theorem 5.39), we get (5.70) (T (z ); 0; 1; 1) = (z; x1 ; x2 ; x3 ); i.e. T (z ) =

(z (z

x1 )(x2 x3 ) x3 )(x2 x1 )

(If one of the xj 's is 1, then use the limit pro ess). Sin e the required map

261

5.7 Prin iple of Symmetry and Mobius Maps a∗

L a C a z0

a∗

Figure 5.15: Symmetri with respe t to a line/ ir le.

must send H + onto itself, Im T (i) > 0. This gives Im T (i) = Im whi h shows that (x1 in the form



i x1 x2 i x3 x2 x3 )(x2

x3 )(x2

T (z ) =

we see that

ad b = (x2 = (x2







x3 x x x x = 1 23 2 3 > 0 x1 1 + x3 x2 x1 x1 ) > 0. Now, rewriting (5.70)

az + b

z + d

x3 )[(x2 x1 )( x3 )℄ [(x2 x3 )(x2 x1 )(x1 x3 ) > 0

x3 )( x1 )℄(x2

x1 )

and this ompletes the proof.

5.7 Prin iple of Symmetry and Mobius Maps Let L be a line in C . Two points a and a in C are said to be symmetri with

respe t to L if L is the perpendi ular bise tor of [a; a ℄{the line segment

onne ting a and a . Clearly, every ir le or line passing through both a and a interse t the line L at a right angle, see Figure 5.15. For example, two points z and z  are symmetri with respe t to the real axis pre isely when z  = z . Similarly, two points z and z  are symmetri with respe t to the imaginary axis pre isely i z  = z. When we do say two points z and z  are symmetri with respe t to the line ft(1 i) : t 2 Rg? A Mobius transformation w = T (z ) with real oeÆ ients maps the real axis in the z -plane onto the real axis in the w-plane and, z and z onto the points w and w, respe tively, whi h are again symmetri with respe t to the real axis. This observation motivates us to formulate the de nition of symmetri points with respe t to a ir le in C 1 . Suppose that K is a ir le jz z0 j = r in C . Two points a and a are said to be symmetri with respe t to the ir le K (or inverse points with

262

Conformal Mappings and Mobius Transformations a ∗ = Jk (a) a

R

2

r

R

z0

Figure 5.16: The re e tion/inversion map JK .

respe t to the ir le K ) i

(5.71)

ja z0 j ja z0 j = r2 and Arg (a z0 ) = Arg (a z0 ):

That is, a and a lie on the same ray emanating from the enter z0 of K , and the produ t of their distan es from the enter of the ir le K is equal to the square of the radius of the ir le. If we let ja z0 j = R, then a = z0 + Rei for some  2 R so that (5.71) is equivalent to

a = z0 + Rei and a

z0 =

r2 ei : R

That is,

r2 r2 (a z0 ) z0 ) = r2 or a = z0 + = z0 + a z0 ja z0 j2 : It follows from (5.71) that if a approa hes the ir le, then a also approa hes the ir le. In other words, a = a i a 2 K . If a approa hes the enter z0 , then the point a moves away to in nity. This fa t is expressed by saying that z0 and 1 are symmetri with respe t to the ir le. This allows us to de ne the symmetri point a of a with respe t to the ir le K in C by the map JK : C 1 ! C 1 : (5.72) (a z 0 )(a

(5.73)

a = JK (a) =

8 > >

> :

1 z0

r2 a z0

if a 6= z0 ; 1 if a = z0 if a = 1:

The map des ribed by (5.73) is often alled the re e tion/inversion in the

ir le K = fz : jz z0 j = rg or re e tion with respe t to the ir le, and the pair of points a and a = JK (a) are said to be symmetri with respe t to K , see Figure 5.16. For example, if we let z0 = 0 then K = fz : jz j = rg

5.7 Prin iple of Symmetry and Mobius Maps

263

and so, we have

r2 JK (0) = 1; JK (1) = 0; and JK (a) = for a 6= 0; 1 a 2 Then, we all the map a 7! r =a, a re e tion/inversion in the ir le jz j = r. For r = 1, this gives a = 1=a and so a and 1=a are symmetri points with respe t to the unit ir le jz j = 1. In parti ular, 0 and 1 are symmetri points with respe t to the unit ir le jz j = 1. Moreover, it is easy to see that if K = L [ f1g for a line L with the equation z + z + = 0 (0 6=  2 C ; 2 R); then the re e tion map JK in a line K in C 1 is obtained by repla ing z and z respe tively by a and a : ( 

a if a 6= 1 (5.74) JK (a) =   1 if a = 1:

By (5.73) and (5.74), JK xes every point of K and JK Æ JK is the identity transformation of C 1 so that JK 1 = JK . If we let M to be the set of all transformations of the form Az + B (A; B; C; D 2 C ; AD BC 6= 0); Cz + D then we note that every re e tion JK de ned by (5.73) and (5.74) belongs to M . Note that

M = f(T Æ J )(z ) : T

2 M and J (z ) = zg;

where M denotes the set of all Mobius transformations. The set M shares many properties of M in ommon. For example, if C is a ir le in C 1 , then, for f 2 M , f (C ) is again a

ir le in C 1 and the omposition of a pair of two transformations in M produ es a Mobius transformation.

5.75. Theorem. (Symmetry Prin iple for Mobius Maps) Let T be a Mobius transformation and K be a ir le in C 1 with K 0 = T (K ): Then (5.76) T Æ JK = JK 0 Æ T; i.e. T (JK (a)) = JK 0 (T (a)) for ea h a 2 C 1 :

In parti ular, T maps any pair of symmetri points with respe t to K onto a pair of symmetri points with respe t to the image ir le K 0 under T .

Proof. Consider the auxiliary fun tion g de ned by g = JK 0 Æ T Æ JK : Then g, being a omposition of Mobius maps, is a Mobius map and so, g is one-to-one on C 1 . Also, we know that JK (a) = a and JK 0 (b) = b for ea h a 2 K and b 2 K 0

264

Conformal Mappings and Mobius Transformations

whi h gives that T (z ) = g(z ) for ea h z shows that g = T so that

2 K . The uniqueness prin iple

T = JK 0 Æ T Æ JK ; i.e. T Æ JK 1 = JK 0 Æ T: Sin e JK is its own inverse, (5.76) follows. Now, we let a and a be two points that are symmetri with respe t to the ir le K . Then, by (5.76), the image points b = T (a) and b = T (a ) satisfy JK 0 (b) = JK 0 (T (a)) = T (JK (a)) = T (a ) = b so that b and b are symmetri with respe t to the image ir le T (K ).

5.77. Remark. Theorem 5.75 may also be proved by using the following two fa ts: (i) Two points are symmetri with respe t to a ir le in C 1 if every ir le

ontaining the points interse t the given ir le orthogonally. (ii) Mobius transformations are onformal and preserve ir les, and so preserve the orthogonality; hen e they preserve the symmetry ondition. 

5.78. Theorem. Let K be a ir le passing through three points in

C 1 . Then the re e tion JK satis es the relation (5.79)

(JK (a); z1 ; z2 ; z3 ) = (a; z1 ; z2; z3 )

for a 6= z1 ; z2 ; z3 . Conversely, if

(5.80)

(a ; z1 ; z2 ; z3) = (a; z1 ; z2 ; z3 )

then a and a in C 1 are symmetri with respe t to the ir le K .

Proof. De ne T (z ) = (z; z1; z2 ; z3). Then T maps z1 to 0, z2 to 1, and z3 to 1, and so T (K ) = R [ f1g. By the invarian e property of the

ross-ratio, one has T (JK (a)) = (JK (a); z1 ; z2 ; z3 ): If a 2 K then, as JK xes ea h point of K and T (K ) is real, we have

T (JK (a)) = (a; z1 ; z2 ; z3 ) = (a; z1 ; z2; z3 ): If a 62 K then, by the prin iple of symmetry, T (a) and T (JK (a)) are symmetri with respe t to the ir le T (K ) = R [ f1g. That is, by (5.76), we have T (JK (a)) = T (a) = T (a) whi h is indeed (5.79).

5.7 Prin iple of Symmetry and Mobius Maps

265

Conversely, suppose that a and a are a pair of points in C 1 su h that (5.80) holds. Case (i): Let K = L [f1g for a straight line L in C . Then, we hoose z3 = 1 and the ondition (5.80) gives a z1 a z1 = z2 z1 z2 z1 and so, ja z1 j = ja z1 j. But, sin e z1 is arbitrary, it follows that a and a are equidistant from the line L. Moreover,      a z1 a z1 Im = Im z2 z1 z2 z1 showing that a and a lie in dierent half-planes determined by L. This is obviously a re e tion with respe t to L. Case (ii): Let K = fz : jz z0 j = rg in C and let K pass through z1 ; z2 ; z3 2 C , i.e. jzj z0 j = r for j = 1; 2; 3. A symmetri use of the invarian e property of the ross-ratio under Mobius transformation gives (a; z1 ; z2; z3 ) = (a z0 ; z1

z0 ; z2 z0; z3 z0) (f (z ) = z z0 )  r2 r2 = a z0 ; ; ; z1 z0 z2 z0 z3 z0 ((zj z0 )(zj z0 ) = r2 )     r2 r2 = ;z z ;z z ;z z f (z ) = a z0 1 0 2 0 3 0 z   r2 ;z ;z ;z : = z0 + a z0 1 2 3 

r2

Hen e, in view of (5.80) (as the ross-ratio (z; z1 ; z2; z3 ) = f (z ) is univalent in C 1 ), we nd that

a = z0 +

r2

or (a z0)(a z 0 ) = r2 : a z0 This means that a and a are symmetri with respe t to K . A pra ti al appli ation of the symmetry prin iple is to nd a Mobius transformation w = T (z ) whi h maps a given ir le onto another ir le. Next we present an example; many similar problems may be solved using the same idea.

5.81. Example. Suppose we wish to present an alternate proof of Theorem 5.59 (see also Theorem 6.45). To do this we onsider (z ), a general Mobius transformation, whi h maps the unit disk
onto itself. Then there must exist a point a 2
su h that (a) = 0. If a and a

266

Conformal Mappings and Mobius Transformations

are symmetri with respe t to the unit ir le 
, then (a) and (a ) are symmetri with respe t to (
) = 
. As (a) = 0, we have (a ) = 1 (be ause 0 and 1 are symmetri with respe t to jwj = 1). As aa = 1, we have a = 1=a so that (z ) has the form

w = (z ) = 



z a 1 az



for some  2 C . Also, as (1) is a point on the unit ir le jwj = 1, we have j(1)j = 1 whi h gives that jj = 1, i.e.  = ei for some  2 R. Consequently, (z ) has the desired form. Finally, if jaj > 1 then (z ) maps jz j  1 onto jwj  1 so that (z ) maps jz j  1 onto jwj  1.  We end this se tion with some remarks. We know that every Mobius transformation T is onformal and it maps a ir le C1 in C 1 onto a ir le C2 in C 1 . Moreover, T an be found by requiring three points z1 ; z2 ; z3 in C1 to map onto three pres ribed points w1 ; w2 ; w3 in C2 . Is it possible to nd a Mobius Transformation su h that a1 2 C1 maps onto b1 2 C2 , and a2 2= C1 onto b2 2= C2 ? Yes, it is. Su h a transformation is given by (w; b1 ; b2 ; b ) = (z; a1; a2 ; a ) 2

2

where a2 is symmetri to a2 with respe t to C1 while b2 is symmetri to b2 with respe t to C2 .

5.8 Exer ises 5.82. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) The fun tion f (z ) = sin z is not onformal on the in nite strip = fz 2 C : jRe z j < =2g. (b) The omposition of two Mobius transformations is again a Mobius transformation. ( ) The set M of all Mobius transformations Tab d(z ) = (az + b)=( z + d) with ad b 6= 0, forms a group with respe t to the omposition as a binary operation. (d) The subset M1 of all Mobius transformations Tab d(z ) with ad b > 0, forms a subgroup in the group M of all Mobius transformations. (e) The subset M2 of all Mobius transformations given by   1 1 z 1 z M2 = z; ; 1 z; ; ; z 1 z z z 1 forms a group with respe t to the omposition as a binary operation.

5.8 Exer ises

267

(f) The set M of all Mobius transformations does not have the ommutative property. (g) Any two Mobius transformations that have the same xed points are

ommutative. Note: If S and T are two Mobius transformations whi h ommute (i.e. S Æ T = T Æ S ), will they have the same xed points? (h) If a Mobius transformation T arries z1 and z2 into a same number w1 , then either z1 = z2 or else T is a onstant map. (i) Every Mobius transformation Tab d(z ) = (az + b)=( z + d) su h that j j = jdj arries the unit ir le 
onto a straight line. (j) A Mobius transformation f : C 1 ! C 1 taking 0; 1; 6 into 2; 3; 4, respe tively, is f (z ) = [z + (2=3)℄=[z + (3=2)℄: (k) Given three distin t points z; z1; z2 and a number  2 C su h that  62 f0; 1; (z z2 )=(z2 z1 )g, there always exists a unique z3 with (z; z1; z2 ; z3) = . (l) If z1; z2 ; z3 are three distin t points in C 1 su h that z; z 0 satisfy (z; z1; z2 ; z3) = (z 0 ; z1 ; z2 ; z3), then z = z 0. (m) The ross-ratios (z4 ; z1; z2 ; z3 ) and (w4 ; w1 ; w2 ; w3 ) are equal i there exists a Mobius transformation f su h that f (zj ) = wj for j = 1; 2; 3; 4. (n) Given a pair of (generalized) ir les, there always exists a Mobius transformation arrying one ir le onto another. (o) If a ir le C is mapped under the inversion w = 1=z onto another

ir le C 0 , then the enter of the ir le C need not be mapped onto the enter of the ir le C 0 unless the enter of the ir le C is zero. (p) A Mobius transformation, whi h maps the upper half-plane fz : Im z > 0g onto itself and xing 0; 1 and no other points, must be of the form T (z ) = z for some  > 0 and  6= 1. (q) A Mobius transformation whi h maps the upper half-plane fz : Im z > 0g onto itself whi h xes 1 and no other points must be of the form T (z ) = z + for some 6= 0 with Im > 0. (r) Let T be a Mobius transformation su h that 1 2 Fix (T ). Then T

arries R onto itself i T (z ) = z + for some  2 R n f0g, 2 R. (s) There exist trans endental entire fun tions having no xed points. (t) A Mobius transformation whi h arries the upper half-plane H + onto the unit disk
su h that z = 2i is mapped onto w = 0 while z = 1 is mapped onto w = 1 is pre isely (2i z )=(z + 2i). (u) A Mobius transformation takes R into R i it an be represented with real oeÆ ients. (v) A transformation whi h arries an in nite se tor of angle =n (n 2 N ) onto the unit disk
is (z n i)=(z n + i).

268

Conformal Mappings and Mobius Transformations

(w) The Ja obian of a Mobius transformation T is identi ally equal to 1 i T (z ) = ei z + b. (x) If a; b 2 C and r; R > 0 are xed, then a Mobius transformation that maps the disk
(a; r) onto C 1 n
(b; R) is given by T (z ) = b + Rr=(z a): (y) The transformation w = (1 iz )=(z i) maps jz j = r, where r < 1, into a ir le in the w-plane, whose enter is on the imaginary axis. (z) A mapping whi h transforms = fz : 0 < Arg z < =6g onto the unit disk
is given by f (z ) = ei' (z 6 )=(z 6 ), Im  > 0:

5.83. What is the angle between images of the urves x2 + y2 = 1 and y = x at their point of interse tions, under the map f (z ) = z 2? Answer the same if the ir le is repla ed by the ellipse x2 + 4y2 = 4. 5.84. Determine points where ea h of the following mappings is onformal: (i) z + e z

5 (ii) zez

3 +1

(iii) os z (iv) z + az 2 (v) z + az 3:

5.85. Determine points where ea h of the following mappings fails to be onformal: (i) z 5 + 1 (ii) z 2

exp(z 2 ) (iii) osh z (iv) sinh z:

5.86. Determine a; b; ; d su h that the Mobius transformation T (z ) de ned by (5.15) oin ides with its inverse given by (5.20). 5.87. Find all Mobius transformations that map the unit disk
onto the left half-plane H = fw 2 C : Re w < 0g. 2z (2 + i)z 2 z ; S (z ) = ; and S3 (z ) = : 5.88. Set S1 (z ) = 2z 1 2 3z 1 z+i Show that 0 and 1 are the xed points for S1 as well as for S2 , whereas 1 + i and 1 i are the xed points of S3 , and S3 is loxodomi .

5.89. Find the image of the ir le jz j = r (r 6= 1) under the Cayley mapping w = f (z ) = (z i)=(z + i); see Example 5.51. What happens when r = 1? 5.90. Using the invarian e property of the ross-ratio, nd a Mobius transformation f in ea h of the following ases:

f1; i; 0g onto f1; i; 1g (b) f0; 1; 1g onto f0; 1; 1g (a)

( ) f1; i; 1g onto f2i; 2; 2ig (d) f0; 2 2i; 4g onto f 3=4; 11i=4; 1g

Chapter 6

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

In Se tion 6.1, we derive the Maximum and the Minimum modulus prin iples/theorems as an important appli ation of the Cau hy integral formula. The Maximum modulus prin iple is a powerful tool in obtaining an expli it estimate for the size of the absolute value of an analyti fun tion. In Se tion 6.2, we use the Maximum prin iple to derive Hadamard's three lines/ ir les theorem. Se tion 6.3 is devoted to S hwarz' lemma whi h is one of the most important onsequen es of the Maximum prin iple and the power series expansion. Later in this se tion, we use the lassi al form of S hwarz' lemma to hara terize the onformal self-maps of the unit disk in the form of the S hwarz-Pi k lemma whi h is a basi tool in the introdu tion of the hyperboli metri and hyperboli geometry in the unit disk. In Se tion 6.4, we dis uss Liouville's theorem and its various generalizations whi h give rise to fas inating and surprisingly pra ti al results su h as the

elebrated dis overy of Gauss, the so alled fundamental theorem of algebra (see Se tion 6.6).

6.1 Maximum Modulus Prin iple When we deal with a fun tion of one variable we frequently speak about the on ept of maxima and minima. On the other hand, we annot speak of maxima and minima of a omplex fun tion f sin e C is not an ordered eld. However, it is meaningful to onsider maximum and minimum values of the modulus of the omplex fun tion f , the real part of f and the imaginary part of f .

6.1. De nition. Let D be a subset of C . A omplex fun tion de ned on D is said to have a (lo al) maximum modulus at a 2 D if there exists

270

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

a Æ > 0 su h that
(a; Æ)  D and jf (z )j (lo al) minimum of jf j is similarly de ned.

 jf (a)j for all z 2
(a; Æ); a

For example, on the losed disk jz j  r, we have jez j = eRe z  er and at the boundary point z = r, ez = er . Thus, ez attains its maximum modulus at z = r. Consequently, M (r; ez ) = maxjzjr jez j = er : Analogously, for jz j  r, we easily see that

iz

iz iz y y r r  je j +2je j = e 2+ e  e +2e = os(ir) whi h shows that M (r; os z ) = maxjzjr j os z j = osh r: Although, in real variable theory, many fun tions su h as sin x, os x are bounded on R, we will see that neither sin z nor os z are bounded in C . iz j os z j = e +2 e



6.2. Theorem. (Maximum Modulus Prin iple) Suppose that f is analyti in a domain D and a is a point in D su h that jf (z )j  jf (a)j holds for all z 2 D. Then, f is a onstant. For the proof of Theorem 6.2, we need the following basi fa t from Real Analysis:

6.3. Theorem. Let h() be a R ontinuous real-valued fun tion on [a; b℄ with h()  0 for all  2 [a; b℄. If ab h() d = 0; then h() = 0 for all  2 [a; b℄. Proof. Geometri ally, the proof of this theorem is obvious. Alternately, we x t 2 [a; b℄. Then we have 0  H (t) :=

Z t

a

h() d 

Z b

a

h() d = 0

whi h gives H (t) = 0 on [a; b℄ so that 0 = H 0 () = h() on [a; b℄:

Proof of Theorem 6.2. Sin e a 2 D and D is open, there exists an r su h that
(a; r)  D. Then, f is analyti inside and on the ir le

= 
(a; r). So, by the Cau hy integral formula (see Theorem 4.66), (6.4)

f (a) =

Z

1 f ( ) 1 d = 2i  a 2

Z 2

0

f (a + rei ) d:

By hypothesis, jf (a + rei )j  jf (a)j and, by (6.4), we note that

jf (a)j  21

Z 2

0

jf (a + rei )j d  21

Z 2

0

jf (a)j d = jf (a)j

271

6.1 Maximum Modulus Prin iple

so that (6.5)

Z

 1 2  j f (a)j jf (a + rei )j d = 0: 2 0 Sin e the integrand in (6.5) is ontinuous and non-negative, Theorem 6.3 applies and (6.5) implies that

jf (a + rei )j = jf (a)j for 0    2: This equation holds on all ir les j aj = s; 0 < s  r and therefore, jf (z )j is onstant on
(a; r). By the Uniqueness theorem, f is onstant on the whole of D.

6.6. Remark. It is important to note that Theorem 6.2 does not ne essarily hold for open sets. Further, one ould also prove Theorem 6.2 without using Theorem 6.3. For example, as f 2 H(
(a; r)), we have f (z ) =

X

an (z a)n for jz aj < r0 with r0 > r:

n0

In parti ular, for ea h ir le jz

aj = s with 0 < s  r, we have

f (a + sei ) = Now, as jf (a + sei )j whi h gives X

n0

jan j2 s2n = 21

n0

an sn ein :

 jf (a)j = ja0 j, we see that jf (a + sei )j2  ja0 j2

Z 2

0

X

f (a + sei )f (a + sei ) d 

1 2

Z 2

0

ja0 j2 d = ja0 j2 ;

that is, an = 0 for ea h n 2 N . Thus, f (z ) = a0 on every ir le j aj = s in
(a; r). Consequently, f (z ) = a0 = f (a) on
(a; r). By the Uniqueness theorem, f is onstant on the whole of D. 

6.7. Example. Consider f (z ) = ez for z 2
(a; r). Then,

i = eRe a+r

max jez j = max ea+re 02 jz ajr

and the maximum modulus of ez is attained at the boundary point z = a+r. Similarly, to determine M = maxjzj1 jz 3 + 3z 1j, we may let z = ei so that p

jz 2 + 3z 1j = je2i + 3ei 1j = jei + 3 e i j = 9 + 4 sin2 : p Thus, we have M = 13 and the maximum is attained at z = i, i.e. when  = =2. 

272

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

It is interesting to observe that the minimum value of jf j may be attained at an interior point of D without f being onstant. For instan e,

onsider f (z ) = z for z 2
r . Then, jf (z )j = jz j  0 = jf (0)j and so the minimum valuepof jf (z )j is attained at the origin. The maximum value of jf (x + iy)j = x2 + y2 is attained at the boundary points jz j = r. Note that f has a zero at the origin. The Maximum modulus prin iple is often used in the following form known as the Maximum modulus theorem:

6.8. Theorem. (Maximum Modulus Theorem) Suppose that f is analyti in a bounded domain D and ontinuous on D . Then, jf (z )j attains its maximum at some point on the boundary D of D. Proof. Re all that a ontinuous fun tion on a ompa t set attains a maximum. So, by hypothesis, f is bounded on D and the maximum value of jf (z )j is attained at some point of D. By Theorem 6.2, it annot be in D, so it must be on the boundary D. 6.9. Example. Consider the fun tion f (z ) = z 2 de ned on the losed disk D = fz : jz 1 ij  1g:pLet us show that the maximum value of jf (z )j is attained at z = (1 + 1= 2)(1 + i). To do this, set z = 1 + i + ei = (1 + os ) + i(1 + sin );  2 [0; 2):

Then, jf (z )j = 3 + 2( os  + sin ): It follows that the maximumpvalue of jf (z )j is attained at  = =4 and the maximum value is 3 + 2 2. The maximum is attained at z = 1 + i + ei=4 . 

6.10. Corollary. Suppose that f is analyti in a bounded domain D and ontinuous on D. Then, ea h of Re f (z ), Re f (z ), Im f (z ) and Im f (z ) attains its maximum at some point on the boundary D of D. Proof. Let u(x; y) = Re f (z ) and g(z ) = ef (z) . By the Maximum modulus theorem, jg(z )j = eu(x;y) annot assume the maximum value in D. Sin e eu is maximized when u is maximized, we obtain that u(x; y) annot assume its maximum value in D. The remaining ases follow similarly. 6.11. Remark. The on ept of a lo al maximum at a point a 2 S  D is meaningful only if a is a limit point. If a is isolated, we do not have jf (z )j  jf (a)j in neighborhood of a. Thus, we observe that an interesting appli ation o

urs when D is a losed region. Therefore, another way of stating Theorem 6.8 is that \if f is analyti inside and on a losed urve C , then jf (z )j attains its maximum value only on the boundary C ."  Another dire t appli ation of Theorem 6.8 is the following orollary whi h is helpful in pra ti e for identifying the maximum modulus of fun tions by their boundary values.

273

6.1 Maximum Modulus Prin iple y C3

C(0, 2π ) C4

B(2π, 2π ) C2

O

C1 A(2π, 0) x

Figure 6.1: [0; 2 ℄ [ [2; 2 + 2i℄ [ [2 + 2i; 2i℄ [ [i2; 0℄:

6.12. Corollary. Let f be analyti on
R and ontinuous on its

losure
R . If jf (z )j  M for some M > 0 on 
R , then jf (z )j  M on
R . 6.13. Example. Let D = fz = x + iy : 0 < x; y < 2g. We wish to nd maxz2D j os z j. To do this, we rst note that q

j os z j = sinh2 y + os2 x: By the Maximum modulus theorem, the maximum is attained on the boundary (see Figure 6.1)

D = [0; 2℄ [ [2; 2 + 2i℄ [ [2 + 2i; 2i℄ [ [i2; 0℄: For z = x + i0 with 0  x  2, j os z j has the maximum value 1 at zq= 0; 2. For z = 2 + iy with 0  y  2, j os z j has the maximum 1 + sinh2 (2) at z = 2 + 2i, sin e sinh y is an in reasing fun tion of y. q

For z = x + 2i with 0  x  2, j os z j has the maximum 1 + sinh2 (2) at the points z = 0 + 2i;  + 2i. Finally,qfor z = 0 + iy with y 2 [0; 2℄, the orresponding maximum is seen to be 1 + sinh2 (2). Hen e, max j os z j = z2D

q

1 + sinh2 (2) = osh 2:



6.14. Theorem. (Minimum Modulus Theorem) If f is a non onstant analyti fun tion in a bounded domain D and f (z ) 6= 0 on D then, jf (z )j annot attain its minimum in D. Proof. Suppose that f 2 H(D) and f (z ) 6= 0 in D. Then, 1=f (z ) is analyti throughout D. The assertion now follows on applying Theorem 6.8 to 1=f (z ) .

274

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

Theorem 6.14 is often stated in the following form: \Suppose that f is analyti in a domain D, f (z ) 6= 0 on D and a is a point in D su h that jf (z )j  jf (a)j holds for all z 2 D. Then, f is a onstant."

6.15. Example. Suppose that f is an analyti fun tion in a neighborhood of the losed unit disk
su h that there exists an M > 0 with jf (z )j > M for jz j = 1 and f (0) = a + ib with ja + ibj < M . Under these assumptions, we wish to show that f has a zero in
. Suppose on the ontrary that f has no zeros in
. Then, by assumption, 1=f would be a non-vanishing analyti fun tion in the neighborhood of
and therefore, would attain its maximum value on the ir le jz j = 1. The Maximum modulus theorem gives that j1=f (z )j < 1=M on jz j  1: In parti ular, 1=ja + ibj = j1=f (0)j < 1=M; i.e. M < ja + ibj; whi h is a ontradi tion. Thus, f must have a zero in
.



Theorem 3.31 and the Uniqueness theorem (see Theorems 3.75 and 4.103) show that a non- onstant analyti fun tion in a domain annot map an open set into a point or an ar . From these two observations, we note that the Open mapping theorem (see Theorem 12.1) about mapping properties of analyti fun tions is a onsiderable strengthening of these fa ts arising out of the Maximum modulus theorem.

6.16. Corollary. (Maximum/Minimum Modulus Theorem for Harmoni Fun tions) Suppose that u(x; y ) is a real-valued non- onstant harmoni fun tion on a bounded domain D. Then, u(x; y ) annot attain its maximum or minimum value in D. That is, if there exists a point (x0 ; y0 ) 2 D su h that u(x0 ; y0) = supz2D u(x; y) or u(x0 ; y0 ) = inf z2D u(x; y); then u(x; y) is onstant on D. Proof. We shall prove only the maximum ase as the proof for the minimum ase follows by simply applying the maximum ase for u. First, we assume that D is simply onne ted. Then (see Theorem 3.39) there exists an f 2 H(D) with u = Re f . The desired on lusion follows if we apply the Maximum modulus theorem to g(z ) = ef (z) (see Corollary 6.10). Now we assume that D is a multiply onne ted domain. Suppose on the ontrary that u(x; y) attains its maximum value at some point z0 = (x0 ; y0 ) 2 D. Sin e D is open, there exists a losed disk
(z0 ; Æ)  D. Then, in
(z0 ; Æ), u(x; y) attains its lo al maximum at the interior point (x0 ; y0 ) whi h is a ontradi tion to the previous ase. Consequently, u(x; y)

annot attain its maximum value on D. 6.17. Example. Consider u(x; y) = 2(x2 y2 ) + 3 for jz j  2. Clearly, u is harmoni on the disk jz j  2. To nd max u and min u on

6.2 Hadamard's Three Cir les/Lines Theorems

275

jz j  2, it suÆ es to nd these on the boundary jz j = 2. Setting x = 2 os  and y = 2 sin , we have

u = 8 os 2 + 3 (0    2) and so, maxjzj=2 = 13 and minjzj=2 = 5. Note that the maximum o

urs when  = 0 and  while the minimum o

urs when  = =2 and 3=2. 

6.18. Example. Suppose that f and g are analyti on the losed unit disk jz j  1 su h that (i) jf (z )j  M for all jz j  1 (ii) f (z ) = z ng(z ) for all jz j  1=3 and for some n 2 N . We wish to use the Maximum modulus prin iple to nd the maximum value of jf (z )j on jz j  1=3. To do this, we pro eed as follows. On jz j = 1, we have M  jf (z )j = jz n g(z )j = jg(z )j and so, jg(z )j  M for jz j  1. Now, for jz j = 1=3, we have

jf (z )j = jz n g(z )j = jz nj jg(z )j = 3 njg(z )j  3 n M: It follows that jf (z )j  3 nM for all jz j  1=3.



6.2 Hadamard's Three Cir les/Lines Theorems We noti e that the hypothesis that D is bounded in the Maximum modulus theorem (see Theorem 6.8) annot be dropped. Therefore, the Maximum modulus theorem is not always true on unbounded domains. To illustrate this, we present three dierent examples. (i) De ne f (z ) = e iz on D = fz : Im z > 0g. Then jf ( )j = 1 on the boundary D = f : Im  = 0g; the real axis. But for z = x + iy 2 D,

jf (x + iy)j = ey ! 1 as y ! +1;

that is, f itself is not bounded. Note also f is a periodi fun tion of period 2. In parti ular, for z = 2 + iy 2 D, Re f (2 + iy) = ey ! 1 as y ! +1;

Similarly, if g(z ) = iz on D = fz : Im z > 0g then Re g(z ) = 0 on D, yet g is not onstant on D. Note that for z = x + iy 2 D, Re g(z ) = y ! 1 as y ! +1: 2

(ii) De ne f (z ) = e iz on D = fz : Re z > 0; Im z > 0g or D = fz : Re z < 0; Im z < 0g. Then, jf ( )j = 1 on the boundary D: But for x + iy 2 D, jf (x + iy)j = e2xy so that f itself is not bounded on D.

276

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

z (iii) Finally, we onsider f (z ) = ee for z 2 D = fz : jIm z j < =2g: Then, for a + ib 2 D = f : jIm  j = =2g, eai=2 iea = e jf (a + ib)j = e = 1:

Again thex on lusion of Theorem 6.8 fails sin e for z = x 2 R  D, f (x) = ee ! 1 as x 2 R; x ! +1: The failure of the Maximum prin iple on ertain unbounded domains raises the following

6.19. Problem. If f (z ) is bounded in an unbounded domain , an we then on lude that supz2 jf (z )j = supz2 jf (z )j?

Here we speak of sup jf (z )j rather than max jf (z )j be ause, although both and  are losed, they are not ompa t. The following result provides an aÆrmative answer in the ase of parallel strips.

6.20. Theorem. (Phragmen-Lindelof Theorem) Let = fz : 0 < Re z < 1g. Suppose that f is analyti and bounded on , and ontinuous on its losure . Then, supz2 jf (z )j = supz2 jf (z )j: In parti ular, if jf (z )j  M for z 2  then jf (z )j  M for all z 2 .

Proof. Set M1 = supz2 jf (z )j and M2 = supz2 jf (z )j < 1. Then, M1  M2 . We laim that M1  M2 . We shall show that jf (z )j  M1 for all z 2 . To do this we x a 2 R \ . Then, 0 < a < 1. Let  > 0 be given. De ne an auxiliary fun tion g de ned by f (z ) g (z ) = 1 + z and onsider the re tangular domain R = fz 2 : jIm z j < Ag, where A > 0 will be hosen later. Then, as j1 + z j  Re (1 + z ) = 1 + Re z  1 for z 2 , we have  g is ontinuous on , analyti on and jg(z )j  jf (z )j for z 2 . In parti ular, jg(z )j  jf (z )j  M1 on the verti al sides of R.  for z on the horizontal sides of the losed re tangle R, we have z = r  iA with 0  r  1 and so p

p

j1 + z j  j1 + (r  iA)j = (1 + r)2 + 2 A2  1 + 2 A2 : Therefore, as jf (z )j  M2 on , jg(z )j  p jf (z )j  p M2

1 + 2 A2 1 + 2 A2 whi h holds for all z on the horizontal sides of the re tangle R.

277

6.2 Hadamard's Three Cir les/Lines Theorems

p

Now hoose A large enough that M2 = 1 + 2 A2 < M1 whi h is possible as M2 is nite. Thus, jg(z )j  M1 on R . We apply the Maximum prin iple to g on R to get jg(z )j  M1 on R. Therefore, jf (z )j  M1j1 + z j for z 2 R. Allowing  ! 0, the last inequality leads to jf (a)j  M1 . Sin e a was an arbitrary point of , jf (z )j  M1 holds for all z 2 .

6.21. Theorem. (Hadamard's Three Lines Theorem) Let = Suppose that f is analyti and bounded on , and

ontinuous on its losure . Suppose that there exist two onstants M0 and M1 su h that

fz : 0 < Re z < 1g.

jf (z )j  M0 for z = 0 + iy 2  , and jf (z )j  M1 for z = 1 + iy 2  : Then, jf (z )j  M01 Re z M1Re z for all z 2 . Proof. Without loss of generality we may assume that M0 > 0 and M1 > 0. De ne an auxiliary fun tion F : ! C by F (z ) = ez f (z ), where  is a real number to be xed later. This fun tion satis es the hypotheses of Theorem 6.20, and sup jF (z )j = max

z2

sup

jF (z )j;

sup

jf (z )j;

z=0+iy2

(

= max

Choose  su h

(

sup

z=1+iy2

e

sup

jF (z )j

)

)

jf (z )j

z=0+iy2

z=1+iy2

   max M0; e M1 : that M0 = e M1, i.e.  = ln(M0 =M1). By Theorem

6.20,

jF (z )j  e M1 for z 2

whi h means that eRe z jf (z )j  e M1 on . Substituting  = ln(M0 =M1 ), this gives jf (z )j  M01 Re z M1Re z for all z 2 . 6.22. Corollary. Let = fz : a < Re z < bg. Suppose that f is analyti and bounded on , and ontinuous on its losure . Suppose that there exist two onstants M0 and M1 su h that

jf (z )j  M0 (a)

for z = a + iy 2 

and jf (z )j  M1 (b) for z = b + iy 2  : Then, for all z 2 ,

jf (z )j  [M0 (a)℄1  [M1 (b)℄ ;  = Reb z a a : Proof. De ne w = (z ) = a + (b a)z . Then,  takes the verti al strip

fz : 0 < Re z < 1g into fz : a < Re z < bg. Now, apply Theorem 6.21 to F = f Æ .

278

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

A striking onsequen e of the Maximum modulus prin iple is the following result whi h an be easily proved using the last orollary and the fa t that ew maps the strip fw : ln a < Re w < ln bg onto the open annulus fz : a < jz j < bg and the boundary maps into the orresponding boundary. However, be ause of its independent interest, we in lude a dire t proof without referring to the orollary.

6.23. Theorem. (Hadamard's Three Cir les Theorem) Let =

fz : a  jz j  bg, f 2 H( ) and M (r) = maxjzj=r jf (z )j: If a < r < b, then ln(r=a) M (r)  [M (a)℄1  [M (b)℄ ;  = : ln(b=a) Here  depends on a; b and r, but is independent of f (z ), and is between 0 and 1.

Proof. We examine the behavior of g(z ) = z  f (z ), where  is a real number to be xed later. Unless  is an integer, g(z ) is learly a multiplevalued fun tion in the annulus (as z  is multiple-valued). So we annot

laim that g 2 H( ) and, as a onsequen e, we annot apply the Maximum modulus prin iple dire tly. But if D = n[ b; a℄, then g(z ) be omes analyti on D as the prin ipal power fun tion given by z  = e Log z be omes analyti on D. Consequently, max jg(z )j o

urs on this new boundary. Note that, as  2 R, there is no maximum for jz j at any interior point on the interval ( b; a) on the real axis. It follows that

jz j jf (z )j  maxfa M (a); b M (b)g and so,

r M (r)  maxfa M (a); b M (b)g: Ignoring the trivial ase f (z )  0, we see that M (a), M (r) and M (b) are all positive, and we may hoose  uniquely (whi h is at our disposal) su h that   ln(M (b)=M (a)) a M (a) = b M (b); i.e.  = : ln(b=a) We then have  a  M (r)  M (a) r = M (a) expf  ln(r=a)g   ln(r=a) = M (a) exp ln(M (b)=M (a)) ln(b=a) 

 ln(r=a)

M (b) ln(b=a) = M (a) M (a) 1 = [M (a)℄  [M (b)℄ :

279

6.3 S hwarz' Lemma and its Consequen es ln M(r) ln M(a) ln M(r) ln M(b)

a r

ln a

O

ln r

ln b ln r

b

Figure 6.2: Geometry of Hadamard's three ir les theorem.

This may be rewritten as [M (r)℄ln(b=a)  [M (a)℄ln(b=r) [M (b)℄ln(r=a) ; or equivalently as ln M (r) 









ln b ln r ln r ln a ln M (a) + ln M (b): ln b ln a ln b ln a

The last inequality shows that we may simply express Hadamard's three

ir les theorem as ln M (r) is a onvex fun tion of ln r (Figure 6.2). .

6.3 S hwarz' Lemma and its Consequen es In this se tion we start with a simple but one of the lassi al theorems in

omplex analysis, namely, S hwarz' lemma, whi h states that if f is analyti and satis es jf (z )j < 1 in
and f (0) = 0, then jf (z )j  jz j for ea h z 2
with the sign of equality i f has the form

f (z ) = ei z for some  2 R. Furthermore, jf 0 (0)j  1 with the equality i f has the form (6.24). This is referred to as an in nitesimal version (or simply a

lassi al version) of S hwarz' lemma and is interesting on its own sake. This result has an important role in the proof of the Riemann mapping theorem whi h is an important theorem on erning the onformal equivalen e of two simply onne ted domains (see Se tion 12.4). Let us now begin by proving the sharp version of the lassi al S hwarz lemma whi h plays a signi ant role in geometri fun tion theory. In fa t, there are now many extensions of this result. The following form is an important appli ation of the Maximum prin iple. (6.24)

6.25. Theorem. (S hwarz' Lemma) Let f :
having a zero of order n at the origin. Then

!
be analyti

280

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

(i) jf (z )j  jz jn for all z 2
, (ii) jf (n)(0)j  n! and the equality holds either in (i) for some point 0 = 6 z0 o

urs i f (z ) = z n with jj = 1.

2
or in (ii)

Proof. Let f :
!
be analyti on
and has n-th order zero at the origin. Then, we have f (0) = 0 = f 0 (0) =


= f (n 1)(0) and f (n) (0) 6= 0. So, we an write 1 X f (z ) = ak z k = z ng(z ) for z 2
; k =n

P k n . The fun tion g (z ) = where ak = f (k) (0)=k! and g(z ) = 1 k=n ak z n f (z )=z has a removable singularity at the origin so that if

g(z ) =



z n f (z ) for z 2
n f0g an for z = 0;

then g is analyti in
nf0g and ontinuous on
. Referring to the Riemann removability theorem (see Theorem 4.88), we on lude that g is analyti in
. (i) We laim that jg(z )j  1 for all z 2
. Now, for 0 < r < 1, (a) g is analyti on the bounded domain
r = fz : jz j < rg (b) g is ontinuous on its losure
r = fz : jz j  rg. Therefore, the Maximum modulus prin iple is appli able. As jf (z )j  1 for every z 2
, it follows that for j j = r

jg( )j = jfj(jn)j  r1n : By the Maximum modulus prin iple, jg(z )j  r n for all z with jz j Sin e r is arbitrary, by letting r ! 1, we nd that jg(z )j  1; that is (6.26)

 r.

jg(z )j  1 for all z 2


and this is same as (i). Equality in (i) holds for some point z0 2
n f0g implies that jg(z0)j = 1. It follows that g a hieves its maximum modulus at an interior point z0 . Consequently, by the Maximum modulus theorem, g must redu e to a onstant, say . Then f (z ) = z n, where jj = 1. (ii) Note that jg(z )j  1 throughout the disk
. Sin e jan j = jg(0)j; (6.26) implies that jg(0)j  1 and so (ii) follows. Again, if jf (n) (0)j = n! then jg(0)j = 1 showing that g a hieves its maximum modulus 1 at the interior point `0'. Consequently, g is a onstant

281

6.3 S hwarz' Lemma and its Consequen es

fun tion of absolute value 1 and as before, this means that f (z ) = z n, where jj = 1.

6.27. Remark. Note that the ase n = 1 of Theorem 6.25 is the original form of S hwarz' lemma stated at the outset of Se tion 6.3.  For instan e, if f 2 H(
) with jf (z )j  1 and f (0) = 0 then what kind of fun tion is f when f (1=3) = 1=3? It must be none other than the identity fun tion be ause the equality in Theorem 6.25(i) holds with n = 1 and z = 1=3 2
. If f is known to satisfy the onditions of Theorem 6.25 in
R instead of the unit disk
, the original form of the theorem an be applied to the fun tion f (Rz ) (see also Exer ise 6.82). More generally, Theorem 6.25 immediately yields the following result.

6.28. Corollary. If f is analyti and satis es jf (z )j  M in
(a; R)

and f (a) = 0, then

(i) jf (z )j  M jz aj=R for every z 2
(a; R), (ii) jf 0(a)j  M=R with the sign of equality i f has the form f (z ) = M(z

onstant  with jj = 1.

a)=R for some

Proof. Use S hwarz' lemma with g(z ) = f (Rz + a)=M; jz j < 1: Does S hwarz' lemma hold for the ase of real-valued fun tions of a real variable? Consider 2x u(x) = 2 : x +1 Then u is in nitely dierentiable on R. In parti ular, u0 (x) is ontinuous on [ 1; 1℄, u(0) = 0 and ju(x)j  1. But ju(x)j > jxj for 0 < jxj < 1.

6.29. Example. Let ! = e2i=n be an n-th root of unity, where n 2 N is xed. Suppose that f :
!
is analyti su h that f (0) = 0. We wish to apply S hwarz' lemma to show that

jf (z ) + f (!z ) + f (!2 z ) +


+ f (!n 1z )j  njz jn and equality for some point 0 = 6 z0 2
o

urs i f (z ) = z n with jj = 1. To do this, we de ne F :
!
by (6.30)

F (z ) =

1 nX1 k f (! z ): n k=0

282

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

Clearly, F is analyti on
, F (0) = 0 and, for 1  m  n 1,

F (m) (z ) =

1 nX1 k m (m) k (! ) f (! z ) n k=0

so that (as !n = 1)

F (m)(0) =





1 nX1 m k (m) f (m) (0) 1 (!m )n (! ) f (0) = = 0: n k=0 n 1 !m

By S hwarz' lemma (see Theorem 6.25), it follows that jF (z )j  jz jn for all z 2
whi h is the same as (6.30). The equality in this inequality for some point z0 6= 0 o

urs i F (z ) = z n with jj = 1, or equivalently nX1

(6.31)

k=0

[f (!k z ) z n ℄ = 0:

We

laim that the above equation implies that f (z ) = z n. If we let f (z ) = P1 a z m, then (6.31) be omes m=1 m

1 X m=1

am

nX1 k=0

!km

!

z m = nz n:

In view of the identity nX1



if m is a multiple of n ; !km = n0 otherwise k=0

the last equation implies that an =  and a2n = a3n =


= 0. On the other hand, as jf (z )j < 1 on
and jan j = 1, we have Z 1 X 1 2 lim j f (rei )j2 d  = jam j2  1 r!1 2 0 m=1 whi h shows that all the Taylor's oeÆ ients of f (ex ept an ) must vanish and so, f (z ) = ei z n .  Our rst appli ation of S hwarz' lemma is the following theorem whi h gives an interesting relationship that exists between the maximum modulus of an analyti fun tion and the maximum of its real part. This result will be used to generalize Theorem 6.60 (see Exer ise 6.89).

6.32. Theorem. (Borel Caratheodary) Let f be analyti in jz j  R.

For ea h 0 < r < R, let

M (r) = max fjf (z )jg and A(r) = max fRe f (z )g: jz j = r jz j = r

283

6.3 S hwarz' Lemma and its Consequen es Then

R+r jf (0)j + R2r r A(R): R r Proof. The result learly holds if f is onstant. Indeed, if f (z ) =  (a omplex onstant), then M (r) = jj; A(r) = Re ; jf (0)j = jj: Substituting these values in the inequality (6.33), then it be omes jj  Re  whi h is trivially true. For a non- onstant f , we rst assume f (0) = 0. Then, A(R) > A(0) = Re f (0) = 0: De ne M (r) 

(6.33)

F (z ) =

f (z ) : 2A(R) f (z )

Then, F is analyti in jz j  R and F (0) = 0. Also, Re [2A(R) f (z )℄ 6= 0; otherwise f would be a real-valued fun tion of omplex variable ontradi ting the analyti ity of f . Further, we also note that 2A(R) + Re f (z )  Re f (z )  2A(R) Re f (z ): This means that

2 jF (z )j2 = [2A(R) Rejff ((zz)℄)j2 + [Im f (z )℄2  1; and so we have jF (z )j < 1 in jz j < R. Thus, by S hwarz' lemma (take a = 0 in Corollary 6.28), we have jF (z )j  jz j=R for jz j = r < R: From the

de nition of F (z ), we have

f (z ) =

2A(R)F (z ) 2A(R)jF (z )j and jf (z )j  1 + F (z ) 1 jF (z )j

 R2r r A(R)

whi h gives (6.33) for the ase f (0) = 0. If f (0) = 6 0, then applying the result to f (z ) f (0), we nd that

jf (z )j jf (0)j  jf (z ) f (0)j  R2r r max fRe (f (z ) f (0))g jz j = R  R2r r [A(R) + jf (0)j℄; that is,

jf (z )j 





2r R+r A(R) + jf (0)j : R r R r

Now we state and prove the invariant form of S hwarz' lemma.

6.34. Lemma. (S hwarz-Pi k Lemma) Suppose that f is analyti on the unit disk
and satis es the following two onditions

284

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem φa

a

φb

f b

O

O

φb ◦ f ◦ φa

Figure 6.3: Illustration for S hwarz-Pi k Lemma.

(i) jf (z )j  1 for all z 2
(ii) f (a) = b for some a; b 2
. Then

(6.35)

2 jf 0 (a)j  1 1 jfj(aaj2)j :

Moreover, for a pair of elements a; a0 in
, the following inequality holds:

(f (a); f (a0 ))  (a; a0 ) where (z; a) = j(z a)=(1 az )j for z; a 2
. Equality is obtained in (6:35) at a point a 2
, or equality is obtained in (6:36) for a pair of points a; a0 with a 6= a0 in
, i f 2 Aut (
); otherwise, there is stri t inequality in jz j < 1 (6.36)

Proof. First, we observe that if jf (z )j = 1 for some point z 2
, then f is a onstant and (6.35) as well as (6.36) are trivial. Thus, we may assume that jf (z )j < 1 for all z 2
. For a xed  2
, we re all ertain fa ts that are already familiar to us (see Figure 6.3):   de ned by  (z ) = ( z )=(1 z ) is analyti on C n f1=g if  6= 0, and 0 (z ) = z . In parti ular, for ea h  2
,  is analyti in a neighborhood of
with  () = 0.  one he ks that j1 z j2 j z j2 = (1 jj2 )(1 jz j2 ) 1 j (z )j2 = j1 z j2 j1 z j2 so that (i) j (z )j = 1 if jz j = 1. (ii) j (z )j < 1 if jz j < 1. Indeed, the fa t that  (
) = 
immediately implies that j (z )j < 1 on
, by the Maximum modulus theorem. (iii)  is one-to-one and onto, sin e  ( (z )) =



  (z ) = 1  (z ) 1





 z 1 z    1 zz

=z

285

6.3 S hwarz' Lemma and its Consequen es

so that  is invertible and  itself is the inverse for  . Note that, sin e  1 =  2 H(
),  2 Aut (
). Now, we de ne g = b Æ f Æ a and apply S hwarz' lemma for g. First, we observe that g satis es the hypothesis of S hwarz' lemma. Thus, by S hwarz' lemma, we on lude that jg0 (0)j  1 and jg(z )j  jz j on
: Now, we ompute (6.37) g0 (0) = 0b (b)f 0 (a)0a (0): Sin e





1 jj2 ;  (1 z )2 it follows that 0 (0) = (1 jj2 ) and 0 () = 1=(1 in (6.37), we nd that

0 (z ) =



1 g0 (0) = 1

jj2 ): Using these

jaj2  f 0 (a) jbj2

and therefore, sin e jg0 (0)j  1 and b = f (a), the on lusion (6.35) follows. For the proof of (6.36), we use the se ond ondition jg(z )j  jz j whi h is equivalent to the inequality

j(b Æ f Æ a )(z )j  jz j; z 2
: Sin e  ( (z )) = z for ea h z 2
, setting z for a (z ), it follows that j(b Æ f )(z )j  ja (z )j; z 2
: In parti ular, if we take z = a0 then the last inequality is equivalent to the inequality jb (b0 )j  ja (a0 )j whi h, by the de nition of  , is the same as the inequality (6.36). By S hwarz' lemma, equality in (6.35) or in (6.36) holds if and only if g(z ) = z =: I (z ), jj = 1; that is, f (z ) = b 1 Æ I Æ a 1 2 Aut (
).

6.38. Remark. Clearly, (6.35) may be obtained dire tly from (6.36). Indeed, rewriting the inequality (6.36) in the form f (a) a





f (a0 ) 1 f (a)f (a0 )  a0 1 aa0

and letting a0 ! a, we obtain (6.35). Thus, every analyti fun tion w = f (z ) from
into
satis es the ondition 2 jf 0 (z )j  1 1 jfj(zzj2)j ; z 2
;

286

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

or equivalently,

jdwj  jdz j (w = f (z ); z 2
) jwj2 1 jz j2 and the equality holds i f 2 Aut (
). 1

Also, we observe that the inequality (6.36) for analyti fun tions f :
!
reveals the ontra tion property of f with respe t to the metri  (in fa t, it is a non-expansive map with respe t to the metri ). Further, f is an isometry, i.e. (f (a); f (a0 )) = (a; a0 ), i f 2 Aut (
). 

6.39. Example. Suppose that f :
!
is analyti su h that f (a) = 0 and f (a 1) = b for some a 2 (0; 1) and b 2
. Then, by (6.36), jbj  1 a(1a 1) = 1 + a1 a2 : In parti ular, this observation implies the following assertions:

(i) There exists no fun tion f that is analyti in
su h that jf (z )j  1, f (1=3) = 0 and f ( 2=3) = 5=6: (ii) There exists no analyti fun tion f :
!
su h that f (1=4) = 0 and f ( 3=4) = 17=20: 

6.40. Example. Suppose we are given that f :
!
is analyti and f (a) = 0 for some point a 2
. Suppose that we are asked to nd an estimate for jf (b)j for some b 2
. To do this, by (6.36), we see rst that (sin e f (a) = 0), jf (z )j  ja (z )j ; a (z ) = 1a azz for z 2
. In parti ular, jf (b)j  ja (b)j and note that the maximum is a hieved when f (z ) = a (z ): For instan e, if f (1=2) = 0 then the estimate for f (1=4) is given by (1=2) (1=4) jf (1=4)j  1 (1=2) (1=4) = 27 and the maximum is attained for 1=2 (z ):  6.41. Example. Suppose that f :
!
is an analyti fun tion su h that f (0) = a for some a 2
. We wish to verify whether there exists su h an f with the property that f 0 (0) = for a omplex onstant . If so, under what ondition on , su h a fun tion exists? To do this, a

ording to (6.35), we rst observe that jf 0 (0)j  1 jaj2 whi h means that must satisfy the inequality j j  1 jaj2 : For instan e, onsider the a (z ) de ned above. Then, a (0) = a, 0a (0) = (1 jaj2) and so, the fun tion f de ned by f (z ) = ka (z ) with jkj  1 does the job. 

6.3 S hwarz' Lemma and its Consequen es

287

6.42. Example. Suppose that f :
!
is an analyti fun tion, jf (z )j  1 on jz j = 1, f (a) = 0 and f ( a) = b for some a 2 (0; 1) and b 2 (0; 1℄. Then, by (6.36), it follows that jbj  2a=(1+ a2): This observation is useful to on lude the following:

(i) There exists no analyti fun tion f :
!
su h that

jf (z )j  1 on jz j = 1; f (1=2) = 0 and f ( 1=2) = 19=20: Note that the existen e of su h a fun tion is guaranteed, for example, if we repla e the ondition f ( 1=2) = 19=20 by f ( 1=2) = 4=5. (ii) There exists an analyti fun tion f :
!
su h that f (1=3) = 0 and f ( 1=3) = 3=5: Su h an f is given by 1=3 (z ). 

6.43. Example. Let f 2 H(
) with f (a) = 0 and jf (z )j  jeiz j for all jz j = 1. How large an jf ( a)j be? To do this, we rewrite the given

ondition on f as jF (z )j  1 for jz j = 1, where F (z ) = e iz f (z ). De ne g by g = F Æ a ; where a (z ) is de ned as above. Then, g(a) = 0 and jg(z )j  1 for jz j  1 (by the Maximum modulus prin iple) so that, by S hwarz' lemma,

jg(z )j  jz j; i.e. je

ia (z) f ( (z ))j  jz j; a

or equivalently,   iw a w w e for all jwj  1 ; i.e. j f ( w ) j  je iw f (w)j  1a aw 1 aw

and the equality holds if f (w) = ei eiw a (w) for some .



6.44. Example. A nite Blas hke produ t is de ned to be a rational Q fun tion of the form B (z ) = ei nk=1 1akakzz ; where a1 ; a2 ; : : : ; an are in
and  2 [0; 2℄. If f is analyti in the unit disk
, ontinuous on
and jf (z )j = 1 for jz j = 1, then it an be easily shown that f is a nite Blas hke produ t. If, in addition, f is entire then f (z ) = ei z n for some  2 R and n 2 N. In fa t, to ex lude the trivial ase, we assume that f is non- onstant. Sin e f is analyti in the unit disk jz j < 1 and f is ontinuous in jz j  1 and jf (z )j = 1, it an have only nitely many zeros; otherwise, the limit point of the zeros must lie on jz j = 1 (sin e the zeros are isolated), whi h is a ontradi tion. Denote these zeros by a1 ; a2 ; : : : ; an (multiple zeros being repeated). De ne F (z ) = f (z ) /

n Y

!

a z ak (z ) ; ak (z ) = k : 1 ak z k=1

288

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

Then, F is analyti for jz j < 1, ontinuous on jz j  1, jF (z )j = 1 for jz j = 1 and F (z ) 6= 0 in
. The same on lusion holds for 1=F (z ) also. Applying the Maximum modulus prin iple, both to F (z ) and 1=F (z ), we get jF (z )j  1 and j1=F (z )j  1 for jz j  1. Thus, jF (z )j = 1 on jz j  1 and so,

f (z ) = ei

n Y

k=1

ak (z ):

If f is entire, then ak = 0 for ea h k, and the se ond part follows.



We have seen a number of variants of S hwarz' lemma whi h are useful. In identally, this result an be used to hara terize all onformal self mappings (i.e. one-to-one, onto and analyti ) of the unit disk
and whi h map the unit ir le 
onto 
(see also Theorem 5.59).

6.45. Theorem. Every univalent analyti fun tion f from
onto itself that has an analyti inverse must be of the form z z0 ; 1 z0z where z0 is a omplex number, jz0 j < 1, and 0    2 . Proof. Let w = f (z ) be su h an analyti univalent fun tion, and let f (a) = b. Then, a

ording to (6.36), we have z a : (w; b)  (z; a); (z; a) = (6.46) 1 az Applying the same argument to the inverse f 1 , for whi h f 1(b) = a, we obtain (6.47) (f 1 (w); f 1 (b))  (w; b); i.e. (z; a)  (w; b) for all z 2
. Equations (6.46) and (6.47) imply that for ea h z 2
f (z ) = ei

f (z ) f (a) = 1 f (a)f (z )

z 1



a ; az

that is

[(z a)=(1 az )℄ + f (a)e i for some real  1 + f (a)e i [(z a)=1 az )℄ whi h has the desired form.

f (z ) = ei

An equivalent formulation of Theorem 6.45 is the following:   z + Aut
= : ; 2 C ; jj2 j j2 = 1 z +    z z0 = ei : z0 2
; 0    2 : 1 z0 z

289

6.3 S hwarz' Lemma and its Consequen es

From Theorem 6.45, it is lear that Aut (
R ) =



R2 ei



z z0 : z 2
R ; 0    2 : z 0 z R2 0

For a xed , f (z ) = ei z is a onformal self-map of
that xes the origin. We now show that these are the only onformal self-maps that x the origin.

6.48. Corollary. Every automorphism f :
!
with f (0) = 0 is

given by f (z ) = ei z .

Proof. The orollary follows if we hoose a = 0 and b = 0 in Theorem 6.45. Alternatively, S hwarz' lemma applied to f and f 1 , be ause ea h of f and f 1 maps
onto itself su h that f (0) = 0 and f 1 (0) = 0, yields

jf (z )j  jz j and jz j = jf 1(f (z ))j  jf (z )j: Hen e, jf (z )j = jz j; that is, f (z ) = ei z . 6.49. Theorem. Let p be analyti in
with p(0) = 1 and Re p(z ) > 0 in
. Then, jp0 (0)j  2 and 1 jz j 1 + jz j

 jp(z )j  11 + jjzz jj ; z 2
:

Equality holds in ea h of these inequalities for p(z ) = (1 + z )=(1

z ).

Proof. De ne (w) = (w 1)=(w + 1): Then,  maps fw : Re w > 0g

onformally onto
and so f =  Æ p maps
onformally onto itself with f (0) = 0: From S hwarz' lemma it follows then that (i) j(p(z ))j = jf (z )j  jz j for z 2
(ii) j0 (p(0))p0 (0)j = jf 0 (0)j  1. The assertion (i) implies that jp(z )j 1 ; 1 jp(z )j   p(z ) 1  jz j; i.e. 1 jz j  jp(z )j  1 + jz j : p(z ) + 1 jp(z )j + 1 1 + jp(z )j 1 + jz j 1 jz j As j0 (p(0))j = j0 (1)j = 1=2, (ii) gives that jp0 (0)j  2.

6.50. Remark. If, in Theorem 6.49, p(0) =  + i ,  > 0, we may apply the above result to the fun tion (p(z ) i )=.  In R, for a dierentiable fun tion f of one variable to have maximum or minimum at x0 it is ne essary that f 0(x0 ) = 0. However, the analog of this does not hold when we deal with fun tions of a omplex variable. This

290

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

surprising fa t in the behavior of an analyti fun tion at a point where it assumes its maximum modulus will be re e ted in the next theorem.

6.51. Theorem. Suppose that f is analyti in a neighborhood of
and z0 2 
satis es jf (z0 )j = max jf (z )j. Then, f 0 (z0 ) 6= 0 unless f is z2


onstant. Proof. Suppose that f is a non- onstant analyti fun tion. Without loss of generality, we suppose 1 = z0 = jf (z0)j, multiplying f by a onstant, if ne essary. Then, by the Maximum modulus prin iple, we have f (
) 
. Assume rst that f (0) = 0. Then, for all t 2 (0; 1), applying the triangle inequality and S hwarz' lemma, we nd that j1 f (t)j  1 jf (t)j  1 t and so f (1) f (t)  1 for 0  t < 1: 1 t Thus, we obtain jf 0 (1)j  1 by making t ! 1 . Suppose f (0) = a 6= 0 and

onsider the fun tion g de ned by g(z ) = (a Æ f )(z ), where a (z ) =

a z ; 1 az

jaj < 1:

Then, g :
!
and g(0) = 0. Applying the pre eding argument to g, we get jg0 (1)j  1. A dire t al ulation shows that

g0 (z ) = so that

f 0(1) =

(1 jaj2 )f 0 (z ) [1 af (z )℄2

g0 (1)[1 af (1)℄2 g0 (1)(1 a)2 = : 2 1 jaj 1 jaj2

Sin e jg0 (1)j  1, this gives that

2 jf 0 (1)j  (11 jjaajj)2 = 11 + jjaajj > 0:

This ompletes the proof. From the proof of Theorem 6.51, be ause of its independent interest, we noti e the following:

6.52. Corollary. If f is analyti on
, f (0) = 0, jf (z )j < 1 in
and if f is analyti at z = 1 with f (1) = 1, then jf 0 (1)j  1.

6.4 Liouville's Theorem

291

6.4 Liouville's Theorem We wish to address, in detail, the following questions:

 Whi h entire fun tions are bounded?  Whi h entire fun tions omit two distin t omplex values? Re all that a fun tion f : ! C is said to omit a value a if a 2 C n f ( ). For 

example, the entire fun tions ez and 1+ ez omit 0 and 1, respe tively. Whi h meromorphi fun tions omit three distin t omplex values? For example, the meromorphi fun tion 1=(1 e2iz ) in C omits two dierent values, namely 0 and 1.

These questions have elegant and omplete answers in the form of Liouville's theorem and Pi ard's little theorem. In Pi ard's great theorem, we shall a tually answer a more general question. Let us dis uss the rst question and the remaining will be done in Se tion 12.7. Re all that a fun tion is entire i it is analyti on the whole omplex plane C . The simplest examples of entire fun tions are polynomials. Entire fun tions whi h are not polynomials (e.g. ez , sin z , os z , osh z , sinh z exp(sin z ) et .) are alled entire trans endental fun tions. Suppose now that f is entire and a 2 C is arbitrary. Then, f admits a Taylor expansion around a: 1 X f (n) (a) f (z ) = an (z a)n ; an = : n! n=0

If an = 0 for all n  1, then f (z ) = f (a) = a0 ; that is, f is a onstant fun tion. Note that a non-zero onstant fun tion is the only (rational) entire fun tion having no zeros. If ak 6= 0 for some integer k  1 but an = 0 for all n > k, then f is a polynomial of degree k  1. Finally if ak 6= 0 for an in nite number of values of k, then f be omes an entire trans endental fun tion. Re all that entire fun tions are just the Taylor's series about any point a with in nite radius of onvergen e. For instan e, the power series su h as 1 zn 1 1 zn 1 zn X X X X n2 z n ; (  > 0) ; e ;  n n ; n=0 (n!) n=0 n=0 2 n! n=0 2 n!

whi h are onvergent for all z 2 C . In on lusion, a non- onstant entire fun tion is either a polynomial of degree n so that it has a pole of order n at in nity; or else an entire trans endental fun tion in whi h ase it will have an essential singularity at in nity (see Chapter 7). With this observation, elementary examples of the last type are the exponential fun tion ez , the trigonometri fun tions sin z , os z , and the hyperboli fun tions sinh z ,

osh z (whi h are, of ourse, well-known simple ombinations of exponential fun tions). From our earlier results, it an be easily observed that the

292

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

sum, dieren e and produ t of a nite number of entire fun tions are also entire; and the quotient of two entire fun tions is also entire provided the denominator is nowhere zero. Now, our aim is to examine the behavior of entire fun tions for suÆ iently large jz j. Let us start with the following simple problem.

6.53. Problem. Are there non- onstant fun tions of a real variable x whi h are both bounded and dierentiable on R? We shall rst see that familiar fun tions on R su h as ex, sin x, os x and log(1 + x2 ) exhibit surprising behavior when we allow x to be omplex rather than just real. To do this, we shall look at three dierent types of simple examples before stating the theorem. Let us look at the fun tion f : R ! R given by f (x) = ex. Then  f is non- onstant and dierentiable on R  f is unbounded on R and f (n) exists on R for ea h n 2 N  f is one-to-one on R. In any subje t, just with one example, it may not be possible to on lude that there exists no fun tion with the desired property as in Problem 6.53. Note that f (z ) = ez is not one-to-one on C . Now, we look at the se ond example. Consider g : R ! R de ned by g(x) = sin x or os x for x 2 R. Clearly,  g is non- onstant and dierentiable on R  g is bounded by 1 and g(n) exists on R for ea h n 2 N  g is not one-to-one on R (be ause it is a periodi fun tion). As the third example, let us onsider f :

R ! R given by

fa (x) = (x2 + a2 ) 1 ; where a is a xed non-zero real number. Then, 0 < fa (x)  1=a2 for ea h x 2 R. However, we see that neither of these fun tions, namely, fa (x), os x and sin x, is bounded if \ omplex values are permitted for the variable x." Indeed, for 1 fa (z ) = 2 2 ; z +a the absolute value of fa (z ) approa hes 1 whenever z approa hes ia, a 2 R n f0g. Note that fa(z ) is not dierentiable on the whole of C . On the other hand, sin x and os x are non- onstant bounded fun tions of real variable and ea h of them is dierentiable on R. Observe that os z and sin z are the well-known non- onstant entire fun tions de ned by

os z =

eiz + e iz eiz e iz ; and sin z = : 2 2i

293

6.4 Liouville's Theorem

It follows that for z = iy, y 2 R,

os(iy) =

e y + ey je y ey j and j sin(iy)j = 2 2

whi h implies that both j os z j and j sin z j in rease inde nitely when z approa hes in nity along the imaginary axis. Thus, the entire fun tions

os z and sin z are unbounded on C ; i.e. no K an exist su h that j os z j  K on C . Similarly, no K an exist su h that j sin z j  K on C . In parti ular, we say that the range of ea h of the fun tions exp z , os z and sin z is an unbounded set in C . As a onsequen e, it follows that both sin z and os z are bounded for jz j < 2004 whereas both sin z and os z are unbounded for jz j > 2004. More generally, for suÆ iently large R, both sin z and os z maps jz j > R into any pres ribed neighborhood of in nity.

6.54. Example. Let , , and Æ be some xed real numbers. Then, we have the following:

 if = fz 2 C : Im z > g, then the entire fun tion f (z ) = e  

iz

is unbounded on the half-plane (note that whi h approa hes +1 as y ! +1); if = fz 2 C : Re z > g, then the entire fun tion g(z ) = ez is unbounded on the half-plane ; 2 the entire fun tion h(z ) = e z is unbounded on any open set whi h

ontains either part of the imaginary axes fiy : y > g or fiy : y < 2 Æg. Note that h(iy) = ey . 

f (iy) = ey

Finally, for n 2 N , we onsider fn(z ) = z=(1 + njz j) or sin(jz jn). Then, for ea h n, fn (z ) is ontinuous on C and jfn (z )j < 1 on C . This observation shows that there exist non- onstant bounded ontinuous fun tions on C . Thus, it is natural to expe t a simple hara terization for fun tions of omplex variable that are both bounded and analyti in C . This is pre isely given by Liouville's theorem in the following form whi h implies that the range of every non- onstant entire fun tion is an unbounded set in C .

6.55. Theorem. (Liouville's Theorem) A bounded and entire fun -

tion is onstant.

We have already observed through examples that this result is quite dierent from any that ould possibly hold for real-valued fun tions of a real variable. We remark that Liouville's theorem10 an be viewed as a statement about the range of a non- onstant entire fun tion. Therefore, 10 Liouville

(1809-1882), a Fren h mathemati ian, is known for his work on analysis and dierential geometry.

294

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

an equivalent formulation of Liouville's theorem is that \the range of a bounded and entire fun tion is a singleton set".

As an immediate onsequen e of Liouville's theorem, we have the following simple appli ations:

 

Non- onstant entire fun tions must be unbounded. For example, as we have already seen, the entire fun tions sin z and os z are unbounded unlike their real ounterparts. Every analyti fun tion in the extended omplex plane is ne essarily

onstant. In fa t if f is analyti at z = 1, then limjzj!1 f (z ) is

nite. Let this limit be L. This means that, given  > 0 there exists an R > 0 su h that

jf (z )j jLj  jf (z ) Lj <  whenever jz j > R; and so, in parti ular, f is bounded for jz j > R and thus, by the

ontinuity of f on the ompa t set fz : jz j  Rg, f is bounded on the whole of C . Hen e, by Liouville's theorem, f is onstant. Equivalently, we say that the only fun tion whi h is analyti on the



Riemann sphere is the onstant fun tion. If f is entire and if there exists an M > 0 with jf (z )j > M for all z 2 C , then f is onstant. This is be ause the given onditions imply

that f 0 (z ) exists and f (z ) 6= 0 in C so that 1=f (z ) is analyti on C and j1=f (z )j < 1=M for all z 2 C . Now applying Liouville's theorem to 1=f (z ), we on lude that f is onstant.

6.56. Problem. Do we really need the boundedness ondition in the hypotheses of Liouville's theorem? For instan e, an we repla e the `boundedness' ondition in Liouville's theorem by a ondition su h as the following?

 Re f (z ) or Im f (z ) is bounded on C  Re f (z ) or Im f (z ) lies in a half-plane. The answer is `yes' under ea h of the above onditions. For example, if f is entire and Re f (z )  M for some xed M 2 R, then f is bounded. Indeed,

f (z ) is entire =) =) =) =) =) =)

(z ) = ef (z) is entire j(z )j = jef (z) j = eRe f (z)  eM for all z 2 C (z ) is onstant, by Liouville's theorem, 0 (z ) = ef (z) f 0 (z ) = 0 for all z 2 C f 0 (z ) = 0 in C , sin e ef (z) 6= 0 in C f (z ) is onstant:

295

6.4 Liouville's Theorem

Alternatively, it suÆ es to observe that if f is entire and Re f (z )  M then g(z ) = 1=[1 + (M f (z ))℄ is entire and bounded by 1 so that g (and, hen e, f ) is onstant. The other ases may be handled in a similar fashion. However, it is important to remark that all these examples follow as a simple onsequen e of Pi ard's theorem whi h we shall dis uss in Se tion 12.7. Note also that the above dis ussion (together with the fa t that a harmoni fun tion in C possesses a harmoni onjugate in C ) learly shows the following:

6.57. Theorem. A fun tion whi h is harmoni and bounded in

must be onstant.

C

This statement an be viewed as the harmoni analog of Liouville's theorem.

6.58. First proof of Liouville's theorem. Sin e f is bounded on C , there is a nite M su h that jf (z )j  M for z 2 C : Let a be an arbitrary omplex number and let = fz : z = Rei , 0   < 2g, where jaj < R < 1. Then, a

ording to the Cau hy integral formula, we have f (a) f (0) =

1 2i

Z 

1

z a



1 a f (z ) dz = z 2i

Z

f (z ) dz z (z a)

so that for ea h xed a, we have

  f (z ) j aj 2R  M jaj jf (a) f (0)j  2 jmax zj=R z (z a) R jaj whi h approa hes zero as R ! 1. Thus, f (a) = f (0) for ea h a 2 C and

hen e, f is onstant.

6.59. Se ond proof of Liouville's theorem. By hypothesis, there exists a nite M > 0 su h that jf (z )j  M for jz j < R and for any R > 0. Equivalently, jf (Rz )j  M for jz j < 1: In parti ular, jf (0)j  M . Set f (Rz ) f (0) ; jz j < 1: 2M Then g satis es the hypotheses of S hwarz' lemma for ea h R > 0, sin e g(0) = 0 and g (z ) =

jg(z )j  jf (Rz )2jM+ jf (0)j  M2+MM = 1:

Hen e, we have jg(z )j  jz j for jz j < 1. In other words,

jf (Rz ) f (0)j  2M jz j for jz j < 1;

296

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

or equivalently,

jf (z ) f (0)j  2RM jz j for jz j < R:

Again remember that M is a xed onstant, whereas R is at our disposal, and an be hosen as large as we please. Thus, restri ting z in the unit disk jz j < 1 and letting R tend to in nity in the inequality, we nd that f (z ) = f (0) for jz j < 1, whi h, by the Uniqueness theorem, gives f (z ) = f (0) for all z 2 C ) and so f is onstant. Third proof of Liouville's theorem is a tually an immediate onsequen e of Cau hy estimate. So we leave this as a simple exer ise. Our fourth proof is ontained in the following theorem whi h shows that if f is an entire fun tion su h that jf (z )j in reases slower than some power of jz j as jz j ! 1, then the fun tion must be a polynomial.

6.60. Theorem. (A Generalized Version of Liouville's Theorem). An entire fun tion f , whi h satis es the inequality jf (z )j  M jz j for some   0, M > 0 and all suÆ iently large jz j, redu es to a polynomial of degree n where n is the largest integer su h that n  . Proof. Note that the ase  = 0 is Liouville's theorem. Let f be entire. Then, f admits a Taylor expansion around 0: Z 1 X f (z ) 1 dz for any R > 0: f (z ) = an z n ; ak = 2i jzj=R z k+1 n=0

Hen e, taking R  M and using the standard estimate for integrals and the given growth ondition on f , we have

jak j 



1 MR 2 Rk+1




2R = MR k :

Sin e R an be made arbitrarily large, it follows that for k >  we must have ak = 0. We on lude that f an only be a polynomial of degree not greater than . For example if f (z ) is entire su h that f (z )=z n is bounded in C , then f (z ) = z n for some onstant . For a generalization of Theorem 6.60, we refer to Exer ise 6.89.

6.61. Remark. We remark that it is possible that a real dierentiable fun tion f an map C onto the unit disk
= fz : jz j < 1g, as the example z 7!

z p ; i.e. f (x; y) = 1 + jz j2

!

x y p ;p ; 2 2 1+x +y 1 + x2 + y2

6.5 Doubly Periodi Entire Fun tions

297

shows. Note that this has an inverse given by

w 7!

!

w

u v p ; i.e. f 1(u; v) = p ;p : 2 2 2 1 jwj 1 (u + v ) 1 (u2 + v2 )

However, an immediate appli ation of Liouville's theorem shows that this is not the ase with analyti fun tions as \every analyti fun tion f : C !
is onstant." In other words, there an be no bianalyti (i.e. a bije tion whi h is analyti together with its inverse) mapping of the unit disk
onto the whole omplex plane C or of the upper half-plane onto C .  Moreover, as a onsequen e of Theorem 6.60, we on lude that if f is entire and jf (z )j  A + B jz j for all suÆ iently large jz j and for some xed onstants A; B and 0   < 1, then f is onstant. This observation reminds us that we need not assume that jf (z )j is bounded in Liouville's theorem, only that its growth is suÆ iently slow.

6.62. Example. Suppose that f = u + iv is an entire fun tion whi h is blessed with the property that uy vx = 2 for all z 2 C . What an you say about the fun tion f ? Can it be a onstant fun tion? Clearly not! Can this be a polynomial of higher degree? The given ondition shows that this is not the ase (how?). Let us now try to nd the pre ise form of this fun tion. By the C-R equation uy = vx , the given ondition is the same as vx = 1 whi h, by the fa t that f 0 (z ) = ux + ivx, is equivalent to Im f 0 (z ) = 1 for z 2 C : This observation implies that f 0 (z ) is a onstant, say a, so that f has the form f (z ) = az + b 0with Im a = 1. Alternatively, as Im f 0 (z ) = 1, h(z ) de ned by h(z ) = eif (z) is entire and jh(z )j = e 1 shows that h (and hen e f 0 (z )) is onstant, by Liouville's theorem. 

6.5 Doubly Periodi Entire Fun tions

Re all that a fun tion f has a period ! if f (z + !) = f (z ) for all z 2 C . To obtain another appli ation of Liouville's theorem, we rst re all two familiar periodi fun tions of a real variable x: 

sin(x + 2) = sin x ; for all x 2 R :

os(x + 2) = os x

Does the property hold when we allow x to be omplex rather than just real, for ea h of these fun tions? By the Uniqueness theorem, yes it is. Thus, sin z and os z are periodi fun tions with period 2. Is ex periodi if x 2 R? No. Indeed, this fun tion is a stri tly in reasing one-to-one fun tion

298

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

on R. However, in view of the periodi ity property of sin z and os z , we have

os(z + 2) + i sin(z + 2) = os z + i sin z for all z 2 C ; that is ez+2i = ez for all z 2 C showing that the omplex exponential fun tion ez is periodi with period 2i, whi h is a purely imaginary number. So, 2 is a period for the entire fun tions eiz , os z and sin z . It is therefore natural to ask whether there exists a non-trivial doubly periodi entire fun tion f (i.e. f with two independent periods !1 and !2 su h that f (z ) = f (z + !1 ) = f (z + !2 ) for all z 2 C ). Equivalently, we raise the following

6.63. Problem. Does there exist an entire fun tion f having both a real and an imaginary period? More pre isely, are there two non-zero real numbers , su h that 

f (z + ) = f (z ) f (z + i ) = f (z )

for all z 2 C ?

Let us rst show that every entire fun tion f su h that

f (z ) = f (z + 1) = f (z + i) on C is ne essarily a onstant. To do this, we onsider the solid square

S = fx + iy : 0  x; y  1g: Sin e f is ontinuous on S , there exists an M su h that jf (z )j  M on S . Now, let z 2 C be arbitrary. Clearly, we an nd two integers m and n su h that z + m + in 2 S and therefore, jf (z + m + in)j  M for z 2 C . Further, the hypotheses imply that

f (z ) = f (z  1) =




= f (z + m) = f (z| +{zm} +i) =


= f (z| +{zm} +in)

showing that f is bounded in C as the behavior of f (z ) over C is ompletely

hara terized by its behavior on the ompa t set S . Therefore, by Liouville's theorem, f is a onstant. More generally, a simple modi ation of this dis ussion gives the following result whi h answers Problem 6.63.

6.64. Theorem. If f :

C !C

is analyti and

f (z ) = f (z + z1 ) = f (z + z2 ) for all z 2 C ; where z1 and z2 are the two non-zero omplex numbers su h that z1 =z2 62 R, then f is onstant.

6.6 Fundamental Theorem of Algebra

299

Proof. Sin e z1 =z2 is not real, ea h z 2 C an be written in the form z = 1 z1 + 2 z2 where 1 ; 2 2 R. But 1 and 2 may be written as 1 = t1 + n1 and 2 = t2 + n2 , that is

z = (t1 + n1 )z1 + (t2 + n2 )z2 = (t1 z1 + t2 z2 ) + (n1 z1 + n2 z2 ) for some integers n1 ; n2 and some 0  t1 ; t2 < 1. Obviously if z1 and z2 are the periods of f , so is m1 z1 + m2 z2 for any integers m1 and m2 and hen e, we must have f (z ) = f (t1 z1 + t2 z2 ): Thus, the behavior of f is now entirely

hara terized by its behavior on the parallelogram

ft1z1 + t2 z2 : 0  t1 ; t2 < 1g: From the analyti ity of f , it follows that jf j is ontinuous on the losed parallelogram D = ft1 z1 + t2 z2 : 0  t1 ; t2  1g and so, f is bounded for all z 2 D. Consequently, f is bounded on C . We then on lude, by Liouville's theorem, that f is onstant.

6.6 Fundamental Theorem of Algebra Does every analyti fun tion have a zero in C ? Clearly, the answer is

no as the familiar exponential fun tion ez shows. Thus, a trans endental entire fun tion may have no zeros in C . Does every polynomial in x 2 R with real oeÆ ients have a zero in R? Again the answer is no as the equation x2 + 1 = 0 has no zeros in R. Does every polynomial in x 2 R with rational oeÆ ients have a rational zero? Observe that the equation x2 3 = 0 has no rational zeros. How about a non- onstant polynomial with omplex oeÆ ients? The answer to this question is given by the fundamental theorem of algebra. Liouville's theorem an be used to provide a natural and short proof for the fundamental theorem of algebra whi h asserts that every non- onstant polynomial with omplex oeÆ ients has at least one omplex zero and hen e has exa tly n zeros. In fa t, from the observations made below Li-

ouville's theorem, it is straightforward to see that if p is a polynomial of degree n  1, then p(z ) = 0 has a zero in C ; be ause if it did not have, then the fun tion 1=p(z ) would be bounded (how?) and analyti in C and would therefore be onstant, whi h is a ontradi tion to the hypothesis. It remains to prove that if p does not have any zeros, then it is bounded in C . To do this, without loss of generality we an suppose that the polynomial (with omplex oeÆ ients) has the form

p(z ) = a0 + a1 z +




+ an 1 z n 1 + z n (a0 6= 0, n  1):

300

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

Then intuitively, for large z , we an expe t that p(z ) should behave like z n , sin e the largest power dominates the other ones. Indeed, for jz j  1 (so that jz jn  jz jn 1 


 jz j), we have





jp(z )j = z n zan0 + z na1 1 +


+ anz 1 + 1 o n  jz jn 1 zan0 +


+ anz 1 ; by the triangle inequality;     jz jn 1 jjzaj0nj +


+ jajnz j 1 j    jz jn 1 jz1j (ja0 j +


+ jan 1 j) : Hen e, for suÆ iently large jz j (e.g. jz j = R  R0 = maxf1; 2(ja0j +

jan 1 j)g), we have (6.65) Then, for jz j  R,




+

jp(z )j  jz2j : n

1 p(z )

 jz2jn  R2n :

On the ompa t set
R = fz 2 C : jz j  Rg, the fun tion 1=p(z ), being

ontinuous on
R , is bounded in the disk by some M = maxjzj=R j1=p(z )j. Therefore, j1=p(z )j is bounded above for all z 2 C by maxfM; 2R ng. Thus, 1=p(z ) is a bounded entire fun tion and hen e, must be onstant whi h is absurd. Therefore, p(z ) = 0 has a zero. Here is a pre ise formulation of the fundamental theorem of algebra. P

6.66. Theorem. Let p(z ) = nk=0 ak z k be a non- onstant polynomial of degree n  1 with omplex oeÆ ients. Then, p has n zeros in C . That is, there exist n omplex numbers z1 ; z2 ; : : : ; zn , not ne essarily Q distin t, su h that p(z ) = an nk=1 (z zk ). Proof. If a 2 C , then by the `division algorithm' there is a polynomial q of degree n 1 su h that p(z ) = (z a)q(z ) + R; where R is a onstant. Clearly, R=0

() p(a) = 0 () z a is a fa tor of p(z ):

Sin e there exists a z1 su h that p(z1) = 0, z z1 is a fa tor of p(z ) with no remainder term. By the `division algorithm' there is a polynomial pn 1 of degree n 1 su h that p(z ) = (z z1 )pn 1 (z ); be ause

p(z ) p(z1 ) = a1 (z z1 ) +


+ an 1 (z n = (z z1 )pn 1 (z ):

1

z1n 1) + (z n z1n )

301

6.7 Zeros of ertain Polynomials

This shows that p has a linear fa tor z z1. Thus, if n > 1, then by applying the same prin iple we on lude that there is another omplex number, say z2 , su h that pn 1(z2 ) = 0 and so pn 1 has a linear fa tor z z2. Pro eeding in this manner, we an express p uniquely as a produ t of linear fa tors:

p(z ) = an

n Y k=1

(z

zk )

where z1 ; z2 ; : : : ; zn are (not ne essarily distin t) the zeros of p(z ). Observe that there is no analogue of the fundamental theorem of algebra in the ase of real numbers. This an be easily seen by onsidering the quadrati polynomial p(x) = 1 + x2 whi h has no real zeros. The following result is referred to as an abbreviated statement of the fundamental theorem of algebra.

6.67. Corollary. Every polynomial p(z ) of positive degree omits no values, i.e. p(C ) = C . Ea h w 2 C is the image of exa tly n points in C . Proof. If p(z ) is a polynomial of degree n, then q(z ) = p(z ) a is also a polynomial of degree n for ea h xed a 2 C . By Theorem 6.66, p(z ) has n-zeros. In other words, for ea h a 2 C there are n points z1 ; z2 ; : : : ; zn su h that p(zj ) a = 0 for j = 1; 2; : : : ; n. Thus, p(C ) = C .

6.7 Zeros of ertain Polynomials We start with a simple result before we move on to a dis ussion on the lo ation of the zeros of ertain polynomials.

6.68. Theorem. If a polynomial p(z ) with real oeÆ ients has a zero at  su h that Im () 6= 0, then the omplex onjugate  is also a zero of p(z ). Indeed, if  is a zero of order k then  is also a zero with order k. Proof. Set p(z ) = a0 + a1 z + a2z 2 +


+ anz n , where a0 ; a1 ; a2 ; : : : ; an are all real and an 6= 0. Sin e  is a zero of p(z ), we have p() = 0: Taking

onjugation on both sides, we have p() = 0. Hen e,  is also a zero of the polynomial p(z ). To prove the se ond assertion we note that if p(z ) has a zero at  of order k, then p() = p0 () =


= p(k 1) () = 0 and p(k) () 6= 0 whi h would then imply that p() = p0 () =


= p(k

1) () = 0

and p(k) () 6= 0:

Thus, omplex zeros o

ur as onjugate pairs with the same multipli ity.

302

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

The onvex hull of a set D  C is the interse tion of all the onvex sets

ontaining D. The losed onvex hull of D is the interse tion of all the losed

onvex sets ontaining D. In fa t, it an be seen that the onvex hull of points z1 ; z2 ; : : : ; zn 2P C is the set of all linear ombinations z = Pnj=1 j zj with 0  j  1 and nj=1 j = 1: The on ept of onvex hull helps us to dis uss the lo ation of the zeros of ertain polynomials.

6.69. Theorem. (Gauss's Theorem) Suppose that p(z ) is a polynomial of degree n  1. Then, every zero of p0 (z ) lies in the onvex hull of the set of zeros of p(z ). Proof. Let p(z ) be a polynomial of the form p(z ) =

n Y k=1

(z

zk );

where z1 ; z2 ; : : : ; zn are (not ne essarily distin t) the zeros of p(z ). Thus, by logarithmi dierentiation, it follows from the above representation that for z 6= zk n n X p0 (z ) X 1 (z zk ) (6.70) = = p(z ) k=1 z zk k=1 jz zk j2 so that  X   0 n  1 p (z ) = 2 (z zk ): p(z ) k=1 jz zk j If 2 C is su h that p( ) 6= 0 and p0 ( ) = 0, then the above equation be omes Pn zk j zk j 2

= Pk=1 n j z j 2 : k k=1 Pn Hen e is of the form = k=1 k zk , where

j =

j zj j 2 ; j = 1; 2; : : : ; n: 2 k=1 j zk j

Pn

This shows that if z1 ; z2 ; : : : ; zn are the zeros of p(z ), then for every zero of p0 (z ) there are non-negative numbers 1 ; 2 ; : : : ; n su h that

=

n X k=1

k zk ; with

n X k=1

k = 1:

The above onstru tion uses the fa t that p( ) = 6 0. If p( ) = 0 = p0 ( ), then we simply take 1 = 1 and = 1
. As an example to Theorem 6.69, we on lude that if all the zeros of a polynomial p(z ) have negative real parts, then all the zeros of p0 (z ) have negative real parts.

303

6.7 Zeros of ertain Polynomials

6.71. Alternate proof of Gauss's Theorem. Re all that every halfplane H an be de ned by the inequality (see 5.42)   z a Im >0 b for some Q omplex onstants a and b (b 6= 0). Consider the polynomial p(z ) = nk=1 (z zk ). Assume that zk 2 H for ea h k = 1; 2; : : : ; n. Then, 

zk

a



> 0: b If 2 C n H , then p( ) 6= 0. Therefore, Im (( a)=b)  0 and      

zk

a zk a = Im Im 0 for ea h k = 1; 2; : : : ; n. Finally, by (6.70), we see that  0  X   n p ( ) b Im b = Im > 0 for 2= H p( )

zk k=1 Im

and onsequently, bp0( ) 6= 0 or p0 ( ) 6= 0 whenever 2= H: p( ) We omplete the proof of Gauss's theorem. As an immediate appli ation of Gauss's theorem, we have

6.72. Theorem. (Lu as's Theorem) If all the zeros of a polynomial lie in a half-plane, then the zeros of its derivative also lie in the same halfplane. If we apply Lu as theorem with referen e to half-planes determined by ea h side of a onvex polygon, we obtain the following

6.73. Corollary. If all the zeros of a polynomial lie in the smallest

onvex polygon, then the zeros of its derivative also lie in the same polygon. The rst equality in (6.70) immediately yields the following result. Later we shall see that this is a onsequen e of a general result.

6.74. Theorem. If N is the number of zeros ( ounted a

ording to multipli ity) of the polynomial p(z ) in
(a; R), then Z

p0 (z ) dz = 2iN: jz aj=R p(z )

304

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

6.8 Exer ises 6.75. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) There exists an analyti fun tion f on
, f (0) = 1 + i, jf (z )j > 2 for jz j = 1, and having a zero in
. (b) For ea h n 2 N , one has maxjzjr jz n + bj = rn + jbj and the maximum is attained at rei(Arg b+2k)=n , k 2 Z. ( ) Let f (z ) = z m=(z n + 2p), where m; n 2 N are xed, and p is real su h that p > 1=2. Then, maxjzj1 jf (z )j = 1=(2p 1): p (d) If = fz = x + iy : 0  x; y  1g, then maxz2 jz 2 2z j = 5. (e) Suppose that f 2pH(
R ) and satis es the onditions jf (z )j  2 on 
R and f (0) = 3 + i. Then, f is a onstant on
R . (f) Suppose that f is an entire fun tion and has zeros at z = 2i. If M = maxjzj=3 jf (z )j, then jf (z )j  (M=5)jz 2 + 4j for jz j < 3. (g) Let f be an entire fun tion and has n zeros at !k = e2ki=n (k = 0; 1; 2; : : : ; n 1), the n-th roots of unity. If M = maxjzj=3 jf (z )j, then jf (z )j  M (3n 1) 1 jz n 1j for jz j < 3. (h) If f 2 H(
) su h that f has a zero of order n at z = 0, and M = max jf (z )j, then jf (z )j  M jz jn for jz j  1. jzj=1 (i) If w = 'a (z ) is a Mobius map of
onformally onto itself, f 2 H(
) and g(w) = f (z ), then (1 jwj2 ) jg0 (w)j = (1 jz j2 ) jf 0 (z )j : Note: What happens when g(w) = w? (j) There exists an analyti fun tion f of
onto itself su h that f (0) = 1=2, f (1=2) = 1=3, and f (1=3) = 1=4. (k) There exists an analyti fun tion f :
!
su h that f (1=2) = 0 and jf 0 (1=2)j  4=3: (l) There exists an analyti fun tion f :
!
su h that f (0) = 1=2 and f 0 (0) = 3=4. (m) There exists no analyti fun tion f :
!
su h that f (1=2) = 3=4 and f 0 (1=2) = 3=4: (n) If f is entire and jf 0 (z )j  jz j for all z , then f is of the form a + bz 2 with jbj  1=2. (o) If jai j  1, i = 1; 2; : : : ; n 1, then ea h zero of the polynomial p(z ) = z n + an 1 z n 1 +


+ a2 z 2 + a1 z + 1 lies in the annulus D = fz : 1=2 < jz j < 2g. (p) Ea h polynomial of the form p(z ) = a0 + a1 z +


+ an 1 z n 1 + z n satis es the inequality supfjp(z )j : jz j  1g  1:

6.8 Exer ises

305

(q) If f is entire and Re f (z ) is bounded as jz j ! 1, then f is onstant. (r) Suppose f is entire whi h takes real z into real and purely imaginary into purely imaginary. Then, f is an odd fun tion. (s) If f is entire su h that f omits a non-empty disk, then f is onstant. (t) An entire fun tion f whose imaginary part is the square of the real part is onstant in C . (u) If k 6= 1 is a xed onstant and f is entire su h that f (z ) = f (kz ) for all z 2 C , then f is onstant on C . (v) An entire fun tion f su h that f (z ) 6= 0 in C and limz!1 f (z ) 6= 0 is ne essarily a onstant. (w) An entire fun tion f su h that jf 0 (z )j  2jf (z )j must be of the form beaz for some onstants a and b with jaj  2. (x) If f = u + iv is entire and u2  v2 + 2004 on C , then f is a onstant. (y) If u(z ) = u(x; y) is harmoni in C satisfying the ondition u(z )  aj ln jz j j + b for some positive onstants a, b and for all z 2 C , then show that u is a onstant. (z) If u(z ) = u(x; y) is harmoni in C su h that u(z )  jz jn for some n 2 N and for all z 2 C , then u(z ) is a polynomial in x and y.

6.76. De ne f (z ) = 1 z for jz j  1. Show that jf (z )j attains its maximum value when z = 1. If f (z ) is repla ed by g(z ) = (1 z )2 or 1 z 2, does the same on lusion hold? 6.77. If f 2 H(
) is su h that jf (z )j < 1 for z 2
and f xes two distin t points of
, then show that f is the identity fun tion. 6.78. Does there exist an analyti fun tion f :
!
with f (0) = 1=2 and f 0(0) = 3=4? Either nd su h an f or state why there does not exist su h a fun tion f . Answer the same question when f (0) = 1=2 and f 0 (0) = 4=5. 6.79. Suppose that f is analyti and Re f (z ) < 0 in
. Find an estimate for jf 0 (0)j. 6.80. If f 2 H(
), f (0) = 1 and jRe f (z )j < 1 for z 2
, then show that jf 0 (0)j  4=. 6.81. If f 2 H(
), f (0) = 0 and if there exists a onstant  > 0 su h that Re f (z ) <  for all z 2
, then show that jf (z )j  2jz j=(1 jz j) for z 2
and jf 0 (0)j  2.

306

Maximum Prin iple, S hwarz' Lemma, and Liouville's Theorem

6.82. Suppose that f is analyti on
R with jf (z )j  M < 1 for all z 2
R , and f (z0) = w0 for some z0 with jz0 j < R. Show that



f (z ) w0  R jRjz2 zz0 jz j for jz j < R: M 2 w0 f (z ) 0 Interpret the ase z0 = 0 = w0 geometri ally and show that in this ase the equality is a hieved for some 0 2
R n f0g i f is of the form f (z ) = Mei z=R, where  is real. M

6.83. Suppose f is analyti and bounded by M for z 2
and has QN z zj zeros at the points z1; : : : ; zN 2
. Prove that jf (z )j  M j=1 1 zj z for all z 2
. Is this an improvement over the hypothesized inequality jf (z )j  M ? What an you say about f if equality holds for some z 62 fz1; z2 ; : : : ; zN g? 6.84. If f :
!
is analyti su h that f (0) = f (1=3) = f ( 1=3), then show that jf (1=4)j  7=572. Show also that the bound 7=572 annot be made smaller. 6.85. Prove or disprove the following: there is no bianalyti mapping of the right half-plane U = fz : Re z > 0g onto the whole omplex plane. 6.86. Let f be entire su h that jf (z )j you say about f ?

 eRe z for z 2 C . What an

6.87. Find the set of all entire fun tions f su h that jf (z )j  jz j5=2 + for z 2
. Justify your answer with a proof.

jz j9=2

6.88. If p(z ) = a0 + a1z +


+ an 1z n 1 + z n (n  1), then show that there exists a real R > 0 su h that 2 1 jz jn  jp(z )j  2jz jn for jz j  R: 6.89. If f is an entire fun tion whi h for some real numbers  and satis es Re f (z )  jz j for all z with suÆ iently large jz j, then f is a polynomial of degree not greater than  (see also Exer ise 10.35). 6.90. In ea h ase given below, determine whether or not there exists a non- onstant entire fun tion f (z ) satisfying the following onditions. If there is, give an example. If not explain why not. (i) f (0) = ei and jf (z )j = 1=2 for all z 2 
(ii) f (ei ) = 3 and jf (z )j = 1 for all z with jz j = 3 (iii) f (0) = 1, f (i) = 0, and jf (z )j  10 for all z 2 C (iv) f (0) = 4 3i and jf (z )j  5 for all z 2
(v) f (z ) = 0 for all z = n, n 2 Z.

Chapter 7

Classi ation of Singularities

Consider the fun tions 1 1 1 ; x sin ; exp( 1=x3 ); : x2 x x(x2 + 2) Then we see that the point x = 0 is a singular point for ea h of these fun tions in the sense that the fun tion is de ned in a deleted neighborhood of 0. The problem of lassifying singularities is not satisfa tory for fun tions de ned only on R. On the other hand the situation is quite dierent for fun tions de ned on domains in C . In Se tion 7.1, we start our dis ussion on isolated and non-isolated singularities and lassify an isolated singularity as a removable singularity, or a pole, or an essential singularity. In Se tion 7.2, we dis uss removable singularities and present Riemann's removable singularity theorem for hara terizing whether an isolated singularity is removable. Se tion 7.3 is devoted to a dis ussion on poles. In Se tion 7.4, we show that isolated singularities an be lassi ed in a simple way using Laurent's series. In Se tion 7.5, we introdu e the notion of an isolated singularity at 1. Analyti fun tions with only poles as singularities play a prominent role in fun tion theory with a spe ial name, meromorphi fun tions. In Se tion 7.6, we dis uss some aspe ts of meromorphi fun tions through a number of examples and hara terizations. As a onsequen e, we present an analog of Liouville's theorem for meromorphi fun tions. In Se tion 7.7, we dis uss the errati behavior of fun tions near an essential singularity via Casorati-Weierstrass theorem whi h provides a basis for understanding the importan e of Pi ard's little theorem.

7.1 Isolated and Non-isolated Singularities A point z = a is alled a regular point for a omplex-valued fun tion f if f is analyti at a. Point a is alled a singular point or a singularity, of f , if f is not analyti at a but every neighborhood of a ontains at

308

Classi ation of Singularities y

a5 N (a) a3 a a2 an a4 a1

t ch cu bran O DA

x

Figure 7.1: Des ription for a non-isolated singularity at a.

least one point at whi h f is analyti . A singular point a is said to be an isolated singular point or an isolated singularity of f if f is analyti in some deleted neighborhood of a. Otherwise, we say a is a non-isolated singular point of f (see Figures 7.1 and 7.2). Equivalently, we say that a point a is a non-isolated singularity of f i a is a singularity and every deleted neighborhood of a ontains at least one singularity of f . For example, z 1 has an isolated singular point at z = 0 while the fun tion 1= sin(z ) has isolated singularities at every integer point n, n 2 Z. On the other hand, every point on the negative real axis (in luding the point z = 0) is a nonisolated singularity of Log z , thepprin ipal bran h of the logarithm fun tion. How about the singularities of z, the prin ipal square root fun tion? Note that the on ept of singularities of a fun tion f very mu h depends upon the domain of the fun tion f . For instan e, f (z ) = z 2 on C n f0g has an isolated singularity at z = 0 whereas the restri tion g(z ) = z 2 on C n
does not have a singularity at z = 0. However, this ambiguity will be removed during the dis ussion on analyti ontinuation. Entire fun tions have no singular points but an have zeros, for example f (z ) = z n ez , n 2 N . Rational fun tions p(z )=q(z ); where p(z ) and q(z ) are polynomials, have isolated singularities at points where q(z ) = 0. For example, the rational fun tion z+1 on C n f0; 3g z 2 3z has isolated singularities at 0; 3.

7.1. Example. What are the singular points of ea h of the fun tions z; Im z; Re z; z Im z; z Re z; jz j2 ? Observe that ea h of the fun tions listed here is nowhere analyti . This does not mean that every point of C is a singularity. In fa t, as there exists no neighborhood (about any point of C ) whi h ontains a point at whi h

7.1 Isolated and Non-isolated Singularities

309

a1 N (a1 )

a3 N (a3 )

a2 N (a2 )

Figure 7.2: Des ription for isolated Singularities at a1 ; a2 and a3 .

the fun tion is analyti , none of the fun tions listed above has singularities in C . 

7.2. Example. Let us dis uss the singularities of g(z ) = 1=f (z ), where f (z ) = sin(1=z ): So the dis ussion of singularities of g follows immediately from the zeros of sin(1=z ). Thus, the singularities of g are at points zn = 1=n for n 2 Z and at their limit point z = 0. This shows g has isolated singularities at zn = 1=n for n 2 Z and a non-isolated singularity at the limit point z = 0. Similar dis ussion shows that (z ) = 1= os(1=z ) has isolated singularities at the points zn = 2((2n+1)) 1 ; n 2 Z: Note that, every neighborhood N of z = 0 ontains singular points dierent from z = 0. So, z = 0 is a non-isolated singularity of (z ). In both the ases, for ea h zn , there are neighborhoods (say ir les of suÆ iently small radius Æ) whi h ontain no other singularity. This means that these singularities are isolated.  Let us now look at the following fun tions de ned for z 2 C n f0g: sin z 1 ; f2 (z ) = n (n 2 N ); and f3 (z ) = e1=z : z z Clearly, ea h of these fun tions is analyti in C n f0g. Is it possible to de ne ea h of these fun tions at the origin so that the resulting fun tion in ea h

ase be omes ontinuous at the origin? We observe the following:

f1 (z ) =

(1) As zlim !0 f1 (z ) = 1, there exists an entire fun tion F1 (z ) su h that F1 (0) = 1 and F1 (z ) = f1 (z ) for z 2 C n f0g. (2) Sin e jf2 (z )j is large when jz j is small, f2(z ) is unbounded near 0. So, there is no way we an de ne f2 (z ) at 0 so that the resulting fun tion be omes ontinuous at the origin. Also, limz!0 f2(z ) = 1, m lim z n f2 (z ) = 1 and zlim !0 z f2 (z ) = 0 for all m > n:

z !0

310

Classi ation of Singularities

(3) Finally, (a) if z = x > 0, then for x near 0, jf3 (z )j = e1=x is large (b) if z = x > 0, then for x near 0, jf3 (z )j = e 1=x is small ( ) if z = iy 6= 0, then jf3 (z )j = je i=y j = 1 for ea h y 6= 0. Thus, zlim !0 f3 (z ) fails to exist (in luding the possibility of the limit being in nity). In parti ular, there exists no entire fun tion F3 (z ) su h that F3 (z ) = f3 (z ) for z 2 C n f0g. Moreover, there exists no n n 2 N su h that the limit zlim !0 z f3 (z ) exists. Indeed, with x > 0 and a xed n 2 N , we have t n e1=z = lim e = 1 lim z n z=x!0 1=x=t!1 t and t t n lim e = 0 n e1=z = lim e = ( 1) lim z t!1 tn t!1 ( t)n z= x!0 t>0 t>0 so that neither f3 (z ) nor z n f3 (z ) is bounded near 0 for any n 2 N . The examples illustrated above motivates one to lassify isolated singularities of f into three ategories. Suppose that f has an isolated singularity at a point a. Then exa tly one of the following holds: (a) f (z ) is bounded near a (b) f (z ) is unbounded near a but (z a)n f (z ) is bounded near a for some n2N ( ) there exists no n 2 N su h that (z a)n f (z ) is bounded near a, i.e. neither (a) nor (b) holds. Thus, the above three situations are lassi ed respe tively as follows: (1) Removable singularity, whi h upon loser examination reveals that this is not a tually onsidered to be a singular point at all. More pre isely, an isolated singularity z = a of f is said to be a removable singularity for f if limz!a f (z ) exists in C . (2) Pole arises from the re ipro al of an analyti fun tion with zero. More pre isely, an isolated singularity z = a of f is said to be a pole for f if limz!a f (z ) = 1. (Note that f (z ) is de ned if z is near enough to a and z 6= a.) (3) Essential singularity, whi h is neither a removable singularity nor a pole. More pre isely, an isolated singularity z = a of f is said to be an essential singularity for f if limz!a f (z ) does not exists in C 1 . The behavior of a fun tion in a neighborhood of an essential singularity is des ribed by the Casorati-Weierstrass theorem (Theorem 7.40). We dis uss ea h of these ases in detail, and prove a few relevant theorems in ea h ase.

311

7.2 Removable Singularities

7.2 Removable Singularities An isolated singularity z0 of f is alled removable or that f has a removable singularity at z0 if f an be de ned at z0 so that it be omes analyti at z0 .

7.3. Example. Consider the following fun tions: f1 (z ) = sin z (z 6= 1);

f4 (z ) =

 z e

z

1

2

(z 6= 0)

z 2(1 os z ) f2(z ) = z (z 6= 0); f5 (z ) = (z 6= 0) e 1 z2 sin z Log (1 z ) (z 6= 0); f6 (z ) = (z 6= 0): f3 (z ) = z z All of these fun tions ex ept f1 have removable singularities at 0; and f1 has a removable singularity at z = 1: These singularities an be removed by letting f1 (1) = sin 1 and fj (0) = 1 (2  j  6), respe tively.  If f is analyti on an open set D and z0 2 D, then the fun tion F de ned by f (z ) f (z0 ) F (z ) = ; z 2 D nfz0g z z0 is analyti on D nfz0g and limz!z0 F (z ) = f 0 (z0 ): Thus F has a removable singularity at z0 , whi h an be removed by letting F (z0 ) = f 0 (z0 ).

7.4. Example. Let f be de ned by z 2 + a2 ; z 6= ia: f (z ) = z + ia Clearly, f 2 H(C n f iag). For z 6= ia, we have f (z ) = z ia and therefore, lim f (z ) = 2ai: Thus, f has removable singularity at z = ia. Now we z! ia set 8 2 2 < z + a for z 6= ia g(z ) = z + ia : 2ai for z = ia: Then, g be omes analyti everywhere in luding at z = ia.  Suppose that f has a removable singularity at a point z0 , say. Then there is a fun tion g analyti at z0 su h that f (z ) = g(z ) for all z in a deleted neighborhood of z0 . In parti ular, f is bounded near z0 . The onverse of this statement provides a useful riterion for determining whether an isolated singularity is removable.

7.5. Theorem. (Riemann's Removable Singularity Theorem) If f has an isolated singularity at z0 , then z = z0 is a removable singularity i one of the following onditions holds:

312

Classi ation of Singularities

(i) f is bounded in a deleted neighborhood of z0 , (ii) zlim !z0 f (z ) exists, (iii) zlim !z (z z0 )f (z ) = 0. 0

Proof. Suppose that f has an isolated singularity at z0 . Then f is analyti in some deleted neighborhood of z0 , say on 0 < jz z0 j < Æ. For proving `(iii) implies z0 is a removable singularity', we introdu e (

h(z ) =

z0 )f (z ) for 0 < jz z0 j < Æ 0 for z = z0 :

(z

By the hypothesis (iii), h is ontinuous at z0 . Sin e h, like f , is analyti in the deleted neighborhood 0 < jz z0 j < Æ, it follows that h is analyti at z = z0 (see Corollary 4.88). This means h is analyti throughout the neighborhood jz z0 j < Æ. For z 6= z0 , let g be de ned by

h(z ) h(z ) h(z0 ) = : z z0 z z0 As limz!z0 g(z ) exists and equals h0 (z0 ) and g = f for z 6= z0 , we de ne f (z0 ) = h0 (z0 ). Thus, when (iii) holds, f an be extended to be analyti in jz z0 j < Æ. This observation, by de nition, shows that the singularity at z = z0 is removable. The remaining parts are dedu ible from (iii), as limz!z0 (z z0 )f (z ) = 0 holds under (i) or (ii). Indeed, (ii) implies (iii) is trivial. Finally, it remains to show that (i) implies that f has a removable singularity. Now, we suppose that (i) holds. Then, jf (z )j  M for 0 < jz z0 j  r and for a small r > 0. The Laurent oeÆ ient ak gives g(z ) =

jak j =

Z 1 2i jz z0 j=r



f (z ) dz M  (z z0 )k+1 rk

whi h approa hes zero as r ! 0, when k < 0. We dedu e that ak = 0 when k 2 N . Consequently, limz!z0 f (z ) exists and therefore, z0 is a removable singularity of f . From the proof of Theorem 7.5, it follows that if f has an isolated singularity at z = z0 and satis es the ondition

jf (z )j  jz Mz j5=6 0 in a neighborhood of z = z0 , then z = z0 must be a removable singularity of f . Is it the ase if we repla e 5=6 by an  with  < 1? Also, Theorem 7.5

313

7.2 Removable Singularities

infers that there exists no fun tion analyti at z0 with f (z )  (z z0 )  whenever  2 (0; 1). Next we illustrate by an example that a result similar to Theorem 7.5 is not available for fun tions of a real variable. Consider f : R n f0g ! R de ned by f (x) = x 1=3 . We see that jxf (x)j = jxj2=3 ! 0 as x ! 0 but f (x) annot be extended to be (real) dierentiable near 0. Here is another example. Consider g(x) = jxj for x 6= 0. Then limx!0 g(x) = 0 but g(x)

annot be extended so that it is (real) dierentiable at x = 0.

7.6. Example. Consider the fun tion f (z ) = os(1=jz j) or sin(1=jz j); z 2 C

n f0g:

Then f is ontinuous (in fa t real dierentiable in nitely often) and bounded on C nf0g, yet f annot be extended to be ontinuous on any neighborhood of the origin. Thus, the situation for (real-valued) C 1 -fun tions is dierent from that of the ase of analyti fun tions. 

7.7. Remark. Let us look at the statement (ii) of Theorem 7.5. Then the following situations o

ur: (a) f may not be de ned at z0 (b) f may be de ned at z0 but f (z0 ) may not be equal to limz!z0 f (z ) ( ) f may be de ned at z0 and f (z0 ) is equal to limz!z0 f (z ). In the last ase, f is not singular at z0 . Thus if f has a removable singularity at z0 then either (a) or (b) holds, as dis ussed in the introdu tion. 

7.8. Remark. If f and g are analyti and if both have a zero of order n at z0 , then we may write f (z ) = (z z0 )n f0 (z ); and g(z ) = (z z0 )n g0 (z ) where f0 and g0 are both analyti and non-zero at z0 , and hen e

f (z ) f0 (z0 ) lim = z!z0 g (z ) g0 (z0 ) exists. This means that f (z )=g(z ) has a removable singularity at z0 . For instan e ea h of the fun tions fj (j = 1; 2), where

f1 (z ) =

ez

1

and f2 (z ) =

z has a removable singularity at z = 0.

z sin z ; z sin z



314

Classi ation of Singularities

7.9. Example. Consider the fun tion f (z ) = (1 z 2 ) s (z ): Then f is analyti on C n Z, and so f an be expressed as a Laurent series about 0. Considering the lo ation of the singularities of f (z ), we see that the Laurent series about 0 is valid for 0 < jz j < 1. Clearly, f extends to be analyti both at z = 1 and z = 1. This observation shows that the largest open set on whi h the Laurent expansion for f (z ) about 0 onverges is 0 < jz j < 2. Moreover, for fun tions su h as g(z ) = s (z ), there exist in nitely many Laurent series about the origin. For the fun tion g, the Laurent series about 0 is valid in ea h of the annuli

Dk = fz 2 C : k < jz j < k + 1g (k = 0; 1; 2; : : :):



7.3 Poles We have seen that if f is bounded in a deleted neighborhood of an isolated singularity at z0 , then z0 is a removable singularity of f . Thus if z0 is not a removable singularity of f then, by Theorem 7.5, f is not bounded near z0 . We might then ask whether (z z0 )n f (z ) is bounded near z0 for some n 2 N . In this ase, we say that the point z0 is a pole of f and the smallest positive integer n su h that (z z0 )n f (z ) is bounded near z0 is alled the order of the pole at z0 . A pole of order one is alled a simple pole. For example, the fun tion f (z ) = z 8 =(z 1)2 has a double pole at z = 1.

7.10. Example. The fun tion f (z ) = 1=(1 + eiz ) has simple poles at z = 1 + 2k (k 2 Z) and z = 1 is the limit point of these poles, sin e the

ondition 1 + eiz = 0 implies that eiz = ei e2ki ; k 2 Z.  7.11. Example. Consider the fun tion f (z ) =

z os(z=2a) ; (z a)(z 2 + b2 )7 sin5 z

where a and b are distin t non-zero real numbers. Then we see that f has poles of orders: 7 at ib, 4 at z = 0, 5 at k (k 2 N ) and a removable singularity at z = a.  Observe that the ondition limz!z0 (z z0 )n f (z ) 6= 0 is equivalent to the ondition that n is the smallest positive integer su h that (z z0 )n f (z ) has a removable singularity at z0 . This expresses the following

7.12. Theorem. If f is analyti in a deleted neighborhood of z0 , then f has a pole at z0 i there exists an n su h that (z z0 )n f (z ) is

315

7.3 Poles

bounded near z0 . More pre isely, f has a pole of order n i limz!z0 (z z0 )n f (z ) 6= 0, and (z z0 )n f (z ) has a removable singularity at z0 , i.e. limz!z0 (z z0 )n+1 f (z ) = 0:

7.13. Remark. Let f be analyti in a domain D and z0 2 D. Re all that the fun tion f has a zero of order n at z0 i f (k) (z0 ) = 0 for k = 0; 1; 2; : : : ; n 1 and f (n) (z0 ) 6= 0. From Taylor's expansion, it follows that f has a zero of order n i f (z ) = (z z0 )n g(z ) where g is analyti at z0 and g(z0 ) = f (n) (z0 )=n! 6= 0. Thus, we see that f (z ) has a zero of order n at z0 i 1=f (z ) has a pole of order n at z0 . For instan e let f (z ) = sin(z 2 ). Thenpwe see p that f (z ) has a double zero at z0 = 0 and simple zeros at zk =   k, k = 1; 2; : : : , be ause f (zk ) = 0 and f 0 (zk ) = 2zk os(p zk2 ) 6= 0 for k 6= 0, and f (z0 ) = f 0 (z0 ) = 0, f 00 (z0 ) 6= 0 (Here the notation  k is de ned as in item 1.3). Thus, 1=f (z ) has a double pole at z0 and simple poles at zp k , k = 1; 2; : : : . Note that Laurent's expansion for 1=f (z ) in 0 < jz j <  ontains even powers of z only.  In general, the following theorem tells us how we may determine poles by their behavior in a deleted neighborhood.

7.14. Theorem. Let N be a deleted neighborhood of z0 su h that f is analyti in N . If f has an isolated singularity at z0 , then z0 is a pole of order n i there are positive onstants C1 and C2 su h that C2  j(z z0 )n f (z )j  C1 for some deleted neighborhood N0 of z0 su h that N0  N .

Proof. Suppose that the inequalities hold. Then the right hand side of the above inequality shows that (z z0 )n f (z ) is bounded for points near z0 so that (z z0 )n+1 f (z ) ! 0 as z ! z0 whereas the left inequality gives j(z z0 )n f (z )j  C2 as z ! z0 : So, by Theorem 7.12, z0 is a pole of order n for f . Assume that f has a pole of order n at z0. Then, (z z0 )n f (z ) is bounded for points near z0 . By the removability theorem, there exists a fun tion g whi h is analyti at z0 su h that (z z0 )n f (z ) = g(z ) in some deleted neighborhood N0 of z0. If g were su h that g(z ) = (z z0 )ge(z ) with ge analyti at z0 , then g (z ) e

= (z

z0 )n 1 f (z )

316

Classi ation of Singularities

in N0 and is bounded near z0. This ontradi ts the fa t that z0 is a pole of order n. This fa t shows that (z z0 )n 1 f (z ) is unbounded near z0. That there are onstants C1 and C2 satisfying the required inequalities is then a

onsequen e of boundedness or unboundedness as the ase may be. Note that Theorem 7.14 hara terizes poles of f through the behavior of the values of f near z0 in the following form:

7.15. Theorem. If f (z ) has an isolated singularity at z0 , then f (z ) has a pole at z0 if and only if limz!z0 f (z ) = 1: Proof. Suppose that f has a pole of order n at z0 . Then f (z ) = (z z0 ) n g(z ); where g is analyti and is non-zero at z0 . So, by the ontinuity of g at z0 (see for example the proof of Theorem 2.10( )), there is a neighborhood N of z0 on whi h f is de ned (ex ept at z0 ) and

jg(z )j  jg(2z0)j for z 2 N:

Hen e, jf (z )j  jg(z0 )j jz z0j n =2 for z 2 N nfz0 g; i.e. limz!z0 jf (z )j = 1: Conversely, suppose that f has an isolated singularity at z0 su h that limz!z0 f (z ) = 1: Then, for a given  = 1, there exists a Æ > 0 su h that

jf (z )j >  = 1 for 0 < jz z0 j < Æ:

Consequently, the fun tion 1=f (z ) is analyti and bounded on the pun tured disk 0 < jz z0 j < Æ with limz!z0 1=f (z ) = 0: By Riemann's removability theorem, it follows that 1=f (z ) has a removable singularity at z0 . De ne ( 1=f (z ) for 0 < jz z0 j < Æ g (z ) = 0 for z = z0 : Then, g 2 H(
(z0 ; Æ)). Clearly, g is not identi ally zero on
(z0 ; Æ) but g(z ) has a zero at z0 . It follows that there exists an n 2 N su h that

g(z ) = (z z0 )n (z ); where 0 < jz

2 H(
(z0 ; Æ)) and (z0 ) =6 0. As (z0 ) =6 0 and g(z ) =6 0 on z0 j < Æ, we have 1 (z ) f (z ) = = g(z ) (z z0 )n



1 (z ) = (z )



n where (z ) is analyti at z0. Hen e, zlim !z0 (z z0 ) f (z ) = (z0 ) 6= 0 showing that f has pole of order n at z0.

7.4 Further Illustrations through Laurent's Series

317

The following theorem is an equivalent formulation of Theorem 7.14 in the language of zeros a useful hara terization of zeros.

7.16. Theorem. If f is analyti in a deleted neighborhood N of z0 , then z0 is a zero of order n i there are nite positive onstants C1 and C2 su h that C2  j(z z0 ) n f (z )j  C1 for some deleted neighborhood N0 of z0 su h that N0  N . 7.17. Example. Let us now dis uss the singularities of z 1 i f (z ) = 2 : z (4 + 3i)z + (1 + 5i) First we obtain that the poles of f , if any, are determined by

z 2 (4 + 3i)z + (1 + 5i) = 0; that is (with t2j = 3+4i for j = 1; 2), we have 2z = 4+3i+tj = 4+3i(2+i): Therefore, we write z  f (z ) = (z )(z ) with  = 1 + i and = 3 + 2i: Thus, f has a simple pole at z = and a removable singularity at z = .  In view of Theorems 7.15 and 7.5, we arrive at a hara terization of isolated essential singularities.

7.18. Theorem. If f has an isolated singularities at z0 , then f (z ) has an essential singularity at z0 i limz!z0 f (z ) fails to exist either as a nite value or as an in nite limit.

7.4 Further Illustrations through Laurent's Series An obvious tool in hara terizing singularities is the Laurent series expansion of a given fun tion about its isolated singularities. If f has an isolated singularity at z0 , then we have the unique representation 1 1 X X (7.19) f (z ) = ak (z z0 )k + ak (z z0 )k ; 0 < jz z0 j < r; k= 1 k=0 where (7.20)

Z

1 f ( )d ak = 2i C ( z0 )k+1 and C an be any ir le with enter at z0 and radius less than r. If z0 is the only singularity of f (z ) in C , then (7.19) is valid with r = 1. Note also that the integral over C is taken in the positive dire tion and has the

318

Classi ation of Singularities

same value on any positively oriented urve whi h en loses z0 but no other singularity of f (see the Cau hy deformation theorem). Then, a

ording to (7.20), Z

1 (7.21) f (z ) dz: a 1= 2i C The oeÆ ient a 1 of (z z0) 1 in the Laurent expansion (7.19) of f about z0 , whi h is of spe ial signi an e, is alled the residue of f at z0 . We use the notation a 1 = Res [f (z ); z0℄ to denote the residue of f at z0 . Equation (7.21) provides a onvenient way of evaluating ertain integrals (see Chapters 8 and 9) of a fun tion f around an isolated singularity, sin e its value is simply the produ t of 2i and the oeÆ ient of (z z0 ) 1 in its Laurent expansion (7.19). Note that, throughout the above dis ussion, we have supposed that f is de ned for all z near z0 , but not ne essarily at z0 itself. Thus, the

lassi ation of the isolated singularity of f at z0 depends only on the lo al behavior of f in a deleted neighborhood of z0 . The rst part in (7.19), i.e. the series with negative powers of (z z0 ), is alled prin ipal part whereas the se ond part in (7.19), i.e. the series with non-negative powers of (z z0 ), is alled the regular/holomorphi /analyti part. It is the rst part whi h plays an important role in deriving the hara ter of the singularities of f at the isolated singularity z0 . The following three ases are mutually ex lusive. Case 1. No prin ipal part. In this ase (7.19) takes the form 1 X f (z ) = ak (z z0 )k ; 0 < jz z0j < r: k=0

Thus if we de ne f (z0 ) = a0 = limz!z0 f (z ); then f be omes analyti at z0 and so, on the entire disk jz z0j < r. This observation shows that z0 is a removable singularity. Similarly if f has a removable singularity at z0 , then ak = 0 for k  1; for, sin e we may write f (z ) = g(z ) for z 6= z0 ; the Laurent series expansion for f must oin ide with the Taylor series expansion for g near z0 . In other words, \f has a removable singularity i, in (7:19), a k = 0 for k  1." For example, onsider the following fun tions:

f1 (z ) = z f2 (z ) = 1 f3 (z ) = z

z3 z5 + 3! 5! z2 z3 + 2! 3! z2 z3 + 2 3




(jz j > 0)


(jz j > 0)


(0 < jz j < 1):

7.4 Further Illustrations through Laurent's Series

319

Then ea h of these fun tions have a removable singularity at z = 0, be ause fj (j = 1; 2; 3) an be de ned at z = 0 in su h a way that fj (j = 1; 2; 3) be omes analyti at z = 0.

7.22. Example. Let f 2 H(
n f0g) su h that jf (z )j  ln(1=jz j) for z 2
n f0g. Then, z = 0 is a removable singularity of f . Indeed, the Laurent oeÆ ients an (n 2 Z) of f are given by an =

Z

1 f (z ) dz 2i jzj=r z n+1

so that jan j  r n ln(1=r). When n < 0, letting r ! 0 gives that an = 0, and when n = 0, letting r ! 1 implies that a0 = 0. This observation shows that the Laurent series expansion of f about the origin does not have the prin ipal part, and so z = 0 is a removable singularity of f . 

Case 2. Finite prin ipal part. In this ase, a

ording to Theorem 7.14, (7.19) be omes f (P z ) = (z z0 ) n g(z ) where g is analyti and non-zero k at z0 . Hen e if g(z ) = 1 k=0 bk (z z0 ) ; then we may write 1 X f (z ) = ak (z z0 )k ; bn+k = ak ; k= n

(a n = b0 = g(z0 ) 6= 0). The uniqueness of f (z ) is obvious from the uniqueness of g(z ). The onverse assertion is lear. In other words, \f has a pole of order n i, in (7:19), a n 6= 0, but a k = 0 for k  n + 1." Moreover, the dis ussion here is equivalent to

7.23. Theorem. An isolated singularity at z0 of f (z ) is a pole of z0 ) n g(z ), where g is analyti at z0 and g(z0) 6= 0.

order n i f (z ) = (z

Consider the following fun tions: ez f1 (z ) = (z 1)4 e e e = + + +


; jz 1j > 0; (z 1)4 (z 1)3 2!(z 1)2 (1 z n)ez 1 f2 (z ) = = m +


; jz j > 0 (m > n): zm z Then, it is lear that f1 has a pole of order 4 at z = 1 and f2 has a pole of order m at z = 0. Case 3. In nite prin ipal part. The fun tion f has an essential singularity at z = z0 exa tly when neither Case 1 nor Case 2 prevails. In other words, \f has an essential singularity i, in (7:19), a k 6= 0 for in nitely many k  1."

320

Classi ation of Singularities

In this ase, limz!z0 f (z ) fails to exist (in luding the possibility of the limit being in nity) as we have seen with the fun tion f (z ) = e1=z . Here e1=z has an essential singularity at z = 0. Similarly, it is easy to see that the fun tion f (z ) = ez=(z b) = eeb=(z b) (b 6= 0) has an essential singularity at z = b. Consider the Laurent series 1 1 zk X X (7.24) : zk + 2k+1 k= 1 k=0 Observe that the rst series onverges for jz j > 1, while the se ond for jz j < 2. Then the ombined series onverges to f (z ) for 1 < jz j < 2, where

f (z ) =

1=z 1=2 + = 1 1=z 1 z=2 (z

1 2)(z

1)

:

Observe that the Laurent series (7.24) has an in nite number of negative powers of z . But, sin e the region of onvergen e of the given Laurent series does not in lude a deleted neighborhood of the origin, it would not be orre t to on lude that the origin is an essential singularity of f . In fa t, the limit fun tion f has simple poles at z = 1 and z = 2. This remark is to aution the reader when dealing with a series of the form (7.24).

7.25. Example. Let f (z ) = (z a)2004 sin (1=(z b)) : Then, we may rewrite f (z ) as 1 X 1 1 f (z ) = (z b + b a)2004 ( 1)k 1 (2k 1)! (z b)2k 1 k=1 whi h implies that z = b is learly an essential 2 singularity of f . Similarly, it is easy to see that ea h of the fun tions ze1=z , e 1=z , sin(1=z ), and os(1=z ) has an essential singularity at z = 0. 

7.5 Isolated Singularities at In nity In the earlier se tions, we have dis ussed singularities of a fun tion whi h lie in C . In this se tion we shall be on erned with fun tions that have singularities at the point at in nity as it is sometimes useful to think about 1 just like any other point in C . The role of the point at in nity is understood through the inversion w = z 1 as it allows us to pass ba k and forth between the neighborhoods of 1 and the neighborhoods of 0. This fun tion is de ned for every z 6= 0 and maps ea h point z in C ex ept z 6= 0, into a point in the w-plane. For instan e, the ir le jz j = R is mapped onto jwj = 1=R. If we assign the point at in nity in the extended w-plane to z = 0 and w = 0 to the point at

7.5 Isolated Singularities at In nity

321

in nity in the extended z -plane, then the inversion is one-to-one from the extended z -plane to the extended w-plane, i.e. C 1 onto C 1 , a fa t whi h was dis ussed in Chapter 5. Let f (z ) be analyti for jz j > R, for some R with 0  R < 1. By putting z = 1=w in f (z ), we obtain (7.26) F (w) = f (1=w): Then, F (w) is analyti in the deleted neighborhood fw : 0 < jwj < 1=Rg of zero. The nature of the singularity of f (z ) at z = 1 (the point at in nity) is de ned to be the same as that of F (w) at w = 0. That is, f (z ) has an isolated singularity at 1 i F (w) = f (1=w) has an isolated singularity at w = 0. In parti ular, \an analyti fun tion f (z ) on jz j > R is said to have a removable singularity, a pole of order n, or an essential singularity at 1 i the fun tion f (1=z ) has a removable singularity, a pole of order n, or an essential singularity at 0, respe tively". For instan e, the fun tion f de ned by f (z ) = z n (n 2 N ) has a pole of order n at z = 1, sin e the orresponding F de ned by F (w) = f (1=w) = w n has a pole of order n at w = 0. More generally, every polynomial of degree n has a pole of order n at 1. Similarly, the fun tion f de ned by f (z ) = z 2 + z m has a pole of order two at z = 0 and a pole of order m at in nity. On the other hand, the fun tion f de ned by f (z ) = ez has an essential singularity at z = 1, sin e F de ned by F (w) = f (1=w) = e1=w has an essential singularity at w = 0. In the same way, it is easy to verify that ea h of the fun tions eiz ; e z ; sinh z; osh z; sin z; os z has an essential singularity at z = 1. Also, we see that every non onstant and nonvanishing entire fun tion on C ne essarily has an isolated essential singularity at 1. Suppose that f (z ) has an isolated singularity at z = 1 and R is the distan e from the origin to the farthest singular point of f (z ). (If point at in nity is the only singularity, then R may be hosen as an arbitrary large positive number). Sin e we may write f (z ) = f (1=w) = F (w) (w = 1=z; jz j > R); F (w) has an isolated singularity at w = 0 and so the nearest singularity of F (w) from w = 0 is at a distan e 1=R. Therefore, F (w) has a Laurent expansion about w = 0: 1 X (7.27) f (1=w) = F (w) = an wn ; 0 < jwj < 1=R: n= 1 Thus, 1 0 1 X X X f (z ) = a nzn = a n z n + a n z n ; jz j > R: n= 1 n= 1 n=1

322

Classi ation of Singularities

Then the following ases arise. Case 1. Suppose that, in (7.27), a n = 0 for all n  1. Then, we de ne F (0) = a0 so that the orresponding F (w) is analyti at w = 0. Hen e, F (w) has a removable singularity at w = 0 i, in (7.27), a n = 0 for all n  1. That is, f (z ) has a removable singularity at z = 1 i its Laurent expansion about z = 1 has the form 1 X f (z ) = an z n; jz j > R; n=0

that is, i the Laurent expansion of f (z ) on jz j > R has no positive power of z with nonzero oeÆ ients. Case 2. Re all that F (w) has a pole of order k at w = 0 i (7.27) takes the form 1 X a a F (w) = kk +


+ 1 + an wn (a k 6= 0; 0 < jwj < 1=R): w w n=0 Hen e, f (z ) has a pole of order k at z = 1 i its Laurent expansion has the form 1 X f (z ) = a k z k +


+ a 1 z + an z n (a k 6= 0; jz j > R); n=0

that is i the Laurent series of f (z ) for jz j > R has only a nite number of positive powers of z with nonzero oeÆ ients. Here we de ne the prin ipal part of f (z ) at z = 1 to be the polynomial

a k z k + a k+1 z k

1+




+ a 1 z:

Case 3. If an in nite number of a n for n  1 in (7.27) do not vanish, then F (w) has an essential singularity at w = 0. In terms of f (z ), we say that f (z ) has an essential singularity at z = 1 and we have 1 X f (z ) = an z n ; jz j > R; n= 1 where an in nite number of a n 's are non-zero for n  1. Case 3 gives rise to the following interesting observation. \An entire fun tion f (z ) is trans endental i f (1=z ) has an essential singularity at 0." To see this we suppose that f (z ) is an entire trans endental. Then, we have 1 X f (z ) = an z n for all z 2 C : n=0

7.5 Isolated Singularities at In nity

323

If 0 is not an essential singular point of f (1=z ), then there exists a k (possibly for a suÆ iently large k) su h that 1 X z k f (1=z ) = an z k n ; jz j > 0; n=0

has a removable singularity at 0. This means that an = 0 for all n  k + 1, that is f (z ) is a polynomial of degree at most k, ontrary to our assumption that f (z ) is entire trans endental. To prove the onverse part, suppose f (z ) is not trans endental. Then, by the de nition of trans endental fun tion, f (z ) is a polynomial and so we write g(z ) = f (1=z ) = a0 + a1 z 1 +


+ an z n: But then g has either a pole (if n > 0) or a removable singularity (if n = 0) at z = 0. The proof of the observation is omplete. The above dis ussion an be equivalently stated as follows: \an entire

trans endental fun tion has ne essarily an essential singular point at in nity", for, otherwise it would have no singularities at all and would, by

Liouville's theorem, redu e to a onstant. Let f be de ned in some open set of C 1 , and let f have isolated singularities at a1 ; a2 ; a3 ; : : :. Also, let a 2 C 1 be a limit point of the set of these singularities. Then, for any neighborhood N of a, there is a singularity (dierent from a) inside N , i.e. f annot be analyti in any deleted neighborhood (
(a; Æ) nfag if a 2 C and jz j > Æ if a 2 1) of a. So f has no Laurent expansion about a. In other words, a is neither a regular point nor an isolated singularity. Thus, a is a non-isolated (essential) singularity of f . For instan e, ot(1=z ) has poles at z = 1=k; k = 1; 2; : : : . Hen e the limit point z = 0 of these poles is a non-isolated (essential) singularity of this fun tion.

7.28. Example. Consider f (z ) = 1= sin z for z 6= k, k 2 Z: Let us dis uss the singularity of f at 1. For this we de ne g(z ) = f (1=z ) = 1= sin(1=z ) and dis uss the orresponding singularity of g at 0. Note that g is not analyti at the origin, sin e every neighborhood of the origin ontains a singularity. It follows that g(z ) has poles at the points z = 1=(k) (k 2 Z). Note that z = 0 is not an isolated essential singularity for g(z ) as z = 0 is a limit point for its poles at 1=(k) (k 2 Z). This observation shows that f (z ) is not analyti in fz : jz j > Rg for any R > 0 and 1 is a non-isolated (essential) singularity for f (z ). Similarly, it is easy to see that 1 is a non-isolated (essential) singularity for f (z ) = (ez 1) 1 . 

324

Classi ation of Singularities

7.29. Example. Consider the fun tion g(z ) = z 1(1 + z ) 1 : Then g(1=w) = w2 =(1 + w) whi h is analyti at w = 0 and hen e, g(z ) is analyti at in nity. Similarly, we may easily verify that ea h of the fun tions 1 z ; ; exp(1=z 2) 1+z 1+z is analyti at the point at in nity. On the other hand, fun tions sin z and os z are not analyti at the point at in nity, be ause both sin(1=z ) and os(1=z ) have an isolated essential singularity at z = 0. 

7.30. Example. The fun tion g de ned by g(z ) = z 4 sin(1=(z + 1)) has a pole of order three at the point at in nity. On the other hand, the fun tion 1=(z 2 sin(1=z )) has a removable singularity at in nity.  Suppose that P f is entire. Then f has a power series expansion of the form f (z ) = n0 an z n whi h onverges absolutely for all z 2 C . Thus,

F (z ) = f (1=z ) =

X

n0

an z n for jz j > 0

and so z = 1 is an isolated singularity of f (z ), sin e 0 is an isolated singularity of F (z ). There are now three possibilities: (i) f (z ) has a removable singularity at z = 1 (ii) f (z ) has a pole of order k at z = 1 (iii) f (z ) has an essential singularity at z = 1. In ase (i), the expansion ontains no powers of z and so f (z ) = a0 . Alternatively, by Liouville's theorem, f (z ) redu es to a onstant. Thus, \an entire fun tion f : C ! C has a removable singularity at z = 1 i f (z ) is onstant". In ase (ii), as dis ussed above, f (z ) has a pole of order k ( 1) at 1 i the expansion of f (z ) ontains only a nite number of positive powers and an = 0 for n > k. Consequently, \an entire fun tion f : C ! C has a pole at z = 1 (of order k ) i f (z ) is a non onstant polynomial (of degree k )". Case (iii) o

urs i an 6= 0 for in nitely many n's. Consequently, f (z ) is a trans endental entire fun tion-a fa t whi h has been on rmed earlier too.

7.6 Meromorphi Fun tions A fun tion analyti in a domain D  C ex ept possibly for poles is alled

meromorphi in D.

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7.6 Meromorphi Fun tions

Suppose that f is meromorphi in D su h that D ontains fz : jz j > rg for some r > 0. Then, the fun tion f is said to be meromorphi at z = 1 if g(z ) = f (1=z ) is meromorphi in a neighborhood of z = 0, i.e. in the usual sense on
1=r . The de nition as stated is equivalent to requiring that f has a pole at 1 but has no poles in fz : jz j > Rg for some R > r. Here are some basi examples of meromorphi fun tions in C . (i) f (z ) = ez =z 2 is meromorphi in C , sin e f is analyti in C n f0g and the singularity at 0 is a double pole. (ii) every analyti fun tion in D  C is obviously meromorphi . Consequently, sums and produ ts of meromorphi fun tions are meromorphi . It is important to remark that the fun tions f (z ) = ez =z 2 and g(z ) = z ez =z 2 are meromorphi in C , but f + g has a removable singularity at z = 0. Here we regard f + g as an extended analyti fun tion in C (by de ning its value at the origin as its limiting value, namely (f + g)(0) = 0) and hen e, we treat f + g as a meromorphi fun tion. (iii) f (z ) = 1= sinh z is meromorphi in C sin e the singularities of f are poles at z = ki; k 2 Z. Further, fun tions 1= sin z , 1= os z ,

os z= sin z , sin z= os z , 1= osh z and 1=(ez 1) are all meromorphi in C . Note also that 1 is the limit point of poles for these fun tions. (iv) any rational fun tion R(z ), where the numerator and the denominator of R(z ) have no ommon fa tor, is meromorphi in C sin e it is analyti in C ex ept at the zeros of the denominator where R(z ) has poles. (Note that ea h of the fun tions involved in (iii) is not rational sin e the set of poles for ea h of these fun tions is ountably in nite). We an express R(z ) as a quotient of polynomials of the form Q

k (z a )ni i R(z ) = A Qli=1 ; ( z b j )mj j =1

(v) (vi) (vii) (viii)

where ai 's and bj 's are all distin t, and A is some onstant. the quotient of a meromorphi fun tion is meromorphi , provided that the denominator term is not identi ally zero. the fun tions (z 2 + 4) 1e1=z and (z 2 4) 1 sin(1=z ) are meromorphi in C n f0g. They are not, however, meromorphi in C , be ause ea h of them has an essential singularity at the origin. every meromorphi fun tion in an open set admits a representation as the quotient of two analyti fun tions in , whi h will be on rmed in Chapter 11. the only singularities of f (z ) = ot z are the simple poles at n (n 2 Z) and hen e, f (z ) is meromorphi in C . It is, however, not meromorphi in C 1 be ause 1 is a limit point of poles of f (z ).

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Classi ation of Singularities

7.31. Remark. A meromorphi fun tion in C an have only a nite number of poles in any bounded subset D of C . For instan e, neither f (z ) = 1= sin(=z ) nor g(z ) = 1= os(1=z ) is meromorphi in C . Note that the set of poles of f (z ) is a bounded set S = f1=n : n 2 Zg and the limit point of S is 0, whi h is a non-isolated (essential) singularity of f (z ). A similar argument shows that g(z ) is not meromorphi in C . On the other hand, both f (z ) and g(z ) are meromorphi in C n f0g. However, the fun tions su h as 1=(ez + 1) and 1=(1 + os z ) are all meromorphi in C .  When we speak of a meromorphi fun tion f without mentioning the domain of de nition for f , it is understood that f is meromorphi in C . We observe that analyti fun tions are, in some sense, a generalization of polynomials while the meromorphi fun tions are then a generalization of rational fun tions. The next two theorems whi h hara terize rational fun tions are simple and elegant. From Liouville's theorem it follows that an entire fun tion on C 1 is onstant. An analog of this result for meromorphi fun tions in C 1 follows.

7.32. Theorem. A fun tion f (z ) is rational i it is meromorphi in the extended omplex plane C 1 . Proof. As noti ed above a rational fun tion is meromorphi in C 1 . Suppose onversely f is meromorphi in C 1 . Note that f (z ) an have only nitely many poles in C 1 sin e otherwise the limit point of the poles would be a non-isolated (essential) singularity whi h is not a pole at all. Let the poles of f (z ) be z1 ; z2 ; : : : ; zm 2 C , of orders k1 ; k2 ; : : : ; km , respe tively. Then, g de ned by g(z ) = f (z )


m Y

j =1

(z

zj )kj

is an entire fun tion. Thus g is a polynomial, sin e otherwise g would ne essarily have an essential singularity at 1. The result follows. For instan e, f (z ) = (z 5 + 3z 2 + 1)(z 2 2z 1) 1 has a triple pole at 1 be ause the fun tion f (1=z ) has a triple pole at the origin.

7.33. Theorem. Let f (z ) be meromorphi in natural number n, M > 0, and R > 0 su h that (7.34)

jf (z )j  M jz jn

C

and there exist a

for jz j > R:

Then, f is a rational fun tion.

Proof. Pro eeding exa tly as in Theorem Q 7.32, we obtain an entire kj fun tion g, where g(z ) = f (z )
p(z ) with p(z ) = m j =1 (z zj ) : Therefore,

7.7 Essential Singularities and Pi ard's Theorem

327

for suÆ iently large jz j, we easily have (see (6.65))

jp(z )j  2jz jN

(7.35)




+ km . Thus, by (7.34) and (7.35), jg(z )j = jf (z )j jp(z )j  2M jz jN +n for suÆ iently large jz j. It follows from Theorem 6.60 that g is a polynomial, where N = k1 + k2 +

and hen e, f (z ) is a rational fun tion.

Note that if, in Theorem 7.33, f is assumed to be analyti in C , then f would be a polynomial. Thus, Theorem 7.33 is learly a natural generalization of Liouville's theorem (see Theorem 6.60).

7.7 Essential Singularities and Pi ard's Theorem We have already proved that sin(C ), os(C ) and exp(C ) are all unbounded subsets in C . Therefore, it is natural to ask: Can we say something more

about entire unbounded fun tions? In fa t, su h questions lead to er-

tain beautiful on lusion about the image domains of non- onstant entire unbounded fun tions. For ertain familiar entire fun tions mu h more is true:

 ea h non- onstant polynomial p(z ) assumes every omplex number as a value; that is p(C ) = C .  ea h of the trigonometri fun tions sin z , os z , and of the hyperboli



fun tions sinh z , osh z assumes every omplex number as a value; that is, sin(C ) = os(C ) = sinh(C ) = osh(C ) = C : the exponential fun tion ez never assumes zero, as exp(C ) = C n f0g.

The rst assertion has been proved in Theorem 6.66 while the other two assertions an be proved by looking at the solutions of the equation f (z ) = for a given 2 C , where f (z ) is one of the fun tions given by sin z; os z; sinh z; osh z; and ez : For example, as sin z = (eiz e iz )=(2i), the solution of sin z = is obtained from p eiz = i  1 2 : p Note that there exists no 2 C su h that i  1 2 = 0. Consequently, for ea h , the equation sin z = has in nitely many solutions given by p

z = i log(i  1 2 ): These observations probably led to \the Pi ard little theorem" whi h we shall dis uss soon. Let us start with the following preliminary result whi h

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Classi ation of Singularities

may be alled a weaker form of the Casorati-Weierstrass theorem (see Theorem 7.40).

7.36. Theorem. The range of a non- onstant entire fun tion is a

dense subset of C .

Proof. Let f be a non- onstant entire fun tion. Suppose on the ontrary that f (C ) is not dense. Then, there would exist a point w0 2 C and a neighborhood
(w0 ; ) su h that
(w0 ; ) \ f (C ) = ;: Then, for all z 2 C , we have jf (z ) w0 j >  so that jg(z )j < 1 for z 2 C , where g(z ) =



f (z ) w0

:

But then, g being a bounded entire fun tion, would be onstant, and hen e f would be onstant, whi h is a ontradi tion. Note that a subset D of D0 is said to be dense in D0 , if for ea h z0 2 D0 and any Æ > 0,
(z0 ; Æ) \ D 6= ;: A

ording to Theorem 7.36, it is almost trivial to show the following: Suppose that f is an entire fun tion satisfying any one of the following

onditions for all z 2 C and for some xed M > 0: (a) jf (z )j  M ; (b) Re f (z )  M or Im f (z )  M or jRe f (z )j > M or jIm f (z )j > M . Then, f is ne essarily a onstant. We have already shown these as a onsequen e of Liouville's theorem. In parti ular, this result demonstrates a simple fa t that \the range of a non- onstant entire fun tion is never on-

tained in a half-plane or in bounded domain or in the omplement of any bounded simply onne ted domain in C ". This pie e of information learly

improves Liouville's theorem but obviously falls far short of Pi ard's little and big theorems. To be a little more pre ise, mu h more than Theorem 7.36 holds: if f is a non- onstant entire fun tion then C n f (C ) ontains at most one point. This result is known as Pi ard's little theorem.11 We may restate it in the following form. For a proof of this result, we refer to p. 540.

7.37. Theorem. (Pi ard's Little Theorem) Every non- onstant entire fun tion omits at most one omplex number as its value.12 In other words, if an entire fun tion omits two values, then it is onstant. 11 Emile Pi ard (1856-1941) was a Fren h mathemati ian who published the proofs of the two famous theorems at the age of 22. He has made a number of ontributions in many other areas. 12 A number 2 C is said to be an ex eptional value of a omplex-valued fun tion f if does not belong to the range of f .

7.7 Essential Singularities and Pi ard's Theorem

329

Now we need some preparations for the proof of Pi ard's little theorem. Pi ard's original proof of his little theorem is of entirely dierent hara ter. The proof whi h we are going to present is due to Landau-Konig. Let us start with a remark that Pi ard's little theorem is an astonishing generalization of the theorem of Liouville as well as the theorem of Casorati-Weierstrass. In 1879, with the aid of ellipti modular fun tions, Pi ard proved a more general and deep result the so alled Pi ard's great theorem: \if f has an essential singularity at a, then the range under f of any deleted neighborhood of a is the whole omplex plane with at most one ex eption". Equivalently, we may state it in the following form.

7.38. Theorem. (Pi ard's Great Theorem) Suppose that f is analyti in
(z0 ; r) nfz0 g and z = z0 is an essential singularity of f . Then C nf (
(z0 ; r) nfz0g) is a singleton set. Examples 7.46 and 7.48 below are spe ial ases of Pi ard's theorem. We shall present the proof of Pi ard's little theorem in Se tion 12.7 as it requires Blo h's theorem. However, let us rst restri t ourselves to the following mu h weaker and simpler result also alled the Casorati-Weierstrass theorem.

7.39. Theorem. (Weaker Form of Pi ard's Great Theorem) Suppose that f 2 H(
(z0 ; R) nfz0 g) and z0 is an essential singularity of f . Then for ea h Æ > 0 with Æ  R, f (
(z0 ; Æ ) nfz0 g) is dense in C .

Proof. Suppose on the ontrarary that the range f (f0 < jz z0j < Æg) is not dense. Then there is a point w0 2 C and a disk
(w0 ; ) = fw 2 C : jw w0 j < g su h that
(w0 ; ) \ f (f0 < jz z0 j < Æg) = ;: For all z with 0 < jz z0j < Æ, we have jf (z ) w0 j >  > 0 and therefore, the fun tion g de ned by  g(z ) = f (z ) w0 is bounded and analyti in the deleted neighborhood
(z0 ; Æ) nfz0g, sin e f (z ) w0 is analyti and non-zero there. From the removability theorem, it follows that g has a removable singularity at z0 . Thus, by de ning g(z0 ) properly, g be omes analyti in
(z0 ; Æ). Clearly, g(z ) 6 0 in
(z0 ; Æ). If g(z0 ) 6= 0,  f (z ) = + w0 g(z ) is analyti at z0 . If g(z ) has a zero at z0 , then z0 is a pole of 1=g(z ) and hen e, the same holds for f (z ). In either ase this ontradi ts the hypothesis that z0 is an essential singularity of f (z ). 7.40. Theorem. (Casorati-Weierstrass Theorem) If f has an essential singularity at z0 and if w0 is a given nite omplex number, then

330

Classi ation of Singularities

there exists a sequen e fzn g with zn ! z0 su h that f (zn ) ! w0 . In other words, f takes values arbitrarily lose to every omplex number in every neighborhood of an essential singularity.

Proof. By hypothesis, f (z ) is analyti throughout a deleted neighborhood of z0 . Suppose that there exists a omplex number w0 and a sequen e fzng with zn ! z0 su h that f (zn ) 6! w0 . Then there exist an  > 0 and a Æ > 0 su h that jf (z ) w0 j   for 0 < jz z0 j < Æ. De ne g(z ) =

1

f (z ) w0

:

Now, pro eeding exa tly as in Theorem 7.39, we would get that f (z ) has either a removable singularity or a pole at z = z0 . In either ase this

ontradi ts the hypothesis that z0 is an essential singularity of f (z ).

7.41. Remark. equivalent.

Note that Theorem 7.40 and Theorem 7.39 are



If 1 is regarded as an isolated essential singularity of g(z ), then 0 is an isolated essential singularity of f (z ), where f (z ) = g(1=z ). In view of this observation, Theorem 7.39 may be rephrased as follows:

7.42. Theorem. If g(z ) is an entire trans endental fun tion, then near 1, the values assumed by g (z ) are dense in C . In other words, g (z ) takes values arbitrarily lose to every omplex number in every neighborhood of 1. Next we re ord a very useful appli ation of Theorem 7.40.

7.43. Theorem. An entire fun tion f (z ) is univalent in C i f (z ) = a0 + a1 z (z 2 C ), where a0 ; a1 are onstants with a1 6= 0; that is, Aut (C ) = ff 2 H(C ) : f (z ) = a0 + a1 z g:

Proof. If f (z ) is entire, then, by Theorem 4.93 (see alsoPCorollary 3.73), f (z ) has a power series expansion of the form f (z ) = n0 an z n whi h onverges absolutely for all z 2 C . There are now three possibilities: (i) f (z ) = a0 , i.e. onstant P (ii) f (z ) = kn=0 an z n ; ak 6= 0 (k  1) P (iii) f (z ) = n0 an z n ; an 6= 0 for in nitely many n.

Suppose further that f (z ) is univalent in C . Then ase (i) annot o

ur. In ase (ii), f (z ) is a polynomial of degree k  1. But, by the fundamental theorem of algebra, f (z ) has k zeros whi h lie inside some ir le

7.7 Essential Singularities and Pi ard's Theorem

331

jz j = R, R large enough. Therefore, f is univalent only when k 6> 1. Hen e, k = 1 and so, f (z ) = a0 + a1 z , where a1 = 6 0.

In ase (iii), f (z ) is trans endental and hen e f (z ) has an essential singularity at 1. Then, by the Casorati-Weierstrass theorem (see also Theorem 7.42), for any given omplex number w, there exists a sequen e fzng su h that zn ! 1 and f (zn) ! w. In parti ular, for w = 0, we have lim z n!1 n

1 1 = nlim !1 f (f (zn )) = f (0) 6= 1:

This ontradi tion shows that f (z ) is a polynomial whi h by ase (ii) yields that f is linear. The onverse part is obvious. This proves that an entire fun tion whi h assumes every omplex value exa tly on e is pre isely the linear fun tion. Pi ard's results mark the beginning of a development whi h eventually

ulminated in the value distribution theory of R. Nevanlinna. In 1896, E. Borel derived Pi ard's little theorem with elementary fun tion-theoreti tools. The theory then took a surprising turn in 1924, when A. Blo h dis overed the theorem named after him. We shall prove Pi ard's theorem using Blo h's theorem (see Se tions 12.6 and 12.7). Now, we observe that an entire fun tion h omits two distin t values, namely a and b with a 6= b, i the fun tion f de ned by

h(z ) a b a is entire and omits 0 and 1. Therefore, h is onstant if and only if f is

onstant. In view of this observation, Theorem 7.37 is equivalent to the following f (z ) =

7.44. Theorem. If f is an entire fun tion su h that 0 2= f (C ) and 1 2= f (C ), then f is onstant. It is natural to see what happens when we repla e \f (z ) is entire" by \f (z ) is meromorphi in C ." To see this, we onsider

(z ) =

1 1 + ez

whi h never assumes 0 or 1 as its value. Clearly,  is meromorphi in C ,  omits 0 and 1, but  is not a onstant fun tion. This example shows that meromorphi fun tions in C an omit two omplex numbers. A value that a meromorphi fun tion does not assume is known as Pi ard's ex eptional value. For instan e, the fun tion (z ) has two ex eptional values, namely 0; 1. Similarly, it is easy to see that tan z is meromorphi in C and has an essential singularity at z = 1 whi h is the limit point of poles of tan z .

332

Classi ation of Singularities

Moreover, tan z assumes every omplex value with Pi ard's ex eptional values i. Meromorphi fun tions of this type are signi ant in view of the following result.

7.45. Theorem. (Pi ard's Little Theorem for Meromorphi Fun tions) Every meromorphi fun tion  in C that omits three distin t values a; b; 2 C is ne essarily onstant. Proof. The desired on lusion follows from Theorem 7.44 as the fun tion f de ned by f (z ) = ((z ) a) 1 is entire and omits two distin t values 1=(b a) and 1=( a). 7.46. Example. We show that e1=z assumes every value in nitely many times with an ex eption of zero (Pi ard's ex eptional value). To do this we re all that exp(1=z ) has an essential singularity at the origin and omits the value zero. By Pi ard's theorem, it therefore assumes all the other values in ea h pun tured neighborhood
(0; r) nf0g. Let us now verify this by a dire t omputation. Consider e1=z = = elog ( 6= 0), or equivalently, (7.47) e1=z = eln j j+i arg where may be taken as a omplex number. If is real and non-negative, then (7.47) implies 1 1 = ln j j + i arg = ln j j + 2ki; i.e. zk = ; k 2 Z: z ln j j + 2ki

Thus, we have a sequen e fzng su h that zn ! 0 and e1=zn = for ea h n 2 N : The general ase for 6= 0 an be handled similarly. Observe that Pi ard's ex eptional value `zero' is a limit point. 

7.48. Example. Let us dis uss one more fun tion whi h has an essential singularity at z = 0. Consider sin(1=z ). Suppose for real and

6= 0, sin(1=z ) = : Then the solution to this equation is given by p 1 1 i p = ar sin = log(i + 1 2 ); i.e. z = : z i log(i + 1 2 ) For k 2 Z, let i p p zk = : ln ji + 1 2 j + Arg (i + 1 2 ) + 2ki Then, we obtain a sequen e fzng su h that zn ! 0 and sin(1=zn) = , n 2 N: In parti ular, if = 1 then sin(1=z ) = 1 is satis ed by in nitely many values, namely, zk = 2=(4k + 1), k 2 Z. However, z = 0 is also an essential singularity of os(1=z ) and there are no ex eptional values (ex ept 1). 

333

7.8 Exer ises

7.8 Exer ises 7.49. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) An entire fun tion f (z ) having z = 1 as a removable singularity is

onstant. Note: See the dis ussion at the end of Se tion 7.5. (b) An entire fun tion f (z ) has a pole of order n at in nity i f (z ) is a polynomial of degree n. ( ) If f (z ) has a pole at z0 , then expff (z )g has an essential singularity at this point. (d) If f (z ) is a non onstant entire fun tion, then expff (z )g has an essential singularity at z = 1. (e) If a is an isolated singularity of f whi h is not removable, then a is an essential singularity of expff (z )g. (f) If f is analyti in C 1 ex ept for a nite number of poles, then the number poles and zeros of f ( ounted a

ording to multipli ity) are equal. (g) If f is non- onstant and analyti at z0, then f (n) (z0 ) 6= 0 for some n  1. (h) Poles are isolated. That is, f has a pole of order m at a i f (z ) = (z a) m g(z ), where g is analyti at a and g(a) 6= 0. (i) Suppose f has an essential singularity at z = a and g has a pole at z = a. Then the produ t fg has an essential singularity at z = a. z4 + 1 (j) The fun tion f (z ) = ; ; 2 C n f0g ( 6= ) (z jz j2)( z jz j2 ) has two simple poles at z = ; and a double pole at z = 0. (k) The fun tion f de ned by the Laurent series

f (z ) =

1 zk z 2k X + k+1 k= 1 ( k )! k=0 2 1 X

has an essential singularity at the origin. (l) If f is meromorphi , then f and f 0 have the same poles and the order of the poles of f 0 in reases by one. (m) The fun tion f (z ) = 1= sin(1=z ) has a singularity at the origin and the Laurent series expansion about the origin does not exist. (n) If z = a is an isolated essential singularity for f (z ), then z = a is neither a regular point nor a pole for g(z ) = 1=f (z ). Further, z = a is not ne essarily an isolated essential singularity for g(z ).

334

Classi ation of Singularities

(o) Suppose f has an essential singularity at z0 . Then there exists a sequen e fzng with zn ! z0 su h that limn!1 jf (zn)j = 1. (p) If f is a trans endental entire fun tion, then there exists a sequen e zn ! 1 for whi h f (zn) ! 0. (q) If f (z ) is analyti at 1, then f 0 (1) = 0. e1=(z 1) (r) The fun tion f (z ) = z has a simple pole at z = 2ki (k 2 Z) e 1 and an essential singularity at z = 1. (s) Every f 2 H(C n f0g) su h that jf (z )j  ajz j1=2 + bjz j 1=2 for some a; b > 0 is ne essarily onstant. (t) Every f 2 H(C n f0g) su h that jf (z )j  jz j2 +jz j 1=2 for all z 2 C n f0g is ne essarily a polynomial of degree at most two. (u) There does not exist a fun tion f 2 H(C n f0g) that satis es jf (z )j  jz j  for all z 2 C n f0g, and for a xed  2 (0; 1). (v) The meromorphi fun tion ot z never assumes i and i and so, i are the Pi ard ex eptional values for ot z . (w) The meromorphi fun tion eiz = os z never assumes 0 and 2. (x) If f and g are two entire fun tions su h that ef + eg = 1, then f and g are onstant fun tions. p p e z pe z , then z = 0 is not a bran h point of f (z ). (y) If f (z ) = sin z p sin z (z) If f (z ) = p , then z = 0 is a removable singularity of f . z

7.50. Suppose that f 2 H(C n f0g) and satis es jf (z )j  ajz j2 + bjz j 2 for all z 2 C n f0g, and for some a; b > 0. If f is an odd fun tion of z , what form must the Laurent series of f have? How about when f is an even fun tion of z ? 7.51. Classify [type (and order where appli able)℄ the isolated singular points (in luding at the point at 1) of the following fun tions: (a) (b) ( ) (d)

sin4 z f (z ) = 8 z (1 + z ) os z f (z ) = z 1 8z 3 f (z ) = 1 4z 2 3 sin(z ) e1=z f (z ) = (z 1)z 3

 

1 1 + 2 z z (2 z ) 1 (f) f (z ) = s z z z2 (g) f (z ) = ; p; q 2 R n f0g (z ip)4 (z + q)3 5 z 10 (z 2 + 1)e1=(z 4) (h) f (z ) = (z )2 sin10 z (e) f (z ) = sin

7.8 Exer ises

335

7.52. Prove or disprove that f (z ) = 1= sin z is not meromorphi on the Riemann sphere. Does f (C ) = C n f0g? Does se (C ) = C n f0g? 7.53. Prove or disprove the following: The point z = 0 is the only singularity of the fun tion f (z ) = sin(1 1=z ) and z = 0 is a simple pole. 7.54. Give an example of a fun tion whi h is meromorphi in C without being meromorphi on the Riemann sphere. 7.55. Suppose that f is any one of (z 3 z )e1=z , z sin(1=z ), e1=z e 1=z and e1=z + e 1=z . Assuming the validity of Pi ard's theorem, de ide whether f (0 < jz j < 1) = C or not. 7.56. Let fj (j = 1; 2; : : : ; 6) be de ned as in Example 7.3. Classify the singularity of ea h of these fun tions at 1. 7.57. Find the general form of a fun tion in C 1 having the following singularities: (i) only simple pole at z0, (ii) one pole of order n at z0 , (iii) one pole of order n at in nity and a pole of order m at z0 .

7.58. Suppose z = a is a singularity of f (z ). Can z = a be a singularity for 1=f (z )? If so what will be the nature of the singularity. Dis uss in details with an example for ea h ase. 7.59. Suppose f and g are analyti in a neighborhood of z0 , f (z0 ) = 0 with multipli ity m, g(z0 ) = 0 with multipli ity n. What is the multipli ity of z0 as a zero of the omposite fun tion f Æ g? 7.60. Constru t a fun tion  whi h is analyti ex ept at the four distin t points zj ; j = 1; 2; 3; 4, where it has the following properties: (i) (ii) (iii) (iv)

simple pole at z1 , simple zeros at z2 ; z3 ; z4, simple pole at in nity, lim z 1 (z ) = 2. jzj!1

7.61. Find the onstant su h that 1

f (z ) = n n 1 + z + z +


+ z2 + z n z 1

an be extended to be analyti at z = 1, where n 2 N is xed.

336

Classi ation of Singularities

7.62. Prove that the fun tion f de ned by f (z ) = (ez 1)=(z (z 2)) has a removable singularity at z = 0 whereas it has a simple pole at z = 2. Prove also that it has an essential singularity at z = 1. 7.63. Prove that the fun tion f de ned by f (z ) = exp(z= sin z ) has a removable singularity at z = 0 whereas it has essential singularities at z = k, k 2 Z nf0g. 7.64. Suppose that f (z ) is analyti on C ex ept pole at P for a double n for jz j < 1: z = ei for some  2 [0; 2). Suppose that f (z ) = 1 a z n n=0 Is fang bounded? Does fang onverge? Justify your answer. How about if we repla e the word `double pole' by `simple pole'? 7.65. Let P f 2 H(
3 nf0g). Suppose that f has a simple pole at z = 2i and f (z ) = n0 an z n for jz j < 2. Is fan g bounded? Does fang onverge? Justify your answer. 7.66. Let f be a meromorphi fun tion in a domain . Then prove that neither the set of zeros nor the set of poles of f an have a limit point in unless f is identi ally zero in . 7.67. Assume that ot(z ) = Find a n for n  1.

P1

n=

n 1 an z in the annulus 0 < jz j < 1.

Chapter 8

Cal ulus of Residues

If f is analyti at a point z = a, then there is a neighborhood N of a inside whi h f is analyti . Let C be a positively oriented losed ontour

ontained R in N . Then the elebrated Cau hy theorem tells us that C f (z )dz = 0: If, however, f fails to be analyti at nitely many isolated singularities inside C , then the above argument fails; whi h means ea h of these singularities

ontribute a spe i ed value to the value of the integral. This motivates us to generalize the Cau hy theorem to fun tions whi h have isolated singularities. This generalization results in the Residue theorem. This result is one of the most important and often used, tools that applied s ientists need, from the theory of omplex fun tions. In Se tion 8.1, we are on erned with the notion of what we all \the residue at an isolated singularity" and dis uss the on ept of residue in detail and illustrate it with a number of examples for nding the residue of a given fun tion at an isolated singularity. In Se tion 8.2, we dis uss the notion of residue at the point at 1. The main result in Se tion 8.3 is Cau hy's residue theorem whi h states that the integral of an analyti fun tion f around a simple losed ontour C equals 2i times the sum of the residues of f at the isolated singular points inside C . Using this important theorem, we shall then develop and illustrate some of the basi methods employed in omplex integration for evaluating omplex line integrals. The residue theorem extends Cau hy's theorem by allowing for a nite number of isolated singularities inside the ontour of integration. Formulae enabling us to do this in lude an alternate proof of the so- alled generalized Cau hy integral formula. When there are no singularities, the residue theorem simply redu es to Cau hy's theorem. The materials presented in Se tion 8.1-8.3, provide a good training ground for the evaluation of omplex integration in the next hapter. In Se tion 8.4, we derive the argument prin iple whi h is another of the most important appli ations of the Cau hy residue theorem. Also, we dis uss several of its onsequen es,

338

Cal ulus of Residues

for example, lo ating the zeros of an analyti fun tion. The argument prin iple provide a tool in the form of Rou he's theorem to see how the number of zeros of analyti fun tions remains onstant under small perturbations. We dis uss a version of Rou he's theorem in Se tion 8.5.

8.1 Residue at a Finite Point We re all that if f has an isolated singularity at z0 , then the residue of f (z ) at z0 is Z 1 (8.1) Res [f (z ); z0℄ := a 1 = f (z ) dz; 2i C where C is any ir le entered at z0 and lying inside a disk about z0 . For instan e, onsider f (z ) = ot z . Sin e sin z = 0 () z = k (k 2 Z), and z k 1 lim (z k) ot z = lim
lim os z = zlim
lim os z = 1; !k os z z!k z!k z!k sin z z!k f has simple poles at z = k, k 2 Z. Suppose we hoose z0 = 0. Then 1 1 X f (z ) = + an z n (0 < jz j < ) z n=0 and, in view of (8.1), this gives, Z

C

ot z dz = 2i:

Here C ould be any ir le around zero in 0 < jz j < . If z0 = k, then 1 X 1 f (z ) = + a (z k)n ; jkj < jz kj < (jkj + 1): z k n=0 n
8.2. Example. We know that the oeÆ ient of z k in (1 + z )n is nk . So, we may write 

n k



= oeÆ ient of z

1

in (1 + z )n =z k+1

= Res [f (z ); 0℄; f (z ) = (1 + z )n =z k+1; Z 1 (1 + z )n = dz; 2i C z k+1 where C is any simple losed ontour en losing the origin (note that f is analyti for all z 2 C n f0g). Similarly, we also see that 

n k



= oeÆ ient of z k in (1 + 1=z )n = onstant term in z k (1 + 1=z )n

339

8.1 Residue at a Finite Point

and therefore, we have n  2 X n k=0

k

= = = =





(1 + z )n k [z (1 + 1=z )n℄ z k+1 Z 1 (1 + z )2n F (z ) dz; F (z ) = n+1 ; 2i C z Res [F (z ); 0℄

oeÆ ient of z n in (1 + z )2n   2n : n

= oeÆ ient of z

1

in



We wish to point out that the most onvenient way to nd the residue is dire tly from the Laurent expansion (if it were already available). We want a tually to develop te hniques for al ulating the residue of a fun tion f without having to nd its Laurent expansion. However if z0 is an essential singularity of f , then, in most of ases, the Laurent expansion of f about z0 will be needed in order to nd the residue at z0 . For instan e, at the essential singular point z = 0 of sin(1=z ) we have sin

 

1 1 = z z

1 1 + 3! z 3

Thus, we see that Res [sin(1=z ); 0℄ = 1; i.e.




; jz j > 0:

Z

Similarly, we obtain Res [z 3 sin(1=z 2); 0℄ = 0; i.e.

C

sin(1=z ) dz = 2i:

Z

C

z 3 sin(1=z 2) dz = 0;

where C is any ir le around the origin in the pun tured plane, C n f0g. The same idea may be used to derive examples of this type (using Taylor's series expansion of the exponential fun tion): ia3 (a) Res [z 4eiaz ; 0℄ = , a 2 C. 3! (b) If a 2 C and n 2 N 0 = N [ f0g, then Res [eaz =z n+1; 0℄ =

an and n!

Z

eaz 2ian dz = : n +1 n! jzj=r z

If a is real and r = 1, then we let z = ei and easily obtain Z 2

0

2an ea os  ei(a sin  n) d = n!

340

Cal ulus of Residues

whi h, by equating real and imaginary parts, gives Z 2

0

and

Z 2

0

Z

ea os  sin(n a sin ) d = 0

ea os  os(n a sin ) d = 

2an : n!

n 1 1 ( ) e1=z dz = Res [f (z ); 0℄ = 01 ifif nn 6= = 1 ; where C is any ir le 2i C around zero in the pun tured plane C n f0g. (d) Res [(z 2) 4 sin(z 2); 2℄ = 1=6! and Res [sin(1=(z 2)); 2℄ = 1.

8.3. Example. De ne f (z ) =

sin(z 2 ) ; a 6= 0: z 2 (z a)

Then the singularities of f are at z = 0 and z = a. Sin e limz!0 f (z ) = 1=a; z = 0 is a removable singularity of f . Thus, Res [f (z ); 0℄ = 0: As f admits a Laurent series expansion about zero in 0 < jz j < jaj, we on lude that, for any r with r < jaj, Z sin(z 2 ) dz = 0:  jzj=r z 2(z a)

8.4. Remark. If f is analyti at z0 , then Res [f (z ); z0℄ = 0: However, this simple fa t holds for other situations as well. For instan e if fn (z ) = bn (z z0) n ; n = 1; 2; : : : ; and b1 6= 0; then Res [fn (z ); z0 ℄ = 0 for n  2: Also, we observe that Res [f1 (z ); z0℄ = b1 but Res [f12 (z ); z0℄ 6= b21 :  In view of Laurent's expansion and Cau hy's prin iple of deformation of ontour, the following result is almost trivial.

8.5. Theorem. If f has a removable singularity at z0 , then we have Res [f (z ); z0℄ = 0: In parti ular, if C is a simple losed ontour ontaining only removable singularities at zk (k = 1; 2; : : : ; n) inside C , then R C f (z ) dz = 0: For instan e, using Theorem 8.5, we have Res [( os z

1)2 =z 2; 0℄ = 0 and Res [z 2= sin2 z ; 0℄ = 0:

8.1 Residue at a Finite Point

341

Note that the fun tions involved here are even. More generally, we have

8.6. Theorem. If f has an isolated singularity at z0 and if f is even z0, i.e. f (z z0 ) = f ( (z z0 )), then Res [f (z ); z0℄ = 0:

in z

Proof. Suppose that f is even in z z0 . Then the Laurent series expansion around z0 annot have odd powers of z z0. Hen e, the assertion follows. Consider the fun tion f de ned by 1 f (z ) = 2 2 2 (a 2 R n f0g): z (z a ) Then the point z = 0 is a double pole and z = a is a simple pole. Observe that f (z ) = f ( z ) and hen e, by Theorem 8.6, we have Res [f (z ); 0℄ = 0: This shows that the residue an be zero, even though f has a non-removable singularity at z = 0 (see Remark 8.4). 2 Similarly, f (z ) = e1=z has an essential singularity at z = 0. Again, as f is even, we have Res [f (z ); 0℄ = 0: By Theorem 8.6, we easily obtain the following: (i) Res [(sin z ) 2 ; k℄ = 0; k = 0; 1; 2; : : : (sin2 (z k) = sin2 z ) (ii) Res [1= sin(z 2 ); 0℄ = 0 (iii) Res [z 3 sin(1=z ); 0℄ = 0 (iv) Res [(sin z z )=z 3; 0℄ = 0 2 (v) Res [e 1=z os(1=z ); 0℄ = 0 (vi) Res [1=[2 + z 2 2 osh z ℄; 0℄ = 0. For evaluating residues in on rete examples, the following theorem is very useful.

8.7. Theorem. If f has a pole of order n at z0 , then 1 dn 1 n Res [f (z ); z0℄ = zlim (8.8) !z0 (n 1)! dz n 1 ((z z0 ) f (z )): The oeÆ ient of (z

z0 ) k in the Laurent expansion is

dn k 1 a k= lim ((z z0 )n f (z )); k = 1; 2; : : : ; n: (n k)! z!z0 dz n k (Case k = n means that we have only to look at (z z0 )n f (z ).) Proof. Suppose that f has a pole of order n at z0 . Then, we have 1 X a n a 1 f (z ) = +


+ + a (z z0 )k (0 < jz z0 j < Æ) (z z0 )n z z0 k=0 k

342

Cal ulus of Residues

for some Æ > 0, where a n 6= 0. Further, limz!z0 (z is non-zero. Note that for z 6= z0 ,

z0 )n f (z ) exists and 1 X

z0 )n 1 +

(z z0)n f (z ) = a n +a n+1(z z0)+


+a 1 (z

k=0

ak (z z0)k+n

and the formula (8.8) follows at on e if we dierentiate the last equation (n 1)-times and then allow z ! z0 . Alternatively, de ne g by (

g(z ) =

(z z0 )n f (z ) for 0 < jz z0 j < Æ n f (z ) for z = z lim ( z z ) 0 0 z !z 0

so that f (z ) = (z z0 ) n g(z ); 0 < jz z0 j < Æ; where g is analyti in jz z0 j < Æ with g(z0 ) = a n 6= 0. This observation, a

ording to the Cau hy integral formula applied to C = fz : jz z0 j = rg for 0 < r < Æ, implies Z

1 f (z ) dz 2i C Z 1 g(z ) = dz 2i C (z z0 )n 1 = g(n 1)(z0 ) (n 1)!

Res [f (z ); z0℄ =

and the result follows at on e. By a similar argument, we have the se ond part. Theorem 8.7 is useful in the ase of rational fun tions. For instan e,

onsider the fun tion z (z 2) f (z ) = ; (z + 4)2 (z 1)2 whi h has a double pole at z = 1. Using (8.8), we obtain

d Res [f (z ); 1℄ = dz





z (z 2) 2 = : 2 (z + 4) z=1 125

At the double pole z = 4, we have Res [f (z ); 4℄ =

d dz





z (z 2) (z 1)2 z=

Similarly, we see that Res [(z 3 + 7)(z

2)

3 ; 2℄ =

4

=

d dz





1 2 (z 1) z=

1 d2 3 (z + 7) = 6: 2 2! dz z=2

4

=

2 : 125

343

8.1 Residue at a Finite Point

8.9. Example. Consider the fun tion f (z ) = e2z = osh z: Sin e

os z = 0 i z = (2k + 1)=2; k 2 Z, and os(iz ) = osh z; we have

() z = (2k +2 1)i : Thus, f has a simple pole at z = (2k + 1)i=2 (k 2 Z). From this we see that if C = fz : jz i=2j = 1=2g, then

osh z = 0

Z

C

f (z ) dz = 2i Res









e2z i ei 2iei ; = 2i = = 2ei:

osh z 2  sinh(i=2) i

Inside the ontour : jz i=2j = 1, f has a singularity only at z = i=2. Therefore, if C is any ir le around i=2 lying ompletely inside then we R have C f (z ) dz = 2ei : 

8.10. Example. De ne (1 + z 2 )n+k : z 2n+1 Then f has a pole of order 2n + 1 at z = 0 and

f (z ) =

Res [f (z ); 0℄ = oeÆ ient of z 1 in f (z ) = oeÆ ient of z 2n in (1 + z 2 )n+k = oeÆ ient of z n in (1 + z )n+k   n+k = : n On the other hand, by Theorem 8.6, we have Res [zf (z ); 0℄ = 0.



8.11. Example. Consider f (z ) =

(z 3

1 : 1)(z + 1)2

Then f has a double pole at 1 and simple poles at 1; !; !2, where ! is a

ube root of unity. We easily see that Res [f (z ); 1℄ = z! lim1





d 1 3z 2 3 = z! lim1 3 = : 3 dz z 1 (z 1)2 4

If a is any one of the ube roots of unity, then we have Res [f (z ); a℄ = zlim !a

(z a) (z + 1) z3 1

2

=

1 1 = 3a2(1 + a)2 3(2 + a + a2 )

344

Cal ulus of Residues

so that Res [f (z ); 1℄ = 1=12, Res [f (z ); !℄ = Res [f (z ); !2℄ = 1=3.



From Theorem 8.7, we have a simple result for omputing residues: \if f has a simple pole at z = z0 , then we have Res [f (z ); z0 ℄ = lim (z z0 )f (z ):" z!z0

From this observation, we also obtain another result whi h is also extremely useful in pra ti e.

8.12. Theorem. If f has a simple pole at z = z0 and if h is analyti at z0 with h(z0 ) 6= 0, then Res [f (z )h(z ); z0 ℄ = h(z0 )Res [f (z ); z0 ℄: Proof. Observe that lim (z

z!z0

z0 )f (z )h(z ) = h(z0 ) zlim !z (z z0 )f (z ) 0

and the result follows immediately. Re all that, f has simple pole at z0 i g(z ) = 1=f (z ) has simple zero at z0 : Thus, in this ase (sin e g(z0 ) = 0 and g0 (z0 ) 6= 0), Res [f (z ); z0℄ = zlim !z (z 0

z z0 1 = 0 : g (z0 ) 0 g (z )

z0 )f (z ) = zlim !z

Hen e we have

8.13. Theorem. Suppose  is analyti at z0 with (z0 ) 6= 0 and g has a simple zero at z0 . Then Res [(z )=g (z ); z0 ℄ = (z0 )=g 0 (z0 ): In parti ular, Res [1=g(z ); z0℄ = 1=g0(z0 ): Consider the fun tion

(z ) f (z ) = n n (a 6= 0, n  1), a +z where  is analyti at z0 su h that (z0 ) 6= 0 for ea h z0 2 C satisfying z0n + an = 0. Then, using Theorem 8.13 (sin e f has simple poles), we have Res [f (z ); z0℄ =

(z0 ) z0 (z0 ) z (z ) = = 0 n0 : n 1 n nz na nz0 0

In parti ular if (z ) = 1, then

zk nan where zk (k = 1; 2; : : : ; n) is the simple pole of 1=(z n + an). If (z ) = z n 1, then  n 1  z 1 Res n n ; zk = ; a 6= 0: z +a n Res [(z n + an ) 1 ; zk ℄ =

345

8.1 Residue at a Finite Point

As the order of poles in reases, the formula for the residue be omes more ompli ated. However, at a double pole, we have the following result whi h an be proved easily.

8.14. Theorem. Suppose  is analyti at z0 with (z0 ) 6= 0, g has a pole of order two at z0 and h has a zero of order two at z0 . Then we have (i) Res [(z )g(z ); z0 ℄ = 0 (z0 )Res [(z z0 )g(z ); z0 ℄ + (z0 )Res [g(z ); z0℄   (z ) 60 (z )h00 (z ) 2(z )h000 (z ) (ii) Res ; z0 = : h(z ) 3[h00 (z )℄2 z=z0

Proof. By hypothesis, there exist r1 , r2 su h that (z ) = a0 + a1 (z z0 ) +


for jz z0 j < r1 ; b 2 b 1 + + b + b (z z0 ) +


g (z ) = (z z0 )2 z z0 0 1 for 0 < jz z0 j < r2 , respe tively. The oeÆ ient of (z z0) is a0 b 1 + a1 b 2 and therefore Res [(z )g(z ); z0 ℄ = a0 b 1 + a1 b 2 = (z0 )Res [g(z ); z0 ℄ + 0 (z0 )Res [(z

1

in (z )g(z )

z0 )g(z ); z0℄

and the se ond part follows similarly.

8.15. Example. Consider f (z ) = (z )=g(z ), with (z ) = 1 + z and g(z ) = os z 1: Then the singularities of f are given by g(z ) = 0. Sin e

os z = 1 () z = 2k (k 2 Z), the singularities o

ur at z0 = 2k, k 2 Z. Further, we note that g0 (z ) = sin z and g00 (z ) = os z and, sin e sin z = 0 () z = n (n 2 Z), we see that g0 (2k) = 0 and g00 (2k) 6= 0 for k 2 Z:

Consequently, g has a double zero at z0 = 2k for k 2 Z. In other words, for ea h su h k, f has a double pole at z0 = 2k. Sin e  and g onsidered above are analyti at z0 with g(z0 ) = 0 = g0 (z0 ) and g00 (z0 ) 6= 0; we ompute that 60 (z )g00 (z ) 2(z )g000 (z ) 2 Res [f (z ); z0 ℄ = = = 2: 00 2 3[g (z )℄

os z0 z=z0  Similarly, Res ez (z



z0 ) 2 ; z0 = ez0 :



8.16. Example. As an immediate appli ation of Theorem 8.13, we obtain (a) to (g) in the following:

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Cal ulus of Residues

(a) Sin e sin z has simple zeros at z = k, k 2 Z, Res (b) ( ) (d) (e)

h os z

sin z

i

; k = Res

h  os z

sin z

i

;k =

os k = 1:

os k

Similarly, we have Res [ osh z ot z ; k℄ = osh k: Res [ s z ; k℄ = Res [ s z ; k℄ = ( 1)k ; k 2 Z: ek Res [ez = sin z ; k℄ = ; k 2 Z:

os k   sinh((2k + 1)i=2) sinh z ; (2k + 1)i=2 = = 1; For k 2 Z, Res

osh z sinh((2k + 1)i=2) where osh z has simple zeros at z = (2k + 1)i=2.   ez ez0 0 For z0 6= z0 , Res ; z = : 0 (z z0 )(z z00 ) z0 z00

e1=z . 1 z (g) For the fun tion f (z ) =  ot z=z 2, we have (using Theorem 8.7) Res [f (z ); k℄ = 1=k2 for k 2 Z nf0g and Res [f (z ); 0℄ = 2 =3; be ause f has a pole of order 3 at z = 0 and a simple pole at k, k 2 Z nf0g.  (f) Res [f (z ); 1℄ = e and Res [f (z ); 0℄ = e 1, where f (z ) =

8.17. Example. Consider f (z ) = (1 + z + z 2 + z 3 ) 1 : As 1 z 4 = (1 z )(1 + z + z 2 + z 3 );

f has a simple pole at k = eki=2 ; k = 1; 2; 3. So (see Theorem 8.13),



1 z Res [f (z ); k ℄ = = 4z 3 z= k

z (1 z )

(1 k ) = k ; k = 1; 2; 3: 4 4 z= k

A simpli ation gives Res [f (z ); eki=2 ℄ =





eki=2 (1 eki=2 ) e3ki=4 k =i sin : 4 2 4



8.18. Example. Consider the fun tion f (z ) = (z sin z ) 1: Then f has a double pole at z = 0 and a simple pole at z = k, k 2 Z nf0g. Note that f is even. By Theorem 8.6, we immediately have Res [f (z ); 0℄ = 0: Sin e the nearest singularity of f to zero is at , we obtain Z

dz =0 z sin z C for any ir le C entered at zero and radius less than .

347

8.1 Residue at a Finite Point

Similarly, we see that the fun tion f (z ) = (z 2 sin z ) at z = 0 and simple pole at z = k, k 2 Z nf0g. Sin e 

1 f (z ) = 3 1 z we obtain



 2 z

z4 +


5!

3!



 1

=

1 11 + + z 3 3! z

1

has a triple pole




for jz j > 0;

Z

1 1 dz i Res 2 ; 0 = ; i.e. = : 2 z sin z 6 jzj=r n =) zlim !z0 g(z ) = 0 (z ) (m) (z0 ) (ii) m = n =) zlim !z0 g(z ) = g(n) (z0 ) (z ) (iii) m < n =) zlim !z0 g(z ) = 1 from whi h the required on lusions follow.

z0j < Æ for some

8.20. Example. Take (z ) = sin z and g(z ) = (1 ez )2 . Then (0) = 0, 0 (0) = 1, g(0) = g0 (0) = 0 and g00 (0) = 2. This implies that the fun tion f de ned by f (z ) = (z )=g(z ) has a simple pole at the origin. z sin z Therefore, Res [f (z ); 0℄ = zlim  !0 (1 ez )2 = 1: 8.21. Example. We wish to onstru t a fun tion f (z ) whi h has the following properties: (i) The only singularities of f (z ) in C 1 are poles of order 1 and 2 at z = 1 and z = 1, respe tively. (ii) f (0) = 0 = f ( 1=2) and Res [f (z ); 1℄ = 1 = Res [f (z ); 1℄.

348

Cal ulus of Residues

To do this, by hypotheses, we observe that the prin ipal part of f (z ) has the form 1 1 b + + P (z ) = z 1 z + 1 (z + 1)2 so that F (z ) = f (z ) P (z ) extends to be analyti in C . As f has no other singularities in C 1 , f is analyti at 1 and so, F (z ) is analyti at 1. Thus, by Liouville's theorem, F (z ) is onstant. Hen e, f (z ) = P (z ) + a for some

onstant a. Finally, as f (0) = 0 = f ( 1=2), it follows that

a + b = 0 and 4b + a = 4=3: Solving these equations imply that b = a = 4=9.



The nal example of this se tion relates to nding the residue of a bran h of a multi-valued fun tion.

8.22. Example. Consider the fun tion f (z ) = (1 + z 2 ) 2 Log (1 + z ): Note that Log (1 + z ) is analyti in C n( 2 at z = i and i. Hen e,

1; 1℄ and f has a pole of order

d Res [f (z ); i℄ = lim [(z + i)2 f (z )℄ z! i dz   d Log (z + 1) = zlim ! i dz (z i)2    1 1 2 = lim
+ Log (1 + z )
z! i 1 + z (z i)2 (z i)3   1 =
1 + Log (1 i)
( 22i)3 1 i ( 2i)2   h p  i  2 1+i 1 =
4 + ln 2 i 4
8i 2 " p #   1  1 ln 2 = + +i + : 8 16 8 4 Res [f (z ); i℄ may similarly be omputed.



8.2 Residue at the Point at In nity First onsider z = 1=w and re all the geometri aspe t of this transformation. If we set z = Me i , then we have w = M 1 ei : This shows that as z des ribes the ir le jz j = M in the z -plane in the lo kwise dire tion, w des ribes the ir le jwj = 1=M in the w-plane in the anti- lo kwise dire tion.

349

8.2 Residue at the Point at In nity

Thus, the point z0 = e i ( > M ) outside the ir le jz j = M orresponds to a point w0 =  1 ei inside the ir le jwj = 1=M . Let f be analyti in a deleted neighborhood of the point at in nity. Then f admits a Laurent series expansion of the form Z 1 X 1 f (z ) f (z ) = ak z k (R < jz j < 1); ak = k+1 dz; 2 i z + C k= 1 with R > 0 suÆ iently large. Here C + denotes the positively oriented ir le jz j = M , where M > R and M is suÆ iently large so that a nite number of singularities in C will be inside C0 , where int C0  int CR = fz : jz j < Rg. De ne C = fz : jz j = M > R; M is suÆ iently largeg where C is traversed in the lo kwise dire tion (so that the point at in nity is to the left of C as in the ase of a nite point). Put z = Me i . Then it follows that (sin e the termwise integration is permissible) Z Z 1 X f (z ) dz = ak z k dz C C k= 1 Z 2 1 X = iak M k+1 e i(k+1) d 0 k= 1 =

ia

1

Z 2

0

d = 2ia 1:

In view of this reasoning, it is natural to de ne Z 1 Res [f (z ); 1℄ = f (z ) dz = a 1 : 2i C In other words, Res [f (z ); 1℄ is the negative of the oeÆ ient of 1=z in the Laurent series expansion of f (z ) with enter at the point at in nity. One should be alerted that a 1 here is neither the residue of f at in nity nor the residue of f at z = 0. Further, we observe that for z = Me i with M = 1=R0, Z 1 Res [f (z ); 1℄ = f (z ) dz 2i C Z 1 2 f (Me i )iMe i d = 2i 0   Z 1 2 1 d(R0 ei ) f 0 i = 2i 0 R e (R0 ei )2   Z 1 1 dw = f 2i C w w2   f (1=w) = Res ; 0 ; w2

350

Cal ulus of Residues

where C = fw : jwj = 1=R0g is des ribed in the anti- lo kwise (positive) dire tion. Alternatively, if we repla e z by 1=z , we see that   1 a 1 a X X 1 1 k k 2 f = = (0 < jz j < 1=R) k+2 k z2 z z z k= 1 k= 1 so that Res



 



1 1 f ;0 = a z2 z

1 = oeÆ ient of z

1

in

 

1 1 f z2 z

and hen e, in both methods, we qui kly have (8.23)

Res [f (z ); 1℄ = Res



 



1 1 f ;0 : 2 z z

Here is an alternate proof of Liouville's Theorem (Theorem 6.55). Let f (z ) beP a bounded entire fun tion. Then f (z ) has a power series expansion n f (z ) = 1 n=0 an z for all z 2 C , so that g de ned by 1 X g(z ) = f (1=z ) = an z n n=0

is analyti for all jz j > 0. Sin e f (z ) is bounded in C , g(z ) is bounded in a deleted neighborhood of 0, and so g(z ) has a removable singularity at 0. Therefore, an = 0 for all n > 0. Thus, f (z ) = a0 is a onstant.

8.24. Example. Consider the fun tion f (z ) = 1 + z 1, z 6= 0. Then F (w) = f (1=w) = 1 + w (w 6= 0); and wlim !0 F (w) = 1: Thus, F (w) has a removable singularity at w = 0 and therefore, the point at in nity is a removable singularity of f (z ). Further, Res [f (z ); 1℄ = 1: From this we also observe that if f has a removable singularity at the point at in nity, then the residue of f at 1 may prove to be non-zero in ontrast to the ase when f has a removable singularity at a nite point. 

8.3 Residue Theorem The ee tiveness of the residue theorem depends, of ourse, on how ee tively we an evaluate residues at various singularities. However, aution must be exer ised to avoid rea hing a hasty on lusion based on appearan es. Having identi ed the type of singularities, we have to hoose a proper ontour. Most often the following theorem will be applied in the next hapter to evaluate dierent types of line integrals.

351

8.3 Residue Theorem

8.25. Theorem. (Cau hy's Residue Theorem) If f is analyti in a domain D ex ept for isolated singularities at a1 ; a2 ; : : : ; an , then, for any

losed ontour in D on whi h none of the points ak lie, we have Z

f (z ) dz = 2i

n X k=1

n( ; ak )Res [f (z ); ak ℄:

Proof. Sin e does not pass through any of aj 's, we an hoose numbers Æ1 ; Æ2 ; : : : ; Æn so small that (i) for every j = 1; 2; : : : ; n no two ir les j : jz aj j = Æj interse t, (ii) every ir le j (j = 1; 2; : : : ; n) lies inside . Sin e aj is an isolated singularity of f , f admits a Laurent series expansion of the form 1 X (8.26) f (z ) = a(nj) (z aj )n ; 0 < jz aj j  Æj ; n= 1 for ea h j = 1; 2; : : : ; n. We denote the prin ipal part of f (z ) at ea h of these isolated singularities by pj (z ) =

1 X

a(nj) (z aj )n :

1 Then for ea h j , fun tion pj is analyti on and outside the ir le j (see Theorems 4.117 and 4.139). Sin e aj lies inside and pj onverges uniformly on j , we have Z

n=

Z

dz (aj lies inside ) z aj

= 2ia(j1) n(aj ; ) = 2in(aj ; ) Res [f (z ); aj ℄; j = 1; 2; : : : ; n:

pj (z ) dz = a(j1)

If we subtra t the prin ipal parts p1 (z ); p2 (z ); : : : ; pn (z ) from f , it follows that the fun tion g de ned by the dieren e

g(z ) = f (z )

(8.27)

n X j =1

pj (z )

is analyti on D nfa1 ; a2 ; : : : ; an g. It follows that all the aj 's are removable singularities of g and, by Cau hy's theorem (see also Theorem 8.5), we have R g ( z ) dz = 0. Consequently, by (8.27),

Z

f (z ) dz =

n Z X

k=1

pk (z ) dz = 2i

n X

k=1

n(ak ; ) Res [f (z ); ak ℄:

352

Cal ulus of Residues

In most of the appli ations will be a simple losed ontour with positive orientation and hen e, in these ases (see Se tion 4.5), 

in the unbounded omponent of C n f g n( ; ak ) = 01 ifif aak is is k inside : Thus, the residue formula be omes elegant for simple losed ontours. Thus if  D is a simple losed ontour with positive orientation, then under the hypotheses of Theorem 8.25 we have the following simple form.

8.28. Theorem. (Residue Formula) Z

f (z ) dz = 2i

X

Res [f (z ); ak ℄:

Here the sum is taken over all ak 's inside .

The proof of this spe ial ase is trivial be ause for the simple losed

ontour in D, by Cau hy's prin iple of deformation of ontour, we have Z

f (z ) dz = 2i

n Z X k=1 k

f (z ) dz = 2i

n X k=1

Res [f (z ); ak ℄:

The general ase an also be proved in the same spirit. The Cau hy integral formula (see Theorem 4.63) an be onsidered as a spe ial ase of the residue theorem. Indeed, if f is analyti in D and a 2 D, then g de ned by f (z ) g (z ) = z a is analyti in D nfag and has the residue f (a) at the simple pole a, by Theorem 8.13. In fa t, the Cau hy integral formula for higher order derivatives

an also be dedu ed as a spe ial ase of Theorem 8.25.

8.29. Remark. In Theorem 8.25, f (z ) an have only a nite number of singularities, be ause otherwise singularities of f (z ) would have a limit point  (possibly at the point at in nity), and so  would not be an isolated singularity of f (z ), ontrary to our assumption.  8.30. Example. Let us evaluate Z Re z I = dz; z  jzj=1 For jz j = 1, we have Re z = z + z

I =

Z

1
=2

jj 6= 1:

so that

1 f (z ) dz; 2 jzj=1

353

8.3 Residue Theorem

where

8 > >
> if  6= 0: :  z  z To ompute the integral, we may either use the Cau hy integral formula or the residue theorem. Note that Res [f (z ); 0℄ = 0 if  = 0.Thus, I0 = 0. When jj > 1, the Cau hy integral formula gives that   Z 1 (z 2 + 1)=(z ) 2i 1 i I = dz = = : 2 jzj=1 z 2   When 0 < jj < 1, we see that 2 + 1 1 and Res [f (z ); ℄ = ; Res [f (z ); 0℄ =   and so, if 0 < jj < 1 then the Cau hy residue theorem gives 2i I = fRes [f (z ); 0℄ + Res [f (z ); ℄g = i: 2 Similarly, for jj 6= 1, we an easily see that 8 Z < 0 if  = 0 Im z dz = = if jj > 1  : jzj=1 z   if 0 < jj < 1: z2 + 1

8.31. Example. For jaj 6= R and C = fz : jz j = Rg, we wish to show that Z jdz j = 2R : I= jR2 jaj2 j C jz aj2 For a proof, we let z = Rei . Then dz = iz d, jdz j = Rd = R dz=iz and jz aj2 = (z a)(z a) = z 1(z a)(R2 az ) so that I takes the form Z R 1 I= f (z ) dz; f (z ) = : i C (z a)(R2 az ) For a = 0, the result is trivial. For a 6= 0, f has two simple poles at z = a and z = R2 =a. If one of them lies inside the ir le jz j = R, then the other lies outside. Finally, the result follows from the residue theorem.  Combining the residue at the point at in nity and Theorem 8.25, we have the\Residue Formula for the Extended Complex Plane" as follows:

C

8.32. Theorem. (Extended Residue Formula) Let f be analyti in ex ept for isolated singularities at a1 ; a2 ; : : : ; an . Then we have

354

Cal ulus of Residues

(i) the sum of all residues (in luding the residue at in nity) of f is zero. Equivalently (by Theorem 8.28 and equation (8.23)), we write Res



 



n X 1 1 f ; 0 = Res [f (z ); ak ℄: z2 z k=1

(ii) if is a simple losed ontour in C su h that all ak 's are interior to

, then     Z 1 1 f (z ) dz = 2iRes 2 f ;0 : z z

The extended residue formula an be used to give another simple proof of Liouville's theorem (see Theorem 6.55).

8.33. Proof of Liouville's Theorem (see Theorem 6.55). Suppose that f is entire and bounded in C . Choose two distin t points, say 0 and a in C , and onsider the fun tion F (z ) =

f (z ) : z (z a)

Then F (z ) has singularities at z = 0; a and possibly at the point at in nity. Sin e limjzj!1 zF (z ) = 0 (as f is bounded in C ), Res [F (z ); 1℄ = 0. Note also that f (a) f (0) Res [F (z ); 0℄ = ; and Res [F (z ); a℄ = : a a In view of Theorem 8.32, we have

f (0) f (a) + =0 a a whi h proves f (a) = f (0) for ea h a 2 C . Hen e, f must be onstant. On most o

asions, al ulation of the residues at many isolated singularities of the integrand is quite diÆ ult. In this situation, we an use Theorem 8.32 to evaluate ertain ontour integrals. Now, we demonstrate this advantageous situation by a number of examples. Consider Res [F (z ); 0℄ + Res [F (z ); a℄ + Res [F (z ); 1℄ =

f (z ) =

(z 2

z 21 : 1)4 (z 4 2)3

Then we see that all the singularities of f lie inside the ir le jz j = 3. Note that f has a simple pole at in nity with Res [f (z ); 1℄ = 4. Consequently, Z

jzj=3

f (z ) dz = 2i Res [f (z ); 1℄ = 8i:

355

8.4 Number of Zeros and Poles

8.34. Example. Let us evaluate the integral Z

1 z 2n+3m 1 f (z ) dz; f (z ) = 2 ; I= 2i jzj=R (z + a)n (z 3 + b)m p

where a; b 2 C n f0g; R > maxf jaj; jbj1=3 g and m and n are xed positive integers. First we note that f has poles of order n at the zeros of z 2 + a, say z1 and z2, and poles of order m at the zeros of z 3 + b, say z3; z4 and z5 . By the onditions on a; b and R, these poles lie inside the ir le jz j = R. Therefore, by the residue theorem,

I=

5 X

j =1

Res [f (z ); zj ℄:

As the al ulation of residues at these poles is quite diÆ ult, to omplete the solution, we make use of Theorem 8.32. A

ording to this, I + Res [f (z ); 1℄ = 0. Sin e 

 







1 1 1=z Res [f (z ); 1℄ = Res 2 f ; 0 = Res ; 0 = 1; z z (1 + az 2 )n (1 + bz 3)m



this gives I = 1.

The o

asional short ut method used in the above two examples should not be missed.

8.4 Number of Zeros and Poles Before we onsider some useful onsequen es of Theorem 8.32, let us de ne the hange in arg f (z ) as z goes around C . This is denoted by
C arg f (z ). Let f be analyti inside and on a simple losed ontour C ex ept possibly for poles inside C and f (z ) 6= 0 on C . As z des ribes C on e in the positive dire tion in the z -plane, the image point w = f (z ) des ribes a losed urve = f (C ) in the w-plane in a parti ular dire tion whi h determines the orientation of the image urve . Sin e f (z ) 6= 0 on C , never passes through the origin in the w-plane. Let w0 be an arbitrary xed point on and let 0 be a value of the argument of w0 . Then, let arg w run

ontinuously from 0 , as the point begins at w0 and traverses on e in the dire tion of orientation assigned to it by w = f (z ). If w returns to the starting point w0 , then arg w assumes a parti ular value of arg w0 whi h we denote by 1 . We de ne
C arg f (z ) = 1

0 :

Note that the dieren e 1 0 is independent of the hoi e of the starting point w0 . Further, we also note that the dieren e 1 0 is an integral

356

Cal ulus of Residues

multiple of 2 and the integer (1 0 )=2, denoted by n( ; 0), is the winding number of around the origin in w-plane as z des ribes C on e in the positive dire tion (see 4.5). If n( ; 0) = 1, then winds around the origin on e in the lo kwise dire tion. If does not en lose the origin, then it is obvious that n( ; 0) = 0. For instan e, onsider the spe ial ase when

f (z ) = z n (n 2 N ) where C = fz : z = ei ; 0    2g. The fun tion f has a zero of order n at z = 0. Then, = fw : w = ein ; 0    2g whi h is the ir le traversed n-times and so, we may de ompose as = [nj=1 j ;



j

= w: w

= ein ;

2(j

n

1)





2j : n

Note that ea h j is des ribed in the positive dire tion and has the origin in its interior. Hen e, we nd that Z Z 1 dw 1 f 0 (z ) 1 = dz: n( ; 0) =
C arg f (z ) = n = 2 2i w 2i C f (z )

8.35. Theorem. If f has a zero of order m at z = a,  0  f (z ) Res ; a = m: f (z ) Proof. If f has a zero of order m at z = a then f (z ) = (z a)m g(z ), where g is analyti at z = a and g(a) 6= 0. It follows that in a deleted neighborhood of a f 0 (z ) (z a)m 1 [mg(z ) + (z a)g0 (z )℄ = f (z ) (z a)m g(z ) 0 m g (z ) = + ; 0 < jz aj < Æ; for some Æ: z a g(z ) As g0 =g is analyti at z = a, the on lusion follows. A proof analogous to that of the above theorem shows that if f has a pole of order n at z = b, then near z = b we have f 0 (z ) n = + an analyti fun tion at z = b. f (z ) z b Hen e at ea h pole z = b of f , f 0 =f has a simple pole at z = b with residue equal to n. One of the important appli ations of Cau hy's residue theorem on erns the number of zeros and poles of meromorphi fun tions.

357

8.4 Number of Zeros and Poles

8.36. Theorem. (Argument Prin iple) Let f be meromorphi in a domain D  C and have only nitely many zeros and poles in D. If C is a simple losed ontour in D su h that no zeros or poles of f lie on C , then Z 1 f 0 (z ) dz = N 2i C f (z )

(8.37)

P

where N and P denote, respe tively, the number of zeros and poles of f inside C , ea h ounted a

ording to their order.

Proof. De ne F (z ) = f 0 (z )=f (z ): Then, the only possible singularities of F inside C are the zeros and poles of f . Therefore13 Z X 1 (8.38) F (z ) dz = Res [F (z ); C ℄: 2i C If aj is a zero of order nj of f and if bk is a pole of order pk of f , then (see Theorem 8.35) it follows that  0   0  f (z ) f (z ) Res ; aj = nj and Res ; bk = pk : f (z ) f (z ) Thus (8.38) be omes Z X f 0 (z ) 1 dz = nj 2i C f (z ) j

X

k

pk = N

P:

8.39. Remark. If, in addition,  is analyti on D, then under the hypotheses of Theorem 8.36 we easily get that Z X X 1 f 0 (z ) (z ) dz = nj (aj ) pk (bk ) 2i C f (z ) j k where aj and bk are the zeros of order nj and the poles of order pk for f , respe tively.  Why is Theorem 8.36 known as an argument prin iple? Let us now restate Theorem 8.36 in terms of the properties of the logarithmi fun tion log f (z ). For this, under the hypotheses of Theorem 8.36, onsider the transformation w = log f (z ): Note that f is analyti on C and f (z ) 6= 0 on C . Hen e, f (z ) 6= 0 in a neighborhood of C . For any analyti bran h log f (z ) of logarithm of f (z ), we have f 0 (z ) d (log f (z )) = dz f (z ) P

use Res [f (z ); D℄ to denote the sum of the residues of f at the singularities aj , where aj belongs to the interior of D. Sometimes, we denote this simply by P Res [f (z ); C ℄ where C is a given losed ontour. 13 We

358

Cal ulus of Residues

and therefore, Z Z f 0 (z ) 1 1 1 (8.40) dz = d(log f (z )) =
log f (z ): 2i C f (z ) 2i C 2i C We refer to this integral as the logarithmi integral of f (z ) along C . Here
C log f (z ) denotes the in rease in log f (z ) when C is traversed on e in the positive dire tion, and we say that the logarithmi integral measures the

hange of log f (z ) along the ontour C . Now, we express log f (z ) = ln jf (z )j + i arg f (z )

where ln jf (z )j is single-valued and hen e,
C ln jf (z )j = 0; as ln jf (z )j returns to its original value when C is traversed. This observation implies that
C log f (z ) = i
C arg f (z ): Therefore, (8.40) yields Z 1 f 0 (z ) 1 dz =
C arg f (z ) 2i C f (z ) 2 where
C arg f (z ) is referred to as the in rease in the argument of f (z ) along C . Thus, the argument prin iple an be restated as follows.

8.41. Corollary. Under the hypotheses of Theorem 8.36, we have 1
arg f (z ) = N P: 2 C 8.42. Corollary. If f is analyti inside and on a simple losed

ontour C and f (z ) 6= 0 on C , then (1=2 )
C arg f (z ) = N: Z 1 f 0 (z ) 8.43. Example. Consider the integral I = dz; where 2i C f (z ) C = fz : jz 1 ij = 2g and z 2 z 2 z2 9 ; (ii) f (z ) = ; (iii) f ( z ) = : z (z 1) z (z 1)2 z2 + 1 Then, we note that 0; 1; 2 and i are inside C and 3; 3 and i are outside C . Thus for (i), I = 1 2 = 1; for (ii), I = 1 3 = 2 and for the last

ase, I = 0 1 = 1.  (i) f (z ) =

8.44. Example. Note that os z = 0 () z = (k + 12 ); k 2 Z; the only zero of os z inside the unit ir le about =2 is at z = =2. Therefore, by the argument prin iple, we have Z Z f 0 (z ) tan z dz = dz = 2i; f (z ) = os z: jz =2j=1 jz =2j=1 f (z )

359

8.4 Number of Zeros and Poles y

v f (z) = z2 − 1

1+i 0

1

2

−1 + 2i

x

3

−1

u

1−i −1 − 2i

Figure 8.1: Mapping w = z 2

1.

From this integral, we also note that Res [tan z ; =2℄ = 1.



8.45. Examples. Let f (z ) = tanh z and C = fz : jz j = 3g. Sin e

osh z = 0 () z = i(k + 12 ) (k 2 Z); we see that osh z has 2 zeros at z = i=2 in the interior of C . Further, we note that Z d ( osh z ) = sinh z; and tanh z dz = 4i; dz jzj=3 by the argument prin iple. Similarly we easily obtain the following: (i) (ii) (iii) (iv)

Z

jz 1j=2

Z Z Z

tanh z dz = 4i

jzj=3 ez jzj=

ez

1

dz = 6i

tan z dz = 12i

dz f 0 (z ) = dz = 2i; with f (z ) = tan(z=2): jzj=1 sin z jzj=1 f (z ) Z



8.46. Example. Consider f (z ) = z 2 1, C = fz : jz 1j = 1g: Using the image points, it is easy to sket h the values whi h w = f (z ) assumes on C : z : 2 1+i 0

#

#

#

w : 3 1 + 2i 1: Sin e f (z ) = f (z ) and sin e the ontour C in the z -plane is symmetri about x-axis, the image urve in the w-plane must be symmetri about u-axis. From Figure 8.1, we easily dedu e that 1
arg f (z ) = 1 2 C

360

Cal ulus of Residues y iR

C Reiθ

θ O

R

x

Figure 8.2: The urve C = fz : jz j = R; 0  arg z  =2g.

as C is traversed on e in the positive dire tion starting from a point and ending at the same point. Further, we also note that f has two simple zeros at z = 1 and z = 1. Only the zero at z = 1 lies inside C . Thus, N = 1. Similarly, it is easy to see that if f (z ) = z n and C is a losed ontour (or ir le) en losing the origin, then 21
C arg f (z ) = n; whi h agrees with the fa t that in the interior of C , the fun tion has a pole of order n at z = 0. 

8.47. Example. Consider f (z ) = z 4 + z 3 + 1: If z = x with x > 0, then (8.48) f (z ) = x4 + x3 + 1 and if z = x with x > 0, then  3 x (x 1) + 1 if x  1; (8.49) f (z ) = x4 x3 + 1 = 4 x + (1 + x + x2 )(1 x) if 0 < x  1: Thus, (8.48) and (8.49) imply that f (z ) has no real roots. Further if z = iy with y-real, then f (z ) = y4 + 1 iy3 whi h shows that f has no purely imaginary roots. On the other hand if C = fz : z = Rei ; 0    =2g, i.e. C is taken round the part of the rst quadrant bounded by jz j = R for suÆ iently large R (see Figure 8.2), then   1 1 4 i 4  3 i 3  4 4 i f (z ) = R e + R e + 1 = R e 1 + i + 4 4i : Re Re If R is suÆ iently large, then the square bra keted term is pra ti ally 1 and so  
C arg f (z ) ! 4 = 2 as R ! 1: 2 On the axis of y,   y3 arg f (iy) = ar tan : 1 + y4

361

8.5 Rou he's Theorem

Here y ranges from 1 to 0 along the positive imaginary axis, the initial and nal values of arg f (z ) are zero. Thus, the total hange when R is suÆ iently large is given by
C arg f (z ) = 2 0 whi h means that there is a root lying in the open rst quadrant. 

8.5 Rou he's Theorem The argument prin iple allows a omparison, under ertain onditions, of the number of zeros of two analyti fun tions.

8.50. Theorem. (Rou he's Theorem) Let f and g be meromorphi in a domain D  C and have only nitely many zeros and poles in D. Suppose that C is a simple losed ontour in D su h that no zeros or poles of f or g lie on C , and, in addition, assume that

jg(z )j < jf (z )j

(8.51)

on C:

Then
C f (z ) =
C (f (z ) + g (z )); i.e. the dieren e between the number of zeros and number of poles is the same for f and f + g :

Nf

Pf = Nf +g

Pf +g :

Proof. By the hypotheses, both f (z ) and f (z ) + g(z ) are nonzero on C . By the argument prin iple (see Corollary 8.41), we have Z Z 1 f 0 (z ) 1 (f + g)0 (z ) Nf Pf = dz and Nf +g Pf +g = dz 2i C f (z ) 2i C (f + g)(z ) so that a straightforward al ulation gives that

Nf

Pf

(Nf +g

Pf +g )  1 f 0 (z )(f (z ) + g(z )) f (z )(f 0 (z ) + g0 (z )) = dz 2i C f (z )(f (z ) + g(z )) Z 1 F 0 (z ) g(z ) = dz; F (z ) = 1 + : 2i C F (z ) f (z ) Z 

In view of (8.51), jg(z )=f (z )j < 1 on C so that, the meromorphi fun tion F (z ) maps C into jw 1j < 1. Thus, as z des ribes C , the point w = F (z ) traverses a losed ontour lying ompletely inside the domain jw 1j < 1 (see Figure 8.3) so that neither passes through the origin nor ontains the origin. It follows that Z Z F 0 (z ) dw dz = =0 F ( z ) w C and so, Nf

Pf

(Nf +g

Pf +g ) = 0. The result follows.

362

Cal ulus of Residues y |z − 1| = 1 C 1

O

w =1+ D

2 x

g(z) f (z)

v |w − 1| = 1 O

1

Ŵ2 u

Figure 8.3: Illustration for the proof of Rou he's theorem.

Sometimes Rou he's theorem is given an equivalent formulation: Instead of assuming jg(z )j < jf (z )j on C , it is assumed that

jg(z ) f (z )j < jf (z )j on C: Then the on lusion with respe t to this assumption is that
C g(z ) =
C f (z ), i.e. Nf Pf = Ng Pg :

8.52. Remark. If f and g have no poles in D, then the on lusion of Theorem 8.50 shows that f and f + g have the same number of zeros inside the ontour C .  Consider, f (z ) = z 6

5z 4 + 7. Then we have

(i) jf (z ) 7j < jz j6 + 5jz j4 < 7 for jz j = 1, and so f (z ) has no zeros in jz j < 1 (ii) jf (z ) ( 5z 4)j < jz j6 + 7 = 26 + 7 < 5jz j4 for jz j = 2, and so f (z ) has four zeros in jz j < 2 (iii) jf (z ) z 6j < 5jz j4 + 7 < jz 6j for jz j = 3, and so f (z ) has all the six zeros in the disk jz j < 3.

8.53. Remark. We next show that the fundamental theorem of algebra follows from Rou he's theorem. Consider f (z ) = a0 + a1 z +


+ an 1 z n 1 + z n and let C = fz : jz j = Rg, R > 1. Then, for z 2 C , we have f (z ) zn



1

 jRa0nj + Rjan1 j1 +


+ janR 1 j




+ jan 1 j℄ R1 (sin e R > 1): Thus, for suÆ iently large R, i.e. for R > maxf1; ja0j + ja1j +


+ jan 1jg, < [ja0 j + ja1 j +

we see that

f (z ) zn



1 < 1; or jf (z ) z n j < jz jn on jz j = R:

1

363

8.5 Rou he's Theorem

By Rou he's theorem, f has n zeros inside this ir le.



8.54. Example. Take f (z ) = 2 + z 2 and g(z ) = eiz . For z = x 2 R, we have f (x) = 2 + x2 > jeix j = 1 = jg(x)j; i.e. jf (z )j > jg(z )j:

On the other hand if z = Rei , 0    , then

jf (Rei )j = j2 + R2 ei2 j  R2 2 and jg(Rei )j = jeiRei j = e R sin  : Therefore, on the upper semi- ir le, we have jf (z )pj > jg(z )j if R satis es

the ondition R2

2 > e R sin  : If we hoose R > 3 then

R2

2 > 1  e R sin  for 0    :

By Rou he's theorem, the number of zeros of f + g in

p

R = fz : jz j  R and Re z  0g

for R > 3 is equal to the number of roots of the equation f (z ) = 2+ z 2 = 0 in R . Thus, in the entire upper half-plane equation 2 + z 2 eiz = 0 has only one root. 

8.55. Example. Let f (z ) = 1+ z 4 and g(z ) = iz 3. Then, on jz j = R, jf (z )j  R4 1 and jg(z )j = R3 so that the inequality jf (z )j > jg(z )j holds on jz j = R if R4 1 > R3: By Rou he's theorem, f and f + g have same number of zeros in jz j < R for whenever R4 1 > R3 : For example, z 4 + iz 3 + 1 = 0 has all the four roots in jz j < 3=2.  8.56. Example. Consider f (z ) = 3 + z + (jjR 3)z n; where jjR 3 > 0 with 0 < R  1 and n = 2; 3; : : : . Then, on jz j = R, we have

j3 + z j  jjR 3  (jjR 3)Rn = j(jjR 3)z nj and therefore, f has exa tly one zero inside the ir le jz j = R. For instan e f (z ) = 3 + z + (jj 3)z n has one zero in jz j < 1 if jj > 3.  8.57. Example. Given  > 1, we wish to prove that e z + z  = 0 has a unique solution in fz : Re z > 0g. To do this, we take f (z ) = z  and g(z ) = e z . Then, for z = iy, we have p

jf (iy)j = jiy j = y2 + 2 > 1 = je iy j = jg(iy)j so that jg(z )j < jf (z )j for z in the verti al line segment onne ting iR and iR. On the other hand if jz j = R, Re z  0, then we have (8.58) jf (z )j  jz j  = R  > 1  je z j = e Re z = jg(z )j

364

Cal ulus of Residues

(sin e Re z  0), provided R > 1 + . By Rou he's theorem, the equation f (z ) + g(z ) = e z + z  = 0 has only one root inside = fz : jz j = R; Re z > 0g for R >  + 1. The root is real be ause the left hand side of the equation e z + z  = 0 for z = x = 0 gives 1  whi h is negative and approa hes +1 as x ! +1. This shows the root must be real. 

8.59. Example. It is easy to prove that for ea h R > 0 there is an integer N = N (R) su h that zn z2 +


+ 2! n! has no zeros in jz j < R for n  N . Indeed, sin e fn (z ) n ! 1, for a given  > 0, there exists an N su h that fn (z ) = 1 + z +

! ez on C as

jfn (z ) ez j <  for all jz j = R; n  N: That is, if we hoose  < minjzj=R jfn (z )j, then the last inequality gives jez fn (z )j < jfn (z )j for jz j = R: Our assertion therefore follows from Rou he's theorem (and the fa t that ez 6= 0 in C ). One an also use Hurwitz' theorem (see Theorem 12.4) to get an alternate treatment of this example. On the other hand, we easily see that if m; n 2 N , then the polynomial

z z2 zn + +


+ + az m 1! 2! n! has m zeros in the unit disk whenever e < jaj. Indeed, as n k X X z 1 1 = e < jaj = jaz m j for jz j=1 ;  k ! k ! k=0 k=0 p(z ) = 1 +

Rou he's theorem gives the desired result.



8.6 Exer ises 8.60. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) If f and g are analyti in a deleted neighborhood of z0 , and if a and b 2 C , then Res [af (z ) + bg(z ); z0℄ = a Res [f (z ); z0℄ + b Res [g(z ); z0℄: (b) If f (z ) has an isolated singularity at a 2 C with nonzero residue at z = a, then the residues of f 0(z ) and (z a)f 0 (z ) at z = a are zero.

365

8.6 Exer ises

( ) If f has an isolated singularity at a, and is a non-zero omplex number, then f ( z ) has an isolated singularity at a= , and Res [f (z ); a= ℄ = (1= )Res [f (z ); a℄: (d) Let f 2 H(
(a; Æ) n fag) for some Æ > 0 and f have a simple pole at a with residue a 1 . Then, for a ir ular ar () of the form a + ei ( 2 [1 ; 2 ℄; 0  1 < 2  2;  < Æ), one has Z

lim f (z ) dz = ia 1 (2 !0  P

1 ):

P

n (e) If f (z ) = n2Zan z n and g(z ) = P n2Zbn z have an isolated singularity at 0, then Res [f (z )g(z ); 0℄ = n2Zan b n 1 . 2 Note: Is this result helpful to ompute Res [e1=z sin(1=z ); 0℄ = 0? (f) If f 2 H(C n f0g and Res [f (z ); 0℄ = a 1 , then there exists z 2 
su h that jf (z ) 1=z j  ja 1 1j. (g) If f is analyti at 1 and has a zero of order n ( 2) at 1, then Res [f (z ); 1℄ = 0. (h) If f is analyti at 1 and has a simple zero at 1, then Res [f (z ); 1℄ = limz!1 zf (z ): (i) If D  C is a domain, M > 0 and fR: D !
(M ; M ) is analyti , then for every losed ontour in D, (f 0 (z )=f (z )) dz = 0: Z 1 (j) dz = 2i. z e 1 2z jzj=1 (k) If f is a rational fun tion su h that the degree of its denominator ex eeds that of numerator by at least two, then the sum of residues at all the poles is zero. (l) If p is a polynomial of degree at most n and jaj < R, then Z

p(z ) n+1 (z a) dz = 0: z jzj=R R

(m) If p(z ) is a polynomial of degree n ( 2), then jzj=R pdz (z) = 0 for large R, and the same is also true if p(z ) is a linear fun tion, i.e. if n = 1. (n) There does not exist an analyti fun tion f de ned on C n f0g su h that f 0 (z ) = 1=z (see also Example 4.89). (o) The fun tion f (z ) = [1 + z + z 2 +


+ z n 1 ℄ 1 ; has simple poles at zk = e2ki=n (k = 1; 2; : : : ; n 1) and Res [f (z ); zk ℄ = 2i





e3ki=n k sin : n n

366

Cal ulus of Residues

(p) If k is a xed integer and C = fz : jz j = (2jkj + 1)=2g, then k X 1

osh z ot z dz =

osh j: 2i C j= k Z

(q) For

f (z ) = ez+1=z ,

Res [f (z ); 0℄ =

(r) For a > 1 and b-real, Z

jzj=1

1 X

1 = Res [f (z ); 1℄: n !( n + 1)! n=0

ebz dz 2 a z2 + 1

=

2i sin(b=a) : a

(s) The equation ez = 2 + 3z has at most one solution in the unit disk jz j < 1. (t) If jz j = 1 and a > 1, then the equation zea z = 1 has exa tly one solution in jz j < 1. (u) All the roots of the polynomial p(z ) = 1 + z + z 2 + z 3 + z 4 have absolute value less than 2. (v) If jf (z )j > m on jz j = 1, f is analyti for jz j  1 and jf (0)j < m, then f has at least one zero in jz j < 1. (w) Let f be analyti in a neighborhood of
= fz : jz j  1g. If jf (z )j < 1 for all jz j = 1, then f has exa tly one xed point in
. P (x) If fn (z ) = 1 + nk=1 kz k 1 and 0 < R < 1, then there exists an N su h that fn has no zeros in jz j < R whenever n > N . (y) All the roots of p the equation z 3 5z 2 + 10 = 0 lie in the annulus fz : 1 < jz j < 2g. (z) If f is a meromorphi fun tion on the Riemann sphere, then Nf = Pf , where Nf and Pf are respe tively the number of zeros and the number of poles of f , ounted with multipli ity.

8.61. Find the residues at ea h of the isolated singularities of the following fun tions in C or C 1 : z (i) 2 z + 3z + 3

(ii)

 2  z +z+1 3

z+1

(iii)

1

: (z 3 + 1)(z + 1)2

8.62. Suppose f and g are analyti in a domain D and f 0 (z ) 6= 0 in D. Let be a losed ontour in D. Then for a 62 , show that Z

g(a) 1 g(z ) n( ; a) = dz: f 0 (a) 2i f (z ) f (a) Apply this result for g(z ) = 1, ez , os z .

367

8.6 Exer ises Z

8.63. Evaluate f (z ) dz , where f (z ) is given by the following jzj=1 fun tions: sin6 z ; (z =6)3 z 4

z z 1 os z (ez e z )2 ; 2 ; z ; : 2 2 6z + 1 (z + 4z + 1) (e 1) sin z z3

8.64. Evaluate the following integrals using the residue theorem: Z Z dz dz (i) 2 (ez e z ) and 2 (ez e z ) z z jzj=1 jzj=4 Z z (ii) dz j z 1j=3 (z 2 1)3 (1 + z 2 ) R (iii) jzj= ot z dz Z e1=(z 1) (iv) dz j zj=3 z 2 Z dz (v) ; where C is any ir le en losing i and i inside. C 1 + z2 8.65. Let Q C be a simple losed ontour en losing the points 0; 1; 2; : : : ; n. If fm (z ) = m k=1 (z k ) for m = 1; 2; : : : ; n, then ompute the integrals Im =

Z

Z

dz fm (z ) and Jm = dz zf ( z ) m C C z

8.66. Suppose that f f is a onstant. 8.67. De ne f (z ) =

for m = 1; 2; : : : ; n:

2 H(
) and su h that f (
)  R. Show that 1 X

1 X 1 zn + : n n 2 n=1 7 n!z n=0 5n + n!

R

Does the series onverge for all z 6= 0? If so, nd the value of jzj=1 f (z ) dz:

8.68. Let f (z ) = (z )=g(z ), where  and g are both analyti around z0 . If (z0 ) 6= 0, g(z0) = g0 (z0 ) = g00 (z0 ) = 0, and g000 (z0 ) 6= 0, show that  00   (z ) 1 0 (z )g(iv) (z ) Res [f (z ); z0℄ = 3 000 : g (z ) 2 (g000 (z ))2 z=z0 8.69. De ne f (z ) = ln jz j + i arg z; z 2 D ; where D = C n fRei : R > 0g,  2 R is xed su h that arg z is the hoi e of arg z in ( 2; ). Assuming  6= =2; 3=2, nd Res [(z 2 + a2 ) 1 f (z ); ia℄:

368

Cal ulus of Residues

8.70. Find the analog of (8.8) when f has a pole at 1. More pre isely, if f has a pole of order n at 1, then show that Res [f (z ); 1℄ = zlim !1

( 1)n n+2 (n+1) z f (z ): (n + 1)!

In the ase when f is analyti at 1, then this formula ontinues to hold if n = 0.

8.71. Let p(z ) and q(z ) be polynomials with no ommon zeros, and with degrees m and n, respe tively. Set f (z ) = p(z )=q(z ). Show that (i) f has a removable singularity at 1 if n  m (ii) Res [f (z ); 1℄ = 0 if n  m + 2.

8.72. Suppose that f is meromorphi on C 1 . Then prove or disprove the following: f has a pole only at 1 i Res [f (z ); 1℄ = 0. 8.73. Suppose that f is analyti on jz j  1, jf (z )j < 1 whenever jz j = 1, and that jj < 1. Find the number of solutions of f (z ) = ((z )=(z 1))2 in jz j  1. 8.74. Let jaj > eR =Rn for a positive integer n. Prove that the equation az n ez = 0 has n solutions ( ounting multipli ity) z satisfying jz j < R. In the ase when R = 1, show that these solutions are simple roots with positive real part in jz j < 1. 8.75. If f (z ) = a0 + a1 z +


+ an 1 z n 1 + z n, and C is a simple

losed ontour en losing all the zeros of f , then show that Z Z zf 0(z ) z 2 f 0 (z ) dz = 2an 1 i and dz = 2i(a2n 1 2an 1): C f (z ) C f (z ) 8.76. Suppose that f (z ) is analyti in the disk jz j < 2 and Z 1 f 0 (z ) In = zn dz: 2i jzj=1 f (z ) Find the value of In when n = 0; 1; 2:

8.77. Let f be analyti for jz j < 2. Show that (

Z f (0) if jaj < 1 1 f (z ) dz = 2i jzj=1 z a f (0) f (1=a) if jaj > 1:

Chapter 9

Evaluation of ertain Integrals

This hapter des ribes some systemati methods to evaluate ertain types of de nite and improper integrals o

urring in Real Analysis. The method of residue al ulus, using the Residue Theorem, is a powerful tool for evaluating su h integrals. Here we illustrate the methods together with a suitable fun tion f and a suitable losed ontour C ; the hoi e, nevertheless, depends on the problem. In Se tion 9.1, we rst dis uss the evaluation of integrals of ertain periodi fun tions over the interval [; 2 + ℄. In the remaining se tions we apply the residue theorem to evaluate various types of real integrals whose integrands have no known expli it anti-derivatives. Let us start with a simple example: Z ez I= f (z ) dz; f (z ) = : z jzj=r An immediate onsequen e of the Cau hy residue theorem (or the Cau hy integral formula) gives I = 2i as f has a simple pole at 0 with Res [f (z ); 0℄ = 1: If we substitute z = rei , then dz = irei d = izd so that 2i = I = i

Z 2

0

exp(rei ) d:

Equating real and imaginary parts, we have Z 2

0

er os  sin(r sin ) d = 0 and

9.1 Integrals of Type

R 2+



Z 2

0

er os  os(r sin ) d = 2:

R( os ; sin  ) d

This se tion provides a method of evaluating integrals of the form (9.1)

I=

Z 2+



R( os ; sin ) d;

370

Evaluation of ertain Integrals

where R( os ; sin ) is a rational fun tion of os  and sin  (with real oeÆ ients) whi h is nite in the range of the integral. Quite often the integrals of the above type an be evaluated by means of some substitutions su h as t = tan , t = tan(=2), et ., but sometimes the evaluation may prove to be diÆ ult or even impossible with the real analyti methods at our disposal. In equation (9.1),  varies between  and 2 + . Sin e  varies over a range of 2, we may onsider  as an argument of a point z on the unit

ir le C = fz : jz j = 1g. Therefore, we may write z = ei so that

z2 + 1 z2 1 dz ; sin  = and d = ; (0    2): 2z 2iz iz Thus, the integral in (9.1) be omes

os  =

I=

Z

C

f (z ) dz =

Z

C

R

 2 z +1



z 2 1 dz ; 2z 2iz iz

where f is a rational fun tion of z that is nite onPthe path of integration C . By the residue theorem, we then have I = 2i nk=1 Res [f (z ); k ℄; where k denotes those poles of f whi h lie inside C and the integral along C is taken in the positive dire tion.

9.2. Remark. As pointed out above, by means of the substitution t = tan =2 or simply by t = tan , we an prove Z =2

d  = (jj < 1) 2 1 2 =2 1 2 sin  +  p Z =2 1 + 2 sin2  (4 3 3) (ii) d = : 1 + 2 os2  6 0 (i)



9.3. Example. Let us evaluate I=

Z 2

0

d ; a + b sin 

where a and b are real with jbj < jaj: First, we note that if b = 0 then I = 2=a. If b 6= 0, then

I=

1 b

Z 2

0

d (a; b real; 1 < ja=bj): (a=b) + sin 

So, it suÆ es to ompute the integral for b = 1. Now, putting z = ei (and assuming b = 1 and a 2 R with jaj > 1), we nd that

I =2

Z

jzj=1

f (z ) dz; f (z ) =

1

z 2 + 2iaz

1

=:

(z

1 )(z

)

:

R 9.1 Integrals of Type 2+ R( os ; sin ) d

371

We see that the only singularities of f are the simple poles at 1 :  Observe that, sin e 1 < jaj and the produ t of the two roots is 1, one root lies inside the unit ir le jz j = 1 while the other lies outside. In fa t  lies in jz j < 1 whenever a < 1, and = 1= lies in jz j < 1 whenever a > 1. Now, p

 = i(a + a2 1) and = i(a

Res [f (z ); ℄ =



1



=

p1 2i a2

1

0

h

d = 2 2i a + sin 

X

a2

1) =

; and Res [f (z ); ℄ =

Therefore, by the residue theorem, we have Z 2

p

i

Res [f (z ); C ℄ =

8 > > < > > :

1 : 

p 22

if a > 1 a 1 p 22 if a < 1: a 1



9.4. Example. In the following examples, we skip some steps. For a > 1; Z 2

0

Z

d 4 z = : f (z ) dz; f (z ) = 2 2 (a + os ) i jzj=1 (z + 2az + 1)2

Observe that f has two poles (ea h of order two) at p

 = a + a2

1 and = a

p

a2 1:

As z 2 + 2az + 1 = (z )(z ), we have  = 1 so that one pole lies inside the unit ir le jz j = 1 while the other must lie outside. Clearly,  lies inside jz j = 1, and therefore, it suÆ es to ompute
d 2 Res [f (z ); ℄ = zlim ! dz (z ) f(z ) d z = zlim ! dz (z )2 ( + ) = ( )3 a = : 2 4(a 1)3=2

Finally, by the Cau hy residue theorem, Z 2

0





d 4 a = 2i 2 2 (a + os ) i 4(a 1)3=2



=

(a2

2a : 1)3=2

372

Evaluation of ertain Integrals y y = f (x)

a

O

x

2a

Figure 9.1: Illustration for an integral by area under a urve.

What happens to the integral if a < 1? Similarly, for a > 1, we an write Z 2

0

where

Z

d 4 = f (z ) dz; 2 (a + sin ) i jzj=1

z z = (z 2 + 2iaz 1)2 (z )2 (z )2 and  and are as in Example 9.3. It follows that f has a double pole at z =  inside the unit ir le, and ( + ) a Res [f (z ); ℄ = = ( )3 4(a2 1)3=2 so that Z 2 d 2a 2 = (a2 1)3=2 ; for a > 1: ( a + sin  ) 0 What is the value of the integral when a < 1?  f (z ) =

9.5. Remark. One may adopt the method of the above example to evaluate su h integrals whose range of integration is not of length 2. In this ontext we often use the following: \If f (x) = f (2a x); then

Z 2a

0

f (x) dx = 2

Z a

0

f (x) dx:"

One an be readily onvin ed of this fa t by interpreting the integrals as areas under a urve (see Figure 9.1). Sin e the urve is symmetri about x = a, the two dierent shaded regions are equal in area. Alternately, it suÆ es to rewrite Z 2a

0

f (x) dx =

Z a

0

f (x) dx +

Z 2a

a

f (x) dx

and use the hange of variable x = 2a t for the se ond integral on the right. An integral over [0; ℄ an also be handled whenever f () is even in  and is 2-periodi , sin e in this situation

R 9.1 Integrals of Type 2+ R( os ; sin ) d Z 

0

f () d =

1 2

Z 



f () d =

1 2

373

Z 2

0



f () d:

9.6. Remark. If a and b are onstants, x a real parameter,  x  d, and R(; x) a ontinuous fun tion with a ontinuous partial derivative with respe t to x for a    b,  x  d, then, a

ording to Leibnitz's rule, we have ! Z Z b b R d R(; x) d = d: dx a a x Leibnitz's rule an be extended suitably to ases where the limits a and b are in nite or dependent on x. Using this rule, we an easily dedu e from Example 9.3 that for 1 < jaj, Z 2

0

d = (a + sin )2

(

2a=(a2 1)3=2 for a > 1 2a=(a2 1)3=2 for a < 1:



Z 2

d 9.7. Example. Set I = 2 2 os  ; 1 6=  > 0: As in the 1 +  0 previous examples, to evaluate this integral, we may rewrite it as Z

i 1 I= f (z ) dz; f (z ) = :  jzj=1 (z )(z 1=) The only singularities of f are the simple poles at z =  and z = 1=. If 0 <  < 1, then z =  is inside jz j < 1 while the other is outside the unit

ir le. Therefore, for 0 <  < 1,

I=

2 2 i f 2i Res [f (z ); ℄g = lim (z )f (z ) = : z !    1 2

Similarly, we dedu e that I = 2=(2



1) for  > 1:

9.8. Example. Let us show that Z 2

Z 2

2(2n)! 22n (n!)2 0 0 Z 2 2(2n)!(a2 + b2 )n (ii) (a os  + b sin )2n d = (a; b are real): 22n (n!)2 0 (i) I =

os2n  d =

sin2n  d =

As usual, let z = ei . Then the rst integral, whi h we have already en ountered in Chapter 4, be omes Z



1 (z 2 + 1)2n 1 1 I = 2n f (z ) dz; f (z ) = 2n+1 = z+ i2 jzj=1 z z z

2n

:

374

Evaluation of ertain Integrals

The only singularity of f is the pole at z = 0 of order 2n + 1. Sin e   2n   2n   2n 2n k 1 k X 2n 2(n k) 1 1X z = z ; jz j > 0; z k=0 k z k=0 k

f (z ) =




= 2nn : Consequently,     1 2 2n 2 (2n)! I = 2n f2i Re [f (z ); 0℄g = 2n = 2n i2 2 n 2 (n!)2

we see that the oeÆ ient of z

1

is a

1

and the integration formula for the rst part follows. Similarly, by onsidering fun tions of the form z k f (z ) (k 2 Z), we an a tually evaluate Z 2

0

os2n () os k d

and

Z 2

0

os2n () sin k d:

However, for the proof of (ii), we may rewrite (as there is nothing to prove if (a; b) = (0; 0)) 



p 2a 2 os  + p 2b 2 sin  a os  + b sin  = a +b a +b and observe that, given a pair (a; b) 6= (0; 0), there exists a unique  [0; 2) su h that p

a2 + b2

p

2

p

os  = a= a2 + b2 and sin  = b= a2 + b2 :

p

Thus, a os  + b sin  = a2 + b2 os( 1 (a2 + b2 )n

Z 2

0

) and

(a os  + b sin )2n d = =

Z 2

0

Z 2

0

os2n ( ) d

os2n  d



so that (ii) follows from (i).

9.9. Remark. If os n or sin n o

urs in the integrand, we may use the formulas zn + z n zn z n

os n = and sin n = 2 2i i where z = e and n 2 Z.  Finally, for a > 1, we write

I=

Z 2

0

Z

sin2  i (z 2 1)2 d = f (z ) dz; f (z ) = 2 ; a + os  2 jzj=1 z (z )(z )

R 9.2 Integrals of Type 11 f (x) dx

where

375 p

p

a2 1  = a + a2 1 and = a are the roots of the quadrati equation z 2 + 2az + 1 = 0. We note that z =  lies in jz j < 1 while z = lies outside the unit ir le jz j = 1. Further, f has a pole of order two at z = 0. Also, it is easy to see that p

Res [f (z ); 0℄ = 2a and Res [f (z ); ℄ = 2 a2 Therefore, by the residue theorem, p in  I = 2i 2a + 2 a2 2 What happens if a < 1?

o

1

= 2(a

p

a2

1: 1):

R 9.2 Integrals of Type 11 f (x) dx

In the previous se tion, we transformed ertain real trigonometri integrals into ontour integrals and then omputed them with the help of residue theorem. This se tion proposes to evaluate ertain types of improper and de nite integrals, but by interpreting the given integral as 2i times the sum of residues at the singularities of a properly hosen analyti fun tion. We start with a simple example of evaluating Z 1 (9.10) I= f (x) dx; 1 where f (x) is a ontinuous fun tion on R. As we know, this is an improper integral of f on ( 1; 1) and has the meaning Z 0 Z S Z 0 Z 1 I = Rlim f (x) dx + lim f (x) dx = f (x) dx + f (x) dx !1 R S !1 0 1 0 provided these two limits exist. We an now write I as

I = lim R;S !1

Z S

R

f (x) dx

with the understanding that R and S have to be allowed Rto run to 1 independently of ea h other; i.e. the existen e of limR!1 RR does not R1 imply the existen e of 1 as the fun tion f (x) = x demonstrates this fa t. However, we next give a pre ise example illustrating this note. Suppose f is ontinuous on interval (a; b) ex ept at a point x0 in (a; b), where f has a singularity (in whi h sense?), i.e. f (x) is unbounded near x0 . Then, Z (9.11)

b

a

f (x) dx

376

Evaluation of ertain Integrals

might not be de ned and so we have to nd a natural way to make a meaningful de nition. For 1 > 0 and 2 > 0, onsider Z x0 1

f (x) dx +

a

Z b

x0 +2

f (x) dx

(note that both the integrals exist for ea h 1 and 2 ) and let 1 ! 0 and 2 ! 0. If both limits exist, we say that the integral (9.11) is onvergent. It may happen that even though this limit does not exist, the limit when 1 = 2 =  with  ! 0 may exist. For instan e onsider f (x) = x 3 , x 2 [ 1; 1℄ nf0g: Then, with x0 = 0, we have Z 0 1



dx 1 = 1 x3 2





Z 1



dx 1 1 3 = 2 2 1 : 1 0+2 x 2 Clearly, the limit of ea h of these integrals does not exist as 1 ! 0 and R R 2 ! 0. This shows that the integrals 01 x 3 dx and 01 x 3 dx are not

onvergent. On the other hand, if we take 1 = 2 = , we nd that 1 21

Z

and

Z



1 dx dx + =0 3 1 x  x3 for all  su h that 0 <  < 1. Thus if we de ne

(9.12)

Z b

a

and (9.13)

f (x) dx = lim 1 !0

(Z

a

2 !0

Z b

a

f (x) dx = lim !0

x0 1

(Z

a

x0 

f (x) dx + f (x) dx +

)

Z b

x0 +2

f (x) dx )

Z b

x0 +

f (x) dx

we see that we may get dierent values for (9.11), depending on whether we use de nition (9.12) or (9.13). Thus if jf (x)j ! 1 as x ! x0 and R the limit in (9.13) exists, then ab f (x) dx is alled a onvergent improper integral. The limit in (9.13), whi h is also of interest to us, is alled the Cau hy's prin ipal value of the integral (9.11). Note also that if the limit exists in (9.12), then it also exists in the sense de ned in (9.13) and hen e, both the limits are equal. When f (x) is nite for all real values, then by R1 RR the Cau hy prin ipal value of 1 f (x) dx we mean limR!1 R f (x) dx R (if it exists). For instan e, sin e RR x dx = 0 for every R > 0, the Cau hy R1 prin ipal value of 1 x dx is zero. In general if f is ontinuous on R ex ept for a nite number of points x0 ; x1 ; : : : ; xn (x0 < x1 <


< xn ) and if lim

!0 R!1

( Z x0 

R

+

Z x1 

x0 +

+


+

Z xn 

xn 1 +

+

Z R !

xn +

)

f (x) dx

R 9.2 Integrals of Type 11 f (x) dx

377

exists and isR nite, then we all this limit the Cau hy prin ipal value of the integral 11 f (x) dx denoted by itself with the additional remark, if ne essary, that the prin ipal value is meant. Some books denote for brevity by Z 1 PV f (x) dx: 1 By (9.13), we see that for a > 0 Z 2a

dx = lim !0 a x

Z 0 

dx + a x

Z 2a



dx = lim !0[ln  ln a + ln 2a ln ℄ = ln 2: 0+ x

On the other hand, using (9.12), we get 

Z 2a



dx  = lim ln 2 + ln 1 ! 0 2 a x 12 !0

(9.14)



:

Clearly, the limit in (9.14) on the right does not exist. This shows that PV

Z 2a

dx = ln 2 a x

R

whereas the integral 2aa dx x does not exist as an improper integral. Similarly, we an easily see that Z 1 Z R xn 1 xn 1 I = PV dx = lim dx R!1 R xn + 1 1 xn + 1 so that I = 0 if n is even. R Now we start with f (x) = 1=(1 + x2 ) and onsider 11 f (x) dx; where the path of integration is the line Im z = y = 0. If we want to use the Cau hy residue formula, we need to onsider an integral along a losed

ontour. This observation suggests that we may have to start with an experiment Z 1 J = f (z ) dz; f (z ) = ; 1 + z2 C where C is the semi- ir ular ontour shown in Figure 9.2. Here we hoose R > 1 so that z = i lies inside C . Therefore, by the residue theorem, we easily get J = 2iRes [f (z ); i℄ = : Write J = J1 + J2 , where

J1 = Sin e

Z R

Z

i

Re dx dz and J = = 2 2 1 + x 1 + z2 i 0 R Re

jJ2 j 

Z 

0

R d R2 1

Z 

0

iRei d : 1 + (Rei )2

! 0 as R ! 1;

378

Evaluation of ertain Integrals y ŴR i R x

−R

Figure 9.2: Contour C = [ R; R℄ [ R . R we dedu e that 11 1+dxx2 = : In parti ular, we have the well-known result Z 1 dx  = : 2 2 0 1+x

Next, we use the same idea and show that Z 1

os ax e am (9.15) dx = (a; m > 0): 2 2 2m 0 m +x Note that if we onsider the most obvious omplex fun tion os az=(m2 + z 2 ), then we will not be able to a hieve the desired result, be ause for z = iR (R large enough) os az m2 + z 2



z=iR

=

eaR + e aR 2jm2 R2j

! 1 as R ! 1:

Nevertheless, as the term eiaz = eiaxe ay is bounded in the upper halfplane, it is natural to onsider

f (z ) =

(z + im) 1 eiaz eiaz = m2 + z 2 z im

and we use the same semi- ir ular ontour as above. Then, we have Z

C

f (z ) dz = 2i Res [f (z ); im℄ =

e am : m

Now, we write (9.16)

Z

C

f (z ) dz =

Z R

R

f (x) dx +

Z 

0

i

iRei eiaRe d = J1 + J2 ; say: m2 + R2 ei2

Sin e sin x is odd and os x is even, we have

J1 =

Z R

R

f (x) dx =

Z R 

 Z R

os ax sin ax

os ax + i dx = 2 2 2 2 2 2 2 dx m +x R m +x 0 m +x

R 9.2 Integrals of Type 11 f (x) dx

379

and (as e aR sin   e0 = 1 for  2 [0; ℄, a > 0 and R > 0)

jJ2 j 

Z 

0

Re aR sin  R d  2 2 2 R m R m2

Z 

0

! 0 as R ! 1:

d

The above observations prove (9.15), in view of (9.16). Further, using Leibnitz's rule, (9.15) readily gives Z 1 x sin ax e am dx = ; a > 0: 2 2 2m 0 m +x In general, problems ofRthis type may be solved using the same idea. Consider the integral I = 11 f (x) dx, where f (x) is a rational fun tion without real poles. Therefore, to evaluate su h an integral, we let

J=

Z

C

f (z ) dz =

Z R

R

f (x) dx +

Z R

f (z ) dz;

where C = [ R; R℄ [ R is the same ontour as in Figure 9.2. Then Z R

R

f (x) dx +

Z R

f (z ) dz = 2i

X

Res [f (z ); C ℄:

As R ! 1, the rst integral on the left tends to I . We shall then have to show that the se ond integral tends to 0. If so, this would then imply that Z 1 X f (x) dx = 2i Res [f (z ); C ℄: 1 R

Thus, we have to nd a suitable ondition under whi h R f (z ) dz ! 0 as R ! 1: Before we establish su h a ondition (see Theorem 9.23) let us rst dis uss a few more examples.

9.17. Example. We wish to prove that Z 1 m 1 x  dx = ; for m; n 2 N with n > m > 0. n 1 + x n sin( m=n) 0 To do this, we let

f (z ) =

zm 1 : 1 + zn

Then f has simple poles at ak = ei(1+2k)=n (k = 0; 1; 2; : : : ; n 1) with Res [f (z ); ak ℄ =

zm nz n

1 1 z=ak

=



z m am = k: n z=ak n

Here we use a dierent ontour to evaluate the integral.

380

Evaluation of ertain Integrals y Reiα γR

ŴR α

O

R

x

Figure 9.3: Contour [0; R℄ [ R [ [Rei ; 0℄.

Note that these poles are simple and the only pole inside C (see Figure 9.3 with  = 2=n) is at a0 , where

C = [0; R℄ [ fz = Rei : 0    2=ng [ fz = rei2=n : 0  r  Rg = [0; R℄ [ R [ R : Now (see Theorem 8.13), by the residue theorem (9.18)

Z

C

Sin e jf (z )j  (9.19)

f (z ) dz =

Z

[0;R℄

+

Z R

+

!

Z

R

f (z ) dz =

2iam 0 : n

jz jm 1  2jz jm 1 as jz j ! 1, we have jz jn 1 jz jn

Z



2 2R f (z ) dz  n+1 m R n R

! 0 as R ! 1:

Next, Z

R

f (z ) dz =

Z R

0

f (rei2=n ) d(rei2=n ) = ei2m=n

Z R m 1 r dr

0

1 + rn

:

Using this equation and (9.19), (9.18) be omes as R ! 1,   Z 1 m 1 x am 0 2 m (1 a0 ) n dx = 2i n 0 1+x and therefore,   Z 1 m 1 x dx 2i  = = : m n m 1+x n sin(m=n) n(a0 a0 ) 0



9.20. Remark. If C is the re tangular ontour with verti es at R and R + 2i (see Figure 9.4) and if f (z ) =

eaz (0 < a < 1); 1 + ez

R 9.2 Integrals of Type 11 f (x) dx

381 y

−R + 2π i

R + 2π i

Ŵ2

Ŵ1 O

−R

x

R

Figure 9.4: Re tangular ontour with verti es at R;

 R + 2i.

then we see that Z R

R

f (x) dx + +

Z 2

0 Z 0

2

f (R + iy) d(R + iy) +

Z

R R

f (x + 2i) d(x + 2i)

f ( R + iy) d( R + iy) = 2if eaig;

as the only pole of f inside C is at z = i. Sin e f (z + 2i) = e2ai f (z ); the above equation simpli es to (1 e2ai )

Z R

R

f (x) dx + i

Observe that

Z 2 f (R + iy) dy =

0

Z 2

0

[f (R + iy) f ( R + iy)℄ dy = 2ieai :

Z 2

0



eaR eaiy 2eaR e(a 1)R dy  = : 1 + eR eiy eR 1 1 e R

Similarly, we nd that Z 2 f(

0



R + iy) dy 

2e aR : 1 e R

As R ! 1, we easily have (sin e 0 < a < 1) Z 1 (1 e2ai ) f (x) dx = 2ieai 1 from whi h we get Z 1 Z 1 eax 2i  (9.21) dx = f (x) dx = ai = : x ai e e sin a 1 1+e 1 Using the new variable ex = t, (9.21) redu es to Z 1 a 1 t  (9.22) dt = : sin(a) 0 1+t

382

Evaluation of ertain Integrals

Further, if we use the transformation t = xn with a = m=n in (9.22), it be omes Z 1 m 1 x  dx = for n > m > 0: n n sin(m=n) 0 1+x However, we an obtain (9.22) dire tly by integrating a suitable fun tion around a suitable ontour.  We may use the same idea as in previous examples to solve more su h problems (in general) by means of the residue theorem. They are mostly dealt with by appli ation of the following

9.23. Theorem. Suppose that f is analyti on C ex ept for a nite number of poles su h that none of its poles lies on the real axis. If there exist two positive numbers M and R0 su h that

jf (Rei )j  RM

(9.24) for some  > 1, then Z

(9.25)

1 1

where H + ; H

f (x) dx =

8
R0 ;

2i

:

2i

X

X

Res [f (z ); H + ℄

Res [f (z ); H ℄;

are respe tively the upper and lower half-planes.

Proof. Let R be arbitrarily large enough so that all the poles of f (z ) are in jz j < R. Let C = [ R; R℄[ R ; where R = fz : jz j = R; 0  arg z  g; the semi- ir ular ar of radius R (see Figure 9.2). Then Z

(9.26)

C

f (z ) dz =

Z R

R

f (x) dx +

Z R

f (z ) dz:

But, by (9.24), the se ond integral on the right of (9.26) is su h that Z

R

f (z ) dz =

Z  i i f (Re )iRe d

0

 RM R = RM  1;

whi h approa hes zero as R ! 1, sin e  > 1. The on lusion now follows from (9.26) by letting R ! 1.

9.27. Corollary. The on lusion of the above theorem holds if f is a rational fun tion, f (z ) = P (z )=Q(z ); where (i) f has no poles on the real axis, and (ii) P (z ) and Q(z ) are polynomials of degree m and m + n, respe tively, with n  2.

R 9.2 Integrals of Type 11 f (x) dx

383

Proof. The ondition (ii) implies that there exist two positive numbers M1 and M10 su h that (see Exer ise 6.88) M 0 jz jm  jP (z )j  M1 jz jm for jz j  R 1

and M2 su h that jQ(z )j  M2 jz jm+n for jz j  R (R  1): Then P (z ) Q(z )

m 1  MMj1zjjzmj +n = MMjz1jn  M
1  M ; say; M jz j2 jz j2 2

2

2

for jz j  R, and Theorem 9.23 applies.

9.28. Example. Let us show that Z 1 sin2 x (1 e 2) dx = : 2 2 1 x +1 To do this, we onsider (sin e sin2 x = (1 os 2x)=2) 1 ei2z : 1 + z2 This fun tion has exa tly two simple poles at i and

f (z ) =

1 ei2z i Res [f (z ); i℄ = lim = (1 e 2 ): z!i z + i 2

Let C = [ R; R℄ [ R ; with R = fz : jz j = R; 0  arg z  g: Then Z R

R

f (x) dx +

Z

Sin e jf (z )j  (1 + e Z

R

f (z ) dz

It follows that Z 1

R

f (z ) dz = 2i

2Im z )=(jz j2

 R2 2 1



1)  2=(R2

Z

R



i (1 e 2 ) = (1 e 2 ): 2 1) for z 2 R , we have

jdz j = R22R1 ! 0 as R ! 1:

1 2 sin2 x i sin 2x dx = (1 e 2 ) 1 + x2 1 1 and equating the real parts yields the desired result.

f (x) dx =

Z



9.29. Remark. Theorem 9.23 may not apply in 2some ases: R p for example to the well-known Gauss Error Integral 11 e x dx = : This 2 is be ause e z does not have the required limiting behavior, as z ! 1. In fa t,

384

Evaluation of ertain Integrals √

y Ŵ√2R

i 2R

(1 + i)R

iR ŴR O

R

√

x

2R

Figure 9.5: Illustration for Gauss Error Integral. 2

(i) je z j = 1 on arg z =  4 and arg z =  34 . 2 (ii) je z j ! 0 faster than re ipro al of any polynomial on the lines arg z =  2 ; 0; . We in lude here a lassi al proof of the error integral. To do this, we let R 2 I = 0R e x dx. Then

I2

=

Z R

0

!

e

x2 dx

Z R

0

y2 dy

e

!

Z RZ R

=

0

0

2 2 e (x +y ) dx dy:

Here we are integrating along a square S in the rst quadrant whose sides have length R. Let R and p2R denote the quarter- ir les in the rst p quadrant entered at the origin having radii R and R 2, respe tively (see Figure 9.5). Evaluating along the ir les in polar oordinates, we have Z =2 Z R Z RZ R Z =2 Z Rp2 2 2 +y 2 ) 2 r ( x e r dr d < e dx dy < e r r dr d; 0

0

0

0

0

and so, for ea h R > 0, we have 2  (1 e R ) < I 2 = 4

Z R

0

e

!2

x2 dx

p Letting R ! 1, we see that I = =2.

0

2  < (1 e 2R ): 4



9.30. Remark. One an evaluate the error integral using the residue theorem dire tly. To do this, we de ne 2

r

e z  f (z ) = ; a = (1 + i) : 2 az 1+e 2 2 2 a ( a ) Sin e = 1 and so a is a period of e 2az . As 2 a =ii, we note that e a e = e = 1, we also note that 2 2 2 e z (9.31) f (z ) f (z + a) = (1 e a 2az ) = e z : 1 + e 2az

R 9.3 Integrals of Type 11 g (x) os mx dx

385

y

√ −R + i π/2

√ S + i π/2

γ2

γ1 O

−R

x

S p

p

Figure 9.6: Re tangular ontour C = [ R; S ℄ [ 1 [ [S + i =2; R + i =2℄ [ 2 .

Note that f (z ) has in nitely many simple poles in C , namely at a=2 + ka, k 2 Z. Therefore, it is not advisable to hoose a ontour that in ludes many poles. Instead, we hoose the re tangular ontour as depi ted in Figure 9.6. Sin e f has only the point a=2 inside the ontour p

p



2

C = [ R; S ℄ [ 1 [ [S + i =2; R + i =2℄ [ 2 ; we have (9.32)

Res [f (z ); a=2℄ =

2

e z i e a =4 = 2 az a2 = 2p : 2ae 2 ae z=a=2

Therefore, by the residue theorem, we have Z S

R

f (x) dx + +

Z

1 Z

2

f (z ) dz +

Z

R S

p

p

f (x + ( =2 + i =2) dx

f (z ) dz = 2i Res [f (z ); a=2℄:

Be ause of (9.31) and (9.32) the above equation be omes Z S

R

2 e x dx +

Z

1

f (z ) dz +

Z

2

p

f (z ) dz = 

R p 2 and letting R; S ! 1, we on lude that 11 e x dx = : R 9.3 Integrals of Type 11 g (x) os mx dx



This se tion dis usses some other important improper integrals. Suppose that f (z ) = eiaz g(z ) for a 2 R and for some analyti fun tion g. Then the

onditions on f in Theorem 9.23 an be weakened; that is in this ase it is enough to assume that

jf (Rei )j  M for R > R0 : R

386

Evaluation of ertain Integrals

Observe that in most of the examples it is suÆ ient that f (z ) ! 0 as z ! 1 in arg z , 0  arg z  . Therefore, Theorem 9.23 and Corollary 9.27 take the following form:

9.33. Theorem. Suppose that g is analyti in C ex ept possibly for a nite number of poles and none of them are real. If there exist two positive numbers M and R0 su h that (9.34) for some  > 0, then

(9.35)

Z

1 1

jg(Rei )j  RM

g(x)eiax dx =

8 < :

2i 2i

for R  R0 ;

X

X

Res [g(z )eiaz ; H + ℄ if a > 0

Res [g(z )eiaz ; H ℄ if a < 0;

where H + ; H are the upper and lower half-planes, respe tively. Further, if g(z ) = P (z )=Q(z ); where P , Q are polynomials su h that

(i) g has no poles on the real axis, (ii) deg Q  1+ deg P () jg(z )j  M=jz j for jz j  R0 , (iii) g is real on the real axis, then and

Z

Z

1 P (x)

os ax dx = Real part of the R.H.S of (9.35) 1 Q(x)

1 P (x) sin ax dx = Imaginary part of the R.H.S of (9.35): 1 Q(x)

In establishing our theorem, we shall make use of Jordan's Inequality (see Figure 9.7).

9.36. Lemma. For  2 [0; =2℄, we have sin   (2=): Proof. Clearly, the inequality is true at the end points. Now,   d sin  () = 2 with () =  os  sin : d   We laim that () =  1 sin  is stri tly de reasing on (0; =2). To do this, it suÆ es to observe that 0 () =  sin  < 0 for  2 (0; =2) so that is de reasing on (0; =2) and thus, () < (0) = 0 whi h shows that 0 () < 0 and hen e, () =  1 sin  is stri tly de reasing on (0; =2). Finally, as sin  2 lim () = lim = ;  !=2 !=2 

R 9.3 Integrals of Type 11 g (x) os mx dx

387

y i

(θ, sin θ )

φ

(π/2, 1)

θ, 2θ π

2 θ π


π/2

O

y = (tan φ)θ =

θ

Figure 9.7: Geometri proof of Jordan's inequality.

we dedu e that () > 2= on (0; =2), as desired.

9.37. Remark. For a geometri proof of Lemma 9.36, we refer to Figure 9.7. Set f () = sin . Then f 00 () = sin  < 0 on (0; =2) so that y = sin  is on ave down on (0; =2). Therefore, the graph of y = sin  lies above the straight lines onne ting the end points (0; 0) and (=2; 1). Note that the equation of the line passing through the point (0; 0) and (=2; 1) is y = (2=): Therefore, the inequality sin   (2=) holds on [0; =2℄.



9.38. Remark. For R > 0, it follows that I=

Z 

0

 e R sin  d < : R

Indeed, if f () = e R sin  then f () = f (

I = 2

 2

Z =2

0

Z =2

0

) so that

e R sin  d (see Remark 9.5) e

(2R=) d

(sin e sin   2  on [0; =2℄)

   e R 1 = 2 = (1 e R ) < 2R= R R 



and the desired inequality follows.

Proof of Theorem 9.33. First onsider a > 0. Choose R large enough so that all the singularities of g in H + lie inside the upper semi

ir le C = [ R; R℄ [ R (see Figure 9.2). By the residue theorem (9.39)

Z R

R

eiax g(x) dx +

Z R

eiaz g(z ) dz = 2i

X

Res [g(z )eiaz ; H + ℄:

388

Evaluation of ertain Integrals y ŴR −γǫ ǫ

−R −ǫ

x

R

Figure 9.8: Contour [ R; ℄ [ (  ) [ [; R℄ [ R .

Now we estimate the absolute value of the se ond integral. Let z = Rei 2 R . Then jeiaz j = e aR sin  . In view of Remark 9.38, taking R  R0 so that jg(z )j  M=R , we obtain Z



R

eiaz g(z ) dz 

M R 1

Z 

M    M e aR sin  d <  1 = : R aR aR

0

Sin e a and  > 0, the R.H.S of the above inequality approa hes zero as R ! 1 and the required result follows from (9.39). As for the ase a < 0, we simply onsider the ontour in the lower halfplane H and pro eed similarly. The proof of the remaining part is similar to the proof of Corollary 9.27. Using the idea des ribed in Theorem 9.33, it is easy to show that Z 1 x sin ax e am dx = (a; m > 0): 2 2 2 0 m +x

9.4 Singularities on the Real Axis We shall now dis uss the ase where f has simple poles on the real axis. Suppose that the only singularity of f on the real axis is a simple pole at the origin. Let C be the indented ontour shown in Figure 9.8. Thus, C

onsists of the line segment [ R; ℄,  < R, the semi- ir le (  ) from  to , the line segment [; R℄, and the semi- ir le R from R to R. Here, the small semi- ir le  is des ribed to avoid the singularity of f at the origin. Assume further that C en loses all the singularities of f in the upper half-plane H + . By the residue theorem Z

C

f (z ) dz =

(9.40)

Z

 R

= 2i

f (x) dx

X

Z



f (z ) dz +

Res [f (z ); H + ℄:

Z R



f (x) dx +

Z R

f (z ) dz

To evaluate integrals of this type we employ

9.41. Theorem. Let  = fz : jz z0j =  and 1  arg(z z0)  2 g.

389

9.4 Singularities on the Real Axis If f is ontinuous on 0 < jz

lim

Z

!0 

z0j   and if zlim !z (z z0)f (z ) = `, then 0

f (z ) dz = i`(2 1 );

where  is positively oriented.

Proof. Write (z z0 )f (z ) = ` + (z ) and hoose  suÆ iently small so that, for any arbitrary  > 0, j(z )j <  if jz z0 j = . Now, Z



f (z ) dz = ` R

Z

Z Z 2 Z 2 dz (z ) + dz = ` i d + i(z0 + ei ) d:

 z z0 1 1

 z z0

As 12 i(z0 + ei ) d  (2 1 ); the last equation implies the desired result.  In parti ular, if 2 1 = 2 then  be omes a positively oriented full

ir le and so, Z lim f (z ) dz = 2i zlim !z (z z0 )f (z ): !0 0



9.42. Remark. If there exists a real number  > 1 su h that jf (z )j  K jz j  as jz j ! 1 in the upper half-plane then, as seen in the proof R of Theorem 9.41, we obtain limR!1 R f (z ) dz = 0: Therefore, (9.40) be omes lim

R!1 !0

(Z



R

f (x) dx + Z

Z R



)

f (x) dx + lim R!1

lim f (z ) dz = 2i !0 

X

Z

R

f (z ) dz

Res [f (z ); C ℄:

From Theorem 9.41 (with z0 = 0), this redu es to Z 1 X f (x) dx + 0 i( 0) Res [f (z ); 0℄ = 2i Res [f (z ); H + ℄ 1 from whi h we get the value of the integral when the only singularity of f on the real axis is a simple pole at the origin. 

9.43. Example. Let us evaluate the integral Z 1 Z sin x 1 1 sin x I= dx = dx: x 2 1 x 0

390

Evaluation of ertain Integrals

We have trouble if we pro eed the way we did in the previous two types of problems as we are now fa ed with a problem at the origin. Again, for z = iR (R large enough), we have sin z eR e R = z z=iR 2R

! 1 as R ! 1:

So, for evaluating the given integral, we have to onsider a suitable fun tion and a suitable ontour whi h avoids the origin. First we rewrite 1 I = Rlim 2i !1 !0

(9.44)

(Z

 eix R

x

dx +

Z R ix e



x

)

dx

and note that jeiz j = e y  1 on the upper half-plane. Thus to evaluate the given integral we onsider Z

C

f (z ) dz; f (z ) =

eiz ; z

where C is the ontour shown in Figure 9.8. Note that C is made up of the (large) upper semi ir ular ontour R = fz = Rei : 0    g, the (small) semi- ir ular ontour  , where  = fz = ei : 0    g, and the real axis inter epted between them, namely the segments [ R; ℄ and [; R℄. Note that f has a simple pole at the origin and there are no other singularities. Sin e z = 0 lies outside C , we have, by the Cau hy theorem, R f ( z ) dz = 0; that is, C (9.45)

Z

 R

f (x) dx

Z



f (z ) dz +

Z R



f (x) dx +

Z R

f (z ) dz = 0:

Sin e zlim !0 zf (z ) = 1; by Theorem 9.41, we note that (9.46)

lim

Z

!0 

f (z ) dz = i( 0):

Alternatively, we an provide a dire t proof. As f (z ) has a simple pole with Res [f (z ); 0℄ = 1, we have f (z ) = 1=z + g(z ) for z near 0, where g(z ) is analyti at z = 0. In parti ular, Z

Z

Z

1 f (z ) dz = dz + g(z ) dz z





 Z Z  iei = d + g(z ) dz i 0 e

 = i +

Z



g(z ) dz:

391

9.4 Singularities on the Real Axis

The integral on the right tends to zero as  ! 0, be ause g(z ) is bounded near 0 and that Z



R

g(z ) dz

 sup jg(z )j ! 0 as  ! 0: z2 

R

Thus, lim!0  f (z ) dz = i. We next laim that limR!1 R f (z ) dz = 0. Note that e sin   e0 = 1 for  2 [0; ℄. But then we annot laim that the last integral approa hes zero as R ! 1. Instead, by Jordan's inequality (see Remark 9.38), we observe that Z

(9.47)



R

f (z ) dz 

Z 

0

 e R sin  d < R

! 0 as R ! 1:

So, if we allow R ! 1 and  ! 0 in (9.45), by (9.46) and (9.47), lim

(Z

R!1 !0

 R

f (x) dx +

Z R



)

f (x) dx = i:



By (9.44), we then have I = =2:

9.48. Example. By onsidering f (z ) = zeiz (z 2 1) 1 (a > 0), and C = [ R; 1 1℄ [ ( 1 ) [ [ 1 + 1 ; 1 2 ℄ [ ( 2 ) [ [1 + 2 ; R℄ [ R ; it is easy to prove that Z 1 x sin ax dx =  os a: 1 x2 1 Indeed, as usual, the Cau hy theorem gives Z

Z

1 1

R

1

+

Z 1 2

1+1

Z

2

+

Z R

1+2

+

!

Z R

f (z ) dz = 0:

As Res [f (z ); 1℄ = eia =2 and Res [f (z ); 1℄ = e ia =2, it an be seen that Z

Z

eia e ia lim f ( z ) dz = i and lim f ( z ) dz = i: 1 !0  2 !0  2 2 1 2 Further, Z

R

f (z ) dz



Z 

0

R2

R2

R2    e aR sin  d  2 ! 0 as R ! 1: 1 R 1 aR

The desired result follows by the limiting pro ess. Similarly, we an easily show that Z 1 sin ax (1 e am) (9.49) dx = (a > 0; m > 0): 2 2 2m2 0 x(x + m )

392

Evaluation of ertain Integrals

Then, by dierentiating both sides of (9.49) with respe t to m (keeping a as onstant) and using Leibnitz's rule, we easily get Z 1 sin ax  am ℄ (a; m > 0): (9.50) 2 + m2 )2 dx = 4m4 [2 (2 + am)e x ( x 0 Similarly, it is easy to see from (9.49) that (by dierentiating (9.49) with respe t to a and keeping m as onstant) Z 1

os ax e am (9.51) dx = (a; m > 0): 2 2 2m 0 x +m One an also give an independent proof for (9.50). In fa t, (9.51) has already been proved in Se tion 9.2. 

9.52. Remark. (i) For 0 < R < 1, integration by parts yields Z R

0

sin x dx = x = = = =

Thus,

Z R=2

sin 2x dx x 0 Z R=2 sin x 2 d(sin x) x 0   Z R=2 2 sin2 x R=2 sin x 2 sin xd x 0 x 0 Z R=2 Z R=2 2 4 sin (R=2) sin x sin2 x 2 d(sin x) + 2 dx R x x2 0 0 Z R=2 Z R=2 4 sin2 (R=2) sin 2x sin2 x dx + 2 dx: R x x2 0 0

Z R

Z

R=2 sin2 x sin x 2 sin2 (R=2) dx = + dx: x R x2 0 0 Sin e limR!1 R 1 sin2 (R=2) = 0; it follows that Z 1 Z 1 sin x sin2 x dx = dx x x2 0 0

and hen e, we have (9.53)

1 sin2 x  2 dx = 2 : x 0 (ii) One

an give an independent proof for (9.53) by onsidering the R integral C f (z ) dz , where

f (z ) =

Z

1 e2iz 1 os 2x 2 sin2 x ; Re f ( x ) = = ; z2 x2 x2

393

9.4 Singularities on the Real Axis

and C is the ontour shown in Figure 9.8. Sin e lim !0 zf (z ) = 2i; we have Z

lim f (z ) dz = i( 2i)( 0) = 2; z!0  where  is positively oriented. Further, j1 e2iz j  1 + e 2y  2 for y  0 and so, Z Z 2 f (z ) dz = 0: f (z ) dz  ; i.e. Rlim !1 R R R As in the above example, we easily dedu e that lim R!1 !0

"Z

 R

f (x)dx +

Z R



#

f (x) dx = 2; i.e.

Z

1 1 e2ix dx = 2: 1 x2

On equating real parts and taking a

ount of the fa t that sin2 x=x2 is an even fun tion we obtain (9.53). (iii) Sin e sin3 x =

 ix e

 e ix 3 1 = Im [(1 e3ix ) 3(1 eix )℄; 2i 4

we have limz!0 zf (z ) = 3, where (1 e3iz ) 3(1 eiz ) : z3 Therefore, pro eeding as in the above example, it is easy to show that Z 1 3 sin3 x  3 dx = 8 : x 0

f (z ) =

9.54. Example. By the error integral, we have p = 2 Z 1 e x2 dx (9.55) 0

Using (9.55), we an evaluate the integrals Z 1 Z 1 sin x2 dx and

os x2 dx: 0

0

2 To do this, we onsider the entire fun tion f (z ) = e z . The ontour C shown in Figure 9.3 (with  = =4) shows that

(9.56)

Z R

0

f (x) dx +

Z R

f (z ) dz +

Z 0

R

f (xei=4 ) d(xei=4 ) = 0:

394

Evaluation of ertain Integrals

R R 2 Note that R0 f (xei=4 ) d(xei=4 ) = ei=4 0R e ix dx and Z

R

f (z ) dz

=

Z =4 i i f (Re )iRe d 0 Z =4

 R

2 e R os 2 d

0

Z

R =2 R2 os  e d = 2 0 Z R =2 R2 sin  = e d ( = =2 ) 2 0 2 R (1 e R ) < ! 0 as R ! 1: 2 2R2 Taking the limit in (9.56), as R ! 1, we obtain Z 1 Z 1 2 i= 4 f (x) dx e e ix dx = 0: 0

So, by (9.55), (9.57)

0

1

Z

2 e ix dx = e i=4



p 

: 2 Separating the real and the imaginary parts in (9.57), we on lude that p Z 1 Z 1

os x2 dx = sin x2 dx = p :  2 2 0 0 0

9.58. Remark. The above 2 method helps us to prove a more general result by hoosing f (z ) = e z and letting C be the ontour shown in Figure 9.3. Then, pro eeding as in Example 9.54, it is lear that Z 1 Z 1 2 f (x) dx = ei e x ( os 2+i sin 2) dx: 0

Using (9.55),

0

1

Z

0

e

x2 os 2 e ix2 sin 2 dx =

pe

2

i

:

Equating the real and imaginary parts, we see that p Z 1 2 os 2 x 2 e

os(x sin 2) dx =

os  2 0 and

1

Z

0

p

2  e x os 2 sin(x2 sin 2) dx = sin : 2

395

9.4 Singularities on the Real Axis



Note that  = =4 yields the Fresnel integrals.

9.59. Example. Finally, by integrating eaz = osh bz ( b < a < b) around the re tangular ontour with verti es at R and R + i=b, we show that Z 1 eax  1 I= dx = :

osh bx b

os( a= 2b) 1 R (a=b)y As I = 1b 11 e osh y dy, it suÆ es to prove the result for b = 1 and a 2 ( 1; 1). So, we let eaz f (z ) = :

osh z Then, inside the re tangular ontour with verti es at R and R + i, f has only one pole, namely, at z = i=2. Further,

i eaz Res f (z ); i = = ieai=2 2 sinh z z=i=2 and therefore, by the residue theorem, we see that h

Z R

R

[f (x) f (x + i)℄ dx +

(9.60)

+

Z 0



Z 

0

f (R + iy)i dy

f ( R + iy)i dy = 2eai=2 :

Consider the se ond integral in (9.60): Z  f ( R + iy ) i dy

0

Sin e



Z 

0

jf (R + iy)j dy =

(R+iy)

j osh(R + iy)j = e

Z 

0

eaR j osh(R + iy)j dy:

+ e (R+iy) eR e R  2 2

R

 e4

and 1 < a < 1, the se ond integral in (9.60) approa hes zero as R ! 1. Similarly, we an show that the third integral in (9.60) approa hes zero as R ! 1. Thus, integrals along 1 and 2 , i.e. the se ond and third integrals in (9.60), approa h zero with in reasing R, so that (9.60) be omes Z 1 Z 1 [f (x) f (x + i)℄ dx = (1 + eia ) f (x) dx = 2eia=2 ; 1 1 that is,  ia=2  Z 1 eax e  dx = 2 = :  ia 1+e

os(a=2) 1 osh x

396

Evaluation of ertain Integrals

9.5 Integrals Involving Bran h Points

Integrals featuring x and log x an in some ases be evaluated using ontour integration. We illustrate this in detail by onsidering a spe ial ase of the real integral Z 1  1 x dx for 0 <  < 1: 1 +x 0 The pro ess used to evaluate su h integrals is often referred to as the integral around bran h point. Clearly, the integral is improper for two reasons as it has a in nite dis ontinuity at the origin and has an in nite limit of integration. Moreover,

x 1 x 1 s x 1 for x near 0, and s x 2 for x near 1 1+x 1+x so that the integral does onverge for 0 <  < 1. Observe that when x is repla ed by z , the integrand be omes z 1 1+z whi h is a multiple-valued fun tion. The origin is a bran h point of f (z ). We onsider the bran h of f (z ) on the slit plane C n[0; 1) so that the positive real axis has been hosen as the bran h ut for f (z ) with z  1 = e( 1)(ln jzj+i arg z) ; 0 < arg z < 2: f (z ) =

This guarantees that f (z ) is single-valued and we an integrate along an appropriate ontour. The fun tion f has a simple pole at z = 1 with residue Res [f (z ); 1℄ = z! lim1(1 + z )f (z ) = z! lim1 z 

1

= e(

1)i

= ei :

Further the residue theorem is appli able only to single-valued fun tions and the origin annot be inside the simple losed ontour C along whi h we integrate. Now, we let R > 1 >  > 0 and set (see Figure 9.9)

C = [ + iÆ; R + iÆ℄ [ R [ [R iÆ;  iÆ℄ [ ( ) so that the inside of C is a simply onne ted domain not ontaining the origin but ontaining the point z = 1. Thus, C onsists of (i) the horizontal line segment + from  + iÆ to R + iÆ (ii) the ir ular ar R of radius R entered at the origin tra ed ounter lo kwise from R + iÆ to R iÆ (iii) the horizontal line segment from R iÆ to  iÆ

397

9.5 Integrals Involving Bran h Points y iR ŴR

−R

ǫ + iδ γ− γ+ R ǫ − iδ

−γǫ

x

−iR

Figure 9.9: Contour for a multi-valued fun tion.

(iv) the ir ular ar (  ) of radius  entered at the origin tra ed lo kwise from  iÆ to  + iÆ. The residue theorem yields Z z 1 (9.61) I= dz = 2iei : C 1+z The value of the integral is independent of Æ, R and , but it depends only on the fa t that z = 1 lies inside C . Therefore, it is natural to write (9.61) equivalently as (9.62)

I=

Z

+

+

Z R

+

Z

+

Z !  1 z



1+z

dz = 2iei:

Our method of approa h will then be to let Æ ! 0, R ! 1, and  ! 0 to obtain the desired value of the given real integral. Note that, despite what is shown in Figure 9.9, it is possible to regard the slit [0; 1) as having an upper side of the positive real axis for whi h arg z = 0 and a lower side of the positive real axis for whi h arg z = 2. For the integrals over R and  , the standard ML-inequality gives Z Z  1  1 z z  1 jdz j  R dz  2R s R 1 ! 0 R 1 jzj=R 1 + z R 1+z as R ! 1 (sin e 0 <  < 1), and Z



z  1  1 dz  2 s  ! 0 as  ! 0: 1 

 1 + z For the remaining two integrals, we pro eed as follows. We have Z

Z

R (x + iÆ ) 1 z 1 dz = dx:

+ 1 + z  1 + (x + iÆ )

398

Evaluation of ertain Integrals

Given  > 0, we an hoose Æ > 0 small enough so that Z +

whi h gives

z 1 dz 1+z



Z R

x 1 dx <   1+x

Z

Z

R x 1 z 1 dz = dx: lim Æ!0 + 1 + z  1+x

As we integrate along (as Æ ! 0),

z

= e(

1

1)(ln jx+iÆj+i arg(x+iÆ))

! e(

1)(ln x+i2)

= x 1 e2i

so that Z Z   1 Z R  1 z 1 x x lim dz = e2i dx = e2i dx: Æ!0 1 + z 1 + x 1 +x R 

Therefore, (9.63)

lim

Æ!0

Z

+

+

!

Z

Z R  1 x z 1 2 i dz = (1 e ) dx: 1+z 1 +x 

Allowing R ! 1,  ! 0 and using (9.62), we obtain (1

e2i )

Z

1 x 1 dx = 2iei 0 1+x

whi h yields the required identity (9.64)

1 x 1 2iei  dx = 2i = : e 1 sin  0 1+x

Z

This identity an also be extended to omplex values of the parameter . To do this, we onsider

F (w) =

1 xw 1  dx and G(w) = : sin w 0 1+x

Z

Then, F and G are analyti on the strip D = fw : 0 < Re w < 1g and oin ides on the interval (0; 1). In view of this observation and the uniqueness theorem, (9.64) holds for all the omplex values of the parameter  with 0 < Re  < 1

399

9.6 Estimation of Sums y (−1 + i)(N +

1 ) 2

(1 + i)(N + 21 ) O

N N +1 x (1 − i)(N + 21 )

(−1 − i)(N + 12 )

Figure 9.10: Square CN .

9.6 Estimation of Sums Convergen e tests for series enable us to verify whether a given series onverges to a nite limit, but they do not give the value of the sum. On the other hand, al ulus of residues permits us to express ertain integrals as a nite sum of the residues of the integrand. Therefore if an in nite sum, su h as 1 1 X X f (n); and ( 1)n f (n); n= 1 n= 1

an be reorganized as a sum of the residues of f , then the al ulus of residues may help us to evaluate it, provided f is a meromorphi fun tion of a fairly simple kind. Thus, for the rst series, we must onstru t a fun tion whose residues are given by ff (n) : n 2 Zg: For this, let f be a fun tion whi h is analyti ex ept for a nite number of poles a1 ; a2 ; : : : ; am (ea h is not an integer). Suppose g is any fun tion having simple poles at z = n (n 2 Z) su h that Res [g(z ); n℄ = 1 (for example, su h fun tions are given by  ot z and 2i(e2iz 1) 1 ): Then for ea h n 2 Z (n 6= ak ; k = 1; 2; : : : ; m), we have Res [f (z )g(z ); n℄ = f (n): Thus if CN is a losed ontour en losing points z = 0; 1; 2; : : : ; N and ak (k = 1; 2; : : : ; m), we have (see Figure 9.10), by the Cau hy residue theorem, Z

where

CN

f (z )g(z ) dz = 2i X=

and

Y=

X

N X n= N n6=ak

X

Res [f (z )g(z ); CN ℄ = 2ifX + Y g

Res [f (z )g(z ); n℄ =

N X n= N n6=ak

Res [f (z )g(z ); at the poles of f in CN ℄ =

f (n)

m X k=1

Res [f (z )g(z ); ak ℄:

400

Evaluation of ertain Integrals

Let us now des ribe this idea by taking g(z ) =  ot z ,  s z and in either ase show that

N

lim

Z

!1

CN

f (z )g(z ) dz = 0:

9.65. Theorem. Suppose f is meromorphi in C , with a nite number of poles a1 ; a2 ; : : : ; am . Suppose, moreover, that there exist two positive numbers M and R su h that for jz j > R

jz r f (z )j  M

(9.66) Then

N

lim

and

N

lim

N X

! 1 nn=6=aN

for a xed r > 1:

f (n) =

k

N X

! 1 nn=6=aN

( 1)n f (n) =

k

m X k=1

Res [ ot zf (z ); ak ℄

m X k=1

Res [ s zf (z ); ak ℄:

Proof.PEquation (9.66) that jf (n)j  Mn r for jnj > R and so P means n the series f (n) and ( 1) f (n) are onvergent sin e the n-th term is dominated by Mn r for large n and r > 1. By hypothesis, for n 6= ak , k = 1; 2; : : : ; m; Res [ ot zf (z ); n℄ = f (n) and Res [ s zf (z ); n℄ = ( 1)n f (n): Let CN be the square with verti es at (N + 1=2)(1  i) en losing all the poles of f (see Figure 9.10), where N is a positive integer (take for instan e N > jak j for all k = 1; 2; : : : ; m). Therefore, by the Cau hy residue theorem, Z N m X X 1  ot zf (z ) dz = f (n) + Res [ ot zf (z ); ak ℄ 2i CN n= N k=1 n= 6 ak

and

Z N m X X 1  s zf (z ) dz = ( 1)n f (n) + Res [ s zf (z ); ak ℄: 2i CN n= N k=1 n6= ak

The result will follow if we an show that Z

Z

lim  ot zf (z ) dz = 0 = lim  s zf (z ) dz: N !1 CN N !1 CN For this we require that there exist two onstants K1 and K2 su h that j ot z j < K1 and j s z j < K2 for all N and for all z on CN . In fa t, we will now prove the following inequalities:

401

9.6 Estimation of Sums

(i) j ot z j < 2 for all z on CN (ii) j s z j < 1 for all z on CN . First we set  = N + 1=2. If z is on the horizontal sides of CN , then we

an write z = x  i, where jxj  . For z = x  i on these horizontal lines, we have iz ix  iz ix  j ot z j = eeiz + ee iz = eeix ee + ee ix ee : By the triangle inequality, jeixe + e ixe j jeixe e ixe j

 

e + e = e + e  ; je e j = e e 

so that 

j ot z j  ee + ee

 

= oth()  oth(3=2) < 2

(sin e the later expression is maximized at  = =2, i.e. at N = 0). Similarly, if z lies on the verti al sides then z =  + iy with jyj  , and so for su h z , we have

j ot z j = j ot ( + iy)j j os() osh y i sin() sinh yj = j sin() osh y + i os() sinh yj j sinh yj < 1: = j tanh yj = p 2 1 + sinh y

Thus, j ot z j < 2 on CN and this proves (i). To prove (ii), we rst note that j sin(x + iy)j2 = sin2 x +sinh2 y: When z = x  i lies on the horizontal sides of CN , 1 1 j s 2 z j = 2  0; a dire t al ulation yields Res [f (z ); ia℄ = and for k 2 Z, Res [f (z ); k℄ =

1 = Res [f (z ); ia℄ 2a sinh a 



1 1 ( 1)k = : 2 2 (k + a )  os k (k2 + a2 )

Thus, for a > 0, we have 







ot z

oth a

ot z Res 2 2 ; ia = = Res 2 2 ; ia : z +a 2a z +a (e) Similarly if a is a non-zero real and not an integer, then, for the fun tion f (z ) = 1=[(z + a)2 sin z ℄, we easily get Res [f (z ); a℄ =  s a ot a and for k 2 Z, Res [f (z ); k℄ = ( 1)k =[(k + a)2 ℄:



9.68. Example. The fun tion f (z ) = z 2m (m 2 N) satis es the

ondition (9.66). Sin e f is even and has a pole of order 2m at 0, Theorem 9.65 gives   1 1 X  ot z 2 + Res ; 0 =0 2m z 2m n=1 n

403

9.6 Estimation of Sums

and 2

1 X

h  s z i ( 1)n + Res ; 0 = 0: 2 m z 2m n=1 n

For m = 1, we nd that (see Example 9.67) 



h  s z i  ot z 2 2 Res ; 0 = ; Res ; 0 = ; z2 3 z2 6

and hen e, we on lude that 1 1 2 1 ( 1)n X X 2 (9.69) = and = : 2 2 6 12 n=1 n n=1 n More generally, sin e (z ) = z ot z has a removable singularity at 0 and lim (0) 6= 0, we note that z!0    ot z Res ; 0 = Res [(z )z 2m 1; 0℄ z 2m Z (z ) 1 = dz; r > 0 is small, 2i jzj=r z 2m+1 (2m) (0) = : (2m)! Pro eeding as above and taking m = 2 we easily derive 1 1 4 X 4 = 90 : n=1 n If we let f (z ) = 1=(z 2 + a2 ); then Theorem 9.65, for a 2= iZ, yields that       1 X  ot z  ot z  ot z Res 2 2 ; ia + Res 2 2 ; ia + Res 2 2 ; n = 0: z +a z +a z +a n= 1 Thus, by Example 9.67(d), we on lude that 1  oth a X 1 (9.70) + 2 + a2 = 0 a n n= 1 whi h gives 1 1 X X 1  1 1 a oth a 1 2 =

oth a ; i.e. : 2 + a2 2 2 + a2 = n a a n 2a2 n=1 n=1 Note that if a is real and positive then, by using uniform onvergen e and making a ! 0, we get the rst equation in (9.69). Dierentiation of the

404

Evaluation of ertain Integrals

last equality with respe t to a immediately yields (sin e the onvergen e is uniform on any ompa t set disjoint from the set fia : a 2 Zg, term by term dierentiation is permitted),   1 X 1 1 2 a2 = + a

oth a 2 : 2 2 2 4a4 sinh2 a n=1 (n + a ) Further, from (9.70) we easily obtain that, for b 2= Z; 1 X 1   ot b : 2 b2 = ib oth( (ib)) = n b n= 1 Thus by grouping the terms with n = m, m = 1; 2; : : : , we have  1  1 1 X 1 1 X 1 1 1  ot b + + 2 ; = 2 b2 = b2 2b m=1 m b m + b b2 m b m=1 that is (9.71)

1 1 X 2b  ot b = + : b m=1 b2 m2

Again, dierentiation of the above result with respe t to b, yields 1 X 1 2 2 (9.72) = Z): 2 =  s b (b 2 ( m b ) m= 1 Similarly, by applying Theorem 9.65, we see that for a 6= 0; i; 2i; : : :       1 X  s z  s z  s z Res 2 2 ; ia + Res 2 2 ; ia + Res 2 2 ; n = 0: z +a z +a z +a n= 1 Thus, by Example 9.67(d), this implies 1 ( 1)n 1 ( 1)n X   s a X + = 0; i.e. = s a: 2 2 2 2 a a n= 1 n + a n= 1 n + a We an also prove (9.72) by onsidering the fun tion f (z ) = (z b) 2 (b not an integer). Indeed, from Theorem 9.65, it follows that (see also Example 9.67( ))     1 X  ot z  ot z Res ; b + Res ; n = 0; (z b)2 (z b)2 n= 1 that is 1 X 1 2 2 = Z: 2 =  s b; b 2 ( n b ) n= 1

405

9.7 Exer ises

Taking b = 1=2 in the last identity, we nd that 1 1 X X 1 1 2 + = 2 2 4 n= 1 (2n + 1) n=0 (2n + 1) so that (9.73)

1 X

1 2 = : 2 8 n=0 (2n + 1)

From (9.73) we an easily dedu e the rst equation in (9.69), be ause 1 1 1 1 1 1 1 X X X 1 2 1 X = + = + : 2 2 2 8 4 n=1 n2 n=0 (2n + 1) n=1 (2n) n=1 n



9.7 Exer ises 9.74. Prove the following integrals Z 2 d 2 1. = p 2 2 (a; b 2 R; jbj < jaj). a + b

os  a b 0 Z 2 d (2a + b) 2. ; a; b > 0. = 2 2 [a(a + b)℄3=2 0 (a + b os ) Z 2

os2 3 (1 + 2 ) 3. d = , 1 <  6= 0 < 1. 2 2 os 2 1  0 1+ Z 2 (4 2 + 1)

os 2 4. d = , 0 < jj < 1. 2 2 os  2 (2 1) 0 1+ Z 2 d 2 5. = p 2 2 2 ; a2 > b2 + 2 . a + b

os  +

sin  a b 0 Z 2 d 2a 6. = p3 2 2 2 ; a2 > b2 + 2 . 2 a b 0 (a + b os  + sin ) 9.75. If a and b are real with jaj > jbj and n is a positive integer, prove that p2 2n Z 2

os n 2b(a a b) p I= d = : 2 a + b

os  a b2 0 9.76. Show the following: Z 3z + 1 1. I = dz = 0 z ( z + 2)(z i)2 jzj=3 Z

os(e z ) 2. dz = 2i sin(1): jzj=1 z 2

406

Evaluation of ertain Integrals

9.77. Prove the following real integrals: Z 1 x2 x + 2 5 1. 4 2 + 9 dx = 12 x + 10 x 1 p Z 1 dx = (2 2) 2. = p (a; b; > 0 and b2 a > 0) 1 a2 + 2b2 x2 + 2 x4 a b2 a Z 1 n

os x e 1 X (2n k)!2k 3. dx = , n 2 N. 2 n+1 n!22n+1 k=0 k!(n k)! 0 (1 + x ) Z 1 dx  4. 2 + m2 )(x2 + n2 ) = 2mn(m + n) ; m; n > 0: ( x 0 Z 1 dx (m + 2n) 5. 2 + m2 )(x2 + n2 )2 = 4mn3 (m + n)2 ; m; n > 0; m 6= n: ( x 0 Z 1

os ax  am ; a; m > 0: 6. 2 + m2 )2 dx = 4m3 (1 + am)e ( x 0 Z 1 x sin ax ae am 7. dx = ; a; m > 0: 2 2 2 4m 0 (x + m ) Z 1 dx  (2(n 1))! 8. 2 )n = 22(n 1) ((n 1)!)2
2 ; n 2 N : (1 + x 0 Z 1

os ax dx (me an ne am ) 9. = ; m > n > 0; a > 0 2 2 2 2 2(m2 n2 )mn 0 (x + m )(x + n ) and hen e dedu e that Z 1 (e an e am) x sin ax dx = : 2 2 2 2 2(m2 n2 ) 0 (x + m )(x + n ) 9.78. Prove the following real integrals:  p Z 1 22 1  1. e (1+i) t dt = for 1    1. 1 + i 2 0 Z 1 x  2. 2 dx = sin  for 1 <  6= 0 < 1. (1 + x ) 0 Z 1 ln x  ln a 3. 2 + a2 dx = 2a for a > 0. x 0

Chapter 10

Analyti Continuation

Analyti ontinuation is an important idea be ause it provides a method for making the domain of de nition of an analyti fun tion as large as possible. Usually analyti fun tions are de ned by means of some mathemati al expressions su h as polynomials, in nite series, integrals et . The domain of de nition of su h an analyti fun tion is often restri ted by the manner of de ning the fun tion. For instan e, the power series representation of su h analyti fun tions does not provide any dire t information as to whether we ould have a fun tion analyti in a domain larger than disk of onvergen e whi h oin ides with the given fun tion. In Se tion 10.1, we dis uss a general method (su h as the power series method). In Se tion 10.2, we present a te hnique for arrying out the ontinuation pro ess and prove the Monodromy theorem. In Se tion 10.3, we develop the Poisson integral formula for harmoni fun tions on the open unit disk (and hen e, for arbitrary disks). We use the Poisson integral to solve the Diri hlet problem for the unit disk and, as a onsequen e, we hara terize harmoni fun tions by the mean value property just as Morera's theorem hara terizes analyti fun tions. Later in Se tion 10.4, we use this hara terization to establish the Symmetry prin iple (due to S hwarz) for harmoni fun tions whi h enables one to nd an analyti ontinuation expli itly under a spe ial situation.

10.1 Dire t Analyti Continuation What do we mean by an \analyti ontinuation"? It is simply a pro ess of extending the domain of analyti ity to larger domains. For example, if is a domain and fj 2 H( ) for j = 1; 2; su h that f1 (z ) = f2 (z ) for all points z in an open subset D  , then, by the uniqueness theorem (see Theorem 4.106), one has f1  f2 on . So, a natural question is the following: Is it always possible to have an extension? Clearly, not. For example, 1 f (z ) = for z 2 D = C n f0g z

408

Analyti Continuation

does not have an extension to C . Similarly, if D = C

ut plane and

n fx : x  0g is the

f1 (z ) = Log z; z 2 D f2 (z ) = z 1=2 = e(1=2) Log z = jz j1=2 ei(1=2)Arg z (  < Arg z < ); f3 (z ) = z 1=2 = e(1=2)( Log z+2i) = e(1=2) Log z (  < Arg z < ); then no extension from D to C is possible in ea h ase. However, if the extension is possible then there are ways to arry out the pro ess of ontinuation so that the given analyti fun tion be omes analyti on a larger domain. To make this point more pre ise, let us start by examining the analyti ontinuation of the fun tion (10.1)

f (z ) =

X

n0

zn:

The series on the right de ned by (10.1), as is well known, is onvergent for jz j < 1 and diverges for jz j  1. On the other hand, we know that the series given by the formula (10.1) represents an analyti fun tion for jz j < 1 and the sum of the series (10.1) for jz j < 1 is 1=(1 z ). However, the fun tion F de ned by the formula 1 F (z ) = 1 z is analyti for z 2 C 1 nf1g = D (sin e F (1=z ) = (1 z 1) 1 = z=(z 1) is analyti at 0, F (z ) is analyti at z = 1). Now

f (z ) = F (z ) for z 2
\ D and we all F an analyti ontinuation of f from
into the domain D, i.e. fun tion f , given at rst for jz j < 1, has been extended to the extended

omplex plane but for the point z = 1 at whi h the fun tion has a simple pole. Thus, it appears that F whi h is analyti globally is represented by a power series only lo ally. Next we onsider another fun tion g de ned by Z 1 (10.2) g(z ) = exp[(z 1)t℄ dt: 0

If Re z < 1, then the integral onverges and Z 1 e(z 1)t 1 1 1 exp[(z 1)t℄ dt = = z 1 = 1 z: z 1 0 0 Thus the integral de ned by (10.2) is onvergent in the half-plane H = fz : Re z < 1g and represents the same fun tion 1=(1 z ) for z 2 H . Consequently, we have F (z ) = g(z ) for z 2 H \ D and hen e we all F the

409

10.1 Dire t Analyti Continuation

ontinuation of g into the half-plane Re z  1 with the ex eption of point at z = 1. Similarly if D1 = C n fz = x : xP 1g, nthen Log (1 z ) is the analyti ontinuation of the power series n1 zn from
into D1 .

10.3. De nition. Suppose that f and F are two fun tions su h that (i) f is analyti on some domain D  C , (ii) F is analyti in a domain D1 su h that D1 \ D = 6 su h that f (z ) = F (z ) for z 2 D \ D1 .

; and D1  D,

Then we all F an analyti ontinuation or a holomorphi extension of f from D into D1 . In other words, f is said to be analyti ally ontinuable into D1 . For a given analyti fun tion f on D, if there exists an analyti ontinuation F of f into D1 , by the uniqueness theorem (see Theorem 3.75), then it is uniquely determined. Thus, we raise

10.4. Problem. When does a power series represent a fun tion whi h is analyti beyond the disk of onvergen e of the original series? One way to provide an aÆrmative answer is by \power series method". Let us start our dis ussion on this method and see how one an use the power series to go beyond the boundary of the disk of onvergen e. A fundamental fa t about a fun tion f 2 H( ) is that for ea h a 2 , there exists a sequen e fan gn0 and a number ra 2 (0; 1℄ su h that 1 X f (z ) = an (z a)n for all z 2
(a; ra ): n=0

To extend f , we hoose a point b other than a in the disk of onvergen e
(a; ra ). Then jb aj < ra and 1 1 X X an (z a)n = an [z b + b a℄n n=0

n=0

= = =

1 X

n=0

1 X

k=0

1 X

k=0

an

n   X n k=0

1 X

k

 

(b

a)n k (z !

b)k

!

n an (b a)n k (z b)k k n=k Ak (z b)k :

410

Analyti Continuation

The inter hange of the summation is justi ed, sin e 1 n   1 X X X n jan j j b ajn k jz bjk = jan j(jz bj + jb aj)n < 1 k n=0 n=0 k=0

whenever jz bj + jb aj < ra : Therefore, the series about b onverges at least for jz bj < ra jb aj. However, it may happen that the disk of onvergen e
(b; rb ) for this new series extends outside
(a; ra ), i.e. it may be possible that rb > ra jb aj. In this ase, the fun tion an be analyti ally ontinued to the union of these two disks. This pro ess may be

ontinued. For example, if f (z ) = 1 1 z then (for z 2
with a = 0; ra = 1) we have X 1 = z n ; z 2
: 1 z n0 Take b = i. In order to get the expression for z 2
(i; rb ), we write 1 1 = 1 z 1 i (z i) 1 X p 1 = (z i)n ; jz ij < j1 ij = 2; n +1 n=0 (1 i) 1 X p (1 + i)n+1 = An (z i)n ; An = n+1 ; jz ij < rb = 2: 2 n=0 P1 P n Thus, np=0 An (z i)n is an analyti ontinuation of 1 n=0 z in
to the P1 1 n disk
(i; 2). Similarly, one an see that n=0 n+1 (z + 1) is an analyti P n from
to the disk
( 2 1; 2).

ontinuation of 1 z n=0

10.5. P Remark. The onvergen e or divergen e of a power series f (z ) = n0 an z n at a point does not determine whether it an or annot be extended beyond that point. For example, onsider X X zn f1(z ) = z n and f2 (z ) = 2: n0 n1 n Re all that the rst series diverges for jz j = 1 and its sum f1 (z ) = 1=(1 z ) de ned by this series is analyti in C n f1g. On the other hand, the se ond series onverges at all points on jz j = 1. However, in both ases, the series

annot be ontinued analyti ally to a domain D with 1 2 D. Note that Z z log(1 t) f2 (z ) = dt and f200 (z ) ! 1 as z ! 1 along reals.  t 0 10.6. Example. As another example of analyti ontinuation, we

onsider the power series (10.7)

f0(z ) =

X

( 1)n z 2n:

n0

411

10.1 Dire t Analyti Continuation y 1 i



√  5 1 ; 2 2

1 −1

O

1 3 2 4

x

1 1

−i

3 5 ; 4 4




Figure 10.1: Dire t analyti ontinuation.

Then (10.7) is absolutely onvergent for z unit disk. Further, if we let

2
and diverges outside the





1 i 1 1 F0 (z ) = = + ; 1 + z2 2 i + z i z

(10.8)

then it is easy to see that F0 is analyti in the extended plane ex ept at z = i and f0 (z ) = F0 (z ) for all z 2
: We also note that F0 (x) = 1=(1+x2) is well de ned for all real values of x, whi h is expandable as a real power series about any point on the real axis. Yet the power series (e.g. about 0) given by (1 + x2 ) 1 = 1 x2 + x4


has the unit interval as the interval of onvergen e and so, this illustrates an interesting fa t about Taylor's series on the real axis. Suppose we want to ontinue f0 on
to a disk about z0 . Then, we write "

i 1 1 F0 (z ) = 2 i + z0 1 + zi+zz00 Thus if (10.9)

F1 (z ) =



!

+

1

i z0 1

1

!#

z z0 i z0

:



iX ( 1)n 1 + (z z0 )n ; 2 n0 (i + z0 )n+1 (i z0)n+1

where R = minfji + z0j; ji z0jg is the radius of onvergen e for (10.9),pthen F1 (z ) is analyti for jz z0 j < R. For instan e if z0 = 1=2, then R = 5=2 and hen e, in this hoi e, we see that (10.9) onverges for some values of z for whi h (10.7) diverges (see Figure 10.1). Therefore, (10.7) and (10.9) both represent parts of (10.8) whi h en ompasses all possible extensions.

412

Analyti Continuation

However, there is no extension whi h is analyti at z = i whi h agrees with F0 in a deleted neighborhood of i, be ause 1  1 + z 2 ! 1 as z ! i: An analyti fun tion f on a domain D will be alled a fun tion element, written as (f; D). If (f1 ; D1 ) and (f2 ; D2 ) are two fun tion elements su h that D1 \ D2 6= ; and f1 (z ) = f2 (z ) for all z 2 D1 \ D2 : Then (f; D1 [ D2 ) is also a fun tion element, where  f (z ) for z 2 D1 f (z ) = 1 f2 (z ) for z 2 D2 : With the observation just made, (f1 ; D1 ) and (f2 ; D2 ) are alled a dire t analyti ontinuation of ea h other, thereby de ning an analyti fun tion in D1 [ D2 . For instan e, if f1 (z ) = a=(a z ) for z 2 C n fag; and

f2 (z ) =

X  z n

n0

a

for z 2 D2 = fz : jz j < jajg;

then (f1 ; C n fag) is a dire t analyti ontinuation of (f2 ; D2 ).

10.10. Example. De ne f0(z ) =

Z

os  1 d 2i j j=1  z

for z 2
: Then, by the Cau hy integral formula, we have f0 (z ) = os z for ea h z 2
: But, sin e os z is analyti in C , ( os z; C ) is a dire t analyti

ontinuation of (f0 ;
).  In general, the following theorem holds:

10.11. Theorem. Let C be a simple losed ontour with interior D and g be an entire fun tion. If, for n 2 N , f0(z ) =

Z

n! g( ) d; z 2 D; 2i C ( z )n+1

then (g (n) ; C ) is a dire t analyti ontinuation of (f0 ; D).

Proof. By the Cau hy integral formula, f0 (z ) = g(n)(z ) for all z 2 D: Sin e g is entire, so is g(n) ; the result follows. 10.12. De nition. Let f be analyti on a domain D. If f annot be

ontinued analyti ally a ross the boundary D, then D is alled natural

413

10.1 Dire t Analyti Continuation

boundary of f . A point z0 2 D is said to be a regular point of f (z ) if f

an be ontinued analyti ally to a region D1 with z0 2 D1 : Otherwise, f (z ) is said to have a singular point at z0 .

For instan e, onsider the power series

f (z ) =

(10.13)

X

k 0

k

z2 :

A dire t onsequen e of the Root test is that the radius of onvergen e of (10.13) is 1 and so, f de ned by (10.13) is analyti for jz j < 1. If jz j  1, n then limn!1 jz 2 j 6=n 0 and therefore, the series diverges for jz j  1. Let  = e2im=2 ,nm = 0; 1; : : : ; 2n 1 (n 2 N ), be the 2n -th root of unity. If z = re2im=2 2
, then nX1 1 k X 2k f (z ) = z2 + z k=0

and so for r ! 1 , we have 1 k X f (r)j  r2 k=n

k =n

1 n X 2k z

k=0



1 X k=n

k

r2

n;

and hen e, for every 2n -th root of unity  , we have limr!1 jf (r)j = 1. Therefore if D is a domain ontaining npoints of
and of its omplement, then D ontains the points  = e2im=2 and so any fun tion F in D whi h

oin ides with f in D \
annot be ontinued analyti ally through  2n = 1 for ea h n 2 N . In other words any root of the equation

z 2 = 1; z 4 = 1; : : : ; z 2n = 1 (n 2 N ) is a singular point of f and hen e any ar , however small it may be, of 
ontains an in nite number of singularities. Thus, f on
annot be

ontinued analyti ally a ross the boundary 
of
. This observation shows that the unit ir le jz j = 1 is a natural boundary for the power series de ned by (10.13). Similarly if X (10.14) f (z ) = z k ! k 0 then, f 2 H(
). Upon taking  = e2im=n ; m = 0; 1; 2; : : : ; n 1; z = r; (where m=n is the irredu ible fra tion), and hoosing r lose to 1 from below along a radius of the unit ir le it an be seen that limr!1 jf (r)j = 1. Hen e, f is singular at every n-th root of unity for any n 2 N . Sin e every point on jz j = 1 is a singular point, f annot be ontinued analyti ally through the n-th root of unity for any natural number n. In

414

Analyti Continuation y

∂1(ζ ; Rζ ) ζ

O

R

x

∂1R

Figure 10.2: Illustration for singularity on ir le jz j = R.

other words, there an be no ontinuation anywhere a ross jz j = 1 and hen e, jz j = 1 is a natural boundary for the power series de ned by (10.14). P

10.15. Theorem. If f (z ) = n0 an z n has a radius of onvergen e R > 0, then f must have at least one singularity on jz j = R. Proof. Suppose, on the ontrary that f has no singularity on jz j = R. Then f must be analyti at all points of jz j = R. This, together with Theorem 3.71, implies that f is analyti for jz j  R. It follows, from the de nition of analyti ity at a point, that for ea h  2 
R there exists some R > 0 and a fun tion f whi h is analyti in
( ; R ) (see Figure 10.2) and f = f on
( ; R ) \
R : In this way, if k and l 2 
R (k 6= l) with G =
(k ; Rk ) \
(l ; Rl ) 6= ; then we have two fun tions fk and fl whi h are respe tively analyti in
(k ; Rk ) and
(l ; Rl ) su h that f = fk = fl on G \
R : Sin e G is onne ted and G \
R is an open subset of G, by the uniqueness theorem, fk = fl on G. Sin e j j = R is ompa t, by the Heine-Borel theorem, we may sele t a nite number of
(1 ; R1 ),
(2 ; R2 ); : : : ;
(n ; Rn ) from the olle tion f
( ; R ) :  2 
R g su h that it overs the ir le 
R . Let

= [nk=1
(k ; Rk ) and Æ = dist (
R ; ): Then, as Rk > 0 for ea h k, we have Æ > 0. Moreover,

fz : R Æ < jz j < R + Æg  and
R+Æ  D =
R [ : Then g de ned by

g(z ) =



f (z ) for jz j < R fk (z ) for jz k j < Rk ; k = 1; 2; : : : ; n;

415

10.1 Dire t Analyti Continuation y e

iα

δ O

r δ+r x 1

Figure 10.3: Existen e of a singularity on the ir le of onvergen e.

is well de ned, single-valued and analyti on D and has same power series representation as f for jz j < R. Thus there exists an analyti fun tion, say , in
R+Æ , whi h oin ides with f on
R . But then by Taylor's Theorem we have the power series expansion

(z ) =

X

n0

bn z n for z 2
R+Æ :

Sin e f = g on
R , by the uniqueness theorem (see Theorem 3.75), we have an = bn for ea h n. This shows that the radius of onvergen e of f is R + Æ, whi h is a ontradi tion. P

10.16. Theorem. If an  0 and f (z ) = n0 an z n has the radius of

onvergen e 1, then (f;
) has no dire t analyti ontinuation to a fun tion element (F; D) with 1 2 D. Proof. For ea h z = rei 2
(0 < r < 1,  2 [0; 2)), we have (10.17)

f (k ) (z ) =

X

nk

n(n 1)


(n (k 1))an z n k

so that (sin e an  0) (10.18)

jf (k) (rei )j 

X

nk

n(n 1)


(n (k 1))an rn k = f (k) (r):

We have to show that 1 is a singular point of f . Suppose, on the ontrary, that 1 is a regular point of f . Then, f an be analyti ally ontinued in a neighborhood of z = 1 and so there is a Æ with 0 < Æ < 1 (see Figure 10.3) for whi h the Taylor's series expansion of f about Æ, namely the series (10.19)

X f (k) (Æ )

k0

k!

(z

Æ)k ;

416

Analyti Continuation

would be onvergent for jz nd that

Æj < r with Æ + r > 1. Now, by (10.18), we

jf (k) (Æei )j  f (k) (Æ) :

k! k! From this, the Root test and the omparison test with (10.19), it follows that the radius of onvergen e of the Taylor series about Æei is at least r. This observation implies that the Taylor series X f (k) (Æei ) (z Æei )k k ! k0 would be onvergent in the disk jz Æei j < r for ea h , with Æ + r > 1. In other words, the Taylors series X f (k) (z0 ) (z z0)k k ! k 0 about ea h z0 with jz0 j = Æ would have radius of onvergen e  r > 1 Æ. Sin e this ontradi ts Theorem 10.15, 1 must be a singular point of f . This

ompletes the proof.

Noti e that the last series is a tually a rearrangement of Indeed, by (10.17), 0

 

1

P

n0 an z

n.

 

n XX n n  an z0n k A (z z0 )k = an z0n k (z z0 )k k k k0 nk n0 k=0 X = an (z z0 + z0 )n n0 X = an z n : n0 X

X

P

10.20. Corollary. If an  0 and f (z ) = n0 an z n has the radius of onvergen e R > 0, then z = R is a singularity of f (z ). Finally, we state the following result whose proof may be found in standard advan ed texts (e.g. [24℄).

10.21. Theorem. Let f (z ) =

P

k0 ak z

nk and

lim inf k!1 nnk+1 k > 1.

Then the ir le of onvergen e of the power series is the natural boundary for f .

By this theorem, it is easy to see that the unit ir le jz j = 1 is the natural boundary for 1 z 3k 1 z 2k X X f1(z ) = and f ( z ) = 2 k k2 : k=0 3 k=0 2

417

10.2 Monodromy Theorem

10.2 Monodromy Theorem

P n We start with a power series 1 n=0 an (z z0 ) that represents a fun tion f with
(z0 ; R) as its disk of onvergen e. A fun tion element of the type (f;
(z0 ; R)) is alled \an analyti germ of f at z0 ", or brie y a \germ at z0 ". Obviously, an arbitrary fun tion element (f; D) determines a germ at ea h point of D. If : [0; 1℄ ! C is a urve with z0 = (0) as its initial point and f (z ) is a germ at z0 , then we say that f (z ) is ontinued analyti ally along if for every t 2 [0; 1℄ there is an (analyti ) germ at (t), i.e. there is a onvergent power series 1 X (10.22) ft (z ) = an (t)(z (t))n

n=0

for z 2 Dt = fz : jz (t)j < R(t)g, su h that (i) f0(z ) is the power series representing the fun tion f (z ) at z0 (ii) for ea h t 2 [0; 1℄, Dt is the disk of onvergen e with enter at (t) (iii) whenever s and t in [0; 1℄ are near to ea h other, then fs (z ) = ft (z ) for all z 2 Ds \ Dt , where Ds and Dt are the disks of onvergen e of fs and ft , respe tively (Note that Ds \ Dt 6= ;); i.e. when s is near t, (fs ; Ds ) and (ft ; Dt ) are dire t analyti ontinuations of ea h other. In this way, we obtain a one parameter family of germs fftg and refer to f1(z ) as the analyti ontinuation of f0 (z ) = f (z ) along , where we regard ft (z ) either as a series or as an analyti fun tion de ned near (t): Clearly, for s near t, the ondition (i) implies that

f (n) ( (s)) an (s) = t n! and so the Taylor oeÆ ients in (10.22) depends ontinuously on the parameter t. Further, in the domain Dt of the germ ft , the radius of onvergen e depends ontinuously on the enter of the expansion of the power series. More pre isely, we have 10.23. Lemma. Let : [0; 1℄ ! C be a urve and f be a germ at z0 = (0). Assume that f (z ) an be ontinued analyti ally along with a onvergent power series given by (10:22). Then either R(t) = 1 for all t 2 [0; 1℄, or jR(s) R(t)j  j (s) (t)j =: jzs zt j whenever s and t are su h that js tj < Æ for some Æ > 0, i.e. the radius of

onvergen e of ft is a ontinuous fun tion of t. Proof. Fix t so that 1 X ft(z ) = an (t)(z zt)n ; jz ztj < R(t); n=0

418

Analyti Continuation

where R(t) is the radius of onvergen e for the series about zt := (t) to whi h ft extends analyti ally. Thus, ft does not extend analyti ally to a larger disk ontaining Dt . If we onsider the power series 1 X an (s)(z zs )n n=0

and hoose zs 2 Dt so that jzs zt j < R(t), then the radius of onvergen e R(s) of this new series is at least R(t) jzs zt j. Consequently,

R(s)  R(t)

jzs zt j; i.e. R(t) R(s)  jzs zt j:

Inter hanging the roles of zs and zt , we see that

R(s) R(t)  jzs

zt j:

From the last two inequalities, we obtain that either R(t) = 1 for all t 2 [0; 1℄, or jR(s) R(t)j  jzs zt j holds for all s and t nearby. In parti ular, the ontinuity of implies that R(t) is a ontinuous fun tion of t. Suppose that f is analyti at z0 , and : [0; 1℄ ! C is a urve with z0 = (0) and z1 = (1), along whi h f has an analyti ontinuation ft . Then the message of Lemma 10.23 is that the radius of onvergen e R(t) of the power series about (t) that represents ft given by (10.22), is a

ontinuous fun tion of t on [0; 1℄. In fa t, R(t) is uniformly ontinuous on [0; 1℄. As R(t) > 0 for ea h t 2 [0; 1℄, we on lude that

R = minfR(t) : t 2 [0; 1℄g > 0 and so, R(t)  R for all t 2 [0; 1℄.

10.24. Example. Consider the prin ipal bran h of log z , where its Taylor series expansion about z = 1 is given by 1 ( 1)n 1 X f (z ) = Log z = (z 1)n ; z 2 D = fz : jz 1j < 1g: n n=1 Note that Log z is analyti in C n fz : Re z  0; Im z = 0g. Let : [0; 1℄ ! C be the urve given by (t) = e2it , starting from z0 = (0) = 1. Then f (z ) a tually has an analyti ontinuation along . In fa t, for ea h t 2 [0; 1℄, there is a olle tion of fun tion elements fft; Dt g su h that (i) (f0 ; D0 ) = (f; D) (ii) for ea h t 2 [0; 1℄, Dt is the disk of onvergen e with enter (t)

419

10.2 Monodromy Theorem

(iii) for ea h t 2 [0; 1℄, and s near t, we have Ds \ Dt 6= ; and fs (z ) = ft (z ) for all z 2 Ds \ Dt ; i.e. (fs ; Ds ) is a dire t analyti ontinuation of (ft ; Dt ). More pre isely, we have 1 ( 1)n 1 e 2int X (z e2it )n ft (z ) = 2it + n n=1 valid for z 2 Dt = fz : jz f0 (z ) = f (z ) and for t = 1

e2it j < 1g: In parti ular, for t = 0, we have

f1 (z ) = 2i + Log z for z 2 D = D1 ;

whi h is a tually another bran h of log z . Thus, we an see that Log z has an analyti ontinuation along any urve in the pun tured plane C n f0g. Note that ( Log z; D0 ) and ( Log z + 2i; D0) are the initial and the nal fun tion elements, respe tively. 

10.25. Theorem. Let (f; D0 ), D0 =
(z0 ; R(0)), be a germ at z0 and be a urve with initial point (0) = z0 : Then any two analyti

ontinuations of f along oin ide in the following sense: if (ft ; Dt ) and (gt ; Ut ), t 2 [0; 1℄, are two analyti ontinuations of (f; D0 ) with (f1 ; D1 ) and (g1 ; U1 ) as the terminal elements of (ft ; Dt ) and (gt ; Ut ), respe tively, then f1 = g1 on D1 \ U1 . Here Dt =
( (t); R(t)) and Ut =
( (t); r(t)) are the disks of de nition for the germs (ft ; Dt ) and (gt ; Ut ), respe tively.

Proof. Set E = ft0 2 [0; 1℄ : ft = gt in Dt \ Ut for all t 2 [0; t0 ℄g. By the de nition, f0 = g0 in D0 \ U0 and so 0 2 E . Thus, E 6= ;. To omplete the proof, we need to show that E is both open (in [0; 1℄) and losed. Then the onne tedness of [0; 1℄ will imply E = [0; 1℄. First we show that E is losed. Take an arbitrary t0 2 E and an in reasing sequen e ftn g of elements of E su h that tn ! t0 . Choose n so large that jtn t0 j < minfR(t0 ); r(t0 )g: Then

(tn ) 2 Dtn \ Utn \ Dt0 \ Ut0 :

Also sin e tn 2 E , ftn = gtn on Dtn \ Utn . Therefore,

ft0 = ftn = gtn = gt0 on Dtn \ Utn \ Dt0 \ Ut0 whi h gives ft0 = gt0 on Dt0 \ Ut0 showing that t0 2 E: Hen e E is losed. To see that E is open in [0; 1℄, we onsider the omplement E = [0; 1℄nE and show that E is losed. As before it suÆ es to show that if ftn g is a

onvergent sequen e in E with tn ! t0 ; then t0 2 E . To see this we note that tn 2= E and so by de nition of E , there exists sn 2 [0; 1℄ su h that tn  sn for all n and (10.26)

fsn 6 gsn on Dsn \ Usn :

420

Analyti Continuation

Passing to the subsequen e if ne essary, we may assume that sn onverges to some s0 : Clearly t0  s0 : Now we easily see that t0 2 E : To on rm this it suÆ es to show that

fs0

6 gs0 on Ds0 \ Us0 :

But if fs0 = gs0 on Ds0 \ Us0 , then the previous reasoning shows that fsn = gsn on Dsn \ Usn for n large enough. This ontradi ts (10.26) and so E is open. We have that E = [0; 1℄. Theorem 10.25 asserts that the analyti ontinuation of a given fun tion element along a given urve is unique if it exists.

10.27. Theorem. (Monodromy Theorem) Let z0 and z1 be two points in a domain D. Let f be an analyti germ at z0 su h that f an be

ontinued analyti ally along every urve in D that begins at z0 and ends at z1 : Assume that (i) 0 and 1 are two urves with initial point z0 and terminal point z1 . (ii) 0 and 1 are homotopi in D. Let f 0 and f 1 be the analyti ontinuations of f along 0 and 1 , respe tively. Then f 0 and f 1 agree in a neighborhood of z1 , i.e. the analyti

ontinuations of f along 0 and 1 produ e the same terminal germ at z1 .

Proof. Let the homotopy onne ting 0 and 1 be F : [0; 1℄[0; 1℄ ! C , and xes the end points of 0 and 1 (i.e. F (0; u) = z1 and F (1; u) = z2 for all u 2 [0; 1℄). We denote by u , the intermediate path asso iated with F : u (t) = F (t; u), where F (t; 0) = 0 (t) and F (t; 1) = 1 (t) for all t 2 [0; 1℄:

Note that

z0 = 0 (0) = 1 (0) and z1 = 0 (1) = 1 (1): Thus, for ea h xed u 2 [0; 1℄, let ft;u denote an analyti ontinuation of f (z ) along the urve u = F (:; u): Then the on lusion of the theorem is that f1;0 = f1;1 (see Figure 10.4): Set E = fu 2 [0; 1℄ : f1; = f1;0 for all  2 [0; u℄g. Sin e 0 2 E , E 6= ;: If we an show that E is both open (in [0; 1℄) and losed, then the onne tedness of E would imply that E = [0; 1℄: In parti ular, 1 2 E so that f1;1 = f1;0 whi h ompletes the proof. To prove that E is open, take an arbitrary u0 2 E: Then, by Lemma 10.23, there exists an R > 0 su h that for ea h t 2 [0; 1℄, the radius of

onvergen e of the power series expansion of fu0 ;t about u0 (t) is at least R: By the uniform ontinuity of F on the ompa t set [0; 1℄  [0; 1℄; there is a Æ > 0 su h that for every t 2 [0; 1℄,

j u (t) u0 (t)j = jF (t; u) F (t; u0 )j < R = 

421

10.2 Monodromy Theorem z0

γ (t)

R(t) z1

Figure 10.4: Illustration for Monodromy theorem. Dt,u γ1 γu

ft,1 z0 f0,u = f0,0

γ0 ft,0

γu (t) z1

γu0(t)

f1,1 = f1,0

Dt,u0

Figure 10.5: Illustration for Monodromy theorem.

whenever ju u0j < Æ and u 2 [0; 1℄: We x u1 2 (u0 Æ; u0 + Æ) \ [0; 1℄ and set T = ft 2 [0; 1℄ : ft;u1 = ft;u0 on Dt;u1 \ Dt;u0 g where Dt;u1 and Dt;u0 are the disks of de nition of ft;u1 and ft;u0 , respe tively. Sin e 0 2 T , T 6= ;: By Theorem 10.25, T = [0; 1℄ and so (see Figure 10.5) f1;u1 = f1;u0 = f1;0 : Thus, every u1 2 (u0 Æ; u0 + Æ) \ [0; 1℄ belongs to E . It follows that E is open in [0; 1℄: To show that E is losed, let fung be a sequen e in E su h that un ! u0 . A similar argument shows that there exists a Æ > 0 su h that if u 2 (u0 Æ; u0 + Æ) \ [0; 1℄ then

f1;u = f1;u0 on D1;u \ D1;u0 : For large n, un 2 (u0

Æ; u0 + Æ) \ [0; 1℄, so it follows that f1;un = f1;u1 = f1;0 :

So u0 2 E and hen e E is losed. We have that E = [0; 1℄:

422

Analyti Continuation

10.3 Poisson Integral Formula For f 2 H(
), the Cau hy integral formula gives Z

1 f ( ) 1 d = f (0) = 2i j j=  2

Z 2

0

f (ei ) d (0 <  < 1):

If, in addition, f is ontinuous on
then f (z ) is uniformly ontinuous on
so that, allowing  ! 1 ,

f (0) =

(10.28)

1 2

Z 2

0

f (ei ) d:

That is, the value of f at the enter of the unit ir le jz j = 1 is the mean value of f on jz j = 1. Our pro edure for omputing f (a) for an arbitrary a 2
will be as follows: onsider a z : a (z ) = 1 az We know that a 2 H(
) and maps the unit ir le jz j = 1 onto itself. Also, a (a) = 0, a (
) =
, a 1 = a ,   1 jaj2 0a (z ) 1 jaj2 0 a (z ) = ; and = : (1 az )2 a (z ) (1 az )(z a) Clearly, F = f Æ a 1 2 H(
), F (0) = f (a) and F is ontinuous on
(as f is ontinuous on
and a 2 H(
)). Therefore, by (10.28), we get (10.29) f (a) = F (0) =

1 2

Z 2

0

F (eiT ) dT =

1 (eiT )
,

1 2

Z 2

0




f a 1 (eiT ) dT:

If we let (T ) = f a then (T ) is 2-periodi and so, we may

hange the variable of integration by setting

a 1 (eiT ) = ei ; i.e. eiT = a (ei ); so that ieiT dT = i0a (ei )ei d. Thus, 0 (ei )ei (1 jaj2 )ei 1 jaj2 dT = a i d = d = i i i a (e ) (1 ae )(e a) je aj2 d: Substituting this into (10.29), we obtain

10.30. Theorem. Suppose that f

Then we have

f (a) =

1 2

Z 2

0

2 H(
)

and ontinuous on
.

1 jaj2 f (ei ) i je aj2 d (a 2
):

423

10.3 Poisson Integral Formula

The fa tor appearing after f (ei ) under the integral sign has a spe ial notation: with a = reit ,  i e



+a 1 jaj2 = i Re i e a je aj2 =: Pr (

t)

where Pr () is known as the Poisson kernel:

Pr () =

1 r2 : 1 2r os  + r2

The extension of Theorem 10.30 for jz j < R follows.

10.31. Theorem. Suppose that f

Then we have

1 f (a) = 2

Z 2

0

f (Rei )

2 H(
R ) and ontinuous on
R .

R2 jaj2 jRei aj2 d (a 2
R ):

Proof. De ne g(z ) = f (Rz ) with a = bR so that jaj < R () jbj < 1 and f (a) = f (bR) = g(b). Further, g 2 H(
) and ontinuous on
. By Theorem 10.30, g(b) =

1 2

Z 2

0

1 jbj2 g(ei ) i je bj2 d (b 2
)

whi h is equivalent to the desired formula.

10.32. Example. If we let f (z )  1 in Theorem 10.31, it follows that Z R2 jaj2 1 1= d ( = Rei ; d = id; jaj < R); 2 j j=R jRei aj2 or equivalently (sin e jd j = jidj = Rd) Z

jd j = 2R if jaj < R: j j=R j aj2 R2 jaj2 How does one handle the problem if jaj > R? (see Example 8.31).



Theorem 10.31 is alled the Poisson integral formula for analyti fun tions in disks, rather than just the unit disk. The integral in Theorem 10.31 is alled Poisson integral for analyti fun tions in
R and an be equivalently written as (10.33)

f (z ) =

1 2

Z 2

0

f (Rei )Pr (R;  t) d (z = reit 2
R );

424

Analyti Continuation

where Pr (R; ) is the Poisson kernel given by

R2 r 2 : R2 2Rr os  + r2 Set f (z ) = f (reit ) = u(z ) + iv(z ) := u(r; t) + iv(r; t): Sin e Pr (R;  t) is real-valued, we may equate the real parts of both sides of (10.33) and obtain the Poisson integral formula for harmoni fun tions in the ir ular domain jz j < R and ontinuous on jz j  R. The Poisson integral formula then reads (with z = reit 2
R ) Pr (R; ) =

u(z ) =

1 2

Z 2

0

u(Rei )

R2 jz j2 1 d = i 2 jRe z j 2

Z 2

0

u(Rei )Pr (R;  t) d

and a similar expression holds for v(z ). Equivalently, we may write 1 u(r; t) = 2

Z 2

0

u(R; )Pr (R;  t) d (0  r < R):

This formula implies that if u is harmoni in
R and ontinuous on
R , then its value u(z ) (= u(r; t); r < R) at an interior point z = reit is

ompletely determined by its boundary values u(Rei ) (= u(R; )) on the

ir le j j = R. Several extensions of this formula are immediate. For example if u(z ) is harmoni in
(z0 ; R) and ontinuous on
(z0 ; R), then for ea h z 2
(z0 ; R) one has 1 u(z ) = 2

Z 2

0

u(z0 + Rei )

R2 jRei

jz z0 j2 d: (z z0 )j2

It means that the value of a harmoni fun tion at a point z an be expressed as 1=2 times the integration of the Poisson kernel with the values of the harmoni fun tion on the boundary of the disk.

10.34. Theorem. Suppose that f = u + iv 2 H(
R ) and is ontinuous on
R . Then for ea h z 2
R we have 1 f (z ) = 2

Proof. Set 

= Rei , Z 2

Z 2

0

u(Rei )

Rei + z d + iv(0): Rei z

z = reit (r < R) so that d = i d, and let Z

 +z 1  + z d d = u( ) :  z 2i j j=R  z  0 It is easy to see that g(z ) is analyti with g (z ) =

1 2

g0 (z ) = lim h!0

u( )

g(z + h) g(z ) 1 = h 2

Z 2

0

u( )

(

2 d: z )2

425

10.3 Poisson Integral Formula

P n Indeed, as ( + z )=( z ) = 1 + 2 1 n=1 (z= ) for jz j < j j, we an perform the term by term integration by the uniform onvergen e of the series (with z 2
R xed) and obtain Z 1 X 1 u( ) n d: g(z ) = b0 + 2 bn z ; bn = 2i j j=R  n+1 n=1

Now, with M = sup ju( )j; j j=R

jbnj 

Z

1 ju( )j jd j  M 2 j j=R j jn+1 Rn

and so,

1 1 lim sup j2bnj1=n  nlim (2M )1=n = R !1 R n!1 showing that g is analyti in
R . Now, sin e 



Z

Z

1 2  +z 1 2  +z u( ) d = u( )Re d; 2 0  z 2 0  z a omparison (see Equation (10.33)) shows that f and g have the same real parts Re g(z ) = Re

Z

1 2 R2 r 2 u(Rei ) 2 d: 2 0 R 2rR os( t) + r2 Hen e, an appli ation of the Cau hy-Riemann equations gives Re f (z ) = Re g(z ) =

Z

1 2 Rei + z f (z ) = g(z ) + i; or f (z ) = u(Rei ) i d + i: 2 0 Re z Setting z = 0, we have  = v(0) (sin e, by the Mean value theorem, the integral on the right equals u(0)), and the result follows. Equating the imaginary part on both sides of the last integral shows that Z 2rR sin(t ) 1 2 v(reit ) = u(Rei ) 2 d + v(0) 2 0 R 2rR os(t ) + r2 or equivalently Z

1 2 2Im (z ) i u( ) 2 0 j z j2 d + v(0) ( = Re ; jz j < R) whi h is of ourse another representation for harmoni fun tions. The quantity Q(; z ) de ned by 2Im (z ) Q(; z ) = j z j2

v (z ) =

426

Analyti Continuation

is known as the onjugate Poisson kernel.

10.35. Example. . We wish to provide an alternate proof of Exer ise 6.89. For jz j = R=2 < R, where R is large enough, Theorem 10.34 gives that   Z 1  + z d f (z ) = Re f ( ) + iIm f (0) 2 j j=R  z i and therefore,   2 2 + jIm f (0)j = 3 R + jIm f (0)j jf (z )j  21 ( R ) RR + R= R=2 so that there exist onstants A and B su h that jf (z )j  Ajz j + B for all jz j suÆ iently large. The desired on lusion follows from the generalized



Liouville's theorem (see Theorem 6.60).

We have shown that a harmoni fun tion on
whi h is ontinuous on
has the property that its values at any point in
an be a hieved by its values on the boundary 
. We wish to work the other way around. First we remark that, the Cau hy integral formula not only reprodu es analyti fun tions but also reates them. Indeed, let us suppose that f ( ) is ontinuous on j j = 1. Then there exists an M > 0 su h that jf ( )j  M on j j = 1. Now de ne g(z ) on
by 1 g (z ) = 2

Z 2

0

Z

f ( ) 1 f ( ) d and let bn = d:  z 2i j j=1  n+1

Then

Z Z 1 f ( ) ! 1 X X f ( ) 1 1 n d = g (z ) = d = z bn z n : 2i j j=1  z 2i j j=1 n=0  n+1 n=0

Note that jbn j  M and lim supn!1 jbn j1=n  limn!1 M 1=n = 1. So, g has a power series representation with radius of onvergen e not smaller than 1 and therefore, g is analyti for z 2
. Also, the fa t that g is analyti in
is a onsequen e of Morera's theorem, sin e 1=( z ) (jz j < 1) is an analyti fun tion of z for ea h  with j j = 1 and g is ontinuous on j j = 1. This observation shows that the Cau hy integral formula reates analyti fun tions. But, in general, there exists no dire t onne tion between f and g. Is it then possible to re over f as the boundary limit of g? The fun tion f ( ) =  = 1 for j j = 1 says `no' as g(z ) = 0. On the other hand, for harmoni fun tions the situation diers. The Poisson integral formula both reprodu es and reates harmoni fun tions. Suppose that f () := f (ei ) is a real-valued ontinuous fun tion of ,  2 (0; 2℄ and de ne a new fun tion

427

10.3 Poisson Integral Formula

u(z ) a

ording to the Poisson integral formula for a harmoni fun tion on
, namely, Z 2

1 jz j2 f (ei ) i je z j2 d if jz j < 1: 0 What property will the fun tion u have? In ontrast to the analyti ase, there is a simple onne tion between f and the reated fun tion u, whi h we shall soon see is harmoni in
. This onne tion is pre isely given by the next theorem whi h is often referred to as the \solution to the Diri hlet problem for disks." What is a Diri hlet problem? Given a bounded domain D and a ontinuous (more generally pie ewise ontinuous) fun tion f : D ! R, the simplest version of the Diri hlet problem is to nd a fun tion u : D ! R su h that u is harmoni on D, ontinuous on D, and equal to f on D. An important onsequen e of the maximum and minimum prin iples is that the solution u to the Diri hlet problem is unique. Indeed, if U is another su h fun tion, then u U is harmoni on D, ontinuous on D, and vanishes on the boundary D, but it takes its maximum and its minimum on the boundary, so it is identi ally zero, i.e. u  U on D. Our spe ial emphasize here on the Diri hlet problem is when D is the unit disk
. 1 u(z ) = 2

10.36. Theorem. (Solution to a Diri hlet Problem) Suppose that f : 
! R is ontinuous. For 0  r < 1, let z = reit and u(reit ) =

1 2

Z 2

0

f (ei )Pr ( t) d =

1 2

Z 2

0

f (ei )Re

 i e



+z d: ei z

Then u is the unique fun tion that is ontinuous on
su h that

(a) u(z ) is harmoni for jz j < 1 and u(z ) = f (z ) for jz j = 1 (b) u(reit ) ! f (eit ) uniformly as r ! 1 .

Proof. Set  = ei , z = reit (r < 1) so that d = i d, and let Z

1 2 i ei + z f (e ) i d: 2 0 e z Then (for example as in the proof of Theorem 10.34), g is analyti in
and so u(z ) = Re g(z ), being the real part of an analyti fun tion in
, is harmoni in
. For a proof of (b), we shall use only the following basi properties of the Poisson kernel Pr ():

g (z ) =

1 r2 : 1 2r os  + r2 (i) Pr () > 0 for  2 R and 0  r < 1. Moreover, Pr () = Pr ( ) and Pr () is a 2-periodi fun tion.

Pr () =

428

Analyti Continuation

(ii) The Poisson integral in Theorem 10.30 for the onstant fun tion 1 shows that Z 1 2 P () d = 1 for 0  r < 1. 2 0 r (iii) If 0 < Æ < , then limr!1 Pr () = 0 uniformly in  for Æ  jj  . This property follows from the following observation. For ea h xed Æ, 0 < Æ < , we see that  2r(1 r2 ) sin   0 for Æ     0 Pr () = (1 2r os  + r2 )2  0 for     Æ and so Pr () is in reasing for     Æ and de reasing for Æ    . This implies that 1 r2 0 < Pr ()  Pr (Æ) = whenever Æ  jj   1 2r os Æ + r2 and the assertion follows. Next, a

ording to (ii), we an write Z 1 2 u(reit ) f (eit ) = P ( t)[f (ei ) f (eit )℄ d 2 0 r Z 1 2 t = Pr ()[f (ei(+t) ) f (eit )℄ d 2 t Z 1  = P ()[f (ei(+t) ) f (eit )℄ d; 2  r be ause the integrand is a periodi fun tion of period 2. It remains to show that lim u(z ) = lim u(reit ) = f (eit ): z2
r !1 z!eit Sin e f is ontinuous on [ ; ℄, it is bounded. So, we may let M = max jf (ei )j. By the uniform ontinuity of f , given  > 0 there exists a Æ > 0 su h that jf (ei(+t) ) f (eit )j <  for all t, whenever jj < Æ. Now, hoose Æ > 0 suÆ iently small to satisfy this ondition and split the interval of integration as f : jj  g = I1 [ I2 := f : 0  jj  Æg [ f : Æ  jj  g so that Z 1 Æ u(reit ) f (eit ) = P ()[f (ei(+t) ) f (eit )℄ d 2 Æ r Z 1 + P ()[f (ei(+t) ) f (eit )℄ d 2 Æjj r =: J1 + J2 :

429

10.3 Poisson Integral Formula

On the set I1 ,

jJ1 j  21

Z

I1

Pr () d 

 2

Z 2

0

Pr () d = :

On the set I2 , we have Æ  jj   and Pr ()  Pr (Æ) so that

jJ2 j  22M Pr (Æ)2 = 2MPr (Æ): We get the estimate

ju(reit ) f (eit )j  2MPr (Æ) +  and, as r is suÆ iently lose to 1, lim sup ju(reit ) f (eit )j  : r!1 Hen e, limr!1 ju(reit ) f (eit )j = 0 and the uniqueness is a onsequen e of the maximum modulus prin iple for harmoni fun tions.

10.37. Corollary. Suppose that a 2 C , R > 0 and h : 
(a; R) ! R is ontinuous. Then there is a unique ontinuous fun tion w(z ) on
(a; R) su h that w(z ) = h(z ) on 
(a; R) and w(z ) is harmoni on
(a; R).

Proof. Consider f (ei ) = h(a + Rei ). By the previous theorem, the fun tion w with the desired properties is given by w(z ) = u((z a)=R), where u is de ned as in Theorem 10.36. The proof of a general ase for pie ewise ontinuous fun tions f uses similar ideas but is slightly te hni al whi h we avoid in this text. So we state the result without proof.

10.38. Theorem. Theorem 10:36 ontinues to hold under a weaker

ondition that f : 
! R is a pie ewise ontinuous (i.e. ontinuous ex ept for nitely many points) bounded fun tion. As an appli ation of Poisson's integral formula, we prove

10.39. Theorem. (Harna k's Inequality for the unit disk) Let u = u(z ) be harmoni on
and ontinuous on
. If u(ei )  0 for all , then for z = rei 2
we have 1 r 1+r u(0)  u(rei )  u(0) (r < 1): 1+r 1 r The estimate is sharp.

430

Analyti Continuation

Proof. Sin e u(ei )  0, by the Poisson integral formula for u, it follows that u(z )  0 for all z 2
. The proof depends on the estimate 2

1 r 1+r

1 r 1+r  1 2r os( = Pr ( t)  for 0  r < 1: 2  t) + r 1 r Multiplying this by 21 u(ei ) ( 0) and integrating, we obtain 

1 r 1 1 + r 2

Z 2

0

u(ei ) d



 u(rei ) 



1+r 1 1 r 2

Z 2

0

u(ei ) d



:

TheR desired inequality is a onsequen e of the mean value property, namely, 1 2 i 2 0 u(e ) d = u(0). The above inequalities be ome an equality for the fun tion u(z ) = Re ((1 + z )=(1 z )) at z = r. A suitable s aling and translation of Harna k's inequality immediately implies the following:

10.40. Corollary. Let u = u(z ) be harmoni on
(a; R) and ontinuous on
(a; R). If u(a + Rei )  0 for all , then for z 2
(a; R) we have u(a)

R r R+r

 u(z )  u(a) RR + rr (0  r < R):

The estimates in this orollary is referred to as Harna k's inequality for arbitrary disks. In parti ular, if u is harmoni and non-negative on
(a; R) then u(z ) 2 [u(a)=3; 3u(a)℄ on
(a; R=2). We know that if u is a harmoni fun tion in a domain D and
(a; r) is

ontained in D, then u satis es the Mean value property (10.41)

u(a) =

1 2

Z 2

0

u(a + rei ) d:

Conversely, it is easy to show that a ontinuous fun tion with the Mean value property is ne essarily harmoni .

10.42. Theorem. Let u : D ! R be ontinuous on a domain D su h that for ea h point a 2 D, (10:41) holds whenever
(a; r)  D. Then u(z ) is harmoni on D. Proof. The fun tion u is obviously ontinuous on
(a; r). Also, if
(a; r)  D, by Corollary 10.37, there exists a fun tion U (z ) harmoni on
(a; r), ontinuous on
(a; r), and equal to u(z ) on the ir le 
(a; r). Sin e U (z ) u(z ) is ontinuous on
(a; r) and satis es the Mean value property, by Theorem 6.16, U (z ) u(z ) attains both its maximum and minimum on the boundary 
(a; r). As U (z ) u(z )  0 on jz aj =

431

10.3 Poisson Integral Formula

r, it follows that U (z )  u(z ) on
(a; r). Hen e, u(z ) is harmoni in a neighborhood of a. Sin e a is arbitrary, the result follows. 10.43. Corollary. A ontinuous fun tion in a domain D is harmoni in D i it satis es the Mean value property at ea h point of D. As an appli ation, we prove the following analog to Weierstrass' theorem for sequen es (see Theorem 4.84) of harmoni fun tions.

10.44. Theorem. If fun(z )g is a sequen e of real-valued harmoni fun tions that onverges uniformly on all ompa t subsets of a domain D to a fun tion u : D ! R, then u(z ) is harmoni in D. Proof. Sin e the limit of a uniformly onvergent sequen e of ontinuous fun tions is ontinuous (see Theorem 2.57), u(z ) is ontinuous on D. Also, if
(a; r)  D, then Z

1 2 u (a + rei ) d 2 0 n for ea h n. By the uniform onvergen e (see the proof of Theorem 4.84),

un(a) =

Z

Z

2 1 2 i ) d = 1 u(a) = nlim u ( a ) = lim u ( a + re u(a + rei ) d: n n !1 n!1 2 0 2 0 So, by Theorem 10.42, u is harmoni on
(a; r) (and hen e on D).

The Harna k inequality allows us to draw some striking on lusion, for example, the following ` onvergen e theorem' for sequen es of harmoni fun tions.

10.45. Theorem. (Harna k's Prin iple/Theorem) Let fun(z )gn1 be an in reasing (i.e. un+1  un on D and for ea h n  1) sequen e of harmoni fun tions de ned on a domain D. Then either un (z ) ! +1 for ea h z 2 D (and uniformly on ompa t subsets of D) or un (z ) onverges to a harmoni fun tion u on D, uniformly on ompa t subsets. Proof. We may assume that u1 (z )  0 (for otherwise, repla e un (z ) by the nonnegative sequen e fun(z ) u1 (z )g). By the monotoni ity property, for ea h z in D either the pointwise limit limn!1 un(z ) exists or limn!1 un(z ) = +1. The fun tion u(z ) an be de ned by this limit as a map u : D ! R. De ne A = fz 2 D : un (z ) ! 1g and B = fz 2 D : fun(z )g onvergesg:

Given a 2 D, hoose a disk
(a; R) Harna k's inequality gives (10.46)

 D. Then, for all z 2
(a; R=2),

(1=3)un(a)  un (z )  3un(a):

432

Analyti Continuation

If a is su h that un (a) ! 1, then the left hand inequality of (10.46) shows that un (z ) ! 1 for jz aj  R=2, and that the onvergen e is uniform in this disk. If a is su h that fun(a)g onverges to a nite limit, then the right hand inequality shows that fun(z )g onverges for jz aj  R=2. Hen e, A and B are both open sets with A [ B = D. Sin e D is onne ted, either A = ; or B = ;. If A = ;, then B = D, and so fun(z )g onverges for all z in D. Next we must show that fun (z )g onverges uniformly on ompa t subsets of D. By using the right hand part of Harna k's inequality (10.46), we nd that (10.47) un+p (z ) un(z )  3[un+p(a) un (a)℄ for jz aj  R=2 and p = 1; 2; : : : . By the Cau hy riterion, this inequality in turn implies that the onvergen e at a point a implies the uniform onvergen e in a neighborhood of a. Sin e every ompa t subset of D an be

overed by nitely many su h neighborhoods, fun (z )g is uniformly Cau hy and so it onverges uniformly on every ompa t subset of D to u(z ). Finally, it follows from Theorem 10.44 that the limit fun tion is harmoni throughout D. The fun tion un (z ) = x + n, or y + n2 , ea h of whi h is harmoni in every domain, satis es the onditions of the theorem.

10.4 Analyti Continuation via Re e tion A domain is said to be symmetri with respe t to the origin if for every z 2 , the point z 2 . For example, disks entered at the origin. Now, we onsider the following two sets of fun tions

F = f1 + z 2; os z; sin z; exp z g; and G = fi + z 2; os(iz ); i sin z; i exp z g: For ea h f 2 F , we have f (z) = f (z ) and f (x) 2 R for all x 2 R. For example, if we allow the three steps for os z we nd that

eiz + e iz e iz + eiz 7! 2 = os z: 2 On the other hand, for ea h g 2 G , we have z 7! z 7!

g(z) 6= g(z ); and g(x) 2= R if x 2 R :

Note that the domain of analyti ity for ea h f 2 F and ea h g 2 G is the whole omplex plane. Further, the re e tion of z with respe t to the real axis of z -plane does not orrespond to the re e tion of g(z ) with respe t to the real axis of the w-plane. We are interested in the following problem:

10.48. Problem. Under what onditions, does an analyti fun tion f on some domain possess the following property: f (z) = f (z ), i.e. the

10.4 Analyti Continuation via Re e tion

433

re e tion of z with respe t to the real axis of z -plane equals the re e tion of w = f (z ) with respe t to the real axis of the w-plane. Our rst result a tually des ribes how it is possible to use two re e tions to onstru t a new analyti fun tion from an old one.

10.49. Theorem. Let be a domain whi h is symmetri with respe t to the real axis. Then, f (z ) 2 H( ) i f (z ) 2 H( ). Proof. De ne f (z ) = u(x; y) + iv(x; y) and F (z ) = f (z). Then, F (z ) = u(x; y) iv(x; y) =: U (x; y) + iV (x; y): Note that u and v belong to C 2 ( ) i U and V belong to C 2 ( ). Further,

Ux (x; y) = ux(x; y); Vy (x; y) = vy (x; y);

Uy (x; y) = uy (x; y); Vx (x; y) = vx (x; y);

whi h, in parti ular, give that u and v satisfy the C-R equations on i U and V satisfy the same property on . Now, the desired on lusion is a onsequen e of Theorem 3.26 and the fa t that the real and imaginary parts of analyti fun tions in are C 1 -fun tions on . Theorem 10.49 leads to a general symmetri prin iple for analyti fun tions.

10.50. Theorem. Let be a domain su h that it ontains a segment (a; b) of the real axis, and is symmetri with respe t to the real axis. Suppose that f (z ) 2 H( ). Then, f (z ) is real on (a; b) i f (z ) = f (z ) for all z 2 . Proof. Suppose that f (z ) is real on (a; b). Consider the dieren e h(z ) = f (z ) f (z): Then, by Theorem 10.49, h 2 H( ). Moreover, h(z ) = 0 for all z 2 (a; b). By the uniqueness theorem, h(z )  0 on . Conversely, if f (z ) = u(x; y) + iv(x; y) and f (z ) = f (z ) on then it follows that u(x; y) + iv(x; y) = u(x; y) iv(x; y): In parti ular, when z = x + iy 2 is real, this equation gives

u(x; 0) + iv(x; 0) = u(x; 0) iv(x; 0) so that v(x; 0) = 0. Therefore, f (z ) is real when z is real.

434

Analyti Continuation y + σ x −

Figure 10.6: Re e tion Prin iple

10.51. Lemma. (Re e tion Prin iple for Harmoni Fun tions) Let be a domain whi h is symmetri with respe t to the real axis and de ne + , , and  as the interse tion of with the upper half-plane fz : Im z > 0g, the lower half-plane fz : Im z < 0g, and the real axis, respe tively (see Figure 10.6). Let v (z ) be a real-valued ontinuous fun tion on + [  , harmoni on + , and zero on  . Then v admits a unique harmoni extension V on all of and the extension satis es the symmetry relation (10.52) V (z ) = V (z ) for z 2 : Proof. First we note that if su h an extension exists it is ertainly unique, so it suÆ es to show that the stated extension de nes a harmoni fun tion in . Se ondly, we note that is a disjoint union of + , and . We extend v(z ) to by setting 8 < v (z ) for z 2 + 0 for z 2  V (z ) = : v(z ) for z 2 : Then, V is ontinuous on , harmoni on + [ , and (10.52) holds. We

laim that V has the mean value property. Fix a 2 . If a 2 + or , then V (z ) has the mean value property for small disks entered at a, sin e V (z ) is harmoni near a. For a 2 , we have Z 



V (a + rei ) d = =

Z 0



Z 

0

v(a + re i ) d + v(a + rei ) d +

Z 

0

Z 

0

v(a + rei ) d

v(a + rei ) d = 0

Thus, V satis es the mean value property on and hen e, it is harmoni on . Lemma 10.51 not only provides a method of extending a harmoni fun tion from a given open set to a larger open set but also gives a proof of a stronger S hwarz's re e tion/symmetry prin iple for analyti fun tions.

10.5 Exer ises

435

10.53. Theorem. (S hwarz's Re e tion Prin iple for Analyti Fun tions) Let , + , , and  be as above. Suppose that f 2 H( + ),

ontinuous on  and f (z ) is real on  . Then f extends to be analyti on

and the extension satis es the symmetry relation f (z ) = f (z ), z 2 D.

Proof. Set f (z ) = u(z ) + iv(z ), z 2 + . By the previous result applied to v(z ) = Im f (z ), v(z ) extends to be harmoni on with v(z ) = v(z ), z 2 . Again, we x a 2  and let
(a; Æ) be a disk that is ontained in

. Sin e v(z ) is harmoni in this disk, whi h is simply onne ted, it has a harmoni onjugate u1 in
(a; Æ) su h that f1 = u1 + iv is analyti in
(a; Æ). Thus, Im (f (z ) f1 (z )) = 0 on + \
(a; Æ): We on lude that f (z ) = f1 (z ) + , a real onstant, on + \
(a; Æ). But f1 is analyti in + \
(a; Æ) and therefore, f is also analyti in
(a; Æ). Consequently, the original fun tion f (z ) extends to be analyti in
(a; Æ). By Theorem 10.49, f (z) is also analyti in
(a; Æ). Moreover, f (z ) = f (z) on the interval (a Æ; a+Æ), and hen e on
(a; Æ) by the uniqueness theorem. Extend f (z ) to by setting f (z ) = f (z) for z 2 . Then the extended fun tion f is analyti on and oin ides with the analyti ontinuation of f (z ) a ross  from + . Hen e, f is analyti on and satis es f (z ) = f (z) on .

We observe that the line of re e tion ould be an arbitrary line, not just the real axis. Thus if is a domain symmetri with respe t to a line and f is de ned and analyti on one side of the line, and real on the line, then f an be extended to be analyti on the whole domain. The proof an be obtained by translating and rotating the domain to the standard ase. Similarly, we an also translate and rotate the image, so it is not ne essary that f (z ) be real on the symmetry line, it is suÆ ient that it maps the line into some other line.

10.5 Exer ises 10.54. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) If f is analyti for jz j < R and satis es the relation f (2z ) = 2f (z )f 0(z ) for jz j < R=2, then f an be ontinued analyti ally into the whole plane. (b) For 0 6=  2 R, the fun tions de ned by the series 1 1 X X (1 )n z n (z )n and ( 1)n (1 z )n+1 n=0 n=0 are analyti ontinuations of ea h other.

436

Analyti Continuation

( ) The series

1 zn X

(e) (f) (g) (h)

1 X

( 1)n

(z

2)n

n n n=1 have no ommon region of onvergen e, but they represent the same fun tion Log (1 z ) in their respe tive disks of onvergen e. P n 1 z n =n on
, and The fun tion f de ned by the series 1 n=1 ( 1) the fun tion F de ned by the series 1 z (1 z )2 (1 z )3 ln 2


2 2
22 3
23 on
(1; 2) are analyti ontinuations of ea h other. P The imaginary axis is a natural boundary for n0 exp( n!z ): P 3k The unit ir le jz j = 1 is the natural boundary for the series 1 k=0 z . P If f (z ) = n0 an z n is analyti for jz j  R, then R annot be the radius of onvergen e of this series de ned by the sum f (z ). The fun tion element (1=(z (1 + z )); C n f0; 1g) is the analyti ontinuation of 1 1 (1 z )n X X f (z ) = (1 z )n n+1 ; for j1 z j < 1 : n=0 n=0 2 n=1

(d)

and i +

P

2n

10.55. Show that the fun tion f (z ) = n1 zn2 is ontinuous for jz j  1, but every point on jz j = 1 is a singular point. 10.56. Let f be analyti in the ut plane D = C n ( 1; 0℄ su h that f (x) = xx for all x > 0. Show that f (z ) = f (z) for all z 2 D. 10.57. Suppose that f (z ) is analyti for jz j < 1 and jf (n) (0)j  3n for all n 2 N: Show that f an be extended to an entire fun tion g su h that f (z ) = g(z ) for jz j < 1: 10.58. Find a fun tion u(x; y) that is harmoni in the region of the right half-plane between the urves xy = 1 and xy = 2 and takes the value 3 when xy = 1, and the value 7 when xy = 2. 10.59. Use S hwarz's re e tion prin iple to show that every fun tion whi h is bounded and analyti in the upper half-plane Im (z )  0, and real on the real axis is onstant. 10.60. Use S hwarz's re e tion prin iple to show that (i) sin z = sin z for z 2 C (ii) z 4 2z + 2 os z = z4 2z + 2 os z for z 2 C .

Chapter 11

Representations for Meromorphi and Entire Fun tions

The entral importan e of this hapter is to study representations of meromorphi fun tions by an in nite de omposition into partial fra tions as well as representations of analyti fun tions by in nite produ ts. These will be done in many dierent ways and we shall also see several appli ations. In Se tion 11.1, we dis uss the Mittag-Leer expansion whi h gives a formula for representing a meromorphi fun tion f as a series involving the prin ipal parts of f at ea h of the poles of f . We begin Se tion 11.2 by presenting a brief introdu tion and basi properties of in nite produ ts of sequen es of omplex numbers and show that their onvergen e properties are similar to those of in nite series. In parti ular, we prove several important tests for the onvergen e of in nite produ ts In Se tion 11.3, we

onsider the onvergen e of in nite produ ts of analyti fun tions whi h will be useful in expressing a non- onstant entire fun tion f as a produ t of the form f (z ) = P (z )H (z ), z 2 C ; where

 P and H are entire fun tions  P and f have exa tly the same zeros with pres ribed multipli ities  H (z ) has no zeros in C . In Se tion 11.4, we present an interesting and useful result alled the Weierstrass produ t formula whi h gives a way of fa toring ertain entire fun tions into an in nite produ t. Clearly, some entire fun tions (su h as polynomials p(z ) and the trans endental entire fun tion p(z )ez ) have only a nite produ t representation. A omparison of produ ts and series expansions provides us a number of interesting identities. In the Weierstrass produ t expansion the prin ipal part of f has no role to play whereas the Mittag-Leer expansion emphasizes the prin ipal parts at the poles of f , but gives no information about its zeros. In Se tion 11.5, we study the

438

Representation for Entire and Meromorphi Fun tions

gamma fun tion as a meromorphi fun tion in C having simple poles at z = 0; 1; 2; : : : . Se tion 11.6 dis usses the zeta fun tion and its various properties. In Se tion 11.7 we prove Jensen's formula. In Se tion 11.8, we ontinue our dis ussion on entire and meromorphi fun tions, and introdu e the on ept of order and the genus of entire fun tions. Also, we study ertain basi properties of entire fun tions of nite order and of nite genus in order to prove the mu h waited Hadamard fa torization theorem whi h provides an interesting relationship between the order and the genus of entire fun tions.

11.1 In nite Sums and Meromorphi Fun tions In this se tion we are interested in the onstru tion of meromorphi fun tions by their poles. Con erning poles there are two possibilities:

 meromorphi fun tions with a nite number of poles  meromorphi fun tions with in nite number of poles. A meromorphi fun tion whi h has a pole of order m at a is (z a) m and a slightly dierent meromorphi fun tion may be obtained by taking linear

ombinations of nitely many su h simpler ones. For example, (z

1

1)2

+

(z

4

3)3

(z

2

2)8

:

Let us now dis uss the onstru tion of meromorphi fun tions in the nite

ase. Suppose that f is a meromorphi fun tion whi h has a nite number of poles at aj (1  j  n) of order mj (1  j  n). Then, by Laurent's expansion of f around ea h aj , there is an asso iated prin ipal part 

mj X



A(jk) Pj = k z aj k=1 (z aj ) 1

whi h may be thought of as a polynomial in the variable 1=(z aj ). De ne

g(z ) =

n X j =1

Pj



1

z aj



:

Clearly, g is analyti in C nfa1 ; a2 ; : : : ; an g. Sin e, for s 6= r, Ps (1=(z as )) is analyti at ar , the prin ipal part of g about ar is Pr (1=(z ar )). Thus, f and g are meromorphi fun tions in C with poles at aj and have the same prin ipal parts at aj , 1  j  n. In parti ular, if we let (z ) = f (z ) g(z ) then  is analyti in C n fa1 ; a2 ; : : : ; an g and has removable singularities at a1 ; a2 ; : : : ; an . Consequently,  an be extended to an entire fun tion. Thus, f (z ) = g(z ) + (z ) where  is entire. Finally, be ause g(z ) ! 0 as

439

11.1 In nite Sums and Meromorphi Fun tions

z ! 1, g has a removable singularity at 1 whi h shows that  and f have the same prin ipal part at 1. The above dis ussion gives 11.1. Theorem. Let f be a meromorphi fun tion with only poles  n). If Pj (1=(z aj )) denotes the prin ipal part of f (z ) at aj (1  j  n), then there exists an entire fun tion (z ) su h that at aj (1  j

f (z ) =

n X j =1

Pj



1

z aj



+ (z ):

In addition,  and f have the same prin ipal part at 1.

11.2. Corollary. Every proper rational fun tion of the form p(z ) p(z ) := ; q(z ) (z a1 )m1 (z a2 )m2


(z an )mn where p and q are polynomials with deg p(z ) < deg q (z ), an be expanded as a sum of polynomials in 1=(z ak ). Here ak 's denote the poles of R(z ) with order mk  1 (k = 1; 2; : : : ; n). (11.3)

R (z ) =

P

n P (1=(z a )) = (z ) is an Proof. By Theorem 11.1, R(z ) j j =1 j entire fun tion. Furthermore,  is bounded sin e R(z ) and ea h of its prin ipal part about ea h aj approa hes 0 as z ! 1. As a onsequen e of Liouville's P theorem, (z ) is equal to a onstant, in fa t, (z )  0 so that R(z ) = nj=1 Pj (1=(z aj )) as asserted.

Suppose that R(z ) is a proper rational fun tion of the form (11.3). Then the partial fra tion expansion of R(z ) may be written uniquely in the form

R(z ) =

(11.4)

n X j =1

mj X

!

A(jk) k : k=1 (z aj )

In general, a meromorphi fun tion f in C may have an in nite number of poles at ak , k 2 N . In that P

ase the series (whose terms are the prin ipal parts of these poles of f ), k Pk ( z 1ak ), need not onverge. It is then ne essary to modify the terms to produ e a onvergent series. For example, suppose we wish to nd a meromorphi fun tion in C having a simple pole at k (k 2 N ) with residue 1 at ea h k. From the pres ription, our fun tion

orresponding to the prin ipal part at z = k is

Pk



1

z k



=

1 1 = z k k(1 z=k)

P 1 but the sum 1 k=1 z k does not onverge in C n N . We need to modify the series suitably so that it be omes onvergent. The onstant term in the

440

Representation for Entire and Meromorphi Fun tions

series expansion of 1=(z k) about 0 is 1=k. So we an try with  X 1 1 1 1   1   1  X X z 1 Pk = + = : z k k k=1 k=1 z k k k=1 k (z k )

This series does onverge uniformly on every ompa t subset of C ex ept at P k, k 2 N (by a omparison with the onvergent series k1 k 2 ). Indeed, z

1

k

 k 1 jz j < k2 whenever jz j < k=2

so that, for jz j  R and R < k=2, we have 1 1 1 1 X X 2 2 k=1 k (z k ) k=1 k

z and hen e, by the Weierstrass M -test, the series 1 k=1 k(z k) represents an analyti fun tion on C n fk : k 2 Ng. Given a meromorphi fun tion f in C with an in nite number of poles at ak , we may always assume that the poles of f are indexed so that fjak jg is non-de reasing. Let us now state and prove a general theorem of MittagLeer although its simple form (namely, Theorem 11.7) suÆ es for our purposes. P

11.5. Theorem. (Mittag-Leer) Let fan gn1 be a sequen e of distin t non-zero omplex numbers su h that jan j < jan+1 j for n 2 N and jan j ! 1 as n ! 1. Then, for a given sequen e of polynomials fPn (z )g without onstant term, there exists a meromorphi fun tion f in C having poles at an with prin ipal part Pn (1=(z an)). Moreover, any other general meromorphi fun tion F having the stated property will be of the form F (z ) =

X

n

Pn



1

z an





Qn(z ) + h(z )

for some polynomial Qn (z ) and for some entire fun tion h(z ). The series

onverges absolutely and uniformly on any ompa t subset of C not ontaining the poles (This series is usually alled a Mittag-Leer expansion of F ).

Proof. As Pn (z ) is a polynomial, we see that n (z ) de ned by     1 1 ( z ) := P = P n n n z an an (1 z=an) is analyti for jz j < jan j and hen e, we may expand it as a Taylor series 1 X ( z ) = A(kn) z k for jz j < jan j: n k=0

11.1 In nite Sums and Meromorphi Fun tions

441

By elementary fa ts about omplex power series, the series on the right

onverges absolutely and uniformly to n (z ) for jz j  12 jan j. Let

Qn (z ) =

n X k=0

A(kn) z k

be the partial sum of the series up to the degree n , where n has been

hosen large enough to satisfy j n (z ) Qn (z )j < 2 n for jz j  12 jan j: Sin e limn!1 jan j ! 1 and jan j < jan+1 j, given any ompa t subset K =
R of C , there exists an N = N (K ) su h that K 
jan j=2 for n  N , i.e. jan j > 2R for n  N . It follows that the series 1 X ( n (z ) Qn (z )) n=N

onverges absolutely and uniformly on K , and therefore represents an analyti fun tion on K . Sin e K is arbitrary, the full series 1 NX1 X 1 ! X (11.6) ( n (z ) Qn (z )) = + ( n (z ) Qn (z )) n=1

n=1

n=N

represents a meromorphi fun tion in C whi h has an as its pole with prin ipal part equal to n (z ) = Pn (1=(z an )). Note that the nite sum in (11.6) is a rational fun tion with pres ribed behavior of poles exa tly at an with jan j < R. The rest of the theorem is trivial. For simpli ation purposes, we prove the following simple version that has wider appli ations and falls under the general theorem of the MittagLeer (see Theorem 11.5).

11.7. Theorem. Let f be meromorphi with only simple poles at a1 ; a2 ; : : : su h that 0 < ja1 j  ja2 j 




 jak j 


;

and bk = Res [f (z ); ak ℄:

Let fCk g be a nested sequen e of positively oriented simple losed ontours (whi h avoid these poles) su h that ea h Ck in ludes only a nite number of poles. Suppose that

Rk = dist (0; Ck ) ! 1 as k ! 1; Lk = length of Ck = O(Rk ); jf (z )j  M for ea h k for z 2 Ck

442

Representation for Entire and Meromorphi Fun tions

( e.g. Ck is a square with verti es Rk (1  i) su h that Rk ! 1 as k ! 1 ). Then, for all z ex ept at these poles, we have the Mittag-Leer series expansion of f  1  1 X 1 f (z ) = f (0) + bk + : z ak ak k=1

Proof. If  is not a pole of f and  2= Ck , then

f (z ) z  has simple poles at , and at ea h ak with Res [g(z ); ℄ = f () and g(z ) =

Res [g(z ); ak ℄ = zlim !a (z

ak )

k

Then

f (z ) b = k : z  ak 

Z

f (z ) 1 dz 2i Ck z  X = Res [g(z ); Ck ℄ X bk = f () + ; k ak 

Fk () := (11.8) P

where k is taken over all the poles of f inside Ck and k is hosen large enough so that  lies inside Ck . Letting  = 0 gives (11.9)

Z

X bk 1 f (z ) Fk (0) = dz = f (0) + : 2i Ck z k ak

Subtra ting (11.8) from (11.9) gives

(11.10)

Z

f (z ) 1 dz 2i Ck z (z )  X  1 1 = f (0) + bk + f ():  ak ak k

Fk (0) Fk () =

Now, for z 2 Ck , jz j  Rk = dist (0; Ck ) and jz j  jz j jj  Rk jj > 0 so that Z (Ck ) jFk () Fk (0)j = 2i1 z (fz (z )) dz  21 R ML ! 0 as k ! 1 ( R k k jj) Ck and therefore, the sequen e fFk () Fk (0)g onverges to zero uniformly on the ompa t set Ck . Allowing k ! 1 in (11.10) gives

443

11.1 In nite Sums and Meromorphi Fun tions

1 X



1



1 f () = f (0) + bk + :  a a k k k=1 Most often Ck is taken to be a ir le or the boundary of a re tangle. In pra ti e, verifying the boundedness ondition, namely jf (z )j  M on Ck , is often a diÆ ult job. Suppose we wish to expand a meromorphi fun tion f whi h has a simple pole at the origin, then the theorem is not dire tly appli able. However, we an apply Theorem 11.5 or 11.7 by onsidering the fun tion f (z ) p0 (z ), where p0 (z ) is the prin ipal part of the Laurent series expansion of f about z = 0. The partial fra tion expansions of fun tions su h as

s z; se z; tan z; ot z; os az; se z os az; s z sin az; se z sin az; (jaj < 1) and of their orresponding hyperboli fun tions are well-known and may be found in some standard texts either as examples or as exer ises. For example, in order to apply Theorem 11.7 for the fun tions  s z and  ot z; we need to onsider

 s z 1=z; and  ot z 1=z; respe tively. Let us now dis uss the situation in detail.

11.11. Example. Let us derive the Mittag-Leer expansion of g(z ) =  ot z: The poles of g(z ) are at n, n 2 Z, ea h pole being simple. As g has a simple pole at the origin, we need to ompute  z  Res [g(z ); 0℄ = zlim !0 sin z os z = 1 and onsider the modi ed fun tion ( 1 f (z ) =  ot z z for z 6= 0 0 for z = 0 whi h is now analyti at the origin and has simple poles only at n, n Z nf0g. It follows that, for ea h n 2 Z nf0g, Res [f (z ); n℄ = zlim !n(z

n)



 os z sin z



1

os n = = 1: z

os n

Choose Ck = fRk (1  i) : k 2 Ng with Rk = k + 1=2 so that

L(Ck ) = 8(k + 1=2) = 8 dist (0; Ck ) ! 1

as k ! 1

2

444

Representation for Entire and Meromorphi Fun tions

and all the other required onditions of Theorem 11.7 are satis ed. Re all that (see the proof of Theorem 9.65) j ot z j < 2 for all z on Ck . Clearly, for z 2 Ck , we have

jz j  dist (0; Ck ) = Rk = k + 1=2 > 1 and therefore,

jf (z )j  j ot z j + jz1j < 2 + 1 for all z on Ck : Finally, it now follows from Theorem 11.7 that

 ot z







n n X X 1 1 1 1 1 = nlim + = lim + !1 n !1 z k k= n z k k=1 z k z + k



k6=0

whi h shows that (11.12)

1 2z 1 X ; z2C  ot z = + z k=1 z 2 k2

n Z:

From this equation, the partial fra tion de omposition of 2 = sin2 z an be obtained by dierentiating (11.12): 1 X 2 1 1 = ; z 2 C n Z: 2 2 z ( z k)2 sin z k= 1 k6=0

Allowing z ! 0 gives that

1 1 1 1 X 2 2 X = ; i.e. = : 3 k= 1 k 2 6 k=1 k2 k6=0

Similarly, if we onsider g(z ) =  s z then it is easy to see that g has a simple pole at n (n 2 Z) with Res [g(z ); n℄ = ( 1)n for n 2 Z. Now, we de ne f (z ) =  s z z 1: Then, applying Theorem 11.7 (sin e f is again uniformly bounded on Ck ), it follows easily that 1 2( 1)k z 1 X  s z = + ; z 2 C n Z:  z k=1 z 2 k2

11.13. Example. We onsider f (z ) =  tan z: Then we see that (i) f has simple poles at (n + 1=2); n 2 Z with Res [f (z ); n + 1=2℄ = 1 (ii) f is analyti at z = 0 with f (0) = 0

445

11.1 In nite Sums and Meromorphi Fun tions

(iii) jf (z )j  2 for ea h z in the square Ck with verti es k(1  i), k 2 N and f is analyti on Ck , k 2 N : By Theorem 11.7, we see that 

k 1 X 1 1 f (z ) = lim ( 1) + k!1 z ( n + 1 = 2) n + 1=2 n= k

=

kX1

z (n + 1=2))(n + 1=2)

lim k!1 n=0 (z

1 X

=



z + n= k (z (n + 1=2))(n + 1=2) kX1 z lim k!1 n=0 (z (n + 1=2))(n + 1=2) kX1

!

z + n=0 z ( n 1=2))( n 1=2)

kX1

=

z lim k!1 n=0 n + 1=2

=

lim k!1 n=0 z 2

kX1



z

1 (n + 1=2)

!

1 z + (n + 1=2)



2z : (n + 1=2)2

Thus, ex ept at the simple poles of tan z , we have

 tan z =

1 X

2z ( n + 1 = 2)2 n=0

z2

; z2C

n f(n + 1=2) : n 2 Zg:

Letting z ! 0 gives 1 1 X tan z X 2 1 2 2 zlim = ; i.e. = : 2 2 !0 z 8 n=0 (n + 1=2) n=0 (2n + 1) Choosing z = 4 shows that 1  X 1 = : 8 n=0 4(2n + 1)2 1 One an also obtain the partial fra tion de omposition of  tan z just by using the identity tan w = ot w 2 ot(2w). 

446

Representation for Entire and Meromorphi Fun tions

11.2 In nite Produ t of Complex Numbers

Let us start by introdu ing the de nition of in nite produ ts. Let fzng1 n=1 be a sequen e of non-zero omplex numbers. We want to dis uss the

onQ vergen e of produ ts of the form 1 k=1 zk : To doQthis, we de ne the partial produ t Pn (or n-th partial produ t) to be Pn = nk=1 zk : The in nite produ t is said to onverge if the sequen e fPn g1 n=1 of partial produ ts onverges to a non-zero ( nite) limit P Qas n ! 1. We symbolize the onvergen e of the produ t by writing P = 1 k=1 zk ; where P 6= 0. If limn!1 Pn fails to exist or if limn!1 Pn = 0, then we say that the in nite produ t diverges. Clearly, some te hni alities are involved if zk = 0 for a nite number of k's. But we would like to allow the in nite produ t to be zero yet able to dis uss the onvergen e of su h produ ts by imposing some onditions. Suppose that zk = 0 for nitely many values of k and let zk 6= 0 for k > m. Then, for k > m, we an write

Pn = (z1 z2


zm )[zm+1 zm+2


zn ℄ := (z1 z2


zm )Pm;n where Pm;n = zm+1 zm+2


zn : In this ase, we say that the in nite produ t Pn onverges to zero provided Pm;n onverges to a non-zero limit Q as n ! 1. Indeed, if limn!1 Pm;n = Q (Q 6= 0; 1) then lim P n!1 n

= (z1 z2


zm )Q = 0:

If Q = 0 or Q =Q1 or if fPm;n g has no limit in C as n ! 1, then the in nite produ t 1 k=1 zk is said to be divergent. More generally, we say that the in nite produ t Q onverges to zero if zk = 0 for a ountable number of k's and that 1 k=1;zk 6=0 zk onverges a

ording to the earlier de nition. For instan e, if

k =

8
1 is xed. We may hoose pn = 0, sin e the series 1 1 1 z pn +1 X X = jz j p p n=1 n n=1 n is onvergent for ea h z , and the desired entire fun tion is then given by 1  Y z f (z ) = z 1 : np n=1 Similarly, if we want to onstru t an entire fun tion f with simple zeros at a0 = 0 and at an = n; n 2 Z (note that these are pre isely the zeros of sin z ). We may therefore hoose pn = 1, sin e the series 1 1 X z pn +1 X z 2 X = 2jz j2 = a a 2 n=1 n n2Znf0g n n2Znf0g n

onverges for ea h z . Again, Weierstrass's theorem gives 1  1  Y z Y z E1 1 + f (z ) = z E1 1 n n=1 n n=1   1 Y z2 = z 1 n2 n=1

465

11.4 Fa torization of Entire Fun tions

whi h is an entire fun tion with the desired properties. Note that the rearrangement of the fa tors in the last expression is justi ed by the absolute

onvergen e of the in nite produ t.  As in Examples 11.45, for ertain sequen es fan gn1 of points, the nonnegative integers in fpn gn1 an be hosen independent of n, say pn = p for all n and for someQ p  0. In this situation, the produ t in Theorem 11.43 takes the form 1 n=1 Ep (z=an ) for some non-negative integer p, and the produ t onverges uniformly on every ompa t subset of C , see Se tion 11.8 for further dis ussion. The Weierstrass fa torization theorem has the following important orollary.

11.46. Corollary. Every meromorphi fun tion in sented as a quotient of two entire fun tions.

C

an be repre-

Proof. Assume that f is analyti on C ex ept for poles say at the points a1 ; a2 ; : : :. Let g be an entire fun tion with zeros pre isely at an su h that the order of the zero of g at an equals the order of the pole of f at an (su h a fun tion g exists by Theorem 11.41). Then, fg has only removable singularities at the points a1 ; a2 ; : : : and thus an be extended to an entire fun tion G with no zeros in C . Hen e G(z ) G(z ) = f (z )g(z ) or f (z ) = ; g(z ) where g and G are entire. We have the following interpolation result for analyti fun tions.

11.47. Corollary. Suppose fangn1 is a sequen e of distin t omplex numbers having no nite limit point. Then for a given sequen e fbngn1 of omplex numbers, there exists an entire fun tion G su h that G(an ) = bn for every n.

Proof. Assume that f is analyti on C ex ept for simple poles at z = an with Res [f (z ); an ℄ = bn =g0(an ), where g is an entire fun tion with simple zeros pre isely at an . Note that su h fun tions exist by Mittag-Leer's theorem and Weierstrass's theorem (In ase bn is zero for some n, then assume that f is analyti at an ). Then fg has only removable singularities at the points a1 ; a2 ; : : : and thus an be extended to an entire fun tion G. Further, we have the expansions for g and f 1 g(k) (a ) X b =g0(an ) n g(z ) = (z an )k and f (z ) = n + f1 (z ); k! z an k=1 where f1 (z ) is analyti in some neighborhood of an . Finally,

G(an ) = zlim !a f (z )g(z ) = bn n

466

Representation for Entire and Meromorphi Fun tions

and we omplete the proof.

11.48. Example. We know that sin z has simple zeros at n 2 Z. Arrange the non-zero zeros of sin z in a su h a way that theyP form a se1 1=n2 quen e with non-de reasing moduli, and observe that the series n=1 P1

onverges whereas n=1 1=n diverges. Therefore, it is onvenient to hoose the number pn guring in Theorem 11.43 equal to 1. Choosing pn = 1, by Theorem 11.43, it follows that Y  z  z=n sin z = eh(z) z 1 e n n2Znf0g 1  Y z  z=n  z  z=n = zeh(z) 1 e 1+ e n n n=1   1 Y z2 = zeh(z) 1 (11.49) n2 n=1 where h(z ) is some entire fun tion. Next we need to show that h(z ) is a

onstant su h that eh(z) = . On e this is done, then we get the produ t representation for sin z in the form  1  sin z Y z2 (11.50) = 1 : z n2 n=1 Now, to nd h(z ), we pro eed as follows: Let (11.51)

Pn (z ) = eh(z) z

n  Y k=1

1



z2 : k2

Then we know that Pn ! sin z (uniformly on dis s) as n ! 1 so that Pn0 (z ) !  os z as n ! 1. Thus, as n ! 1, Pn0 (z ) !  ot z for z 2 C n fm : m 2 Zg: Pn (z ) P 2 Using the lo al uniform onvergen e of 1 k=1 jz=k j in C , we may form the logarithmi derivative on both sides of (11.51) and nd that n Pn0 (z ) 1 X 2z = h0 (z ) + + !  ot z as n ! 1: Pn (z ) z k=1 z 2 k2

A omparison of this with (11.12) shows that h0 (z ) = 0, so h(z ) is a onstant, say, . Therefore, (11.49) be omes 1  z2  Y sin z = e 1 : z k2 k=1

11.4 Fa torization of Entire Fun tions

467

Sin e limz!0 (sin z )=z =  and the right hand side approa hes e as z ! 0, we have e =  and we obtain the desired formula (11.50). 

11.52. Example. Using Example 11.11, we an provide an alternate proof of (11.50). To do this, we start by observing that the series P1 k 2 k=1 z =k onverges uniformly on every ompa t subset of C . Thus, 1  z2  Y g(z ) = z 1 k2 k=1 represents an entire fun tion. Observe that for x 2 (0; 1), g(x) > 0 and 1  x2  X ; ln g(x) = ln x + ln 1 k2 k=1

by the uniform onvergen e of the derived series. Finally, for x 2 (0; 1), 1 2x 1 X d (ln g(x)) = + dx x k=1 x2 k2 =  ot x; by Example 11.12, d = (ln sin x) ; dx whi h shows that      d g(x) g(x) ln = 0; i.e. ln = onstant: dx sin x sin x Therefore, there exists a real onstant su h that 1  x2  sin x 1 Y g(x) = sin x; i.e. = 1 ; x 2 (0; 1): x

 k=1 k2

Taking the limit x ! 0, we get  = 1. Thus, the formula (11.50) follows from the uniqueness theorem. Now, some spe ial ases follow. (i) Substituting z = i in (11.50) gives  1  Y 1 e e  sinh  1+ 2 = = : n 2  n=1 (ii) The produ t representation for the osine fun tion may be obtained dire tly from (11.50). Indeed, by (11.50), we see that 1  4z 2  sin 2z Y = 1 2z n2 n=1

468

Representation for Entire and Meromorphi Fun tions

whi h, by the identity 2 sin z os z = sin 2z; gives   1  4z 2  Y sin z

os z = 1 z n2 n=1  2 !   ! 1 Y 2z 2z 2 = 1 1 2n 2n 1 n=1     1 1 Y z2 Y 4z 2 = 1 1 n2 n=1 (2n 1)2 n=1 so that

os z = Using the formulae

1  Y n=1

1



4z 2 ; z 2 C: (2n 1)2

eiz + 1 = 2eiz=2 os(z=2) and eiz

1 = 2ieiz=2 sin(z=2);

the produ t representation for the entire fun tions eiz + 1 and eiz 1

an be a hieved with the help of the orresponding representations for os z and sin z . Consequently, if we repla e z by iz then we an qui kly establish the formula  1  Y z2 1+ 2 2 ; z 2 C: (11.53) ez 1 = zez=2 4 n n=1 (iii) A omparison of the produ t expansion of an entire fun tion with its power series expansion often leads to interesting on lusions. For example, by equating the produ t expansion (11.50) of sin z with its Taylor's series expansion about 0, we see that  1  Y z2 (z )3 (z )5 1 (11.54) z = z +


: 2 n 3! 5! n=1 Be ause of the uniqueness of the Taylor oeÆ ients, a omparison of the oeÆ ients of z 3 on both sides shows that   1 1 2 X 1 1 3  1+ 2 + 2 +


= ; i.e. 2 = 6: 2 3 3! n=1 n (iv) If z = 1=2, then (11.50) gives the `Wallis produ t formula'  1 1  4k2 1  Y 1 Y 1= 1 = : 2 k=1 4k2 2 k=1 4k2

469

11.5 The Gamma Fun tion

Consequently,     1 2k Y  2k 2
2 4
4 6
6 =
=


2 1
3 3
5 5
7 k=1 2k 1 2k + 1       2
2 4
4 2n
2n 1 = nlim !1 1
1 3
3


(2n 1)
(2n 1) 2n + 1 or equivalently r





 2
4
6


2n 1 p = nlim !1 2 1
3
5


(2n 1) 2n + 1 whi h is often alled the \Wallis identity". This formula an be written in the form r

(

 (2
4
6


2n)2 1 p p = nlim !1 2 (1
3


(2n 1))(2
4


2n) 2n 1 + 1=2n

p

whi h gives  = nlim !1

22n (n!)2 : (2n)!n1=2

)



11.5 The Gamma Fun tion The study of the -fun tion that involves the integral form  Z 1 Z 1 1 x 1 (x) := log (11.55) dt = e ttx 1 dt (x > 0) t 0 0 was rst introdu ed by Euler in 1729. Dierentiating the integral (11.55) gives Z 1 (n) (x) = e t tx 1 (log t)n dt; n 2 N : 0

Clearly (1) = 1 and for x > 0, integration by parts yields the fun tional equation (11.56) (x + 1) = x (x) be ause Z 1 Z 1 1 (x + 1) = tx d( e t) = tx e t j0 + x e t tx 1 dt = x (x): 0

0

For x = n 2 N , the fun tional equation be omes (n + 1) = n! and therefore, the -fun tion is seen as an extension of the fa torial fun tion for x belonging to R n f0; 1; 2; : : : g: Indeed, in view of the fun tional equation (11.56), we have (x) =

(x + 1) (x + 2) = = x x(x + 1)




= x(x + 1) (
x

+(xn+) n 1) :

470

Analyti Continuation

This observation suggests that it is possible to extend the -fun tion on the whole real axis ex ept on the negative integers f0; 1; 2; : : : g. We are interested in dis ussing the following questions:  Does there exist an analog of (11.55) for Re z > 0?  Is it possible to extend (x) de ned by (11.55) to C n f0; 1; 2; : : : g?  What an we say about the points in f0; 1; 2; : : : g?  What is the ounterpart of (11.56) when x takes omplex values?  What are the basi onsequen es of the extended gamma fun tion? There are two simple approa hes through whi h one an answer these questions. Let us rst allow the integral (11.55) to extend the \fa torial fun tion" to the omplex plane. Now,

jtz j = jez log t j = eRe z ln t = tRe z for t > 0 so that if we repla e x in (11.55) by the omplex variable z , the resulting fun tion ( alled the lassi al gamma fun tion) given by Z 1 (11.57) (z ) = e t tz 1 dt 0

is uniformly onvergent for Re z > 0. For this fun tion, the point z = 0 is a singularity be ause Z 1 Z 1 e t t  1 () = e t dt = 1  dt ! 1 as  ! 0: 0 0 t

11.58. Gamma fun tion via produ t representation. We re all that 1 h z  i Y sin z (11.59) zG(z )G( z ) = ; G(z ) = 1 + e z=n :  n n=1 Here G(z ) is entire and it has simple zeros at all negative integers. Clearly, the zeros of G and G(z 1) are the same, ex ept that G(z 1) has a zero at z = 0. It follows from the Weierstrass fa torization theorem that G(z 1) = ze (z)G(z ) or equivalently, 1 z + n 1 Y e (z n n=1

1)=n

= ze (z)

1 Y n=1





z+n e z=n ; n

for some entire fun tion (z ) whi h is to be determined. In order to nd

(z ), we take the logarithmi derivative of the last equation and obtain   1  1  1 X X 1 1 1 1 = + 0 (z ) + : z n=1 z + n 1 n n=1 z + n n

471

11.5 The Gamma Fun tion

  P 1 1 Repla e n by n + 1 on the L.H.S and note that 1 n=1 n+1 n = 1. This gives that 0 (z ) = 0 so that (z ) is a onstant whi h we denote by . Therefore, G(z 1) = ze G(z ): To nd the value of the onstant , we observe that G(0) = 1 and so we may let z = 1. This gives  1  Y 1

e 1=n : 1 = e G(1); i.e. e = 1+ n n=1

Taking the logarithm on both sides   1   X 1 1

= ln 1 + n n n=1 n X









1 1 = nlim !1 k=1 ln 1 + k k        2 3 n 1 1 1 = nlim ln



+ ln 1 + 1 + +


+ !1 1 2 n 1 n 2 n      1 1 1 = nlim 1+ +


+ ; sin e nlim !1 ln n !1 ln 1 + n = 0: 2 n Thus,   1 1 (11.60)

= nlim !1 1 + 2 +


+ n ln n = 0:5772


: The onstant is alled Euler's onstant. The question of whether is rational or irrational seems to be still open. In on lusion, the produ t representation of the gamma fun tion is de ned by  1  z  1 1 e z Y z=n (11.61) (z ) = z = 1+ e : ze G(z ) z n=1 n We have the following onsequen es of the de nition:  In view of (11.61) and taking into a

ount (11.60), we dedu e the formula of Gauss  n  e z Y z  1 z=k (z ) = nlim 1 + e !1 z k=1 k 



n ez ln n Y k = nlim !1 z k=1 z + k n!nz = nlim !1 z (z + 1)


(z + n) whi h is valid for all z 6= 0; 1; 2; : : : .

472

Analyti Continuation

 In view of (11.61), we have (z +1) = z (z ), whi h is alled Riemann's 

Fun tional Relation for the gamma fun tion. By (11.59), we have the identity

(1 z ) (z ) = z ( z ) (z ) =

p

 

1  = : zG(z )G( z ) sin z

In parti ular, (1=2) = . Repeated appli ation of (z + 1) = z (z ) gives (n + 1) = n!. (z ) never vanishes in C as, for z 62 f0; 1; 2;


g, the gamma fun tion is given by a onvergent in nite produ t of non-zero fa tors. Moreover, the representation 1 h z  i Y 1 = ze z G(z ) = ze z 1 + e z=n ; z 2 C ; (z ) n n=1 makes expli it that the gamma fun tion never vanishes and that it has simple poles pre isely at 0; 1; 2; : : : . The fun tional equation (z ) =

(z + 1) = z




= z (z +(z1)+
n

+(z1)+ n)

shows that Res [ (z ); 0℄ = limz!0 z (z ) = limz!0 (z +1) = (1) = 1 and for n 2 N , we have Res [ (z ); n℄ = z! limn(z + n) (z ) (z + n + 1) = z! limn z (z + 1)(z + 2)


(z + n 1) (1) ( 1)n = = : ( n)( n + 1)


( 1) n!



So, (z ) is meromorphi in C . Take the logarithmi derivative of (11.61) to get  1  1 0 (z ) 1 X 1 = + + (z ) z n=1 z + n n so that (11.62)

d dz



1 0 (z )  X 1 = 2: (z ) n=0 (z + n)

As (z ) (z + 1=2) and (2z ) have the same set of poles, we an write (z ) (z + 1=2) = e(z) (2z )

473

11.6 The Zeta Fun tion

for some entire fun tion (z ). In fa t, by (11.62), we see that  0 1 1 0 (z + 1=2)  X X d (z ) 4 4 + = + 2 dz (z ) (z + 1=2) (2 z + 2 n ) (2 z + 2 n + 1)2 n=0 n=0 1 X 1 = 4 (2 z + n)2 n=0  0  d (2z ) = 2 dz (2z ) whi h, upon integration, shows that (z ) (z + 1=2) = eaz+b (2z ) for some onstants a and b. To nd the values p of a and b, wepsubstitute z = 1=2; 1 and obtain (note that (1=2) =  , (3=2) = =2)

p = e(a=2)+b ;

p 2

= ea+b

p

whi h, by dividing one by the other, gives ea=2 = 1=2 and eb = 2 : It follows that a = 2 ln 2 and thus, we obtain the so alled Legendre's dupli ation formula (11.63)

p (2z ) = 22z

1

(z ) (z + 1=2):

11.6 The Zeta Fun tion The zeta fun tion was rst introdu ed by Euler. There are two important formulas whi h de ne the zeta fun tion in two dierent ways either as series form or as an Euler produ t. By the series formula of Riemann, this is expressed as 1 1 X (11.64)  (s) = s ; s =  + it; n=1 n where we use the traditional notation to denote the omplex variable by s =  + it. The fun tion  represented by the series (11.64) is alled the Riemann zeta fun tion or simply the zeta fun tion. This fun tion, whi h is of spe ial interest, provides a link between number theory and fun tion theory. Also, it is of entral importan e in number theory parti ularly in the study of the distribution of prime numbers. First we shall show that the series (11.64) onverges for  > 1. To do this we let   0 > 1. Then, by the de nition of ns ,



jns j = je(+it) log n j = e ln n = n  n0 ; i.e. n1+it  n10 :

474

Analyti Continuation

Further, sin e x10 is de reasing for x > 0, Z n+1

n

Z

n dx dx 1 < < x0 n0 n 1 x0

where the inequality on the left holds for n  1 and that on the right holds for n  2. Hen e, for any natural number N  2, we have N X

1 0 < n=2 n

Z N

1



dx 1 1 =   0 x 0 + 1 N 0



1

1

! 1 1  as N ! 1 0

and so, as the partial sum is an in reasing sequen e bounded above, the series onverges. Indeed, Z 1 1 1 X X dx 1 0 j (s)j  n1+it  n10 < 1 + 0 = 1 + 0 1 = 0 1 : x 1 n=1 n=1 Therefore, the series on the right of (11.64) onverges uniformly and absolutely for   0 > 1. As f (s) = n s = e s ln n , ea h term in the series (11.64) is an analyti fun tion of s and therefore, f 0 (s) = lnnsn : Moreover, the Weierstrass M -test shows that the  fun tion is analyti for Re s > 1 with derivatives 1 (ln n)k 1 ln n X X (k) (s) = ( 1)k and   0 (s) = s s ; k 2 N: n=1 n n=1 n The  fun tion so de ned is related to the study of prime numbers by the following

11.65. Theorem. (Euler's Produ t Formula) For s =  + it,  > 1,

we have

(11.66)

 (s) =

Y

1

1 ps

 1



Y 1 or = 1  (s)



1 ; ps

where the produ t is taken over all prime numbers p. In parti ular,  = 6 0 for Re s > 1. Q

s Proof. By Corollary 11.22, the in nite produ t p; prime (1 p ) onP verges uniformly forP  0 > 1 as the series p prime p s is obtained by s omitting terms of 1 n=1 n whi h onverges uniformly for   0 > 1. Now onsider the series (11.64) for  > 1, and

1 1 1 1  (s) s = s + s + s +


: 2 2 4 6

475

11.6 The Zeta Fun tion

Subtra ting this equation from (11.64), we obtain that 



1 1 1 =1+ s + s + s 2 3 5

 (s) 1




:

Similarly, one an nd that 

 (s) 1

1 2s



1



1 1 1 1 =1+ s + s + s + 3s 5 7 11




:

More generally,


X  (s) 1 2 s 1 3 s


1 pNs = m s = 1 + pNs+1 +






where the sum on the right being over all positive integers that ontain none of the prime Q fa tors 2; 3; : : : ; pN . Therefore, allowing N ! 1, it follows that  (s) p; prime (1 p s ) = 1 and the on lusion follows. The two de nitions of the Riemann zeta fun tion, namely the series form (11:64) and the produ t form (11.66), are equivalent. Either of them may be taken as a de nition of  (s) for Re s > 1. The following relationship between the -fun tion and the  -fun tion may be studied in several forms.

11.67. Theorem. (Integral representation of zeta fun tion) For Re s > 1, we have Z 1 s 1 x 1 (11.68)  (s) = dx: (s) 0 ex 1 Proof. The integral de nition of the -fun tion given by (11.57) together with the hange of variable t = nx imply that for Re s > 1 Z 1 (s) = e nx xs 1 dx: ns 0 Summation over all positive integer n in the equation gives Z 1 Z 1 s 1 1 X x (s) (s) = xs 1 (e x)n dx = x 1 dx 0 0 e n=1 where used the fa t that the partial sums of the geometri seP we have x n ries 1 n=1 (e ) form an in reasing sequen e of fun tions that onverges uniformly on ea h interval [; 1),  > 0. This observation justi es the inter hange of the summation and the integration.

11.69. Global representation of  (s). The de nition of the  fun tion given by (11.64) and (11.66) are valid only for Re s > 1. We wish to extend the  fun tion to C n f1g and show that  is meromorphi in C and

476

Analyti Continuation y

y 2π i

2(n + 1)πi 2nπ i C = Cǫ

O

x

−n

n

x

−2nπ i −2π i

−2(n + 1)πi

Figure 11.1: Contour for the ontinuation of  fun tion.

has a simple pole only at s = 1 with residue 1. A

ording to (11.68), if we an show that the integral in (11.68) is a meromorphi fun tion of s in C then the  -representation given by this integral an be used as the de nition of the  fun tion on C . Unfortunately, in the present form, the integral in (11.68) is improper and is divergent for Re s < 0, be ause as x nears zero, s 1 x Re s 2 : ex 1  x Riemann over ame this diÆ ulty, through a tri k. To dis uss this, we need to represent  (s) as a ontour integral by hoosing a suitable ontour that avoids the origin so that the resultant form be omes entire. For s 2 C xed,

onsider ( z )s 1 e(s 1) Log ( z) f (z ) = z = : e 1 ez 1 This fun tion is analyti in C n [fx 2 R : x  0g [ f2ki : k 2 Zg℄: De ne Z 1 (s) = (11.70) f (z ) dz; 2i C where C = C = C (Æ) is the \Hankel ontour" shown in Figure 11.1 with 0 <  < 2. This integral onverges, and it represents an entire fun tion of s. Indeed, the integrand is an entire fun tion of s implying that (s) is entire. Sin e the region bounded by the ontours C and C ontains no poles of f (z ) for 0 <  <  < 2, by Cau hy's theorem, we see that the value of the integral in (11.70) is independent of .

11.71. Theorem. For Re s > 1, Z (1 s) ( z )s 1 (11.72) dz:  (s) = 2i C ez 1

477

11.6 The Zeta Fun tion

Proof. Sin e the integral in (11.70) is independent of the shape of C as long as C does not en lose any integer multiple of 2i, we are free to allow  ! 0. We also note that, for z = x + iÆ := ei with x > 0, Log ( z ) = Log ( x iÆ) ! ln x i as 0 < Æ ! 0 and, for z = x iÆ, Log ( z ) = Log ( x + iÆ) ! ln x + i as 0 > Æ ! 0: First, as Æ ! 0, we express the integral (11.70) as Z

Z

Z 

( z )s 1 ; ez 1

1

2

3 Z  (s 1)(ln x i) Z Z 1 (s 1)(ln x+i) e e = dx + f (z ) dz + dx x e 1 ex 1 1 jzj=  Z 1 s 1 Z x = 2i sin(s) dx + f (z ) dz jzj=  ex 1

2i(s) =

+

+

f (z ) dz; f (z ) =

be ause

e(s

1)(ln x+i)

e(s

1)(ln x i)

= xs =

h

e(s 1)i e (s xs 1 2i sin(s): 1

1)i

i

Suppose, for the moment, that Re s > 1. As ez 1 has a simple zero at the origin, by the ontinuity of ez 1, we see that for jz j suÆ iently small, jez 1j  jz j=2 and so there exists an M > 0 su h that Z jzj=



Z

2Re s ( z )s 1 dz  M z e 1 jzj= 

1

jdz j = 4MRe s 1 :

Consequently, the integral tends to zero as  ! 0 and be ause Re s > 1. Finally, we may evaluate (s) by letting  ! 0 in the last equation. This gives Z sin s 1 xs 1 (s) = dx  0 ex 1 sin s = (s) (s); by Lemma 11.67;   (s) ; sin e (1 s) (s) = sins : = (1 s) Thus, (11.72) follows. Note that (s) is an entire fun tion of s and the fa tor (1 s) has simple poles at s = 1; 2; : : : . The left hand side of (11.72), on the other

478

Analyti Continuation

hand, is de ned at all su h points ex ept for s = 1. Thus, the equation (11.72) de nes an analyti fun tion ex ept possibly with a simple pole at s = 1. Now, (s) has a simple pole at s = 0 with residue 1 and so, 1 1 (1 s)  = for s near 1: 1 s s 1 Thus, 

Z

1 ( z )s 1 lim ( s 1)  ( s ) = lim ( s 1) (1 s ) dz s!1 s!1 2i ez 1   Z 1 dz = ( 1) 2i ez 1   1 = Res z ;0 = 1 e 1



and we have established the following

11.73. Theorem. The  fun tion is analyti in simple pole at s = 1 with residue 1.

C n f1g and has a

Riemann used the de nition of the  fun tion de ned by (11.72) to derive many interesting relations (su h as the fun tional equation of the  fun tion and the fun tion  (s), below) whi h a tually help to verify the so- alled Riemann hypothesis. Let us rst prove the fun tional equation whi h provides more expli it information about the analyti ontinuation of  to C n f1g.

11.74. Theorem. For all s 2 C , the  -fun tion satis es Riemann's

fun tional equation

(11.75)

 (s) = 2s s 1 sin(s=2) (1 s) (1 s):

In parti ular (as  (1

s) 6= 0 for Re s < 0),  ( 2k) = 0 for k 2 N .

Proof. It suÆ es to prove this theorem for s < 0. De ne Z

1 ( z )s 1 f (z ) dz; f (z ) = z ; n (s) = 2i Cn e 1 where Cn is as shown in Figure 11.1. Note that the re tangle has verti es at n  (2n + 1=2)i. The idea is to relate (s) to n (s) using the al ulus of residues and then to let n ! 1. For z on the sides of the re tangle, we have jez 1j > 1=2 and so for s < 0, (s 1) Log ( z) e jf (z )j = ez 1  2ns 1 :

479

11.6 The Zeta Fun tion

By the ML-inequality (see Theorem 4.9(iii)), jn (s)j  Kns ! 0 as n ! 1. Consequently,

n (s) (s) = 2i  (sum of the residues of f (z ) inside the re tangle):

The poles of f (z ) are simple and they o

ur at zk = 2ki, 1 with residue ( z )s 1 1 lim ( z z ) = lim lim ( z )s 1 k z z!zk z!zk ez z!zk e 1 = (2ki)s 1 = e(s 1)[ln j2kji=2℄ = (2k)s 1 e(s 1)i=2 :

 k  n,

It follows that

n X 1 ( z )s 1 (2k)s dz = 2i CnnC ez 1 k=1 Z

= 2

1



e (s

1)i=2 + e(s 1)i=2



n X

(2k)s 1 os[(s 1)=2℄

k=1

n X 1 = 2s s 1 sin(s=2) 1 s: k k=1

Allowing n ! 1, it follows that (sin e n (s) ! 0 as n ! 1) Z

( z )s 1 1 dz = 2s s 1 sin(s=2) (1 s): 2i C ez 1 If we ombine this with (11.72) and follow the dis ussion presented after its proof, we obtain the fun tional equation for s < 0. Sin e both sides of the fun tional equation are meromorphi and agree on a non-empty open set, it holds for all s by the uniqueness theorem. There are equivalent forms of the fun tional equation (see for example, Exer ise 11.125). If we repla e z by (1 s)=2, then Legendre's dupli ation formula for the gamma fun tion (11.63) be omes 2 s  ((1 s)=2) 1=2 (1 s) = 2 s ((1 s)=2) (1 s=2) = (s=2) sin(s=2) so that 2 s1=2 ((1 s)=2) (1 s) sin(s=2) = : (s=2) In view of this equation, (11.75) is equivalent to

 s=2 (s=2) (s) = 

(1 s)=2

((1 s)=2) (1 s):

480

Representation for Entire and Meromorphi Fun tions

Thus, the fun tional equation takes the form (s) = (1 s); (s) =  s=2 (s=2) (s): Clearly,  has simple poles at s = 0; 1. If we multiply it by s(1 s)=2, we see that the fun tion s(1 s) s(1 s) s=2  (s) = (s) =  (s=2) (s) 2 2 is entire and satis es the relation  (s) =  (1 s).

11.76. Riemann Hypothesis. The fun tional equation of the  fun tion enables us to lo ate the zeros of  (s). Be ause of the produ t representation (11.66), the  fun tion has no zeros if Re s > 1. We know that (s) never vanishes in C and is analyti ex ept at s = 0; 1; 2; : : : . Consequently, both (1 s) and  (1 s) are analyti and non-zero for Re s < 0. The fun tional equation then says that the only zeros of  (s) for Re s < 0 are the zeros of sin(s=2), that is only at s = 2; 4; : : : : These are known as the trivial zeros of the  fun tion. It is not diÆ ult to show that there are no zeros on Re s = 0 and Re s = 1. We on lude that all the non-trivial zeros of the  fun tion lies in the strip fs : 0 < Re s < 1g, whi h is alled the riti al strip. It is known that there are in nitely many zeros on the line s = 1=2 + it; t 2 R. This line in the s-plane is alled the riti al line. Again, sin e 1 1 (1 21 s ) (s) = 1 s + s


> 0 2 3 for 0 < s < 1 and  (0) 6= 0,  (s) has no zeros on the real interval (0; 1). This observation implies that all possible zeros of  (s) in the riti al strip are

omplex numbers. The Riemann hypothesis asserts that \All the nontrivial zeros of  fun tion lie on the riti al line Re s = 1=2". Although this has been shown to be true for more than one billion non-trivial zeros, the onje ture remains open, despite the eorts of some of the greatest analysts sin e Riemann's time. It is the most famous unsolved problem onfronting 21-th entury mathemati ians, espe ially after the proof of Fermat's last theorem. No al ulation had ever ontradi ted the hypothesis. The Clay Mathemati s Institute of Cambridge, Massa husetts, oers one million US dollars for a proof of the Riemann hypothesis.

11.7 Jensen's Formula Suppose that g 2 H(
) and that it is zero-free on
. Then g is a nowhere

vanishing analyti fun tion on
R for some R > 1, and so there exists an h 2 H(
) su h that g(z ) = eh(z) . In parti ular, g admits an analyti logarithm on
R . The Cau hy integral formula applied to log g(z ) yields that Z 1 2 log g(0) = log g(ei ) d: 2 0

481

11.7 Jensen's Formula

Equating the real parts gives ln jg(0)j =

(11.77)

1 2

Z 2

0

ln jg(ei )j d

(we observe that the right hand side of (11.77) is an improper integral if g has a zero on the ir le jz j = 1). For example, if jaj < 1 and one onsiders g(z ) = 1 az then g 2 H(
) and is zero-free there so that (11.77) gives (11.78) 0 = ln 1 =

1 2

Z 2

0

ln j1 aei j d =

1 2

Z 2

0

ln j1 ae i j d:

More generally, we prove

11.79. Lemma. For a 2 C with jaj < r, we have 1 2

(11.80)

Z 2

0

ln jr

ae i j d = ln r:

(Note that r = 1 gives (11:78)).

Proof. We may rewrite jr ae i j as rj1 (a=r)e i j = rj1 (a=r)ei j so that ln jr ae i j = ln r + ln j1 (a=r)ei j and 1 2

Z 2

0

ln jr

ae

i j d = ln r +

1 2

Z 2

0

ln j1 (a=r)ei j d:

Note that the integral on the right vanishes with the same reasoning as above (by onsidering g(z ) = 1 (a=r)z ). Therefore, (11.80) follows. Alternatively, for jbj < 1, we obtain that 1 2

Z 2

0

ln j1

bei j d



Z



R 2

ln j1 ei j d; then I = 0.

1 2 = Re Log (1 bei ) d 2 0 " # Z Log (1 z ) 1 = Re dz 2i jzj=b z = ln 1 = 0; by the Cau hy integral formula,

and therefore, (11.80) holds.

11.81. Lemma. If I = 21

0

Proof. Sin e sin '  0 on [0; ℄, we write

j1 ei j = je

i=2

ei=2 j = 2 sin(=2)

482

Representation for Entire and Meromorphi Fun tions

for  2 [0; 2℄. Therefore, ln j1 ei j = ln 2 + ln sin(=2) so that

I=

Z 2

0

ln j1 ei j d = 2 ln 2 +

Z 2

= 2 ln 2 + 2

0Z

ln sin(=2) d



ln sin t dt ( = 2t):

0 R

To omplete the proof it suÆ es to show that 0 ln sin t dt =  ln 2. Noti e that this integral is improper, but its onvergen e is obvious. Now, we nd that Z  Z =2 Z =2 J= ln sin t dt = 2 ln sin t dt = 2 ln os s ds 0

0

0

(use the hange of variable t = =2 s in the se ond integral) and we also see that

J = =

Z 

Z0 

0

ln[2 sin(t=2) os(t=2)℄ dt ln 2 dt +

=  ln 2 + 2

Z 

0 Z =2 0

ln sin(t=2) dt +

ln sin(t) dt + 2

=  ln 2 + 2J:

Z 

0 Z =2 0

ln os(t=2) dt ln os(t) dt

Thus, J =  ln 2 and so, I = 0. We will now investigate what happens to (11.77) in the presen e of zeros as well as poles in
. Also, we remark that there is a similar generalization of the Poisson integral formula (see Exer ise 11.119).

11.82. Theorem. (Jensen's Formula for the losed unit disk) Let f be meromorphi in
. Suppose that (i) 0 is neither a zero nor a pole of f (ii) ai (1  i  m) and bj (1  j  n) denote the zeros and poles of f in
( ounted as many times as its order of multipli ities), respe tively. Then we have

(11.83)

1 2

Z 2

0

ln jf (ei )j d

= ln jf (0)j + ln

Qn

j =1 jbj j Qm i=1 jai j

!

:

11.84. Observations. There are a number of interesting observations we an make from (11.83).

483

11.7 Jensen's Formula

(i) Equation (11.83) is alled the Jensen formula for meromorphi fun tion in the losed unit disk
. First we shall obtain a general result (see Corollary 11.85) as a onsequen e of this result. (ii) The se ond sum on the right hand side of (11.83) will not appear when f has neither zeros nor poles on jz j  1 (in this ase, f is a tually in H(
) with f (0) 6= 0; ompare with (11.77)). This fa t will be lear in the proof. (iii) Equation (11.83) yields Jensen's inequality: "

ln jf (0)j

Qn

j =1 jbj j Qm i=1 jai j

!#

 sup ln jf (ei )j: 0 0,


jf (z )j  exp jz j2 + for jz j  r0 : This implies that
M (r) = max jf (z )j  exp r2 + for jz j = r  r0 jzj=r

so that for large r, we have ln ln M (r)  (2 + ) ln r; i.e.

ln ln M (r) ln r

 2 + :

On the other hand, by de nition,
ln ln M (r) M (r) > exp r2  ; i.e > 2 ln r for an in nite number of r's, r ! 1. Thus,

2 = lim sup r!1



ln ln M (r) ln r

and hen e, 1 = 2 . Finally, 1 = 1 i 2 = 1. Indeed, if 2 = 1 then for any given A > 0, ln ln M (r) M (r) > exp(rA ) or >A ln r for some values of r suÆ iently large. Thus, 2 = 1 i 1 = 1. We have the following (i) the number (f ) = 1 = 2 is alled the order of f (z ). When there is no onfusion, we may simply use the notation  instead of (f ) as we did in the proof of the above theorem. So, 0   < 1 as we work here only with fun tions of nite order. (ii) Clearly, if 2 < 0, then f (z ) is bounded on C and so redu es to a onstant, by Liouville's theorem. This is another reason why we have de ned 2 (f ) in (11.92) through nonnegative real numbers A satisfying the growth ondition.

11.8 The Order and the Genus of Entire Fun tions

489

11.93. Example. For z = rei
(i) j exp(ez )j = j exp(er os  eir sin  )j = j exp er os  os(r sin ) j so that M (r) = exp(er ) and ln ln M (r) r = ! 1 as r ! 1: ln r ln r Therefore,  = 1 and we say that f (z ) = exp(ez ) is of in nite order. Alternatively, as x ! 1 along the positive real axis, we have exp(ex ) > exp(xk ) so that   k for any k  0. Consequently, by (11.92), f (z ) = exp(ez ) is of in nite order. (ii) If f (z ) = exp(z m) for some m 2 N , then jf (z )j = j exp(rm os(m))j so that M (r) = exp(rm ) whi h gives easily that  = m, by (11.91). 2 In parti ular, the order of ez is 1 while the order of ez is 2. More generally, if p(z ) is a polynomial of degree m ( 1) then the order of exp(p(z )) is m. (iii) If f (z ) = os z , then jf (z )j  j(eiz + e iz )=2j  (ey + e y )=2  (ejzj + ejzj)=2 = ejzj so that   1, a

ording to (11.92). But, for z = iy, jf (iy)j = (e y + ey )=2  ejyj=2

whi h shows that  is at least 1. Thus, the order of os z is 1. (iv) If f (z ) = sin z , then for jz j = r iz e



e iz ey + e y  2  ejzj jf (z )j = 2i so that   1, a

ording to (11.92). But, for z = iy jf (iy)j = (e y ey )=2  (ejyj e jyj)=2  ejyj=3 as y ! 1 whi h shows that  is at least 1. Thus, (sin z ) = 1. P (v) If f (z ) = a0 + a1 z +


+ an z n, then, for jz j = r > maxf1; nk=0 jak jg,

jf (z )j  (ja0 j + ja1 jr +


+ jan jrn ) < (ja0 j + ja1 j +


+ jan j) rn and so, jf (z )j < rn+1 , i.e. M (r)  rn+1 . Thus,

ln ln M (r) ln(n + 1) + ln ln r  ! 0 as r ! 1 ln r ln r showing that every polynomial is of order 0.



490

Representation for Entire and Meromorphi Fun tions

Before we turn to the dis ussion on Hadamard's fa torization theorem for entire fun tions, we re all the Weierstrass fa torization theorem (see Theorem 11.43) whi h states that every entire fun tion f an be fa tored in the form Y (11.94) f (z ) = z m eh(z) Epn (z=an) 1n! where (i) h is entire (ii) fang!n=1 forms a sequen e of zeros of f distin t from z = 0, ea h of them listed a

ording to its multipli ity (iii) ! 2 N if the sequen e is nite and ! = 1 otherwise (iv) m = 0 is allowed if f (0) 6= 0; otherwise m is the multipli ity of the zero of f at the origin (v) fpng!n=1 is a sequen e of non-negative integers su h that X 

 r pn +1 < 1 for every r > 0: jan j 1n!

Our main task is to deal with the situation ! = 1. Let f be entire with zeros fan gn1 , listed a

ording to multipli ity and arranged su h thatP0 < ja1 j  ja2 j 


. If p is the smallest nonnegative (p+1) < 1, then f is said to be of rank p. It integer su h that 1 n=1 jan j is trivial to see that p = 0 whenever f has only a nite number of zeros. An f is of in nite rank, i.e. if there exists no p for whi h P1entire fun tion (p+1) < 1. This is possible whenever P1 ja j (p+1) = 1 for j a j n=1 n n=1 n all p  0. IfP f is of nite rank p, i.e., there exists a nonnegative integer p  0 su h (p+1) < 1, then, by (11.94), f (z ) an be rewritten as that 1 n=1 jan j   1 Y z (11.95) f (z ) = z meh(z) Ep =: z meh(z) P (z ): a n n=1 Q1 Here the produ t P (z ) := n=1 Ep (z=an) is alled the Weierstrass produ t or the anoni al produ t asso iated with the sequen e fangn1 . Further, if f is of nite rank p, and p0 is any integer with p0 > p, then 1 X jan j (p0 +1)  jan j (p+1) ; i.e. jan j (p0 +1) < 1 Q1

n=1

and so there is another produ t n=1 Ep0 (z=an) showing that the fa torization for f is not unique. If p happens to be the rank of f , then p is alled the genus of the

anoni al produ t P (z ). Moreover, the produ t P (z ) so de ned is said to be

11.8 The Order and the Genus of Entire Fun tions

491

P p in standard form for f . It is important to remember that 1 n=1 jan j = 1 P1 ( p +1) whereas n=1 jan j < 1. Further (as in Examples 11.45), in the representation (11.95), the fa torization is unique ex ept that h(z ) may be repla ed by h(z ) + 2mi for any m 2 Z. If p is the genus of the anoni al produ t and h(z ) is a polynomial in the representation (11.95), then the fun tion f (z ) is said to be of nite genus and, in addition, if q is the degree of the polynomial h(z ) then the  := maxfp; qg is de ned to be the genus of f (z ). The number q is referred to as the exponential degree of f (z ). If P (z ) is not of nite rank or h(z ) is not a polynomial, then f is said to be of in nite genus. In the sequel, we are interested only on fun tions of nite genus. For example, an entire fun tion of genus zero is of the form  1  Y z m

z 1 an n=1 P 1 with 1 n=1 jan j < 1. By de nition, the anoni al representation of an entire fun tion of genus 1 is either of the form 1 z Y

z me z E1 ; 2 C; an n=1

with

P1

n=1 jan j

2

< 1,

P1

n=1 jan j

z me z with

P1

n=1 jan j

2

1

= 1, or of the form

1  Y n=1

1

z an



< 1 and 6= 0.

11.96. Examples. (i) From the representation of sin z shown in (11.50), we see that sin z is an entire fun tion of genus 1. 2 (ii) ez is of genus 1 whereas ez is of genus 2. (iii) A polynomial p(z ) = a0 + a1 z +


+ an z n is of genus 0. 

11.97. Lemma. Let f be a non onstant entire fun tion of order , with zeros at a1 ; a2 ; : : : , with ounting multipli ities. Suppose that 0 < ja1 j  ja2 j 


. If p is an integer with p + 1 > , then for z 2 C n fa1 ; a2 ; : : : g,   1 X dp f 0 (z ) 1 = p ! p p+1 : dz f (z ) j =1 (aj z )

492

Representation for Entire and Meromorphi Fun tions

Proof. First we assume f (0) 6 0. Let a1 ; a2 ; : : : ; an be the zeros of f in
R so that f (z ) 6= 0 on jz j = R. De ne F (z ) = f ( z )

Y (R) j =1

1  z ;  (z ) = : aj =R (z=R) 1 z

Then F is a non-vanishing analyti fun tion in
R with jF (z )j = jf (z )j on jz j = R and so there exists a g 2 H(
R ) su h that F (z ) = exp(g(z )), z 2
R . This gives

f (z )

Y (R) j =1

R2 aj z = exp(g(z )): R(aj z )

Taking the logarithmi derivative with respe t to z , we obtain

f 0 (z ) f (z )

X (R) 

aj

j =1

1

z

+



aj

R2 aj z

= g0(z )

and so dierentiating p-times yields that "

 

X (R) dp f 0 (z ) = p ! dz p f (z ) j =1 (aj

# apj +1 1 + + g(p+1) (z ); z )p+1 (R2 aj z )p+1

(11.98) p = 0; 1; : : : : As for the se ond term on the right in (11.98), we rst note that for jz j < R=2 and jan j  R,

jR2 aj z j  R2 jaj j jz j > R2 R2 =2 = R2 =2

and so

(R) X j =1



apj +1 Rp+1 2p+1 

( R )  (AR+ + B ) (R2 aj z )p+1 (R2 =2)p+1 R(p+1)

for some onstants A and B . Here, as f is of order , we have used the estimate (see (11.88))

(R)  AR+ + B for R suÆ iently large: As p +1  > 0, if we hoose  so that  +  < p +1 (e.g.  = (p +1 )=2), 2p+1 (R)R (p+1) approa hes zero as R ! 1; that is the se ond sum in (11.98) onverges to zero. As for the last term, we note that Re g(z ) = ln jF (z )j for z 2
R and Re g(z ) = ln jf (z )j for jz j = R. By Theorem 10.34, g has the form

g(z ) = iIm g(0) +

1 2

Z 2  i Re

0



+z ln jf (Rei )j d: i Re z

493

11.8 The Order and the Genus of Entire Fun tions

Thus, for jz j < R (11.99)

g(p+1) (z ) =

(p + 1)! 2

Z 2

0

2Rei ln jf (Rei )j d: z )p+2

(Rei

For ea h xed z with jz j < R=2, the Cau hy integral formula applied to the fun tion (w) = (w z ) p 2 shows that Z

(p + 1)! dw (p + 1)! 0= = 2i jwj=R (w z )p+2 2

Z 2

0

Rei d: (Rei z )p+2

In view of this, for jz j < R=2 Z 2 2Rei i )j ln M (R)
d ln j f ( Re jg(p+1) (z )j = (p 2+1)! i z )p+2 0 (Re Z 2
R  ((Rp + j1)! ln M (R) ln jf (Rei )j d z j)p+2

0

 on a

ount of R

  Z (p + 1)!R 2 p+2 2 (ln M (R) ln jf (0)j) d  R 0

jz j > R R=2 = R=2 and the Jensen inequality 1 2

Z 1

0

ln jf (Rei )j d  ln jf (0)j:

Therefore, sin e ln M (R) < R+ for large R and for a given  > 0, we have  p+2 jg(p+1) (z )j  (p + 1)! R2 (R+ ln jf (0)j)2R = (p + 1)!2p+3 (R+ ln jf (0)j)R (p+1) ! 0 as R ! 1 ( sin e ( + )=(p + 1) < 1): Finally, as R ! 1, (11.99) gives the desired formula whenever f (0) 6= 0.

If f (0) = 0 and zero is of multipli ity m, then we write

f (z ) = z mG(z ); G(0) 6= 0: Then

f 0 (z ) m G0 (z ) = + : f (z ) z G(z ) Dierentiate p times to omplete the argument. 11.100. Theorem. (Hadamard's Fa torization Theorem) Let f be an entire fun tion of nite order . Suppose that a1 ; a2 ; : : : ; are the zeros

494

Representation for Entire and Meromorphi Fun tions

of f (z ) listed with multipli ities and 0 < ja1 j  ja2 j 


Then f has a nite genus  satisfying the inequality   .

 jan j 


.

Proof. Let f have an order  and p = [℄. Then, p   < p + 1 and for any small  > 0 ln M (r)  r+=2 for r suÆ iently large.

If f has a zero of order m at the origin, then for jz j = r ln jf (z )=z mj

 ln(M (r)r m )  r+=2 m ln r  r+ for r suÆ iently large.

So F (z ) = f (z )z m is an entire fun tion of order  with no zero at the origin and F (0) = limz!0 f (z )z m = f (m) (0)=m!. Thus, by Corollary 11.86, 1 (11.101)

(r)  ((2r)+ ln jF (0)j) = Ar+ + B; ln 2 where A = 2+ = ln 2 and B = ln jF (0)j= ln 2, independent of r. Therefore, taking r = jan j, we obtain

n  (jan j)  Ajan j+ + B for large n:

In parti ular, (11.102)



A jan jp+1  n B 1

(p+1)=(+)

for large n

and if we hoose  (e.g.  =P(p + 1 )=2) so that p + 1 >  +  (re all that p + 1  > 0), the series jan j p 1 is dominated by a onvergent series. Therefore, 1 X jan j (p+1) < 1 n=1

and so, f (z ) an be written in the form 1 Y (11.103) f (z ) = z m eg(z) Ep (z=an) =: z meg(z) P (z ); p = [℄; n=1

where m = 0 when f (0) 6= 0. Next, we laim that g(z ) is a polynomial of degree  . Q First we let PN = N j =1 qj (z ) be the N th partial produ t of P (z ). Note that PN (z ) ! P (z ) lo ally uniformly on C and so, PN0 (z ) ! P 0 (z ) lo ally uniformly on C . Consequently, N q 0 (z ) 1 q0 (z ) PN0 (z ) X P 0 (z ) X j j = ! = PN (z ) j=1 qj (z ) P (z ) j=1 qj (z )

495

11.8 The Order and the Genus of Entire Fun tions

where the series on the right onverges uniformly on every ompa t subset of C not ontaining the zeros of P (z ). Also, we note that

z z2 qj (z ) = Ep (z=aj ) = (1 z=aj ) exp + 2+ aj 2aj so that and






zp + p paj

!

qj0 (z ) 1 1 z zp 1 = + + 2 +


+ p qj (z ) aj z aj aj aj   dp qj0 (z ) p! = : dz p qj (z ) (aj z )p+1

Thus   1 X dp P 0 (z ) 1 (11.104) = p ! p+1 ; z 6= a1 ; a2 ; : : : : dz p P (z ) ( a j =1 j z ) Finally, by (11.103), we have f 0(z ) m P 0 (z ) 0 = + + g (z ): f (z ) z P (z ) Dierentiating both sides p times and applying Lemma 11.97 and (11.104) gives g(p+1) (z ) = 0 and g is a polynomial of degree at most p. In parti ular, the genus  of f is less than p. Sin e p  , we have   . The Hadamard fa torization theorem is often stated in the following equivalent form.

11.105. Theorem. (Hadamard's Fa torization Theorem) Let f be an entire fun tion of nite order . Suppose that a1 ; a2 ; : : : ; are the zeros of f (z ) listed with multipli ities and 0 < ja1 j  ja2 j 


 jan j 


. Then there exists a polynomial g (z ) of degree not greater than , and a nonnegative integer p (p  ) su h that f (z ) = z meg(z)

1 Y

n=1

Ep (z=an):

The onverse of Theorem 11.100 is also true. In fa t, we have

11.106. Theorem. If f is an entire fun tion of nite genus , then f is of nite order  and    + 1. Proof. If f is of nite genus , then f (z ) = z meh(z) P (z ); P (z ) =

1 Y n=1

E (z=an)

496

Representation for Entire and Meromorphi Fun tions

where the degree q of the polynomial h(z ) is  . We note that the order of the produ t of the two entire fun tions annot ex eed the order of the ea h of the fa tors. In view of this observation,

(f )  maxf(eh(z) ); (P (z ))g: Sin e f is of nite genus , the order of exp(h(z )) is q whi h is  . Hen e, to omplete the proof, it suÆ es to show that the order of the

anoni al produ t P (z ) is   + 1. Sin e fq; pg  p and the P1 = max (p+1) < 1, it follows

onvergen e of the produ t P ( z ) implies j a j n=1 n P (+1) < 1. First we laim that that 1 n=1 jan j ln jE (z )j  (1 + )jz j+1 ; z 2 C : Note that 1 + jz j  ejzj, i.e. ln(1 + jz j)  jz j for z 2 C . Consequently (11.107)

ln jE0 (z )j = ln j1 z j  ln(1 + jz j)  jz j; z 2 C ; showing that the estimate (11.107) holds when  = 0. Next we prove (11.107) for   1. To do this, we re all Lemma 11.39: 1 + jE (z )j  jE (z ) 1j  jz j+1 for jz j  1 so that jE (z )j  1 + jz j+1 for jz j  1. Consequently, for ea h   1 (11.108) ln jE (z )j  ln(1 + jz j+1 )  jz j+1  ( + 1)jz j+1 for jz j  1: For arbitrary z 2 C with jz j  1, and   1

jE (z )j = j1 z j jez j jez2 =2 j


jez = j  (1 + jz j)ejzjejzj2 =2


ejzj = so that, for all jz j  1,  X jz jk ln jE (z )j  ln(1 + jz j) + 

 jz j +

 X

k=1

k

jz j (be ause jz jk  jz j for k = 1; 2; : : : ; ) k

k=1  ! X

1 jz j  (1 + )jz j k k=1 (1 + )jz j+1 :

 1+ 

Combining the last inequality with (11.108) proves (11.107). The estimate (11.107) gives at on e 1 1 X X ln jP (z )j = ln jE (z=an)j  (1 + )jz j+1 jan j (+1) k=1

n=1

11.8 The Order and the Genus of Entire Fun tions

and so, there exists an  > 0 with ln jP (z )j  jz j+1 for all z follows that P (z ) is at most of order  + 1, i.e.    + 1.

497

2 C . It

Finally, we obtain the following result whi h exhibits the strength of Hadamard's fa torization Theorem.

11.109. Corollary. An entire fun tion of fra tional order assumes every omplex value in nitely often. Proof. Let f be an entire fun tion of order , where  62 N . Clearly, (f ) = (f a) for any onstant a 2 C . Therefore, it suÆ es to show that f has in nitely many roots. Suppose not. Then f has only nitely many zeros, namely a1 ; a2 ; : : : ; an . Then f is of the form f (z ) = eh(z)

n Y

k=1

(z

ak ) =: eh(z) P (z ):

By Hadamard's fa torization theorem, h is a polynomial of degree m  (f ). But then (f ) = (eh(z) ) = m. This ontradi tion proves the orollary. Thus, every entire fun tion of nite order has either in nitely many zeros or else f (z ) = eh(z) P (z ), where h(z ) and P (z ) are some polynomials.

11.110. Convergen e exponent. Let fan gn1 be a sequen e of nonzero

omplex numbers, listed a

ording to in reasing moduli su h that jan j ! 1 as n ! 1. Then the onvergen e exponent of fangn1 is de ned by ( ) 1 X   = inf  > 0 : jan j < 1 : n=1

Thus, for ea h  > 0, one has 1 1 X X 1 1 < 1 and  +    = 1: j a j j a n=1 n n=1 n j For example, if an = n for all n 2 N , then we have  = 1. Also, we have P  = 1 for all  > 0, then we set  = +1 as the (i) If 1 n=1 jan j in mum of an empty set. P  (ii) If 1 n=1 jan j < 1 for all  > 0, then  = 0. (iii) From the proof of Theorem 11.100, we have the following: if f is an entire fun tion of order  and a1 ; a2 ; : : : ; are the zeros of f (z ) listed with multipli ities and 0 < ja1 j  ja2 j 


 jan j 


, then 1 X jan j  < 1 for  > : n=1

498

Representation for Entire and Meromorphi Fun tions

Indeed, let be a number su h that  < < . Then, for small  > 0 and for large n, we have

n  (jan j)  Ajan j + + B; i.e.



A jan j  n B 1

=( +)

:

The on lusion follows, as = > 1. If f is an entire fun tion having fangn1 as its non-zero zeros, then  := (f ) de ned as above is alled the onvergen e exponent for the zerosequen e fan gn1 of f .

11.111. Examples. (i) For > 0, set an = n1= for all n 2 N . Then, as 1 1 X X jan j  = n = < 1 if  > ; n=1

n=1

the exponent of onvergen e  of fn1= g is . Note that 1 1 1 X X X jan j  = jan j = n 1 = 1: n=1

n=1

n=1

(ii) If f (z ) = sin z , then the zeros of f (z ) are an = n (n 2 Z) and so 1 1 1 X X 2 X   jan j = 2 jan j =  n n=1 n=1 n2Znf0g showing that (f ) = 1 = (f ). If g(z ) = os z , then (g) = (g) = 1. (iii) Sin e ez has no zero in C , we set that (ez ) = 0. 

11.9 Exer ises 11.112. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) The meromorphi fun tions sin z=[e2iz + 1℄ and 1=(2i os z ) dier by an entire fun tion. (b) If f is meromorphi and has an in nite number of poles, then every

losed disk jz j  R (0 < R < 1) ontains only a nite number of poles inside it. ( ) There an be no meromorphi fun tion in C with poles at 1=n, n 2 N .

11.9 Exer ises

499

(d) If f is entire and f (z ) 6 0, then either f has a nite number of zeros or has an in nite number of zeros an with an ! 1 as n ! 1. Q (e) An in nite produ t 1 k=1 (1+ak ) is divergent i the sequen e of partial produ ts either diverges to in nity or onverges to zero or os illates. (f) If f is a non- onstant entire fun tion su h that f (z ) 6= 0 in C , then f 0(z ) may or may not have a zero in C . Q P1 (g) The in nite produ t 1 k=1 (1 + ak ) onverges i the series k=1 ak

onverges. Q (h) For 1 < P ak  1 for k 2 N , the produ t 1 k=1 (1 + ak ) onverges i 1 the series k=1 ak onverges. Q (i) The onvergen e of the produ t 1 k=1 (1 + ak ) isPne essary and suÆ ient for the absolute onvergen e of the series 1 k=1ak .   Q1  Q 1 (j) Ea h of the produ ts k=1 1 + 2k+11 1 and 1 1 + k=2 k2 1 onverges. Q 1
(k) For 0 6=  2 C , the in nite produ t 1 n=2 1 n onverges absolutely provided Re  > 1. Q n (l) The in nite produ t 1 an entire fun tion. n=1 (1+a z ), jaj < 1, de nes P1 (m) If R is the of the series n=1 an z n , then the Q1radius of onvergen e n produ t n=1 (1 + an zP) onverges for jz j < R. 1 jz 3n j onverges for jz j < 1, the produ t Note: As the series n=1 Q1 3n n=1 (1 + z ) onverges for jz j < 1. Q 2 n onverges to z whereas (n) For z 2 C n fx + i0 : x  0g, 1 n=1 z Q1 n 3 does not onverge to z . n=1 z (o) P Let p(z ) be a non- onstant polynomial in z (e.g. z , z 2 ). If the series 1 ja j is onvergent (e.g. a = 1=k with  > 1), then the produ t k k=1 k Q1 k=1 (1 + p(z )ak ) is onvergent for all z 2 C and represents an entire fun tion.   Q 2 (p) The produ t 1 k=1 1 + kz +k3 1 represents an analyti fun tion for Re z > 3:   Q n2 and (q) The region of absolute onvergen e of ea h of 1 1 + z n=1 Q1 2
is dierent. 1 + z=n n=1 Q 2 (r) The produ tQ 1 n=1 (1 z=n ) represents an entire Q fun tion whereas 1 z=n is the produ t n=1 (1 z=n) does not although 1 n=1 (1 z=n)e an entire fun tion. (s) There exists an entire fun tion whi h has zeros of multipli ity n at z = n (n 2 N ), and no other zeros. Q z (t) The produ t 1 n=1 (1 + n ) onverges uniformly on every ompa t subset Re z  1 + Æ (Æ > 0) and represents an analyti fun tion in the half-plane Re z > 1.

500

Representation for Entire and Meromorphi Fun tions

  Q 2z (u) The produ t 1 n=1 1 n2 +z represents a bounded analyti fun tion on the right half-plane Re z > 0. (v) The terms of an absolutely onvergent produ t an be rearranged without ae ting the onvergen e or the value of the produ t. (w) There exist entire fun tions with simple zeros at n2 , n 2 N . (x) There exists no non- onstant entire fun tion having zeros at z = 1=n (or 1=n2 or 1=n3), n 2 N . (y) An entire fun tion, whi h has simple zeros at 0, n1=4 (n 2 N ) and no other zeros, is given by   2  1  Y z2 z z4 1 p exp p + z : n n 2n n=1

(z) There exists an analyti fun tion f in the unit disk su h that f (z ) = 0 i z = 1 1=n, n 2 N .

11.113. Determine whether ea h of the following statements is true or false. Justify your answer with a proof or a ounterexample. (a) The order of entire fun tions f (z ) and f (z ) ( 6= 0) are the same. (b) The entire fun tions f (z ) and z nf (z ) (n 2 N ) have the same order, where the point z = 0 is a zero of multipli ity n for f (z ). ( ) The order of the entire fun tion f (z ) = zez is 1. (d) For any two entire fun tions f and g, (f + g)  maxf(f ); (g)g. (e) For any two entire fun tions f and g, (f + g) = maxf(f ); (g)g whenever (f ) 6= (g). (f) If f and g are two entire fun tions su h that (f ) > (g), then (f + g) = (f ). (g) For any two entire fun tions f and g, (fg)  maxf(f ); (g)g. (h) If f is an entire fun tion and h(z ) = f (az ), then (h) = (f ). (i) If f is an entire fun tion and h(z ) = f (z n), then (h) = n(f ). (j) If f is an entire fun tion and h(z ) = z nf (z ), then (h) = (f ). Q (k) The exponent of onvergen e of 1 n=1 E1 (z=n) is 1. Q 1 m (l) The order of z n=1 E1 (z=n) ( m 2 N ) is 1. (m) If (f ) 6= 0 and (g) 6= 0, then (f + g) is not ne essarily a non-zero number. (n) There are in nitely many prime numbers. (o) If R is the re tangle with verti es at n  (2n + 1=2)i, then one has jez 1j > 1=2 for z 2 R.

501

11.9 Exer ises

11.114. Constru t meromorphi fun tions f in properties: (i) (ii) (iii) (iv) (v)

C with the following

simple poles only at an = n 2 N with Res [f (z ); an℄ = n p simple poles only at an = n, n 2 N , with Res [f (z ); an℄ = 1 simple poles only at an = n, n 2 Z, with Res [f (z ); an℄ = 1 simple poles only at an = n(1+ i), n 2 Znf0g, with Res [f (z ); an ℄ = 1 poles only at an = n 2 N of order n.

Q p 11.115. P Set an = ( 1)n+1 = n. Show that 1 n=1 (1 + an ) diverges even though 1 a

onverges. Does this provide an example of a series n=1 n that is onvergent but not absolutely?

11.116. Constru t entire fun tions with the following properties (i) simple zeros at an = n; n 2 N and no other zeros (ii) simple zeros at an = n; n 2 N and no other zeros p p (iii) simple zeros at an = n; n 2 N and double zeros at bn = i n; n 2 N , and no other zeros (iv) simple zeros at an = n5=4 ; n 2 N and no other zeros (v) simple zeros at an = n4=5 ; n 2 N and no other zeros (vi) simple zeros at an = n1=2 ; n 2 N and no other zeros. Q1 11.117. Q1 Suppose that the produ t k=1 (1+ ak ) onverges whereas the produ t k=1 (1 + jak j) diverges. What an we say about the sequen e? Q1 11.118. Show that produ t k=1 (1 + fk (z )) is uniformly onvergent P1 in a domain if the series k=1 jfk (z )j onvergesQuniformly in : Using this result, dis uss the onvergen e of the produ t 1 k=1 (1 + fk (z )) when

(i) (ii) (iii) (iv) (v)

fk (z ) = (k=(k + 1))k z k , z 2
fk (z ) = (1 z )=(1 z 2k ), z 2
fk (z ) = (1 z 2k )=(1 z ), z 2
fk (z ) = z k =k!, z 2 C fk (z ) = z=[k(ln k)2 ℄, z 2 C .

11.119. Let f be meromorphi in
R . Suppose that (i) a is neither a zero nor a pole of f (ii) ai (1  i  m) and bj (1  j  n) denote, respe tively, the zeros and poles of f in
R ( ounted as many times as its order of multipli ities).

502

Representation for Entire and Meromorphi Fun tions

Then show the Poisson-Jensen Formula of the form   Z 1 2 R + ae i Re ln jf (Rei )j d 2 0 R ae i = ln jf (a)j +

m X i=1

2 R ln R(a







n R2 bj a ai a X ln i a) j =1 R(bj a)

a (z ) = (a z )=(1 az ) (Note that if f is free from poles, then the last term on the right does not appear in the formula). Also, dis uss what modi ation is required if a is a zero or a pole of f (z ). 11.120. Give an example of an entire fun tion f (z ) whose order is not an integral number. 11.121. Prove that, whenever  62 Z,  1  Y z z

ot  1+ e z=(n+) : sin (z + ) = e sin  n +  n= 1 11.122. Find the order and genus of the entire fun tion  1  Y z f (z ) = 1 : n(ln n)2 n=2 by

11.123. For a > 0, onsider the Hurwitz zeta fun tion  (s; a) de ned 1 X 1  (s; a) = : ( n + a)s n=0

Show that  (s; a) is analyti for Re s > 1. Also show that Z 1 e (a 1)sxxs 1 1  (s; a) = dx: (s) 0 ex 1

11.124. Prove the Gauss multipli ation formula (2)(n 1)=2

(nz ) = nnz 1=2

nY1  k=0



k z+ : n

11.125. For s 2 C , prove the fun tional equation  (1 s) = 21 s  s os(s=2) (s) (s):

Chapter 12

Mapping Theorems

We shall dis uss a number of interesting results on erning ertain mapping problems between domains. Se tion 12.1 begins with the open mapping theorem whi h is a elebrated result about analyti mappings. In Se tion 12.2, we study some basi results on univalent fun tions. Se tion 12.3 is devoted to a preliminary dis ussion on normal families. In addition, we also prove Montel's theorem for normal families of analyti fun tions. The main result in Se tion 12.4 is the Riemann mapping theorem whi h asserts that every simply onne ted domain of the omplex plane having at least two boundary points an be mapped onformally onto the open unit disk. In Se tion 12.5, we state the elebrated onje ture due to Bieberba h whi h led to the development of a great number of dierent and deep methods that have solved a large number of problems in fun tion theory. Sin e the on rmation of the Bieberba h onje ture by de Branges, one of the outstanding open problem in omplex analysis is that of nding the exa t value of the Blo h onstant. Our nal se tion dis usses this onstant along with the long awaited Pi ard's little theorem and S hottky's theorem.

12.1 Open Mapping Theorem and Hurwitz' Theorem In the subje t of topology, ontinuity of f on  C is equivalent to saying

that the inverse image of every open set in f ( ) under f is open. We are interested now in these fun tions for whi h the dire t image of any open set is open. A fun tion de ned on an open set D is said to be an \open mapping" if for every open subset of D, the image f ( ) is open. Thus if there exists an open set in D whose image under f is not open, then f is not an open mapping in D. Consider fj : R ! R (j = 1; 2; 3) de ned by

f1 (x) = x2 ; f2 (x) = sin x and f3 (x) =

ex + e x ; 2

504

Mapping Theorems

respe tively. Clearly,

f1 (R) = [0; 1); f2 ((0; )) = (0; 1℄ and f3 (R) = [1; 1) showing that for ea h j , ea h of the real-valued fun tions fj of a real variable fj is not an open mapping. Our emphasize will be on plane domains in C , and the above examples show that the following theorem has no analog in R.

12.1. Theorem. (Open Mapping Theorem) If f is a non- onstant analyti fun tion on a domain D, then f is an open mapping, i.e. f (D) is an open set in C . Clearly, the name of this theorem is derived from the property of \openness". What does this theorem onvey? If D is a domain in C and f 2 H(D), then f (D) is either a domain or a single point; i.e. f (D) is a domain or else f is a onstant fun tion. For example, this theorem prohibits a non- onstant C 1 -analyti mapping of a disk onto a portion of a line (see Corollary 12.3). Is there an analogous result for real-valued C 1 fun tions de ned on D  C ? De ne f : C ! C by f (z ) = jz j2 : Then

f (C ) = fx + i0 2 C : x  0g

whi h is not an open subset of C . Note that f is nowhere analyti but is real-dierentiable be ause f (x; y) = x2 + y2 and the partial derivatives of all orders exist and are ontinuous on R2 . On the other hand, f (z ) = z 2 is an open mapping on C . Is there a non- onstant omplex-valued fun tion (need not be analyti ) de ned on a domain that is an open map? How about f (z ) = z, z 2 C ? Proof of the open mapping theorem. We shall show that if f is a non- onstant analyti fun tion in D and is an open subset of D

ontaining a, then f ( ) ontains an open disk about f (a). Sin e zeros of the non-vanishing analyti fun tion f (z ) f (a) are isolated, there exists a disk
(a; r) with
(a; r)  su h that

f (z ) f (a) 6= 0 in 0 < jz aj < r:

In parti ular, f ( ) f (a) 6= 0 for  2 C = 
(a; ) where 0 <  < r. Let 2m = min jf ( ) f (a)j:  2C Then m > 0. Further, for every w 2
(f (a); m), we note that

jf ( ) wj  jf ( ) f (a)j jf (a) wj > 2m m = m > jf (a) wj for all  2 C . Rewrite the last inequality as jf (a) wj = j(f ( ) w) (f ( ) f (a))j < jf ( ) wj for all  2 C:

12.1 Open Mapping Theorem and Hurwitz' Theorem

505

It follows from Rou he's theorem that the fun tions f ( ) w and f ( ) f (a) have the same number of zeros inside the ir le C . But as f ( ) f (a) has at least one zero inside C , f ( ) w has at least one zero inside C . Hen e there exists z 0 2
(a; ) su h that f (z 0 ) = w and so w is in the range of f , w being an arbitrary element of
(f (a); m) the assertion is true. As the onne tedness of D implies the onne tedness of f (D) (see Theorem 2.24), the open mapping theorem is often formulated in the following form:

12.2. Theorem. A non- onstant analyti fun tion maps the domain D onto the domain f (D). Proof. By the open mapping theorem f (D) is open; so we only need to show that f (D) is onne ted. We provide a dire t proof to show that f (D) is onne ted. Let w1 ; w2 2 f (D). Then there exist z1 ; z2 2 D su h that f (z1) = w1 and f (z2) = w2 . Be ause D is onne ted, z1 and z2 an be onne ted by a nite number of line segments that lie entirely within D. The image of ea h line segment under f is always a urve in f (D), sin e f is dierentiable in D. It follows that w1 and w2 an be onne ted by a

urve in f (D). Note that this urve an be approximated by line segments in f (D). 12.3. Corollary. Theorem 3.31(ii) follows from Theorem 12.2. Proof. The hypothesis of Theorem 3.31(ii) shows that, in the w-plane, f (D) is either a subset of a ir le or u = onstant, or v = onstant, or tan 1 (v=u) = onstant, respe tively. However, we note that none of these takes open sets into open sets. It follows from Theorem 12.2 that in ea h

ase, f must be a onstant. This ompletes the proof. Here is another appli ation of Rou he's Theorem.

12.4. Theorem. (Hurwitz' Theorem) Let ffng be a sequen e of non-vanishing analyti fun tions in a domain D whi h onverges to f uniformly on every ompa t subset of D. Then either f (z )  0 or f has no zeros. Proof. Sin e, by assumption, ffng onverges uniformly on D, f is analyti on D. Suppose f (z0 ) = 0 but f (z ) 6 0. Then (as zeros are isolated) there exists a small ir le C = 
(z0 ; Æ) su h that
(z0 ; Æ)  D and
(z0 ; Æ) nfz0 g \ fz : f (z ) = 0g = ;: In parti ular, f (z ) 6= 0 on C . Let f have k zeros in
(z0 ; Æ). Sin e f is analyti on
(z0 ; Æ), jf j attains a minimum value m on C . As f (z ) 6= 0

506

Mapping Theorems

on C , m > 0. Sin e fn ! f uniformly on the ompa t set C , for a given m > 0, there exists an N su h that

jfn (z ) f (z )j < m=2 < m  jf (z )j for n  N and for all z 2 C: By Rou he's Theorem, it follows that fn and f have the same number of zeros on
(z0 ; Æ), and for some n  N . Thus, fn has a zero in
(z0 ; Æ); this is a ontradi tion. Hen e we must have f (z )  0. We note that for ea h n 2 N , fn (z ) = ez =n has no zeros in C , but the limit fun tion f is identi ally zero in C .

12.5. Corollary. Let ffn g be a sequen e of analyti and univalent fun tions in a domain D whi h onverges to f uniformly on every ompa t subset of D. Then either f is onstant or univalent on D. Proof. Assume that f is a non- onstant analyti fun tion that is not univalent. Then there exist two distin t points z1 and z2 in D su h that f (z1 ) = f (z2) = . Choose r so small that
(z1 ; r)  D;
(z2 ; r)  D and
(z1 ; r) \
(z2 ; r) = ;: If f (z ) 6 , then, sin e fn  ! f  uniformly on ompa t subsets of D, Hurwitz' Theorem applied to fn(z )  shows that there is an n su h that fn (z )  has a zero in
(z1 ; r) and a zero in
(z2 ; r). That is, fn (z10 ) = fn (z20 ) for some z10 2
(z1 ; r) and z20 2
(z2 ; r): This is a ontradi tion to the hypothesis that fn is univalent in D. So f (z )  . For example, we note that fn (z ) = z=n, n  1, is univalent in C , and the limit fun tion f is identi ally zero in C .

12.2 Basi Results on Univalent Fun tions We have already en ounted a large number of examples of univalent fun tions. Now start with P P1 n 12.6. Theorem. If f (z ) = z + 1 n=2 an z is su h that n=2 njan j  1, then f is univalent in the unit disk
. P

Proof. Suppose jan j  1. Then, we n2 nP P that P have jan j  1 for all n  2, and hen e n2 jan z nj  n2 jz jn: Sin e n2 jz jn onverges for jz j < 1, by the omparison test, we see that the series represented by f

onverges for jz j < 1 so that f is analyti in
. Let jz0j < 1. We have

507

12.2 Basi Results on Univalent Fun tions

(f (z ) f (z0)) (z =

X

n2

= (z As jz n

1 + z n 2z

0+

z0 )

an (z n z0n ) z0 )

X

n2

an (z n

1 + zn 2z

0+




+ z0n 1 ):




+ z0n 1 j < n for jz j < 1, we have

j(f (z ) f (z0 )) (z z0)j < jz z0 j

X

n2

njan j  jz z0 j:

A

ording to Rou he's Theorem, f (z ) f (z0) and z z0 have the same number of zeros in
, that is f (z ) = f (z0 ) has exa tly one solution.

12.7. Theorem. Let f be analyti at a. Then f is one-to-one in some neighborhood of a i f 0 (a) 6= 0. Proof. Clearly, f is analyti at a i g de ned by g(z ) = f (z + a) f (a) is analyti at 0. Note that f 0 (a) = g0 (0). In view of this observation, it suÆ es to prove the theorem with a = 0 and f (a) = 0. (=: Let f 0 (0) 6= 0, f (0) = 0 and h(z ) = f (z )=f 0(0). Then h0 (z ) = 0 f (z )=f 0(0) whi h is analyti at 0, and h0 (0) = 1. The ontinuity of h0 (z ) at z = 0 shows that there exists an open disk jz j < Æ su h that jh0 (z ) 1j < 1=4 for jz j < Æ: (The univalen y of h in jz j < Æ may be obtained qui kly from Theorem 12.18). In parti ular, for any two distin t points z1 and z2 in this disk, and for = [z1 ; z2 ℄, the line segment joining z1 and z2 , we have

jh(z2 ) h(z1 ) (z2 z1 )j =

Z z2 (h0 (z ) z 1 Z

1) dz

 jh0 (z ) 1j jdz j [z1 ;z2 ℄  (1=4)jz1 z2 j whi h, by the triangle inequality, implies that

jh(z2 ) h(z1 )j  jz2 z1j (1=4)jz1 z2 j > 0: Thus, h and (hen e f ) is one-to-one in a neighborhood of 0. =): Let f be analyti , f (0) = 0 and one-to-one in a neighborhood
R of 0 . Assume the ontrary that f 0 (0) = 0. Then by the Ma laurin series of f around 0, there exists k  2 su h that f (z ) = z k (z ) where  is analyti at 0 and (0) 6= 0. By hypothesis, f is univalent (and so f is

508

Mapping Theorems

not a onstant). As zeros are isolated (and f 0 (z ) is analyti in
R with f 0 (0) = 0), we have f (z ) 6= 0 and f 0 (z ) 6= 0 for 0 < jz j  Æ: Now, m = min jf (z )j > 0. Pi k any omplex number su h that 0 < jzj=Æ 0, we note that this

ase leads to the previous ase.

Thus, we deal only with the analyti fun tions on
. The reason for this is due to the elebrated Riemann Mapping Theorem. So, it suÆ es to pay attention to the last ase be ause, by means of ompositions, we retrieve the properties of the image domain of the analyti fun tions de ned on any arbitrary simply onne ted domain. Avoiding onstant fun tions in our dis ussion, it suÆ es to onsider the family of fun tions F = ff 2 H(
) : f 0 (0) = 1g: The ondition f 0(0) = 1 is to ensure that f is not a onstant fun tion. First we noti e that the existen e of su h a disk in f (
), f 2 F , is ensured by the open mapping theorem. To make the size of su h disks more meaningful, we de ne two onstants. Consider f 2 F . Choose a point w 2 f (
) and nd the radii of all possible disks entered at w su h that they lie ompletely in f (
). Repeat this with every point of f (
). Let these radii be named as r; , where  and belong to some indexed sets  and 0 , respe tively. If r; is the radius of the -th disk entered at w 2 f (
), then we de ne Lf = supfr; :  2 ; 2 0 g: Then the Landau16 onstant is de ned as

L = inf Lf : f 2F Clearly, the de nition of L is related to the size of the image domains of fun tions in F . For L, Ahlfors's [1℄ ultra-hyperboli method produ ed the non-sharp lower bound 1=2. Later Yanahigara [12℄ improved the lower bound for L to 12 + 10 335. Again it is important to mention that, we do not know the exa t value of L. The following bounds were determined by Robinson (1938) and independently by Radema her [10℄, 1 (1=3) (5=6)