Classical Complex Analysis: A Geometric Approach (Volume 2) 9789813101074, 9813101075

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Classical Complex Analysis A Geometric Approach-Voll.. 2

Classical Complex Analysis A Geometric Approach - Voll. 2

1-~iung

Lin

National Taiwan Normal University, Taiwan

1!i» World Scientific NEW JERSE Y • LOND ON • SINGAPO RE • BEIJ IN G • SHA NGHAI • HONG KONG • TAIPEI • CHENNA I

Pre face

Complex analysis , or roughly equivalently the theory of analytic functions of one complex variable, budded in the early ages of Gauss, d )Alembert , and Euler as a main branch of mathematical analysis. In the 19th century, Cauchy, Riemann, and vVeierstrass laid a rigorous mathematical foundation for it (see Ref. [47]). Nourished by the joint effort of generations of brilliant mathematicians, it grows up into one of the rernarkable branches of exact science, and serves as a prototype or model of other theories concerned with generalizations of analytic functions such as Riemann surfaces, analytic functions of several complex variables, quasiconformal and quasiregular mappings, complex dynamics, etc. Its methods and theory are widely used in many branches of ma.thematics, ranging from analytic number theory to fluid mechanics, elasticity theory, electrodynamics, string theory, etc. Elementary complex analysis stands as a discipline to t he whole mathematical t raining. T his book is designed for beginners in this direction, especially for upper level undergraduate and graduate mathematics maJors, and to those physics (or engineering) students who are interested in more theoretically oriented introduction to t he subject rather than only in computational skills. The content is thus selective and its level of difficulty should be then adequately arranged. Beside its strong intuitive flavor, it is the geometric (mapping) properties, derived from or characterized the analytic properties, that makes the theory of analytic functions differ so vehemently from that of real analysis and so special yet restrictive in applications . T his is the reason why I favor a geometric approach to t he basics. The degree of difficulty, as a whole, is not higher than t hat of L .V. Ahlfors's classic Complex Analysis [1]. But I

try my best to give detailed and clear explana.tions to the theory as much as possible. I hope that the presentat ion will be less arduous in order to be more available to not-so-well-prepared or not-so-gifted students and be easier for self-study. Neit her the great est possible generality nor the most up-to-date t erminologies is our purpose. P lease refer to Berenstein and Gay V il

Vlll

Preface

[9] for those purposes . I would consider my purpose fulfilled if t he read ers are able to acquire elementary yet solid fundamental classical results and techniques concerned. Knowledge of elementary analysis, such as a standard calculus course including some linear a lgebr a , is assumed. In many sit u ations , mathem atical

maturity seems more urgent t han purely mathematical prerequisites. Apart from these the \Vork is self-contained except some difficult theorems to which references have been indicated. Yet for clearer and thorough understanding where one st ands for t he present in the whole mathematical realm and for the ability to compare with real analysis, I suggest readers get fam iliar with the t heory of functions of two real variables.

Sk etch of the C ontents

If one takes a quick look at the Contents or read over Sketch of t he Content at the beginning of each chapter and then s/he will have an overall idea about t he book. A complex number is not just a plane vector but also carries by itself the composite mot ion of a one-way stretch a nd rotation , a nd hen ce, is a

two-dimensional "number" . They constitute a field but cannot be ordered. Mathematics based on them is the one about similarity in global geometric sense and is t he one about conformality in local infinitesimal sense. Chapter 1 lays t he algebraic, geometric and point-set fow1dations barely needed in later chapters. Just as one experienced in calculus , we need to know some standard elementary complex-valued functions of a complex variable before complex differentiation and integration are formally int roduced. It is the isolatedzero principle (see (3) and (4) in (3.4.2.9) ) t hat makes many of their algebraic properties or algebraic identit ies simila r to their real counterparts . Owing to t he complex plane C having the same topological structure as the Euclidean plane R 2 , their point-set properties (such as continuity and convergence) are the same as t he real ones, too . It is the geometric mapping properties owned by t hese elementary functions t ha.t distinguish them from t h e r eal ones o,nd make one feel t hn.t complex o,nulysis is not just a copy of t he latter. In particular, the local and global single-valued continuous branches of arg z are deliberately studied, and then , prototypes of Riemann surfaces are introduced. Chapter 2 tries to figure out, t hough loosely and vaguely organized, the common analytic and geometric properties owned by

Preface

IX

these individual elementary functions and then, to foresee what properties a general analytic function might have. A complex valued function f (z), defined in a domain (or an open set ) n , is called analytic if any one of the following equivalent condit ions is satisfied: 1.

2.

f (z) is differentiable everywhere inn (Chap. 3). f (z) is continuous inn and fa.6. f (z)dz = 0 for any triangle ~ contained

inn (Chaps. 3 and 4). 3. For each fixed point z0 En, f (z) can be expressed as a convergent power series L~=O an(z - zo)n in a neighborhood of zo (Chap. 5). An analytic function f(z) infinitesimally, via t he Ca.uchy-Riema~quar tions, appears as a conformal mapping in case f' (z) -=/= 0 and f (z) can be interpreted as the velocity field of a solenoidal, irrotationa.l flow (see (3.2.4.3)). Chapter 3 develops the most fundamental a nd important analytic and geometric properties, both locally and globally, which an analytic function might possess. The most subtle one, among all, is that a function f (z) analytic at zo can always be written as f(z) = f(zo) + (z - zo) 1.

Consequently ,

g(f( z)) = bo

+

t, {J;

bnakn } (z - zo)• D

Take, for instance,

JcOsZ =

1-

~z 2 - ~z 4 - ~z 6 4

96

ezsin z

5760

= 1 + z2 +

-

...

)

lzl < ; ,

where

1 3z4 + .. . ' lzl < 00.

J cosO

= 1;

Sec. 5.1.1

Algorithm of Power Series

7

Section (3) The inve r se function ope rat ion Suppose that 00

1(z) = L an( z - zo r'

Iz - zoI < r (r > 0)

0

satisfies f' (z0 ) = a 1 =I= 0. According to the inverse funct ion theorem (see (3.4.7.1) or (3.5.1.4)) , there are 8 > 0(0 < 8 < r) and p > 0 such that w = l (z) : {lz - zo l < 8} ---7 {lw - wo l < p}, wo = l(zo), is a univalent a nalytic function and the inverse function z = 1- 1 ( w) : {lw - wo I < p} ---7 {Iz - zo I < 8} is an analytic function, too. How can we find the power (Taylor) series expansion of 1-1 at wo? All one needs to do is to let 1- 1(w) = L~=O bn(w - wor, lw - wo l < p , and then , try to apply (5.1.1.5) to 1- 1 o l(z) = z . Vve summarize as

The inverselunction operation. Suppose l(z) = L~ an(z - zo)n has a positive radius r of convergence and ai = j'(zo) =I= 0. Let wo = l (zo) = ao . Then there exists a power series g(w) = Lgo bn(w - wot, lw - wo l < p (p > 0), such that lz - zol < r,

(g o f)(z) = z,

v.rhere t he coefficients bn, n 2: 0, satisfy t he following recursive relations:

g(wo) = bo = zo aib1

= 1,

+ a2b1 = 0, a~b3 + 2a1a2b2 + a3b1 = 0, aib2

in which P11 (a1, . .. , an- I, bi, ... , bn-1 ) is a polynomial in ai, ... , an-1 and bi , ... , bn-1 · (5.1.1.6) For instance,

w = Arc tanz = z z = ta.nw = w

1

1 3

z

3 3

+ 1z 5 -

5

2

5

1 7

z

7 1

7

+ · ··

+ -3 w + -15 w + -7 w + · · · )

lz l < 1, and 1T

lwl < 2·

Chap. 5

8 n

oo

L ;n.

w = ez =

Fundamental Theory n

oo

1

and z = Log(l

+ w)

=

n= 0

L (-1)11-

1

n =l

:__ )

n

lzl < 1,

where Log 1 = 0. 1

00

w

= sinz = L (-1) 2 n+l n =O

z

.

1

1

(2n + 3

z 2'11+l , ·

1) .1

1·3 1

5

lzl < oo, and 1·3 · 5 1

7

= A re sm z = z + 2 . 3z + 2 . 4 . 5z + 2 . 4 . 6 . 7z + . . . '

lzl < 1.

\Vhat is Arccos z? Section (4 ) The double series op e ration This is the ltVeierstrass double series theorem. Suppose 00

fn(z) =

L ank(z - zo)\

k 2:: 0,

k=O

all converge in

lz -

zo l < r(r > 0) for n 2:: 1. Also, suppose 00

L fn(z) n =l

converges uniformly on each compact subset of f(z). Then,

to a function

lz -

zo l < r , and f( z) = I::r=oak(z - zo)k, lz - zol < r, where ak = I::~=l ank = I::~=l

1. f(z) is analytic in

2.

lz - zol < r

tJk>(zo) k!

)

k

0 2:: .

In short, (5.1.1.7) This is a rather special case of a general theorem by W eierstrass, to be formally stated and proved in (5.3.1.1) and (5.3.1.2). See also Exercise A(3) i:: .. 1?~. o f. ec. o

s

Sec. 5.1.l

Algorithm of Power Series

9

Example 1 (J. H. L ambert, 1913). Show that

oo L

oo L d(n)zn,

n

n =l

1~ zn = n = l

lzl < 1,

where d(n) denotes the number of positive divisors of n. The series in the right is called the L ambert series. Solution. For a ny 0

< p < 1, in lzl < p , we have

oo

=? ( since

n

L pn converges) L 1 ~ zn converges loca.lly uniformly in n =l

lzl < 1. On the other hand, n

fn(z) =

oo

z

=

l - zn

~ znk , L

lzl < 1, for n >

1.

k= l

The result follows by (5.1.1.7). Example 2. The E uler rp-function rp(n) is defined to be the number of positive integers, both less than n (a positive integer) and relatively prime ton. Designate n

0

s k l , lz-al 0 so that f (z) # 0 in 0 < lz - al < 8. (2) The interior uniqueness principle: If f (z) has a sequence zk of zeros in lz - zol < r such that Zk , k > 1, has a limit point in lz - zo l < r , then f (z) - 0 throughout lz- zo l < r. In particular, f (z) - 0 if f (z) is identically equal to zero in a nonempty open subset of lz- zol < r. (5.1.2.4) This leads immediately to the follow ing

Principle of (direct) analytic confinuation of power series. Let 00

f(z) = L an(z- zo), 0

00

lz- zol < r;

g(z) = L bn(z- zo),

lz-zo l < r.

0

Then

{:::? {:::? {:::?

{:::?

f(z) - g(z ), lz - zol < r. an = bn, n ~ 0. There is a point a in lz -zo l < r such t hat f (n)(a) - g(n )(a), n ~ 0. There is a sequence zk of points in Iz- zol < r , converging to a point in lz - zo l < r, such that f(zk) - g(zk), k ~ 1. (5) There is a subdomain in lz - zo l < r on which f(z) = g(z) holds.

(1) (2) (3) ( 4)

(5.1.2.5) Proofs are left to the readers. Second, exactly like the process hO\;\' (*1 )- (*4) in Sec. 3.4.4 lead to (3.4.4.1), we have

The max- min modidits principle. Let f(z) be as in (5.1.2.1) . (1) The maximum principle. If f(z) is not a constant function in lz-zo l < r, t hen lf(z)I cannot attain its local maximum at any point in lz-zo l < r. Equivalently, if f(z) is also continuous on lz-zo l ~ r < oo, then lf(z) I attains its global maximum on lz - zol ~ r at a boundary point of the disk. (2) The minimum principle. If f(z) is not a constant function and f (z) # 0 throughout lz - zo l < r, then lf(z)I cannot attain its local minimum at any point in lz - zo l < r . Equivalently , if f(z) is also continuous on

20

Chap. 5

Fundamental Theory

lz- zo J< r < oo, then lf(z)I

attains its global minimum on at a boundary point of t he disk.

lz-zol

~

r

(5.1.2.6)

Let

M(p) =

max

l z - zol~P

lf(z) I and m(p) =

nlin

l z- zol~P

lf(z) l

1

0~ p

< r. (5.1.2.7)

Then M(p) is a nondecreasing continuous function of pin (0, r) and it is strictly increasing if f (z) is nonconstant; while m(p) is a nonincreasing continuous function of p jn (0 1 r) if f (z) :/= 0 on lz - zo I < r and it is strictly decreasing if f(z) is also nonconstant. Refer to Exercise A(6) of Sec. 3.4.4. As a bonus, these results help us to imagine what the graph lf(z)I might look like in the Euclidean three-dimensional space without graphing it practically. For instance, suppose f (z ) is not a constant in lz - zo I < r and fix a point a within. Then If (a)I is neither a local maximum nor a local minimum, if f(a) :/= 0. For any p, 0 < p < r - la - zo l, there are points z 1 and z2 on lz- al = p so that lf(z1) I < lf(a)I < lf(z2) I. By the intermediate value property of a continuous function on a connected set, there is at least one point Zp on lz - al = p so that lf(zp)I = lf(a) I holds. Thus , the level curve lf(z)I = lf(a)I, namely,

{zl lz - zol

0 wherever

26

Chap. 5

Since I:~ an no so that

Fundamental Theory

= 0, li1n,i-.oo A n = 0. For any given s > 0, there exists an IAnl < c,

00

=?

n >no

00

L

L

Anzn
1· Note that there is a point of the form ( = e2lr.i/ 2 '·' (0 :::; l:::; 2k - 1, k > 0), lying on lzJ= 1, that lies in the disk lz - zol < r z0 . For the series ( *8), lim g (r ( ) = g((),

r-+ l

0< r


1

'vVell1 for such a k, sep arate

lzl < 1 and lz - zol < r z0 )

0 < r < 1.

f (z) into two parts as k

f

39

00

cro = 2= r 2n k + 1,

N

r2n

>

L

r2"

> (N -k-

l)r2N

n = k+ l

=? lim f (r() = oo r---> 1

contradicting to t he fa.et t hat g(() is a finite complex number.

D

*Section ( 4) Com1nents on convergence, diverge n ce and r egular , singular points

There is no obvious connection, in general, as might be seen in Example 4, between the convergence or divergence of a po\.ver series at a boundary point (, I( - zol = rzo, and t he regularity or singularity of t hat power series at(. It is even possible that a power series converges uniformly and is infinitely differentiable on its n atural bounda ry (see Exercises A(3) and (20)). A power series could diverge at its regular points except under some additional restriction on its coefficients. For instance, Fatou's theorem. Suppose f(z) = .2:.":~ an z n has radius 1 of convergence and limn~oo an - 0 holds. Then , the series converges at its regular point (if exists); moreover, it converges uniformly on its set of regular points. (5.1.3.10)

See Refs. [58], Vol. II , pp. 404- 408 or [76], pp. 217- 220 for the proof. Lusin constructed a power series satisfying the assumption of Fatou 's theorem yet diverging everywhere on lzl = 1. Generally speaking, the coefficients an for all sufficiently large n determine completely the distribution of singular points on the circle lzl = r of

40

Chap. 5

Fundamental Theory

convergence. For instance, when these an which are not equal to zero are evenly distributed, then z = r turns out, in many cases, to be the only singular point (see (5.1.3.9) and Example 4) . Furthermore, suppose 00

f(z) = L

ankZnk (k,a fix ed positive integer),

ank 2:: 0

0

has the radius 1 of convergence. In this case, the number of the missing terms between two consecutive terms is equal to a constant, namely, k(n + 1) - kn = k, and the k-th roots of 1 are singular points of the series (see "Another proof,, for Example 5) . Even more, if the distribution of nonzero an is badly irregular, some addit ional singular points might be added, as Example 5 shows. As a matter of fact, Example 5 is a special case of the following

Hadamard 's gap theorem. Suppose a seq'l.lence nk; k > O; of positive integers satisfying:

1. no < ni < · · · < nk < nk+l < · · ·, and limk~oo nk = oo; 2. there is a positive n'l.lmber o: such that nk+l - nk 2:: o:nk, k 2:: 0. If the power series f(z) = L~o ankznk has a positive radi'l.ls r of convergence, then lzl = r is the natural bo'l.lndary of the series. In partic1dar, in case or then

lzl =

nk+l = Aknk (A.k, apositiveinteger) , k 2:: 0

r is the natural boundary.

(5.1.3.11)

Sk et ch of proof (Faber, IVIordall) . Replacing f(z) by f(r e·iB z) for some constant we may suppose that r = 1 and z = 1 is a regula r point of the series, and we want to get a contradiction. According to the definition of a regular point (5 .1.3.5) , there is an analytic function F( z) in lzl < 1 and in a neighborhood of 1 so that F(z) = f (z) on lzI < 1. Choose a positive integer p > ~. Let

e,

z

= -1 (wP + wP+l ). 2

Note that lwl :S: 1 is mapped into that z = 1 if w = 1. Then

lzl :S: 1, and into lzl < 1 if lwl

cp(w) = F (

~ (wP + wP+ 1 ))

:S: 1, except

Sec. 5.1. 3

is analytic on

Since p nk

>

Boundary B ehavior of a Power Series

lwl < 1 and

41

has the power series expansion in w as

±implies t hat 1 < ±(n1~~

1

-

1)

< p ( n~! 1

-

1) , namely , pnk +

< pnk+ 1 , we see that every power of w in (*9) occurs at most once and

is t hen obtained by simply rearranging t he t erms in the middle expression. This ser ies ( *9) has its radius of convergence greater than 1. It will converge at a real point ro > 1, and t he series for f (z) would be convergent at the 1 point ~ (r'b + r'b+ ) > 1, a contradiction. This proves (5 .1.3.11 ). 0 A slight modification of this method can be adopted to prove Ostrowski 's overconvergence theorem (see Exercise B (5)). Hadamard >s gap theorem can be used to prove Fatou- P 'olya's theorem, saying t hat ((most'' of the power series with positive radii of convergence cannot be continued analytically beyond their circles of convergence (see Exercise B (8)) . For related information , see Refs. [71], pp. 246- 252 ; [25], pp. 126- 133 1 153- 157; [76], Chap. VII , in particular , pp. 214- 224.

*Section (5) The bounda ry va lue function of a powe r series Case 3 (p > 1) in Example 4 gives a concrete illustration. The most commonly adopted definition for the boundary value at a point on the circle of convergence is provided by t he one stat ed in Abel's limit t heorem (5 .1.3.1 ): Let f (z) = 0 an(z - zo)11, lz - zo l < r (the radius of convergence) and ( be a point on lz- zol = r . Denote by D8 (0 ~ < ~) a Stolz domain with vertex at ( . If the limit

I::,

o

h(() =

lim

zE Da - ~

f( z)

(5.1.3.12)

exists as a finite complex number or oo, then h( () is called the nontangial boundary value of .f (z) at ( as z approaches ( from inside t he circle lz zo I = r . In case h( () is d efined everywhere on z - zo I = r , it is called the nontangential boundary value function off (z). In particular, J

lim f(t()

t---.1 -

= h ((),

0 no for some no . Show that zo is a singular point. (10) An extension of Vivanti- Pringsheim theorem. Let r = 1 be the radius of convergence of both f(z) = L~ a 11 zn and g(z) = L~ (Re an)zn . In case R ean > 0 for all sufficiently large n, show t hat z = 1 is a singular point of f (z) . (11) An extension of Vivanti- Pringsheim theorem.

Lgo

(a) Let f (z) = 2:~ a 11,zn and g(z) = /3nzn be series with real coefficients) both having their radii of convergence not less than 1. If 1 is a singular point of (an + i/3n)zn, show t hat 1 is a singular point of both f(z) and g(z). (b) Use (a) to reprove Exercise (10). (c) Suppose f( z ) = anzn has 1 as its radius of convergence and !Im an l < Jvf Re an, n > 0, for some positive constant M. Use (a) to show that 1 is a singular point of f (z).

Lgc

Lgo

(12) Show that t he set of singular points of a power series is a closed subset of its circle of convergence. (13) Dienes theorem. Suppose f(z) = L~ anzn has t he radius of convergence equal to 1 and JArg anl ~ a < ~ i n > 1, for some constant a > 0. Show that 1 is a singular point of t he series. (14) Let f( z) = anzn be a power series with real coefficients and having 1 as its radius of convergence. Suppose Sn = L~=O ar.: --+ oo or -oo as n --+ oo. Show t hat z = 1 is a singular point of t he series. Yet the result is not necessarily true if the assumpt ion is changed to lsnl --+ oo as n --+ oo. For instance, (H\)3 = 1 - 3z + · · · + (-l)n

Lgo

Chap. 5

46

! (n + l )(n + 2) zn + · · ·, imate value). (15) Let r zo (0 < r zo

Fundamental Theory

lzl < 1, is a case in which lsnl ,. . ,

2

.,~ (approx-

< oo) be the radius of convergence of t he power series

lz -

f( z) = .2:.:: ~ an(z - zo)11-. Then,

zo l = rzo is its natural boundary

if and only if 1 rzo+ !r zo ei9 = 2rzo' where r zo+ ! rzo eie is the radius of convergence of t he T aylor senes expansion of f (z) at zo + ! r zo eiB, 0 ~ B ~ 21T, namely, -1-in1 . -

k~oo

2= en1,;an (1-?rz e )nt = 1, oo

iB

0

~

n= k

(16) Suppose r = 1 is t he radius of convergence of f(z) = .2:.::go anzn. Then z = 1 is a singular point of the series if a.nd only if lim 71, ---4 00

\/TbJ =

2,

""'n

. ifJ . . 1ar porn . t , t h en a1c, w h ere bn = u k = O en k ak; m case z = e is a smgu 0 ~ k < n , in t he expression of bn should be replaced by eikB a k . (17) T ry to use Exercise (15) or (16) t o prove t hat t he power series in Example 4 has only one singular point, namely, z = 1. (18) Let p be a positive integer and hp(t) = t(t - l )(t - 2) · · · (t - p + 1). Try to show t hat tm = hm(t) + ai ,m h m -1 (t) + · · ·+ am-1 ,mh1 (t), m is a positive integer , by induction, where each coefficient aj,m, 1 ~ j ~ m - 1, is a positive integer. Then use this fact or some other method to show that oo

n =l

P n

n z =

Qp(z)

-~--.

(1 - z)P+l'

lzl < 1,

·w here Qp(z) is a polynomial of degree p , with coefficients all positive integers. Finally, show that z = 1 is t he only singular point of t he sen es. (19) Try to use Exercise (16) to redo Exa mple 5. (20) Try to use HadamarcFs gap theorem (5.1.3.11) to show that lzl = 1 is t he natural bounda.ry for each of the following power series. And discuss the convergence (including absolute, conditional a nd uniform)

Sec. 5.1. 3

or divergence on 'oo i 2" (·a·) " L.JO 2n z .

47

Boundary B ehavior of a Power Series

Jz l = 1 for each series.

(b) L~ zn' . (c) L~ 2~. zn ! . 1 nl (d) 'l\'oo L.JO ~ z " 'l\'00 1 2n (e) L.JO nT z . Exercises B

(1) Show that the power series (-1)[vnJ zn, n

L 00

n =l

Jzl < 1

conditionally converges at each point on JzI = 1. (2) Let an E C , n ~ 1, and nan form a bounded sequence. In this case, f(z) = I:~ anzn has its radius of convergence at least equal to 1. (a) Show that f is bounded in the real interval [O, 1] if and only if the sequence I:~=O ak, n ~ 0, is bounded. (b) Show that f is bounded in Jzl < 1 if and only if the sequence I:~=O akz\ n ~ 0, is uniformly bounded on lzl = 1, namely, n

sup sup n2:0 JzJ=l

L ak zk

< oo.

k=O

(3) Consider the power series (see Ref. [44], Vol. I, p. 133)

f(z)

=

~ 1 2n ~1 z ,

o n.

Jzl
,, - /J 000006

F ig. 5.10

Sec. 5.2.1

Analytic Continuation along a Curve

53

o,

2. there is a 0 so that ) if s , t E [O , l] and Is - t l < U z(s), D z(s) ) and (f z(t), Dz(t)) are direct analytic continuation of each other: indeed, = min1~j~n ltj - t.i-11 will work.

o

The detail is left to the readers. We summarize the above as the

Equivalent statements of analytic continuation. Let (Jo , Do) and (f, D ) be two analytic function germs, and .zo and z be t he centers of the open disks D o and D , r espectively. Then ,

{:::?

{:::?

(1) (fo ,D o) and (!, D ) are a nalytic continuation of each other (see Fig. 5.9 and (5.2.2)) . (2) (fo ,D o) and (!, D ) are analytic continuation of each other via a chain of disks (see Fig. 5. 10). (3) There is a cont inuous curve .z = z(t) : [O, l] ~ C connecting .zo = z(O) to z = z(l) and satisfying: 1. to each t E [O , l], there is a germ (fz(t )i D z( t)) at .z(t) so that (fz(O) , Dz(O)) = (Jo, Do) and (fz(l)i D z(1) ) = (f , D ), and 2. (consistence) there is a > 0 so that , for s , t E [O, l ] with ls-tl < O, the germs (fz(s)> D z(s) ) a.nd Uz(t), D z(t)) are direct analytic continuation of each other.

o

Note that z(t) is the center of the disk Dz(t) a.nd, in 2, t he subcurve z l[s,t] lies in the set D z(s) n D z(t) . In the case of (3), {fz(t) }tE[0,1] is called an analytic continuation of Jo along the curve .z = .z(t), and (Jo , D o) and (f, D ) are analytic continuation of each other along the curve or along the chain of disks on the curve (in view of (2)) . See (5.2.1.2) below. (5.2.1.1) Remind that , as a point set, z( [O , 1]) is a cornpact subset of C , and thus, (3) ~ (1) follows obviously . In what follows , we develop basic properties of analytic continuation. F irst , the

Uniqueness. Suppose {fz(t)}tE[0 ,1] and {gz(t)hE[0,1] be two analytic continuations of Jo along the same curve z = z(t) : [O , l] ~ C. Then, on their common domain, fz(t) = 9z(t) for each t E [O , l].

(5.2.1.2)

Chap. 5

54

Fundamental Theory

(}

Fig. 5.11

It is in this sense t hat we are able to say that {Jz(t )}tE [O, l ] is the analytic continuation of the function Jo or the germ ( J o) D o) along the curve. Proof. Let A = {t E [O, l] IJ z(s) = 9 z(s), O ::;; s ::;; t}. 0 E A because J z(O) = 9z(O) = Jo on the common domain of J z(O) and 9 z(O) . Let t o = sup A. Then 0 ::;; t 0 ::;; 1. It is routine to show that t 0 E A and t 0 = 1 hold. Let 61 > 0 be the 6 in (3)2 in (5.2.1.1), corresponding to {9z(t) }tE [0,1]· Denote 60 = min {6 , 6'}. Choose any t E A such that It - to I < 60. Then J z(t) = 9 z(t) on D z(t ) n .6.z(t), where .6.z(t ) is the open disk associated to the germ (9 z(t), .6.z(t)) - See Fig. 5.11. Let D denote the open disk with center at z(to) and radius the minimum of the radii of D z(to) and .6.z(to) . Since both J z(to ) and 9 z(to) are direct analytic continuations of Jz(t) = 9z(t )i it follows that J z(to) = 9 z(to) = =?

(by Eq. (5.1.2.5))

=? J z(to) = 9z(to)

on

f~(t) on D z(t)

f z(to) = 9 z(to) D z( to )

n n .6.z(t)

on

D

D ~ D z(to), .6. z(t-0)

n

.6.z(to) ·

This shmvs that to E A. In case to < 1, choose t so that t o < t < 1 and Ito - tl < 60 (as above). By (3)2 in (5.2.1.1) , both J z(s) and 9 z(s) are direct analytic continuat ions of J z(to ) = 9z(to) for each s, t o ::;; s ::;; t. Adopting argument similar to the last paragraph, then Jz( s) = 9 z( s) for to < s ::;; t; in particular' Jz( t ) = g z( t ) holds. Hence, t EA yet t > t 0 , contradict ing to the definition of t 0 . Consequently, to = 1 should hold. O Remark 1 (The resulted multiple-valued function) . In (5.2.2), it is quite possible that there exist j and k with k - j ~ 2 so that D j n D k =/:

Dz(tn-1)) · For any t, tj-1 s t < tj and 1 s j s n , by assumption r

r

lw(t) - z(tj-i) I s jw(t) - z(t) I + lz(t) - z(tj- 1)1 < 4+4

r

= 2·

This indicates th at w(t) E D z(ti_ 1) holds for t.i-I S t S tj and 1 S j Sn. Use /j to denote the line segment joining z(t j) and w(t1 ) for 1 s j s n. Note that /j lies completely in Dz(tj) for 1 s j s n . The key point in the argument relies mainly on the fact t hat , t he direct analytic continuation to the same point a.long any curve in the open disk of a germ always lead to the same germ (see (5.1.2.2) and (5.1.2.5), or t he implication of (3) in (5.2.1.4)). Now, consider the disk Dz(O) = Do. (Jo, Do) can be analytically continued to U z(t 1 ), D z(t 1 )) a.long z = z(t) from z(O) as well as to (9w(ti), D w(t 1 )) along w = w(t) from w(O). If we continue (9w(ti), Dw(t 1)) from w(t1) to z(t1) along 'Y:l\ t hen we get (Jz(t1 ), D z(t 1 )) because a ll the processes proceed in t he same open disk Do of the germ (Jo, Do). See F ig. 5.12. Then, proceed to t he disk D z(l) . (Jz(ti), D z(t 1 ) ) can be analytically continued to U z(t2 ), Dz(t 2 )) along z = z(t) and (9w(t 1), Dw(t 1 ) ) to (9w(t 2 ), Dw(t2 )) a.long w = w(t) . Based on what we obtained in the last paragraph and ( 5.2.1.5), if we continue (Jz( tt)> D z(ti )) from z(t1) to w(t1)

Sec. 5. 2.1

Analytic Continuation along a Curve

59

0

Fig. 5.12

along 'YI and then to w(t2) along w = w(t) and t hen to z (t2) along 12 1 , we will obtain the same (f z(t2 ), D z(t?J ) . Repeat this process after a finite number of steps and we will prove the statements. D Exe rcises A (1) Let

Ji (z) = Jlzjett

z E D1 = C -

1

[O, oo) with 0 < Arg z = () < 27r

1

and

h(z) = v'fzje~

= C - (-oo, O] with - n < Argz = () < 7r . Then, J1 (z ) - h(z) on Im z > 0, while J1 (z) # h(z) for all z with Im z < 0. So (!11 Di) and (h: D 2 ) are not direct analytic continuation of each other. Try to show that (Ji , Di) and (h , D2) are analytic con1

z

E D2

tinuation of each other by showing some intermediate analytic germs between them. (2) Given two power series oo

n

fi( z ) = ~ :.._ D 'n,

and

n=l

f2(z) = 1fi +

Loo (-l)n n (z - 2)n. n= l

(a) Determine their radii of convergence a.nd show that t heir disks of convergence do not intersect each other. (b) Show that Ji (z) can be analytically cont inued to h (z) by presenting a third analytic function f3(z) so that both fi(z) and f 2(z) are direct analytic continuations of f3(z). (3) Shows t hat t he series 00

L

n=l

(

1 1 ) 1 - z n+ l - 1- zn

Chap. 5

60

Fundamental Theory

represents an analytic function Ji (z) on Jz I < 1 and another one h (z) on lzl > 1, yet fi(z) and h(z) cannot be analytically continued to each other. (4) Do the same problem as Exercise (3) for the series 1 -1-z

oo

+L

n=O

z

2n

z 2n +1 - 1.

(5) Let L:::: =l an be a series with an> 0 for all n 2:: 1. Let r n 1 n > 1, be the sequence formed by all the rational numbers. Show that the series

=" an L......t z-r 00

f(z)

n=l

n

is analytic in Imz > 0 and in Imz < 0) respectively, but the analytical functions then represented cannot be analytically continued to each other. (6) Prove (5.2.1.4) in detail. (7) Prove (5.2.1.5) in detail. (8) Try to use (3) in (5. 2.1.1) to reprove (5.2.1.6).

Exercises B Try to use the concept of analytic continuation to formally define the Riemann surface of the multiple-valued function F( z ) mentioned in Remark 1 and justify that it is coincident with the one give11 by (2.7.2.8). It is beneficial to refer to Ref. [81 L or, in less arduous, to Ref. [75L Chap. 3. 5.2.2

Homotopy and monodromy theorem

Roughly speaking, that a curve is continuously deformed into another curve without leaving a given open set during the process is termed as a homotopy between the two curves. See Sec. 4.2.2 or Exercises B of Sec. 2.7.1 for formal definition. Section 4.2.3 (see also Exercise B (2)(d) of Sec. 2.7.1 ) discussed the homotopic invariance of winding numbers. Here, in this section, we will see that the analytic continuability is another homotopic invariance which is specifically called the monoclromy theorem in t he realm of complex analysis . Our main result is

The homotopic invariance of the analytic contim.wtion or A1onodromy theorem. Let 'Y and :Y be homotopic with fixed endpoints zo and z1 in a domain

Sec. 5.2.2

Homotopy and Monodromy Theorem

61

n, and H: [0,1] x

[O,l]~n

be the homotopy between "/ and ~'· Suppose a germ (f 0, Do) at z 0 can be analytically continued along each curve H(-, T) : [O, 1] ~ n for 0 S T ::; 1. Then the continuation of (.to, Do) along "/ and i' lead to the same germ at Z1 . (5.2.2.1) This result can be restated in terms of covering surface (see (2.7.2.9) and Sec. 7.5.2, in particular, (7.5.2.3)) .

Proof. The proof is based on (5.2.1.6). Let (!, D) be the germ obtained by analytic continuation of (Jo , Do) along / . Let

A = {TE [O,l] I the germ obtained by continuation of (fo ,Do) along t he curve

H (t,T) , 0 0 such that, for any curve a = a(t) : [O, 1] ~ n satisfy ing:

1. a(t) and H(t , To) having the same encl points, namely , a(O) = H (O, To) = zo and a(l) = H (l, TO) = z1; 2. la(t) - H(t , TO) I < c, 0 < t::; 1, then the continuations of (Jo , Do) along a(t) and H(t, TO ) lead to the same germ at z 1 . For this:::, by the uniform continuity of H(t , T) on [O , 1] x [O, 1], there is a > 0 so that

o

IH(t , T)-H(t , To)I < s whenever t E [O , 1] and T, TOE [O, 1] with IT-To i
oo

./1

By Morera's theorem (3. 4.2.13) , J (z ) is thus analytic inn.

Sec. 5.3.1.

Analyticity of the Limit Fimction

65

However, Cauchy 's integral formula provides another proof for this fact and even more: Choose any closed disk lz - zol : :; r containing in n. Then

fn( z) =

r

~

=? (Since

n

--t

lz - zol < r

fn(() d(,

2ni 1 1(- zo f=r ( -

and

~

n

1

Z

fn(z) converges uniformly to f(z) on

lz - zol

Sr.) Letting

oo, thus

~ f

f(z) =

f (() dC

lz - zol < r.

lz - zol
0 and there is an no 2 1 such that

lfn(z) - f (z) I < c for all z

E 0

and n 2 no .

This implies that f n ( z ) =/: 0 on 0 for all n 2 no . Assume the result holds for m - 1, where m > 1. Let zo E 0 be a zero of f( z) . According to (2) in (5.3.2.1), t here is an integer ni 2 1 so that fn (z) has a zero Zn in 0 for each n 2 ni and lim11 _ 00 Zn = zo holds. For each of such n 2 ni, there are analytic functions 9n (z) a nd g(z) on n such that

f n(z) = (z - z 11 )g11 (z ),

n 2 n1,

and

f( z) = (z - zo)g(z) hold throughout n. Since z - Zn converges to z - Zo uniformly in n, by assumption , 9n (z) converges to g(z) locally uniformly in n. (*) says that g(z) =/: 0 along 80 and g(z ) has (m-1) zeros in 0. By inductive assumption, there is an integer no 2 n1 so that 9n (z) has exactly (m- 1) zeros in 0 for n 2 no . It follows that f n(z) has exactly m zeros in 0 for n 2 no.

Sec. 5.3.2

Zeros of the Limit Function

85

To prove (1)::::} (2). Vie may suppose f( zo) = 0 when f( z ) is replaced by f (z) - f (z0). All one needs to do is to show t hat, if f( z) is not identically

equal to zero, then each zero of f (z ) is the limit point of a sequence Zn, where f n(zn) = 0 for each n > 1. Let zo be a zero of f( z) . Choose p > 0 so that lz - zol s; p is conta ined inn and f (z ) =J 0 in 0 < lz - zol < p. Let Pk = 2-kp fork > 0. According to (1 ) in (5.3.2.1 ), t here is an integer nk so that f n(z ) h as a zero Zn,k in lz - zol < Pk for each n > n k and k z 0. l\ila.y choose nk+l > nk for k z 0. Define a sequence Zn in n by Zn = Zn,k' Then Zn

--4

nk < n s; nk+b

k

> 0.

zo and f n(zn) = 0 ~ 0 = f( zo). This finishes t he proof.

D

Even though fn(z ) is univalent in n for each n z 1, t he limit funct ion f( z) in (5.3.2.1) could be identically equal to zero. For instance,

z fn(z) =- , z ED. =C -{O} n is univalent in f2 and f n(z) =j 0 every;vhere in D., yet fn(z ) COnVf)rges to the zero function locally uniformly in n. Vlhile 9n(z ) = ~ =J 0 in C for n 1, and 9n(z) converges t o 0 locally uniformly in C. Anyway, we have

z

the following

Two special cases of Hurwitz 1s theorem. Suppose fn(z) : n ~ C is analytic for n 1 and the sequence fn(z) converges t o f (z) locally uniformly in D.

z

(1) If f n ( z ) does not have zeros in n for each n z 1, t hen the limit function f (z) is either ident ically equal to zero in n or does not have any zero inn. (2) In case each f n(z ) is univalent in D, t hen f( z) is either a constant or is univalent in n. (5.3.2.2) (1) is obvious. As for (2), fix any point Z Q E n. Since fn (z) - fn(zo ) converges to f (z)- f( zo) locally uniformly inn - {zo }, by (1), f( z)- f (z o) does not have zero in D - { zo} in case f (z) is not a constan t . Hence, f (z) =J f( zo) for all z E n and z =J Z Q. This proves t he cla im. vVe illustrate t hree examples. Example 1. Example 2 in Sec. 5.3.1 showed that ez = limn---.oo (l + locally uniformly in C. Try to use t his fact to show that ez =J 0 in C.

;r

Solution. Suppose there exists a point z o E C so t hat ezo = 0. Since f( z) = ez is not a const ant , t here is an R > 0 so that ez =J 0 on 0 < lz - zol s; R.

86

Chap. 5

Fundamental Theory

;r

By (5.3.2.1) , there is a.n integer no 2 1 so that fn(z ) = (1 + has the same number of zero as f( z) does in lz - zol < R for each n 2 no. Yet, for all sufficiently large n such t hat 1-n - zo I = In + zo I > R , fn (z) f:. 0 in lz - zoI < R , a contradiction. Hence, ez f:. 0 for all z E C. Example 2. Let an

> an+l > 0 for n 2 0. Show that 00

is analytic in

lzl < 1 and f( z) =/: 0 there.

a::

Solution. limn-+oo an = p 2 0 exists. In case p > 0, then limn-+oo = 1 1 shows that t he series has its radius of convergence equa.l to 1; in case p = 0, .l. then limn-+oo a1~ ::; 1 and t he radius of convergence is not less t han 1. Hence, 2=~ anzn is analytic in lzl < 1. According to Enestrom- Kakeya theorem (2.5.1.8) , Sn(z ) L:~=O akzk =/: 0 on lzl < 1 for each n > 1. Since Sn(z) converges to f(z) locally uniformly in lzl < 1 and /(0) = ao > 0, by (1) in (5.3.2.2) 1 f (z) =/: 0 throughout lzl < 1. Example 3. Let

R be a strictly increasing continuous 1

f( z )

=lo cp(t) cos ztdt

is an entire function and can only have real zeros.

Solution. Let

fn( z) =

1 (k) k7 2=-

0, there is a 8 > 0 so t hat

IF( z, t) - F (z, t')I < £,

z EK

and

t, t'

E [O , 1] with

It - t' I < 6.

Sec. 5.3.2

Zeros of the Limit Function

Choose n la rge enough so that ~
[~ ]

+ 1).

Then , for

z EK, f( z ) - f n(z) =

Ln 1·k/n rp(t)cosztdt- L (k-1 )/n n

k=l .

=

I

k /n

k=l. (k- 1)/n

(k) cos -

kz

oo - Zk

r - lzk - zo l

lz - zol r

< 1) lz - zol < r,

under the assumption that limk_,, 00 Zk = zo . In this case, it follows that the sequence Ln_ 1 (z) will converge to f (z) locally uniformly in z - zo l < r (see (*3) below). Now) suppose fm (z), rn > 1, a.re analytic in Iz - zo I < r and continuous on lz - zo l < r . Also, suppose f m(z), m > 1, are uniformly bounded in lz - zo l < r, namely,

(5.3.3.2)

sup sup Jfm(.z )l = J..lf 2.

Sec. 5.3.3

Some Sufficient Criteria for Local Uniform Convergence

91

Under the additional assumption that lim m ~ oo fm( zk) = 0 for each k > 1, for fixed 1 < k ::; n and s > 0, there is an integer mt: ,k > 0 so that for m

> m e ,k rn-1

n

where L

=2

L

I / ( )I)

k=l Wn Zk

if lz -zo l < p Combining ( *4) and (*5), for 0 < p N = N(p ,s) > 0 so that for 1n > N ,

and

> l max m e k· ::=;k::=;n '

m

< r and s > 0, there is an integer

lfm(z)I ::; lfm(z) - L;~~)1(z)I + IL;:)1(z)I < c zol < p.

if n is large enough and lz -

This means that fm(z) - t 0 locally uniformly in lz In conclusion, we have a

zol < r .

Conve rgen ce theore m (I ). Suppose fn( z) is analytic in lz - zo l continu01.ts on lz - zol < r for each n > 1. Also suppose that,

1, are uniformly bounded (see (5.3.3.2)) in lz - zo l < r; 2. zk, k > 1, are distinct points in lz - zol < r satisfying limk~ oo Zk = zo, and 3. lim n~oo f n (zk) = 0 for each k > L Then fn(z) converges to 0 locally 'ltniformly in lz -

zol < r.

(5.3.3.3)

Section (2) An application of H adamard's three-circles theore m (3.4.4.4) Suppose f(z) is analytic in ma.x

jz-zo l::;2R

lf(z) I =

lz - zol < 4R, where ma.x

lz-zo l=2R

R

> 0. Then

lf(z)I 1

< [ max If(z) 1 !z- zol=R

w h ere a:

~

- a [

max lz- zol =4R

log2R - log R

If (z) ~ a , ~

log 2

= log4R - log R = -. log4

0 bserve that the constant a = l~~ ~ is good for any function analytic in lz - zo I < 4R so that (*6) remains valid. Let f(z) be analytic in a domain D. Fix any closed disk lz - zo l < 4R containing inn and choose any

z1 in an open subset 0 of n. Construct a

92

Chap. 5

Fimdamental Theory

sequence of open disks D(zj , R j) : Jz - Zj l and R,1 = R , satisfying

< Rj, 1 ::;

j

< n,with

Zn

=

zo

1. D (z1, R1) ~ O; 2. the closed disks D(zj, 4Rj) C !1 for 1 ::; j < ri, and 3. D (z.i+l , R j+i) C D(zj, 2Rj) for 1 < j < n - 1.

Then, applying ( *6) to R1 < Jz that supzEn Jf( z)I = 1, we have

z1

J ::; 4Ri , and under the assumption

1

sup lf(z) I::; max lf( z) I ::; [ max lf(z) l - a: [ max Jf(z) la: D(z2 ,R2) Jz- z1J=2R1 Jz- z1J=R1 ~ Jz- z1J=4R1 ~

< [ sup lf (z)l l 1-a: 1, be

1. fn(z) converges uniformly in a nonempty open sitbset 0 of n, and 2. fn(z), n > 1, are uniformly boimded in each compact subset of n.

Then fn(z) converges locally uniformly in

n.

(5.3.3.5)

Proof. To prove t his , choose any com pact set K inn and a closed disk D in 0 . Construct a bounded domain U so that D u K c U (compact) c n. Apply (5 .3.3.3) to U with c = c(U, D , K) . T hen sup z EJ(

IJ~(z) -

fn(z) i S [sup lfm(z ) z ED

f~i(z)i] l - c [sup lfm(z) -

fn(z) i] c

z EU

says everything needed. By t he way, via the concept of subh armonic function (refer to Sec. 6.4, if needed), one can prove A generalized maximum principle (A . Ostrowski ( 1922/23)) . Let r 1 and f 2 be two nondegenerated complimentary arcs on izJ= 1 (namely , f 1 =I= , f 2 =/= , f 1 n f 2 = , and f 1 U f 2 is the circle izi = 1) . Let J(z) be an analytic function in Izi < 1 satisfying

-

lim If (z) I s

Z---+(

·where 0 < Nfi >.(r) , 0 < >. (r)

{Nii. , Nh,

s

NI2 < oo. Then, for each r , 0 < r < 1, there is a. number < 11 so t hat sup

lf(z)i S

.NI{(r) J\,!{;->-(r) .

(5.3.3.6)

lzl ~r

T his >.(r) is valid for a.ny such an alytic function f(z) . T hus, it follows easily t he following Convergence theorem (III) (P . Montel, 1927). Let f 1 and f 2 be nondegenerated complimentary arcs on lzl = 1. Suppose fn(z), n > 1, are analytic in lzl < 1 and continuous on the set {lzl < 1} U f 1. Also, suppose 1. fn( z ), n 2: 1, are uniformly bounded in lzl < 1; 2. fn( z ) converges locally unifomily on f 2.

Then fn(z) converges locally 1miformly in

izl < 1.

(5.3.3.7)

Chap. 5

94

Fimdamental Theory

Sectio n (3) How can Schwarz's le mma h e lp? Fix a point zo, lzo l < 1. Then

z - zo 1 - zoz

lzl + lzo l

.,

s 1 + lzol lzl (see (3.4.5.0)) , lzl < 1 lzol . 1 - lzol lzl) . 1 + lzo lIzl < 1 - ( 1 - lzl) 1 + lzl 1-

= 1 - (1 -

1- lzl}

< exp { -(1- lzo l) 1 + lzl (using 1- t

< e-t, t > O), lzol < 1 and lzl < 1.

(*s)

Suppose f( z) is a bounded a nalytic function in lzl < 1, say lf(z)I :::; M there. Let zi, ... , Zn be n distinct points in lzl < 1 so that lf(zk) I < a for 1 :::; k < n . Let Ln-1(z) be Lagra.nge's polynomial given by (*1) . Observe that Ln-1(z) - J(z) has zeros at z1, ... , Zn and

Ln-1(z ) - f( z) Bn(z)

n

where Bn(z) =

zII 1. :._ 7

k

ZkZ

k =l

is a nalytic in lzl < 1 (see (3.4.2.17) ). Apply the maximum principle to this function on lzI < r < 1 and we have

L n-1(z) - f (z ) Ln-1 (z) - f( z) :::; sup B n(z) Bn(z) lz l ~r = sup lzl=r ~Letting r

--7

1 and observing that

Ln- 1(z) - f (z ) Bn(z)

IBn(z)l --7 1 as lzl

Ln-l (z) - f( z ) < sup ILn- 1(z) - f(z) I, Bn(z) lzl

Then f n (z ) converges locally uniformly in D. (2) (P. JVlontel, 1907) Suppose fn(z), n 2: 1, are locally ·u .niformly bounded inn. Then f n(z) has a subsequence fnk(z), k 2: 1, converging locally uniformly inn. (5 .3.3.10)

(2) is still valid even if n is only a nonempty open set. In what follows , we try to prove the validity of (1) via (5.3.3.9) (see Exercise A(2) for a second proof), and then the equivalence of (1) and (2). For another proof of (2) via the concepts of equicontinuity and normal family, see Sec. 5.8.1. Exercises A(3) and (4) will give two applications of (5.3.3.10). To prove (1). Let A = {z E Dlfn(z) ,n 2: 1,converges}. Then E ~A. Let Do be the set of the limit points of A in D. Then, no =f. c/; holds. Suppose zo E Do and the closed disk lz - zol :::;; r is contained in n. Choose a sequence zk of distinct points in A such that lzk - zo I < r for k 2: 1 and limk~oo Zk = zo. In this case, 2::: ~ 1 (1 - lzkl) = oo holds and it follows by (5.3.3.9) that f n(z) converges locally uniformly in lz - zol < r. Consequently, {lz - zo l < r} ~ no n A holds and thus no is a relatively open set in n. Since Do is relatively closed in D which is a connected set, hence Do = D and then {lz - zol < r} ~ A. Therefore no = D ~ A and, indeed, A = D holds. By using (5.3.3.9) again, f n(z) converges locally uniformly inn. To prove (1) ::::} (2). Vve may suppose that E = {z 1 : z2 , ... ,zk, .. . } is a countable set. Since {fn(z1 )}~=l is bounded , it contains a convergent subsequence f nk( 1) (z1), k 2: 1. Similarly, the boundedness of f nk(1) (z2), k > 1, implies that it has a convergent subsequence f (2} (z2 ), k 2: 1. Repeat this pronJ.:

cess. Then, each sequence fnul (z1+i), k 2: 1 1 has a convergent subsequence k

Sec. 5.3.3

f

(1+1}

, nk

Some Sufficient Criteria f or Local. Uniform Convergence

(z) .,

k

97

> 1, converges on {z1, . .. , z1+1} for l 2 0.

By Cantor's diagonal process, the sequence

f mk (z),

k 2 1, defined by

is a subsequence of t he original sequence a.nd converges point wise on E . By ( 1), it follows t h at f mk (z), k ;::: 1, converges locally uniformly in n. Note t hat ea.eh open set 0 in C can be expressed as a countable union of connected open set s . A second usage of the Cantor diagonal process will extend t he known result on each component t o t he whole set 0. To prove (2) ~ (1 ) . Suppose there is a point zo E n so t hat f n(zo), n > 1, diverges. Since {fn(zo) } ~=l is bounded , there are a, f3 E C , a =/= /3, and t wo subsequences f n(1 ) (zo), k 2 1, and f.nc2}(zo), k 2 1; of it such t hat ~

lim f

k --oo

k

(1)

nk

(zo) = a

and

lim

f

(2)

k-too nk

(zo) = /3.

Since t he two subsequences f nk(1) (z), k > 1, and f n,,(2J(z), k 2 1, a re locally uniformly bounded in n , by (2) , it follows that each h as a subsequence f 1n,,(l)(z) --7 1 1, of distinct points from E , converging to a point inn. Then, by assumption 2 in (1), lim {f

k -too

( 1)

mk

(zt) - f. (2)(z1)} = 0 = j (l )(zt) - j (2)(z1) for l 2 1. m ~,

By the interior uniqueness t heorem , j (l) (z) _ j ( 2)(z) inn, a contra.diction. Therefore, the sequence f n(z) converges everywhere inn. And, finally, D (5.3.3.9) guarantees the desired result . Section ( 4) Finally , Caratheodory ine quality

applications

of

H adamard- Bore l-

Reca.11 t he inequality (see Exa mple 3 in Sec. 3.4.5) : Suppose f (z) is a.nalyt ic in izl < R and cont inuous on izl s R (or, Re f (z) is just assumed to be bounded a bove in izl S R). Then

R-r

2r

.

supRe f (z)< Re f (O)+ sup Re j (z); lzl=r R +r R+ r lzl =R

Chap. 5

98

sup

lzl=r

R +r

lf(z)I SR

- r

Fundcimental Theory

lf(O) I + R

2r

sup Ref(z ), - r lzl=R

0 Sr< R.

(5.3.3.11) A direct application of (5.3.3.11) is the following Convergence theorem (VI). Suppose f n ( z ), n 2:: 1, are analytic in a domain Sl. Let zo E S1 be a point so that the sequence f n (zo) converges. Then7 any one of following conditions will imply that the seq'ltence fn (z) converges locally 'ltniformly inn: 1. Refn(z) , n

> 1, converges locally uniformly inn .

2. Refn(z) s Refn+ 1(z) on n for each n 2:: 1. 3. lfn(z )I < lfn+1(z)I on n for each n > 1 and limn--oo f n(zo) =I=

o holds. (5.3.3.12)

Proof. Let

A = {(

E S1I

0 = {( E

.Sll

fn(() converges},

and

fn(z) converges uniformly in an open neighborhood of(}.

(*9) Since zo E A , A =f. 0 so that lz - (I s R is contained in n. Apply (5.3.3.11) to Un - fm)(z) , n > m 2:: 1, in lz - ( I < R. \"f\Te have, for n > m 2:: 1, su~R lfn(z ) - fm(z)I lz- ( 1- "r

R+R/2

< R- R/ 2 lf n (() - fm(() I

+ ~

sup l z -(1~~

lf n(z) -

2 · R/2

/ sup Re(fm (z ) - f n (z)) R - R 2 1z- ( l=R

f m(z) j S 3lfn(() - fm(()I + 2 sup

Re (f~(z)

- f n(z )),

(*10)

lz -( J ~ R

by t he maximum modulus principle. Condition 1 will show, by (* 1o) , that fn(z) converges uniformly in lz - ( I S ~·Hence

Sec. 5.3.3

Some Snfficient Criteria for Local Uniform Convergence

99

whenever ( E A and lz - ( I ::; R is contained inn. Similarly) Condition 2 implies that suplz-C1s 4 lfn(z)-j~(z)I::; 3lfn(()-fm(01 holds because Re fm( z) ::; Re fn (z) if n m 1. In this case, (*n) is still t rue. In conclusion 1 both cases show that A = 0 =f. 0 so that fi(z) =f. 0 in 0 < lz - (o l ::; R which is assumed to be contained inn. Then {lz - (o l = R} ~ n 0 . By the result from the last paragraph, fn(z) converges uniformly along the circle lz - (o l = R and then, uniformly in the disk lz - (ol ::; R by using

(5.3.3.1). This ends up the proof.

D

For another application of (5.3.3.11), we set up a preliminary inequality as follows: Suppose f (z) is analytic in Iz I ::; R and f (z) =f. 0 t here. Then 2r

sup lf(z) I ::; lf(O)liFf. [sup lf(z) ll R+r ) lzl=r lzl=R

0 ::; r::; R.

(5.3.2.13)

Chap . 5

100

Fundamental Theory

To see this, choose p > R so that f(z) is analytic in lz l < p on which f( z) =I= 0. Choose a bra.nchg(z) of logf(z) in lzl < R, namely, eg(z) = f(z). Applying (5.3.3.11) to g(z) , we have 2r R-r sup Reg(z) < R sup R eg(z) + R Reg(O), where lzl = r ~ R. lzl =r + r lzl=R +r Observe that et is an increasing function oft and that inequality follows easily. (5.3.3.13) can be used to obtain the following

If (z) I = eRe g( z) . The

C onvergen ce theor e m (VII) (Monte l, 1912) . Sitppose fn(z), n are analytic in a domain

n

:;::=:

1,

and satisfy:

1. fn(z) =I= 0 inn for each n :;::=: 1; 2. fn(z) , n :;::=: 1, are locally uniformly bounded; and 3. There is a point zo ED so that limn-too fn(zo) = 0.

Th en f n (z) converges to 0 locally 1..lniformly inn.

(5.3.3.14)

This result can provide (2) in (5.3.2.2) another proof. See Exercise A(5). The proof of (5.3.3.14) is exactly the same as that of (5.3.3.12). Sk etch of proof. Suppose ( E n so that f n(() converges to zero. Choose R > 0 so that lz - (I ~ R is contained in n. Let NI = supn'2:1 suplz-( l::;R lfn (z) I. Note that 1\lf < oo . Chooser = ~ in (5.3.3.13) and we have sup lz-(1:::; ~

lfn(z)I

~ lfn(()I~+~~~ { sup lfn(z)I } lz-(l=R l

l

~Nl 3 lfn(()l 3 ,

ii!Jh

n2".l.

This means that f n(z) converges to 0 uniformly in lz to finish the proof.

(I
1, converges to f (z) locally uniformly. 9 (b) limn-+oo ank = ak , for each k ;:::: 0. In this case, '\"00 L.Jk =O

f (z )

k

akZ .

(5) Try to use (5.3.3.14) to prove (2) in (5.3.2.2). (6) (Osgood, 1901- 1902) Let f 11 (z), n > 1, be analytic in an open set n a.nd converge pointwise to a function f (z) in n. Then, there is an open dense subset n o of n on which fn(z) converges locally uniformly to f( z) . Hence , j(z) is analytic in n o. (7) Let f(z) be analytic and bounded in the angular domain -Bo< Arg z < Bo (Bo > 0) . Suppose limx ___.,of (x) = a holds. Show that , for each x>O

0
1; 3. indifferent if lf' (zo) I = 1; more precisely, rational neutral if lf'(zo) I = 1 and f'(zo) = 1 for some integer n , and irrational neutral if lf'(zo)I = 1 and f'(zor "f. 1 for any integer n . (5.3.4.1)

1. attractive if

11

Two ftmctions f : n --t n and g: O(open) --t 0 are said to be (conforma.lly) conjugate to each other if there is a univalent mapping : n --t 0 so that g = of o - 1 , i.e.)

(f(z)) = g((z)) ,

zE0

(5.3.4.2)

referred to as Schr"oder)s equation (1871). In this case, gn = o Jn o - 1 and thus, and gn are also conjugate for each n > 1, and so are 1- 1 and g- 1 , ·when both are defined. i\!Ioreover, maps fixed points of f to fixed points of g and vice versa, and the multipliers at the corresponding fixed points are equal. This plays as a coordinatizing map, specifically called the confagation, and f and g can be regarded as the same function viewed in different coordinate systems related by this . In what follows, we will sketch some results about t he existence of fixed points and the univalence of an analytic function, mostly in terms of the iterates of the function via the concept of local uniform convergence. For detaiLis, see Ref. (12]. The study of the behavior of analytic functions under iteration constitutes the content of complex dynamics, a field originated vvith the Nev,rton- Ra.phson iteration method for approximating roots (see, for example, Ref. [7], Chap. 11 , and Sec. 5.8.5 for some fundamental results) and of current research interest. One may refer to Ref. [15] and the references therein for its ana.lytic aspects. The following is divided into four subsections.

rn

Section (1 ) Linear fr action al t r a n s forma tion s as pre limina r y e x a mples Here, readers a.re supposed to be familiar with the content of Sec. 2.5.4 and the exercises t here, in particular, Exercises A(9) and B(l ). vVe will have no hesitation to adopt materials presented there. Also, the families of Steiner's

Chap. 5

104

Fitndamental Theory

and Appollonius 's circles introduced in Sec. 1.4.4 are helpful, too , in the understanding of this subsection. Given az + b w = , with normalized ad - be = 1. cz +cl ' w = w( z) has either two distinct fixed points z1 and z2 or one coincident fixed point zo except the identity mapping w = z . Th e case that c = 0 and z1 = d~a, z2 = oo: Then

w =

az+b cl

(*1)

ad = 1

{:::} w - z1 = k(z - z1), with k = ~ = w'(z1) and k- 1 = w' (oo) , the multipliers. ( *2)

{:::} (canonical form) 'rJ = k(, where ( = z - z 1 and 'rJ = w - z1.

Note t hat ( = (z) = z- z 1 acts as the conjugation of w = f( z) = azfb to 'r/ = g(() = k( in the sense (5.3.4.2): g(() = of o - 1 ((). The advantage of (*2), namely, 'rJ = g(() , over (*1) , namely, w = f( z), is that it is easy to write down the iterates gn:

ry = kn(,

n 2:: 1.

In case w = f( z) is hyperbolic (k > 0 but k =I= 1) or lox odromic (k = lklei8 , lk l > 0, lkl =I= 1, and =I= 2n1r for integers n), gn(() converges to 0 or oo uniformly in any bounded neighborhood of 0 or oo, respectively . Consequently, so does f n(z) to z 1 or z2 as n ~ oo. In case gn(() = g(() for some integer n 2:: 2, then it is necessary t hat kn = 1, and thus, w = J(z) is elliptic (lkl = 1 and k =I= l ). Th e case that c =I= 0 and z 1 =I= z 2 : Then

e

w = {:::}

az+ b cz + d '

(a-cl) 2 +4bc =I= 0

ad-be = 1 and

w - z1 z - z1 . ,( ) a - cz1 and w , ( z2 ) = k- 1 , =k ' wit h w z1 = k = w - z2 z - z2 a- cz2 the multipliers.

{:::} (canonical form) 'rJ = k(, where ( =

z - z1 Z -

Z2

and rJ =

w - z1 W -

Z2

.

(*4)

Hence ( = !l>(z) = ~= ~; is the conjugation of w = f( z) = ~;$~ to 'r/ = g( () = k(. Similar explanations are still valid in this case as in the last paragraph.

Sec. 5.3.4

An Application

105

The case of one coincident fixed point z o: If c = 0, t hen zo = oo a.nd w = w(z) reduces tow = z±b) b =I= 0. If c =I= 0, then zo = 0 : ; / . In this case, az +b w - cz+ d '

1 w -zo

{:::;> - - -

{:::;>

1 - -

z- zo

ad-bc = l ,

and

± c (it is c if a+ d

(*5)

a + d = ±2

= 2 and - c if a+ d = -2).

(canonical form) rJ = ( ± c, c =I= 0) where ( =

1 z- zo

1 and rJ = - - w - zo

Now, ( = 1. gn( () converges to oo locally uniformly (according to the spherical chord metric) in C . As a consequence, the original fn (z) converges to zo = 0 2cd locally uniformly in C . Observe tha.t zo is a parabolic fixed point. For the sake of reference and completion, we list the following three types of popular geometries and leave the details or proofs to the readers or refer to Appendix B. Recall that the set of all linear fractional transformations forms a group under composite of mappings (see (1) in (2.5.4.13)) and is called the (general) linear group on the plane. \i\Then the symmetric mapping z ~ z is added, the resulted one is called the generalized linear groitp. P arabolic (or Euclidean) geometry. The group of rigid motions

Gp = { eie z

+ blB E R , b E C}

is a subgroup of the linear group. It is composed of the following two fundamental motions: I. rotation (elliptic type) : w = eie z; 2. t ranslation (parabolic type): w = z +b.

The distance lz1 - z2I between two points z1 a.nd z2 in C is in'variant under Gp . Note. Sometimes, GP is replaced by ~~~ group GP of Euclidean motions, generated by adding z ~ .Z to Gp. Gp is a subgroup of the generalized linear group. (5.3.4.3) Elliptic (or spherica0 geometry. The imitary group

Ge = {

~;~ ;ia,bE C

and

lal 2 + lbl2 =

1}

Chap. 5

106

Fimdamental Theory

is a subgroup of the linear group. Elements in Ge are always of the elliptic type: zo l+zow W -

iB Z - ZQ

---= e

Ima+ J(Im a) 2 b

wh ere zo =

l+zoz'

+ lbl2

. i

and

() E R

(see Exercise B(2) of Sec. 1.6). Consequently,

z1 - z2 1 + Z2Z1

z1, z2

E C*

(

it is

1

lzi I

if

z2

=

oo)

is an invariant under Ge and so is ? -1 ~tan ')

~tan

-1

-z2 II' z1, z Ilz1 _ 1 + Z2Z1 _1_

Iz1I,

z1 E

2E C and z 2 =

C

oo

Under de(,) , the Riemann sphere or C * is a complete metric space with the great circles (or parts of it) as geodesics. (5.3.4.4) Hyperbolic (or non-Euclidean or Lobachevski) geometry. The group of nonEuclidean motions Gh = { e

i8

z- a l _ ()E R , - az

and

aEC

is a subgroup of the linear group (sometimes, z in G h are classified into three types:

~

and

la!
0 so t hat f (z) is analytic in lz - zo l ~ r . Then, for any integer k > 1' -

~ l(fn)(k)(zo)I ~ k~iVl, r

~ (setting k

'l!vhere J\!1 =

ma.x lf(z)I

lz- zo l:S:r

) / n = 1) I (f ·111 ') (zo I = If (zo) I

~

NI r

Chap. 5

108

~ lf'(zo)I ::;

Fundamental Theory

z/¥

~(letting n --too)

lf'(zo)I < 1.

This is the desired condition for f(zo)

(5.3.4.6)

= zo to be true.

Section (2) Attractive and repuls ive fixe d points

Suppose zo is an attractive fixed point for f. Choose r > 0 so that f is analytic in lz - zo I < r on which If' (z) I < a < 11 where a > 0 is a constant. By mean-value theorem (3.4.1.1) or just writing J(z} = f(zo) + fz: f'(()d( in Jz - zol < r, vve have

lf(z) - zo l < alz - zol, lz - zo l < r ~

lfn(z) - zo l < a lfn-l(z) - zol < · · · < o:nlz - zol, lz - zol < r for n > 1.

As a consequence, t he iterate sequence Jn (z) converges to zo uniformly in lz - zo l < r . Conversely, suppose Jn(z) converges to z 0 locally uniformly in lz - zo I< r . In particular, fn(zo) = f o Jn-l (zo) --t zo. By continuity of f at zo, it follows that f(zo) = zo . On the other hand, Weierstrass's theorem (5.3.1.1) says that (fn(z))' = (j'(z)) 11 --t 0 locally uniformly in lz-zol < r; in particular, (j'(zo))n --t 0 as n --t oo. It follows that lf'(zo)I < 1 holds and thus z0 is an attractive fixed point of f. As a matter of fa.et, ( *8) says geometrically that the closure of the image of the open disk lz - zo I < r, under f 1 is contained in a smaller closed disk lz - zo I ::; a:r . This suggests that we are able to relax the condition (*8) a little in order to determine whether there is a fixed point for f. To see this, now suppose f : n (a domain) --t n is analytic such that f(f2) is a. conipact set (of course, contained in n). Let On = f 11 (f2), n 2 1, with Do = n. Note that f21 C f21 1, converges tog uniformly in f21, (refer to Exercise B(4) of Sec. 5.3.1)

~

w E g(D1) ~ g(D),

~

K

~

g(D).

Combining together, g(D) = K is a nonempty compact set in n and thus, g(D) = {zo} should hold ; otherwise, the nonconstantness of g(z) would imply that g(D) is an open set in n , a contradiction (see (1) in (3.4.4.4)) . Consequently, by Exercise A(3) of Sec. 5.3.3, f n(z) converges to zo locally uniformly in n and thus, f( zo) = zo holds . vVhat a re the conjugation and the canonical form for f(z) near such an attractive fixed point zo? It is the loca.l factorization formula f (z) = zo + f'( zo) (z - zo) + O((z - zo )2 ) in a neighborhood of zo (see (5.1.2.3)) that enables us to solve t his problem. After the substitution of z - zo by w, we may suppose that zo = 0, for simplicity. For the sake of later reference, we consider a little general sit uation

where g(z) is analytic in a n eighborhood of 0, g(O) = 1, and ak induction,

i=

0. By

fn( z) = (f'(O)tz+akzkgn(z), where 91(z) = g(z) ,

and

9n+1(z) = f'(O)gn(z)

+ [(f'(O)t + akzk-l gn(z)Jkg(fn(z)),

n 2:: 1.

(*10) It is understood henceforth that, if (*9) holds in lzI < r = ri, then, since fn(O) = 0, it is possible to choose 0 < rn+l < rn so that Jn( { lzl < rn+d) ~ {lzl < rn} and 9n+l (z) is analytic in lzl < rn+l for n 2:: 1.

110

Chap. 5

Fundamental Theory

For this moment , let k = 2 in (*9) and suppose o setting

=? 1, converging locally uniformly inn to an analytic function h( z) . F ix any point WOE g(D). There exists a point zo En so thatwo = g(zo). Since fn j (zo) ---+ g(zo) = wo , so does fn j" (zo) as k ---+ oo; hence, lim

jmJk

oj

nik (zo) =

k-+oo

lim fi jk+ 1 (zo) k-+ oo

= WO

(*13)

On the other hand , choose r > 0 so that lz - zol :s; r is contained in n. Since fmi (z) , j > 1, converges uniformly to h(z) on the compact set nzc~ 1 fn k ( { lz - zo l :s; r}) =I= 0) it follows that lim rmjk

k-; oo

0

fn jk (zo) = h(wo)

118

Chap. 5

Fundamental Theory

holds. Combining together,

h(w) = w,

w E 9 (n) ~ n.

By the interior uniqueness theorem, h(w) = w throughout on n. Yet (3) in (5.3.4.11) says, under this circumstance 1 that f( z ) is both univalent and onto n , a contradiction to our assumption. In conclusion, under our assumption, any convergent subsequence of f n(z), n > 1, should converge to a constant function. In addition to the previous assumption, suppose that f (z) has a fixed point z0 in n. Since fn(z 0 ) = z 0 holds for n > 1, by the last paragraph, any convergent subsequence of fn(z) , n > 1, ·will always converge locally uniformly in n to the point zo. So does t he original iterate sequence .fn(z), n > 1, as indicated in Exercise A(3) of Sec. 5.3.3. As a matter of fact , the boundedness of n can be replaced by the assumption that the complement C - n contains at least two distinct points. Under this circumstance, fn(z), n > 1 1 form a normal family (see (5.8.3.1)) which means that any subsequence of fn(z) contains another subsequence converging locally uniformly in n. In order to prove this claim, observe that the a.fore-mentioned fmi (z), j ~ 1, has a subsequence .fmh, (z), k > 1, converging locally uniformly in n to oo or to an analytic function h(z). Since, for a fixed point zo E n, the set K = {h(zo) u {fn j (zo)lj ~ 1} is a compact set in n, the functional relation fmik 0 f n j k = fnh+i shows t hat fm j,, (z), k > 1, cannot converge to oo in K but converges to h(z) instead. Consequently, fmh, (z ), k ~ 1, converges locally uniformly in n to h(z). And finally, the same process from (*13) to ( *14) will prove the claim. Summarize the above as Th e cases that f (z) fails to be 'l.tri,ivalent or onto. Let n be a domain in C whose complement C - n contains at least two distinct points. Suppose f : n --t n is analytic, neither univalent nor onto. Then,

(1) any convergent subsequence of the iterates fn(z), n > 1, converges pointwise (and hence, locally uniformly) to a constant function inn; (2) in case f(z) has a fixed point zo inn, the iterates fn( z ), n > 1, themselves converge to z 0 locally uniformly inn. (5.3 .4.12) Recall that (4) in (5.3.4.11) is an obvious consequence of results stated here. More concrete result can be obtained if the domain n is an open disk. \Ve state as Th e case (5.3.4.12) where n is the open unit disk. Let < 1} be analytic, neither univalent nor onto. Then 1

f : {lzl < 1} - {lzl

Sec. 5.3.4

An Application

119

(1) either f(z) has only one fixed point zo in lzl < 1, and the iterate sequence f 11 (z) converges to zo locally uniformly in lzl, (2) or f(z) does not have any fixed point in lzl < 1, and fn(z) converges to a constant locally uniformly in izj < 1 such that lfn(z)I --+ 1 locally uniformly, too. (5.3.4.13)

(1) is an easy consequence of the uniform boundedness of f 11 (z), n 2 1, (5.3.3.10) and (5.3.4.12). A complete proof of the former part in (2) needs to u se the concepts of L ebesgue measure zero and Fatou's radial limit theorem (see (5.1.3.4)) and one may refer to Ref. (12] and, Craig and Macintyre: Inequalities for functions regular and bounded in a circle. Pac. J. Math. 20

{1967), 449- 454. For the later part of (2), see Exercise A(9) . Via Riemann mapping theorem (see Cha.p. 6), these results are still valid in a simply connected domain n ~ C. For extension to multiply-connected domains, see M. Heins: On the iteration of functions which are analytic and single-valued in a given multiply-connected region. Am. J. Math. 63 {1941), 461 - 480. E xercises A

(1) Let f(z) =

{~a~ 1

- 1 < a < 0. Show that f has an attractive fixed

point at 1 with mult iplier i:!:~ and a repulsive fixed point at -1 wit h multiplier i+~· Also, show that fn(z) --+ 1 in lzl < 1 by the following

two methods:

:+i

:+i

1. Rewrite w = f(z) as = i~~ (see (*3) and (*4)). z -a h d a+an - 1 > 2 . Show th at f n (z ) = l-a n z , 'vV ere ai = a an an = I + aan - i, n _ 2, and then try to see if limn ~oo an exists. 11

(2) (3) (4) (5)

Prove (5.3.4.3) in detail. Prove (5.3.4.4) in detail. Prove (5.3.4.5) in detail. Let f(z) = z 2 - 2. (a) Show t hat f(z) has a superattractive fixed point at oo . Also, (() = ( + ~ (1(1> 1) conjugates f(z) to g(() = ('2 so that the basin of attraction of oo for f(z) is t he set A(oo) = C * - [-2) 2]. (b) The set of nonnormality, namely, ~'f(f) = { z EC*

I the

iterates f 11 (z) , n > 1, do not form a

normal family}, is called t he Julia set of an analytic function f; and the complement of it is called the Fatou set F(f ). F(f) is open while 0, and a + /3 = 1. Set

t

{J

f( z) = az + -; '

z

E ft =

C - {0}.

Shovv f n(z), n 2 1, converges locally uniformly to 1 in Rez > 0 a nd to -1 in Re z < 0. Determine the basins of attraction , Julia set, and Fatou set of f. In case f (z ) is redefined as az + ~' where a and /3 are nonzero complex numbers (refer to Exercise B(l) of Sec. 2.5.5) , what happens to results obtained in the previous case? (8) Let n - {zllzl < 1, TI.ez > 0 and Imz > O}i the pa.rt of lzJ< 1 that lies in the first quadrant. Show that w = f(z) = e ~1l'iZ maps n into n a nd has a fixed point in n. Use this to show that in converges as n --t oo, where i1 = i and in+l = ii" = e 1 1l''iin, n z 1. (9) Prove that lfn (z)I --t 1 locally uniformly in JzI < 1 as stated in (2) of (5.3.4.13) . Exe rcises B (1) Let n be a domain whose complement C - n contains at least two distinct points. Let a E n be a fixed point. Set Gn = {an analytic function f mapping

n onto itself univa.lently

and preserving a fixed}. (a) Show that Ga is a group under composition and is group isomorphic to a subgroup of the multiplicative group Jzl = 1. Hence, Ga is abelian . (b) Suppose fn(z) E Ga , n 2 l , and converges to a function f(z). Show that f( z) E Ga . (c) In case f n(z) E Ga, n 2 1, and f/i(a ) converges, t hen f n(z ) converges to a function f( z) E Ga. (d) Define a mapping : G a --t {Jzl = 1} by i!>(f (z)) = J'(a) .

Then is continuous (namely, once f n ( z) --t f (z) locally uniformly, then so is f~(a) --t J'(a)) and so is - 1 : (Ga) --t Ga. Consequently,

Sec.

5.4

Meromorphic Fimctions

121

G a can only be isomorphic to

1. eit her a finit e cyclic subgroup of 2. or t he whole unit circle lzl = 1.

lzl =

11

Not e. By Riemann mapping t heorem (see Chap. 6), in case n is a simply connected domain whose boundary has at least two distinct points, G a is necessary isomorphic to lzl = 1; on the contrary, Aumann and Carath eodory (1934) I)roved t hat Ga is isomorphic to a finite cyclic subgroup if n is not so. (2) (Shields, 1964) Let ~ be a family of functions f(z) satisfying the following properties:

f(z): {lzl ::; 1} ~ {lzl ::; 1} is continuous; 2. f( z) is analytic in lzl < 1, and 3. f o g(z) = go f(z), lzl ::; 1, for all such f (z) and g(z). 1.

Show that members in ~ have a common fixed point. (3) Let f : n ~ n be analytic, where n is a domain whose comple1nent c - n contains at least two distinct points . Suppose there is a point zo E n so that the sequence f 11 ( zo) converges to a point WO in n. Then 1. either f is univalent and onto n, 2. or f(w 0 ) = w 0 and t he iterate sequence f n(z) converges to w0 locally uniformly in n.

5.4

M eromorphic Functions : Mittag-Leffie r's Partial Fractions Theore m

It might be constructive and instructive to learn Sec. 5.5.2 (in particular , (5.5.2.10) ) in advance rather t han t he material in t his section, at least, for logical reasons . But for some technical problems in handing the infinite product representations concerning entire function with simple zeros (see Examples in Sec. 5.5.2 and the content of Sec. 5.4.2), it is harmless, for pedagogical purpose, to sketch a head some propert ies of meromorphic functions. A meromorphic function f(z) in the 'vhole plane can be classified as one of t he following three types according to its behavior at oo: 1.

f (z) h as a pole or removable singularity at oo. {::} f (z) = ~f~~, the quotient of two relatively

prime polynomials (see Example in Sec. 2.1 , t he rational functions in Sec. 2.5.3 and Example 6 in Sec. 4.10.2).

Chap. 5

122

2:J=

Fundamental Theory

9jC ~a)' where g(z) and 9j(z) , 1 ~ j < k , are polynomials (see the partial fractions expansion in (2.5.3.5) and Example 6 in Sec. 4.10.2).

f( z) = g(z) + c +

?

1

2. f(z) has an essential singularity at oo (a transcendental one, and hence, has only finite many poles in C).

J(z) = g(z) + 2::;=l 9jC.!a_), where g(z) is a t ranscendental entire function; 9j(z) , 1 ~ j ~ k~ are polynomials.

?

3. f( z) has oo as the limit of its poles (a transcendental one).

(5.4.1)

The main theme in Sec. 5.4.1 is to expand fun ction f(z) of type 3 into a convergent series whose terms are principal parts Rn ( z-1a,, ) at poles an, n 2 1, such as (5.4.2) where g( z) is entire; Pn (z), n > 1, are suitably chosen polynomials acting as convergence-producing terms. Under some constrained condit ions on the growth of f (z), Cauchy used the residue methods to show that g(z) in (5.4.2) indeed is a polynomial. This is shown in Sec. 5.4.2, in addition to some illustrative examples.

M ittlag- L effler's partial f ractions exp ansion f or meromorphic f unctions

5.4 .1

Given a sequence an, n 2 1, of complex numbers with limn_,00 an = oo. Let R n ( z), n 2 1, be polynomials without constant term. We try to construct a meromorphic function in C with poles at an, n 2 1, and the corresponding singular or principal parts R,i C .!aJ, n 2 1. Suppose an =I= 0 for some fixed n . It is well-known already (see (3.3. 2.4)) that 1

--- =

z - an

1

00

an

k= O

z - -2=(- )k converges absolutely an

and locally uniformly in ~

( _1 z

a 11

lzl < Ian I·

(*1)

)l can be represented as a power series converging absolutely and

locally uniformly in lzl < lanl for l 2 1. ~ Rn( z_!an) can be represented as a pmver series converging absolutely and locally uniformly in lzl < lanl·

Sec. 5.4.1

123

Mittlag- Leffier's Partial Fractions Expansion

Let Pn(z) be the first kn-t h partial sums of the power series expansion 1 of R n( -z-an - ) so t hat

IR n (

I

1

z - an

1-< ~ 2

) - p n (z) I

n

Try to show that this Pn(z) will work as claimed in (5.4.2). To see this) let R > 0 be given. There is an integer N

~

1 so that

l an l > 2Rfor n~N .

Now, consider t he closed disk Iz I n ~ N . By (*1) and (*2))

< R on which Iz I
1 -

i

Sec. 5.4.1

where

B 2k

Mittlag- L effler's Part-ial Fractions Expansion

129

are Bernoulli 's numbers. In particular, deduce that

Exercises B Generalized Cauchy 's integral formula (2.9.13) (or (4.11.1.8)) can be used to prove a weak form of the .fVIit tag- Leffier's theorem (5.4.1.1). In (1)- (3) that follows, readers a re supposed to be familiar wit h the content of (2.9.13) and its meaning or its proof. (1) (One -dimensional B-problem) Let

JzI (remind that z is a fixed point in Int /in) . Under this circumstance , _1

{

2rri ./1 n

f(() d'"' < .2_. Nfd~L('Yn) (P+l (( - z) "I - 27r ~,+ 1 (dn - Jzl)


1, namely , supn~ l SUPze rn lf( z)I < oo.

Sec. 5.4 .2

Cauchy's R esidue Method

145

Then

L rng(a + n);

lim S m = 1n~oo

n

where Sm denotes the sum of residues of g(z)f(z) at its singularit ies in Int 'Ym, and L n denotes t he sum over these n for which a + n is a simple pole of .f (z) yet g( z) is analytic at a + n . (2) Imitate :Method 2 in Example 2 and Exercise Example 1. (3) Set Cn : Jz J = n

+~

(1) to redo

for n 2:: 1 and redo Example 2.

(4) Try t o use at least two distinct methods (say1 model after Example 2) to prove the following expansions.

+ L.., ~= n =l

1- = l (a) -tanz z

(b)

=

_ 1_ cos z

7r ~ oo

L..,n = l

.

00

sen es

(- l)n

( -1r -

L n =l

2z . [ (2n 2' 1) ro J2-z2 2n-1

z2- [( 2 n ;

1 )1r

=

L~=l z 2 +(c:?:;il"J2 .

(e)

= ~ + 2z L~=l z2~n2.

coth 7rZ

1 ( f) Sl'i1IiZ

~=

1

= z + L..,n = l

7r 2

~00

(g)

cos2 rr z

(l1 )

_ z_ _ e -" - 1 -

cot ZU: ? 2

\i\Th at is

~ sec

7r{ ? Sum the

· 4 ~ 00 1 ;: w n =l z+( -1)"'(2n -1) ·

(d) tanh z = 7r

. ]2 ·

2L 2

1

2n - 1

( c) tan 7r2z I sec rr2z

What is

( -1)"2z z 2+n 27r2 ·

1

= L..,-oo ( z - l - n ) 2 · 2

1_

~

2

2 + L..,n ~ oo 2z =l z 2+ 4n:trr 2 ·

(5) Show that 1

1

sin z sinh z = z2

+

~ (-l)n4n7rz 2 L..t (z4 - n47r4) sinh n7r ·

n= l

And 1 then , 1, be a syst em of circles or squares, with centers at 0 and passing no poles, such that 1. ''in~ lnt / 'n+l, for n > 1, and 2. Int 'Y"i contains only poles z1, ... , Zn but not others .

In case t here are constants }.If > 0 and a: < 1 so that If (z) I ::; J\lf Izla along 'Yn for all sufficient ly large n, then

1) ,

00

f( z) = f(O) + L Res(f (z); Zn) n =l

+-

1

(

Z -

Zn

Zn

Then, try to prove the following expressions, where - 1

(a)

7re'""~ _ '\"'oo ( - l)"(zcosn7ra- nsin n7ra) si nh 1l"Z - LJn =-00 z2+n2 .

7rc_osh 11a.z s111h7rz .

= '\"'oo

(c)

( - l)nzcosn7ra w n =-oo z2 +n2 . '\"'oo (- l)''nsjn n7ra 1f sin 1faZ CSC 7f Z = w n =- oo z2· - n2 ·

(d)

1f COS11aZCSC 1fZ

(b)

'\"'oc

= w n =- oo

(- l )nzcosn7ra z 2-n2 ·

< a
1, Zn # Q for n > m, and if the partial products Zm+l · · · Zn (n > m + 1) converges to a nonzero finite number p as n - 00 . In this case) rr ~=l Zn is said to converge to z1 · · · Zrn P ) call t he product of z1, z2, ... ) Zn, .. ., and denote it a.s 00

II

Zn =

z1 . . . Zm P

n=l Note t hat rr~=l Zn = 0 if and only if some of its terms is equal to zero. On the contrary, if Zm+ l · · · Zn - 0 as n - oo or does not converge to a finite limit, then rr~= l Zn. is called divergent. Suppose IT~ Zn is convergent. Let rn be as above. For n > m + 1) Zn

=

Zm+l · · · Zn-lZn Zm+l · · · Zn-1

-

P l = p

-

as n - oo .

So it is preferable to rewrite (5 .5.1.1) in t he form 00

II (1 +an)

n=l and an -

or

II (1 + a

11 )

(5.5.1.2)

0 becomes a necessary condition for convergence.

Section (1 ) Con ver gence a nd a b solute conve r gen ce of infinite product o f number s

It is convenient, in practice, to use tests of convergence for series (see Sec. 2.3) to check if a given infinite product is convergent or not. \¥ithout loss of its generality, v,re assume henceforth that each an # -1 for n > 1 in (5.5.1.2) during the process of proofs in the following. First, let P,i = (1 + al) ·· · (1 +an ) a nd Sn = ~~=l Log(l + ak), where, for n ~ 1, Log( l +an) with -7r < Arg(l +an) ~ 11 , is the principal branch

Sec. 5. 5.1

Infinite Products

151

of log(l + an) · Then

Pn =

eSn

shows that. if S 11 - Sas n - oo) then Pn - e8 and fl(l +an) converges to the product e8 . Is the conver e t rue'? To see this, let limn-.oo Pn = P (# 0, oo). Choose the principal branch Log P ; -n < Arg P < 7f. It is possible (see (1.7.3)) to choose arg Pn so t hat Arg P - 11 < arg P11 < Arg P + 7r if n is large enough. For such n, then n

S n=

L Log (1 + a.1,;) = logPn + 2h n1fi,

logPn = log IPn l + i argPn,

k= l

where hn i::> an integer to be determined later on. Hence Log (1 + an+l) = STI+l - Sn = log Pn+I - logPn =? Arg

(1 +an+ I) = arg Pn+l - arg Pn

+ 2(hn+l

+ 2(hn+I -

hn) rri

- hn)7r.

Since Log (1 + an+1) - Log l = 0 and logPn+l) logPn - Log ? , there is an integer no > 1 so that IArg (1 + an+1) I < 23) I arg Pn+I - Arg PI < 23, and Ia rg Pn - Arg PI < 2; if n >no . Consequently)

2l(h..., 1 1

-

h ... )111" < 3 ·

2

7r

3

= 2n, n >no

I

< 1. n >no . Since h are integers, h,1 +1 = hn for n > no. Let h = h =? (hn-.I - hn) I

11

11 ,

n > no. Then

limn~ Sn = Log P + 2h11i and l:~=l Log (1+ zn) converges (to t his li1nit) . Combining together, we have proved the following

Equivalent for convergence of infinite product. 00

IJ (1 +an) converges (to P ;if 0). n=l 00

~ the series

L Log (1 +an) converges (to Log P + 2h7ri n=l

for some integer h ). where each Log (1 +an) is t he principal branch of log (1 +an), n > 1. (5.5.1.3) Ca n (5.5.1.3) be ftuther reduced to the convergence of 2::~ 1 a11 '? Si1nple examples say t hat it is negative in general. For instance,

g n:

1 = 1, while

I; (n:

1

- I) diverges to - oo:

Chap. 5

152

Fundamental Theory

.~ rtoo=l (r\ ( -

diverges if 0 < p < 1 , while L

2

ln)Pn+ 1 )

converges (see Exercise A(3)(e));

!!

oo (

1+

(

-

l)n+l) n

converges

oo ( ( -

and L

n=2

1 In+ i '

)

n

converges (to what sum?) . However , in t he realn1 of absolute convergence, the answer is in the affirn1ative. To justify t his claim, note that

. Log( l + z) 1nn = 1 z

z --+0

~

For

i::

> 0, there is a o= o(s) > 0 so t hat, for 0 < lz l < o,

(1 - s )lz l < ILog (1 + z)I < (1 + s) lz l. Applying this inequality to each Log (1 + Zn), n > 1, it follo'vvs irnm~diately t hat 2::= lznl and 2::= ILog (1 + Zn) I converges or diverges si1nultaneously. Also, the inequality 1+ x

~ la1I


0

+ · · · + lan l < (1 + la1 I) · · · (1 + Ian !)< elail+·+lanl ,

n

>

1.

(*2)

Hence, L lan l and fl (l + lanl) converges or diverges sin1ultaneously, too. \Ve smnmarize the above as Th e absolute convergence of infinite prod'uct.

fl(l + lan l) converges. {:} 2. L lanl converges. {:} 3. 2::= ILog(l + an) I converges. l.

In this case, fl (l +an) is said to be absolutely convergent. Moreover, absolute convergence has the following properties.

(1) Absolute convergent infinite product does converge. (2) The order of terms in an absolute convergent infinite product can be altered arbitrarily without affecting the absolute convergence a.nd the product. (5 .5.1.4) The last two statements can be proved by known results from series and the very definitions introduced above.

Infinite Products

Sec. 5. 5.1

= fl~=l (1 + ak)

Sketch of proof. (1) let Pn Then

153

and P~

= fl~=l (1 + lak I).

"'"""' (~i - P~_ 1 ) = n-li1noo P~ converges.)

=> (Since L

n=l

°"'(Pn - Pn-1) = n---.oo li1n Pn = P converges (absolutely). L

n=l

= 1 and E lanl converges,

To see P =f. 0, observe t hat, since li1nn---.oo( l+ an)

=?

(by statements 1 and 2, a nd t he result just proved)

II (1 n=l

an

=

)

1 +an

l"Im rr n n-oo k = l

(i

-

ak

)

1 + ak

=

l"lffi n-

nn (11 + k=l

ak)

1

= p (=f. 0, oo) converges. Hence P =f. 0 holds. (2) Let 7r: {1,2, ... ,n .... } - t {1.2, ... ,n., ... } be a permutation (namely, one-to-one and onto n1ap) . Since I: lanl < oo =?I: lc7r(n)I < oo holds, by (1), then fl (l + a7r(n)) = P' converges. Fix any n, choose 'm large enough so that p~i = n~:l (1 + a11(k)) contains all the factors appearing in Pn . Then m > n and

< elaa.1 l+ .. +la"'k I - 1, < el::~n-1 la, I (and hence rn

=? -P' p

= 1 or

This fini hes the proof.

P'

=

by ( *2)

1 -+ e0 - 1 = 0 -t

a n.

-+

oo

)

P. 0

154

Chap. 5

Fundamental Theory

vVe illustrate two exan1ples. Example 1. If L n=l absolutely.

lznl 2

converges, t hen f1 n=l coszn converges

-!·

Solu t ion. Note t hat lin1z_.Q cosz1-1 = lin1z~O -;~1 z = Hence I.: lznl 2 and I.: I cos z11 - II converge sin1ultaneously. It follows by (5.5.1.4) t hat the product converges absolutely. Example 2. The product

oo (

II n=l

1+

(-l) n- 1) nz

converges in the half-plane Rez

n z

'

= ez logn

t and absolut ely in Rez > 1.

>

Solution. Using Log (1+ z ) = z - t z 2 + · · ·+ (- l )n- l ~ zn+ · · · , lzl Log ( 1 + Zn

=

(- l )n-1 )

= Zn + R,i(z),

nz

(- l )n-1

and

Ptn(z)

>

k

=L

(- 1)1•- 1 ; ,

k= 2

if n is chosen so t hat In case Re z

where

00

nz

< 1, let

lzn I < 1.

~. then

lznl =

1

1

nRez

=> IRn(z)I
4

f

~lzn lk = lznl2 f ~lznlk- 2

k=2

k=2

1 + 1 + · · · ) < 2lzn l2 if n > 4. < lzn I2(1 + 2 22

Hence, 00

L Zn and L lznl n=l

2

converge if Rez

> ~;

n=l

00

L lznl converges if Rez > 1. n= l

oc

=>

L IRn(z)I < 2 L lznl n=4

2

< oo if Rez >

~·

n=l

The claimed results fo llow fro111 (*4), (5.5.1.3), and (5.5.1.4).

Sec. 5. 5.1

Infinite Products

155

S ection (2) (Local) Uniform conver gen ce of TI (1 + fn(z)) Let us turn to infinite product of complex-valued functions. Suppose fn (z) : n (domain) ~ C . for n > 1. If Zo E n such that f1 1 (1 + fn(zo)) converges (see Example 2 above), then t he infinite product

II (1 + fn(z))

or

n=l

II (1 + fn(z))

(5.5.1.5)

is said to converge at z 0 . The set of points in n where (5.5.1.5) converges is not necessarily open or connected, in general, and is called the domain of convergence of t he product. The prodttct J(z) defined on this domain of convergence (if exists) is a function and is denoted by 00

J(z)

=II (1 + tn(z)) n=l

or

II(l + !n(z)).

(5.5.1.6)

If the partia l product Pn(z) = TI ~=l (1+ f k(z)) converges un ifor111ly to f (z) on a subset of n , t hen (5.5.1.6) is said to converge uniformly to f(z) on the set; and converge locally uniformly to f (z) in the domain n if it converges uniformly to f(z) on each compact subset of n. (5 .5.1.3) and (5 .5.1.4) have obvious counterparts for (local) uniform convergence (namely. these results are still valid if convergence in (5.5.1.3) and absolute convergence in (5.4.1.4) are replaced by uniform convergence, respectively). See Exercise A(8) . Anyway, we have a concise N ecessary and sufficient condition for absofote and uniform convergence.

(1) TI ~=l (1 + f n(z)) converges absolutely and locally uniformly m a domain n ~ (2) E :=l fn(z ) converges absolutely and locally uniformly inn. Let

f (z) = f1 (1 + f n.(k ))

inn, in t his case. Then ,

1. z 0 E n is a zero of j (z) ~ z0 is a zero of at least one factor 1 + f 11 (z) in

the product; a nd 2. The order of zero off (z) at z0 is equal to the sum of the orders of zeros of the finitely many factors having z0 as their common zeros. It is tacitly understood henceforth that absolute convergence and local uniform convergence a re considered only on subsets of n where t he zeros of f(z) are deleted. (5.5.l. 7)

156

Chap. 5

Fundamental Theory

In case lfn (z) I is sufficiently s1nall (if n is large enough), then ( *1) sa.ys

(1 - c) Ifn (z) I < I Log (1 + f n (z)) I < (1 + c) If n (z) I n+p

~ (1 - s)

n+p

L lfk(z)I < L

k=n

n+p

ILog (1

+ fk( z)) I < (1 + s) L lf k(z) I, p > 0.

k=n

k=n

And (5.5.1.7) follows obviously. Alternatively, one can imitate (*3) to present another proof. The details are left to the readers . What is important for our purpose is the following

Infinite product of analytic functions . Let fn(z) : D (do1nain) --t C be analytic for n > 1. Suppose 1 f n(z) converges absolutely and locally uniformly in D.

I:.:

Then, converges absolutely and locally unifonnly in n to an analytic function f( z) (recall that the behavior of convergence is performed on the set D - {z lf(z ) = O} ). (2) Also,

(1) rt~=l (1

+ fn(z))

f' (z)

=

f (z)

f .+f~ ( ~

n =l

1

z ) absolutely and locally uniforn1ly in fn z

D- {z if(z) = O} .

(5.5.1.8)

Sketch of proof. (1) is an easy consequence of (5.5 .1.7) and vVeierstrass's theore1n (5.3.1.1) . 00

Set f(z)

=

IT (1 + f n(z)) = (1 + f1( z )) · · · (1 + fn(z) )

Fn(z),

where

n =l 00

F,i(z) =

IT

(l + f1;;( z) ),

z ED.

k=n+ l

Choose any co1npact set J( c D - { z jf (z) = O}. By logarithn1ic differentiation (see Exercise A(8) of Sec. 3.3.1) , t hen

f'(z) ~ f{( z) f(z) = ~ 1 + fk(z)

F~(z)

+ Fn(z) ' z

E ]{.

Since Pm(z) = (1 + fn+1(z)) · · · (1 + f m(z)) converges to F,i(z) uniformly on J( as m --t oo, so does P:n(z) to F~(z) on J( , again by (5.3.1.1) .

Sec. 5. 5.1

Infinite Products

157

Then

F~(z)

P~i(z)

"°"'

~ f~ (z) j~(z) Fn(z) - m~ Pm(z) = m~11 . ~ 1 + fk(z) = ~ 1 + j,,(z)

r

1

li

k=nTl

~ =n+l

unifonnly on J(. =?

~(~j

=

f

~~J~iz) pointwise (and, indeed , locally uniformly)

1

n=l

= O}.

inn - {zl f( z)

\!Vhat remains is to prove t he absolute and local unifonn convergence of the above right series inn- {z lf(z) = O}. Let I< be a closed disk lz -z0 I < r, contained in D - {zlf(z) = O}. There is a n no > 1 ::>O that lfn (z) I < ~ on ]( for n >no; in t urn, ll + fn(z ) I > on J( for n >no . Since f( z ) # 0 on K , so do 1 + f n(z) # 0 for 1 < n < noi and there is some ex> 0 so that ll + fn(z)I >a on J{ for 1 < n Al ;

z E I 1

< Alf lJ+,n' (z)I.'

z E

-

J(.

n > 1.

Try to show that L lf~(z)I converges uniformly on I - z) 2 · 1(-zo l-

1

.

1

dr

lz - zo l
l

and that the series L~ e-::~e: converges if and only if Re z > 0. Hence) the product converges absolutely and locally unifonnly in Re z > O. (5) _Jote that lim

n-

l. · 11

sin z/n sin z/n = lim = 1. z/n2 n~oo z/n

The product converges absolutely and locally uniformly in C.

Sec. 5. 5.1

159

Infinite Products

Exe r c ises A

(1) Suppose an =/= 0 for n > 1 a nd designate -7r < Argan < 7r. Show that IT: =l an and I::: =l Logan converge or diverge simultaneously. \i\That happens if 0 < Arg an < 27r is assumed for each n > 1? If a< Arg an< a+ 27r (a:< O)? (2) Suppose both IT an and IT bn converge. Determine if the following products converge or diverge: (a) IT(an + bn). (b) IT(an - bn)· (C) IT a 11 bw (d) IT ~ · (3) Prove t he following staternents. (a) Let an> 0 for n > 1. Then IT (l + an) converges {:::? I:: an converges; and IT ( 1 - an) converges {:::? I:: an converges. (b) Let 0 < an < 1 for n > 1. Then I:: an diverges =? IT(l - an) diverges to O; but :Z::::: an a.nd f1(1 + an) diverge or converge si1nultaneously . (c) If I:: an , I:: a:1 , . . . , I::a ~- l (k > 2) and I:: lanlk converge, so does IT (l +an) · (cl) Suppose an E R for n > 1 and I:: an converges. Then IT(l + an) converges or diverges to 0 according to IT a~ converges or diverges, respectively. (e) Suppose an E R for n > 1 and I:: an converges, while I:: lan l diverges. Then IT (1 + an) converges {:::? I:: a~ converges. (f) Suppose L Ian 2 < oo. T hen IT (1 +an) converges {:::? Lan converges. 1

(4) (a) Letan = (-l)n-l'* , n> 1. Show that :Z::.::an convergesand I::a~ diverges, while IT (1 + an) diverges to 0. (b) Let a2n-1 = J~~l a nd a2n = J n\ l + ~ + 1 for ri > 1. Show that I:: an a nd I:: a~ diverge, while IT (l +an) converges. (5) Prove the identity sine 2k sin~' IT~=l cos 2~ for k > 1 and then , deduce that IT: =l cos 2~. e. Also, show that 2

2

2

sie

_

72 . J2+h . }2+V2+72 ... -

7r

2·

(6) Show that IT:=i e ~ (1 + ~ )- 1 = ei·, where / = lirn 11,_. 00 (I::~=l ~ logn) is the Euler;s constant. (7) Find the don1ain of absolute and local uniforn1 convergence for each of the following products .

(a) IToo n=l (1 - z n) . (b) IT: =2 [l - (1 - ~ )-nz -n} .

Chap. 5

160

Fundamental Theory

2

(c) rr:=l [ 1 + (1 + ~)n zn] . ( 1, converges absolutely a nd locally uniformly in C. ( *1)

Chap. 5

162

Fundamental Theory

To this end, choose any fixed R > 0. There is an integer N so that lanl > R if n > N (and thus, lan l S R if n < N) and we con ider only these tenns with la 11 I > R. T hen -

z

an

::::;. Log

lzl < R

if

< l

(1- _:an_ )

Pn (z) = _:_ an

(and 1 of cotU'se, n > JV)

lzl < R ,

-pn(z) +rn(z),

=

+ -1 ( _:_ ) 2 + · ·· + - 1 2

an

mn

( z )

where

11ln

a11

,

and

f= ~ (: )

rn(z) = -

k

71

k=·mn+l

The remainder r 77 (z) h as the estin1ate, in absolute value, 1 ( R )

lrn(z)I < k=~+l k lan l ( R )

1

mn

< -

+1

1nn

1(

nn ~

1

1

( R + 1 la11I

)mn-'-1

R - an l

1

111

lanl

< mn + 1

lanl

2

( R )

1

k

n

+

(

R )

k-1

{; Ian I

)-1

if l~il > 2R

for n large enough.

(*2)

It is possible to choose mn so th at t he series

2: mn l + 1 ( -anR ) mn+l 00

-

n=l

1

converges for any fixed R > 0.

(5.5.2. 1)

1

Say, let 1nn = n for n > 1. For any R > 0, since limn-+ lan l = oo, then Ian I > 2R will hold for all sufficient ly large n , say n > k . Then, in this case, 1

~ n+ l

( R ) n+1

lanl

00

0 or

n

n=l

IT (i - ~) : an

n= l

if h = 0, converges absolutely and locally uniformly in C to an entire fun ction.

(5.5.2.5)

Assume that h is the smallest nonnegat ive integer for which the series (5.5.2.3) converges; then t he corresponding expression (5.5.2.5) is called the canonical product 11ssociu.tcd wit h t h e sequence a11 , a.nd this h is t he genus

of the canonical product. Note that the canonical product associated "vith given an , n > 1 is thereby uniquely determined. See Exercise B(l) for a generalized version of this theoren1. Suppose the infinite product in the expression (5.5.2.2) for f (z) is canonical with genus h and g(z) reduces to a polynomial: the function f (z) is then said to be of finite genus and the

genus of J(z) = inax{h: degree of g(z)} . (def.)

(5.5.2.6)

Chap. 5

164

Fundamental Theory

For instance, an entire function of genus zero: genus one:

if Ae°'z

ii (i - aZ?i)

zm

e

1

L

lanl < oo;

*

if L111 - = 00 CLn

n=l

and

L

Aea:z

zm

1

lanl2 < oo ,

ii

or

(1 - ..:_)

n =l

if

an

L

-I

1

an

I

< 00.

(5.5.2.7)

Readers are urged to write out forms of entire functions of genus t hree. Also see Exercise A (11). vVe list two sirnple corollary of (5.5.2.2). As an easy consequence of (5.5.1.8) , we have The logarithmic derivative of a canonical product. Given a canonical product 00

C(z) =

IT En(z)

where En(z) =

n=l

if h > 0 or En(z) Then, in C -

=1-

• • )/t 1- -Z ) eii':i + l( 2 ; - )2 , .. +1i.l( 7i':i an t

(

-

z

an

if h

= 0.

{a1,a2, ... , an, ... } ,

C'(z) __ ~ E~(z) L C(z) n =l En(z)

absolutely and locally uniformly.

(5.5.2.8)

In case iVI(z) is ineromorphic in the whole plane, by (5.5.2.2); there exists an entire function f 1 (z) with the poles of lvl(z) for zeros. The product JVI(z)f 1 (z) has removable singularity at each pole of iVI(z) and hence is an entire function f2(z). It follows that l111(z) = J~t~~. Then we obtain Th e expression for rneromorphic function in C. A function is meron1orphic in C if and only if it is the quotient of two entire functions (with the den01ninator not identically equal to zero). (5.5.2.9) As an easy consequence of (5.4.2.2) and (5.4.2.3)) \•ve have the following counterpart for infinite product. A special case of Weierstrass 's factorization theorem (5.5.2.2) . Suppose an entire function f(z) satisfying /(0) =/= 0 and having si1nple zeros a1 , a2, ... ) an , ... so arranged as 0 < la1 I < la2 I < · · · < lanl < · · · that lin1n ..... oo lanl = 00.

Sec. 5. 5.2

Weierstrass's Factorization Theorem

165

(1) Let l'n be as in (5.4.2.1) with properties 1) 2, and 3 so that t' (() . .LJ.il limn-+ J," ((-z) /(() d( = 0 for z E Int {n - {poles of Kif} . Then

f( z) = f(O)

IT (1- az ) locally uniformly in C- {a a 1.

n=l

2 , .. . }.

n

(2) Suppose ~{:,> satisfies either condit ion 1 or 2 in (5 .4.2.2). Then

.!!

z)e;;;-.

00

£iQl

f(z) = f(O )eT{(J)z

(

1 - an

locally uniformly in C - {a1, a2, ... }. Note that, in t hese infinite product expressions, factors corresponding to zeros between /'n and 'Yn+l , should be considered as one "term'' in t he product. (5.5.2.10) Proof is left to t he readers. See Exercise A(l ) for application. Four examples are illustrated. Exa mple 1. (Ref.

[lL p. 197). Shows that

= 7rZ Tin=l ( 1 - ~) = 1rZ TI '~=-oo (1 - *) ei" locally uniformly in C , where TI '~=-oo means the term n sin 7rZ

absolutely and

= 0 is 01nitted

in the product. Try to deduce: 1.

2.

rv.vallis. pro duct: ~- = lm1n__.oo . 2242 ... (2n- 2)2 ? 3 2. 5 2.. -( 2n - l )'l · :...:rr; oo (i + :;;:'! z2 1rWZ h . _ + z4 ) _ sin 1rWZ1f2·zsin 2 ' w ere w Ti n= 1 n4

3. in parti cular, TI~= l

-

(1 + :;z + :rt:r)

=

e

~. ;s

)

l±c~~~ v17T .

l\/[ore generally, in case a is not an integer , t hen

IT (1+ n+a z

s in7r(z+a)=sin 7raenzcotna:

) e-,,.+o

n=- oo

. 7ra ( 1 + -z ) = Slll

a

II ,oo

n= _ 00

(1+

z

n

+a

) e _ ni

•

Solutio n . M e thod 1 (based on (5.5. 2.7)). Observe t hat sin nz has simple zeros at integers n (n = 0, ±L ... ). Since L ; = oo while L 111\ 2 < oc, according 1 1

Chap. 5

166

Fundamental Theory

to (5.5.2. 7) , there is an entire function g(z) so that (refer to Example 3(3) in Sec. 5.5.1) sin nz = zeg(z)

II (1 - ~ ef.' absolutely and locally uniformly in C. (*3) n~O

Taking the logarithmic derivatives on both sides> we find 7r

cot 7r z = -1 z

+ g (z) + L..t I

'"" (

n~O

1 z -n

+ -n1)

=> (by Exa1nple 2 of Sec. 5.4.2 or adopting the method u8ed in Example 2 of Sec. 5.4.1) g'(z) _ 0

=c, a constant.

=> g(z)

Letting z - 0 in (*3) and observing that lin1z-.Osi ~;z = 1, we obtain the constant g(z) = c = log n and t he result follows.

M ethod 2 (based on (5.5.2.10)). Set f (z) = si~;z, z E C . f(z) is entire and has simple zeros at z = ±n (n = 1 , 2,3, ... ).Note that the derivative of J(z)

f' (z) 1 ----'- = n cot 11z - -

f (z)

z

satisfies Condition 1 in (5.4.2.2) (refer to :\Iethod 1 in Exa1nple 2 of Sec. 5.4.2) . Hence 1 the required expression follows directly from (2) in (5 .5.2.10), namely,

II' (1- -z ) en....

S lll 71" z = 1TZ

n

n=-

Catttion again that ~i~;z cannot be expressed as IT ''.:'.'.' the divergence of this product. Set z = ~ in ( *4 ) and the \~Tallis product follows. Also, by (*4), 2 2

l _ w z

II (

n.=l

n2

)

= sin nwz; 1TWZ

IT (

1_

n =l

2 2

wz n2

)

(1 - *) because of

= sin ~wz. 1TWZ

The absolute convergence of both products permits us to multiply these two expressions termwise together (see (5.5.1.4)) to obtain the claimed identity. Set z = l in the identity and we get the third one.

Sec. 5.5.2

Weierstrass 's Factorization Theorem

167

Exactly the same methods as before still work to express sin 7r (z +a) as t he claimed product. Note that, in this case, In is the square with center at a; and vertice at a:± (n + ~ ) ( 1+i) for each n > 1. For the econd expression for sin 7r(z +a.), just use Example 2 of Sec. 5.4.2 as follows:

(1 + a~) exp z[1f cot7ra -

sin ~(z +a.) = sm 7f a.

x

exp [- z n

+z

-

n

~]

a

·

IT ' (1 + n_z _) +a

_z_l a+n

1

OO I = (l+ -Z )exp z [ 7rcot7ra- - - '°"' a a -oo ~

x

IT

I

00

1+

(

1 1)

(

n+a

- n

l

z ) .;.. exp- " n +a

Example 2 . Show that COS7rZ

=ii (1 n=O

2

(n

z

+ 112) 2

sin 7fZ = nz(l -z)

) ;

J1

(

z - z2 ) 1+ n+n2 ·

Solution. f( z) = cosnz has si1nple zeros at z = +(n+ ~) for n = 0.1. 2 ... . . Let /'n, n > 1, be the square with vertices at ±n(l + i) . Along the vertical side z = ±n + iy, -n < y < n : 2 I cos n z 1

=.;.

!' (z) f (z)

I sm . n z 12

= cos h 2 rry > 1 ;

2 = 7!' 2 tanh2

ny

. h 2 7ry < smr1 . L 2 nn = sm

< 1f2

-

Along the horizontal side z = x ±in, -n < x < n: lcos7rz l2 >cosh2 nrr-l ;

f (z) f (z) I

lsin 7rzl 2 1. Since f(O) = 1 and f' (0) = 0, it follows from (2) in (5.5.2.10) that CO 1T"Z

=

IT n=-oo

(1 -

z

n

+ 1/ 2

) e-z/(n-1/2)

(absolutely by Exan1ple 3(3) of Sec. 5.5. 1)

168

Chap . 5

fi (

1-

Fiindamental Theory

+z:/2)2)

(n

(see also the explanation in (5.5.2.10)). Set ( = z - ~ to replace z in ( *5) and we obtain

!! Il g 00

cos7r( =

(2

(

1-(n + l/2)2

)

= (1- 4(

=?

00

2

n

+n

(n + 1/

(1

2) 2

(

z- z

+n

2

(2

(·

1 -(n +l / 2)2

)

2

1 [( 1) (n + 1/ 2)2 n+ 2

00

=

)1]

2

00

cos 7r( 1 - 4(2 =

2

z-

1)

2

]

)

+n2

·

Letting z = ( + ~ ~ 0 in ( *6) and observing that .

(~~\'& =?

cos 1r( 1 - 4(2 =

.

(~1~i7

sin 7r( 7r rr oo n 2 + n -8(2 = 4 = (n + 1/ 2) 2

-'jT

n =l

;1 . h f f' ( )) sin 7r z 7rz(l _ z) ( set .., = z - 2' in t e 1et o *6

4 n(l - 4(

rr oo

COS7T(

=

2) = n = l

(l + n + n z

Z -

2 )

2

·

Example 3. (Ref. [l L p. 197). Let an be a sequence of distinct nonzero complex nwnbers converging t o oo and An , n > 1, be arbitrary complex numbers. Find an ent ire function f (z) such that f (an) = An, n > 1. Solution. Using (5.5.2.2) , construct an entire function g(z ) with si1nple zeros at an, n > 1, nan1ely,

g(z ) =IJ 00

(

z 1-;-

~

)

1

~

2

l

~

ea:n+z(;m)+ .. + ;nn (an-)

mn

,

z E C.

n

n =l

Then g' (an) =I= 0 for each n > 1. Observe that the function

g(z) g'(an) . Z - a11 An

has a re1nova.ble singularity a.t an and may thus be assigned the value .4n at z = an . It is entire for each n > 1. Try to choose suitable constants r n so that the series

J(z) =

f

n=l

An . g(z) er,.(z -a,,,) g'(an) z - an

S ec. 5.5.2

Weierstrass 's Factorization Theorem

169

converges locally uniformly in C. To see this, fix any R > 0. There is an integer f\T > 0 so that lanl > R if n > fl.T. On lzl < ~ and n > N , lz - anl > R - ~ = ~ holds; consequently, at least one of Re (z - an) and Iin(z - an) should be not less than ~. Novv

X

or

O:n

rn ,n

=

1

< 2 · max lg(z)I · lzI ::; ~

R

Set

·f rn is · rea;1

e r nRe(z -an ) ,

I 9,(;,.5I for

n

>

An g' (an )

e ir n l m(z -a,,.)

'

if irn is real.

1. Suppose that we are able to choose constants

> 1 so t hat A

oo

~

L

n=N

n

G g

'(an )

ern(z-a)


1, and the corresponding principal parts g,(n )· a.u z -lan ) n

>

1, nan1ely ) 00

NI(z) =

L

n=l

(

An

1

9 (an)

1

· Z -

an

-

Pn(z) ) ,

where Pn(z) , n > 1, arf) suitably chosen polynomials (see (5.4.1. 1)). Then the entire function

f(z) = g(z).!Vl(z) satisfies the condition that f(an) = n > 1.

limz -.a.n

g(z)NI(z) = g';;~~~,, = An,

Example 4. (Ref. [1], p. 198) . Suppose f( z ) is of genus 0 or 1 with real zeros, and f(z) is real for real z. Then all t he zeros of f' (z ) are also real , and between any two consecutive zeros of f (z) t here lies precisely only one zero of f'(z) .

170

Chap. 5

Fundamental Theory

This is a well-known result for polynomial with all its zeros real which can be easily shown by using the mean-value theorem for differentiation. The extension to t he entire function of genus 0 or 1 was due to Laguerre. 2 However, the function (z + l)ez , which is of genus 2. shows that it is no more true for entire functions of genus > 2.

P roof. Such an entire function of genus 0 or 1 can be expressed as (see Exercise A(ll))

where m > 0 and an, n > 1, are real. By assumption that f( z ) is real for real z, a within should be real, too . Differentiating (*io) logarithmically or using (5.5.1.8) and (5.5.2.8) directly, we have

f'( z ) _a+ ni

f (z) -

+

z

t( n= 1

1

+ ~)

z - an

an

'

locally uniformly in

C - {an , n > 1} U {0} . =?

f'( z) ( m ~ Im - ( -) = -(Im z) - 2 + ~ I

lzl

f z

n =l

1

z-

)

O,,,i 12

.

Then f'( z) = 0 implies that Im)(~/ = 0 which, in turn , implies that Im z = 0 hould hold. Hence f' (z) = 0 can hold for real z only. Differentiate ( *11) again to obtain

df'( z ) dz f( z )

m

= - z2

-

~

1

~ (z - an.)2'

zE C - {an,n>l } U{O},

which is real and negative for real z . As a strictly decreasing function from +oo to -oo os z increases through real values between two consecutive zeros of f( z ), 5(~}, and thus f'(z) assumes the zero value once between these two zeros. O

Exe r cises A

(1) Try to adopt the following two methods: 1. (5.5.2.10). and 2. the infinite product expansion for sin 7rZ (see Example 1)

Sec. 5. 5. 2

Weierstrass 's Factorization Theorem

171

to prove the follo,ving expansions. Detennine t he genus for each entire function. (a) sinhnz =nz f1n=l

= z 2 IT~ 1

(b) cosh z - cos z

= ze t

(c) ez - 1

(cl)

eaz -

(1+ ~~) -

Tin=l ( 1 +

(

1+

41:'=t~).

2 ; 4 2 -;r2) .

ebz = (a - b)zef Ca +b)z

rr= n=l (1 +

(a-b

)2z2) .

4n i -;r 'i

(2) Find the product function for each of the following infinite products. (a)

rr~=2

(b)

rr~=2 ( 1 -

(c)

f1 n=l (1 + fi:'z 56' )

(1 +

~) .

s) and sho-w that rr~=2 (1 -

00

and shovv that

~) =

1 ) f1 n=l (1 + -:;}! 00

8; (e7l" -

= 12 .

e-71") .

::;inh 71" (cosh 7f ~

- cos J31f). (d)

rr~=l (1 - ~ ).

(3) Let An be the positive root of the equation tan z = z in the interval (n7r. n + t 7r) for n = 1, 2, . ... Show that 00

sin z - z cos z

=

1 ~ ..>Z

II n=l

(

.,.2 )

1 - ~2

A11

(compare to Exercise A(7) of Sec. 5.4.2) .

(4) Show that sin3 z smz

00

--

II n=-oo

4z2 ) + z)2 '

(

1 - (n1f

z

=I=

nn

for n = 0, ±1, ±2, ....

n;

(5) Show that cos 7 - sin = (1 - z)( l + ~)( 1 - ~) .. . . (6) Determine the genus of cos JZ. (7) If f(z) is of genus h, how large and how s1nall can the genus off (z 2 ) be? ( ) Derive the results of Exan1ple 1 in Sec. 5.4.1 directly from vVeier trass's factorization t heorem. (9) Deduce the periodicity of sin z from its infinite product e>...rpansion (refer to Exan1ple 1). (10) Extend Example 3 to the case where the sequence an contains the number zero.

172

Chap. 5

Fundamental Theory

(11) Show that an entire function f(z) of genus h can be represented in the form

f( z) = zmeg(z)

IT (1- ~) e~+! (in-)2T···-r t. ( «-,. )h, n

where m > 0 and g(z) is a polynomial of degree < h. I\ote that the infinite product, in this case, is not necessarily a canonical one since its genus might be strictly less than h. (12) Try to find a.n entire function g(z) so that ;J~~z1 = eg(z) . (13) Deduce i\!Iittag- Leffler's theorem (5.4.1.1) from Weicrstrass's theore1n (5.5.2.5) . Exercises B (1) A generalized version of Weierstrass's factorization theorem (5.5. 2.2). (5.5.2.2) is still valid in a domain (or open set) n in C. l\/Iorc precisely, let an' n > 1) b e a sequence of points in n without limit points in it, and m11, n > 1, be positive integers. Then there exi. t s an analytic function f : n - C with these and only these zeros at an, n > 1, the zero at a 71 being of the order m n for each n > 1. '\Ioreovcr, the quotient of any two uch functions is analytic and zero-free in n. Hint. l\ Ioclel after the proofs of (5.5.2.2) and Exercise B(4) of Sec. 5.4.1. Refer to Ref. [5 L Vol. I , p. 86. (2) Given a domain n in c , show that there is a function analytic inn but cannot be analytically continued outside n. Hint. Cornpare to Exercise B(6) of Sec. 5.4.1.

5.5.3

Hadamard's order theorem

Let f( z) be a nonconstant entire function and fl/l(r) = n1ax1zl=r If (z)I for r > 0. Then l\!I(r) is a strictly increasing continuous function in r and lin1r_, 00 f\lf(r) = oo holds (see Exercise A (6) of Sec. 3.4.4) . In case f (z ) is a polynomial of degree n > 1: since lin1z- f},~l =F 0, there are constants llfi > M2 > 0 so that llhrn

< l\tl(r) < Alirn for r > ro (some positive constant) . logl\I (r) =n r-oo log r .

=> hm

or l\I (r )rvrn

asr-

(*1)

This is to say. the order of growth of l\I(r) is the same as that of r 11 as n -oo.

Sec. 5.5.3

Hadamard 's Order Theorem

173

Well, in case f (z) is a transcendental function: we do have a remarkable fact that log AI (r) . . logAI (r) . 11m = oo. or eqrnva1ent1y 1im r -oo log r 1ogr r-+oo To see this 1 suppose on the contrary that . log AI (r) lim = a< oo 1:=;, log r =? For each fixed s > 0, there is a sequence rn ----? oo so that logNI(rn) c < a + s 1or n > 1. 1ogrn Exactly as in the proof of Example 6 of Sec. 3.4.2, it fo llows that f(z) is a polynomial of degree < [a], contradicting to the assumption that f (z) is trani:icendental. Consequently, no distinction can be made among transcendental entire functions if (*2) or the power funct ion r 11 is chosen as a measure of rate of their growth. Here comes naturally to compare f\l(r) with er" . If there is a nmnber p > 0 so that

l\I(r) < erP

for all sufficiently larger.

(5.5.3.1)

then f (z) is aid to be of finite order: otherwise, of infinite order. Suppose. for t he moment, that f (z) is of finite order. Set .A = inf p where p's are these that satisfy (5.5.3.1).

For any .A1

(5.5.3.2)

> >., there is an ro > 0 so that whenever r > ro, NI(r)
2· ogrn (S ince .A2 < .A < .A1 , and .A1 and .A2 are arbitrary.) , -.- log logN/(r) /\ = 11n1 . r -+oo logr

NI(rn) >

.x 2

ern

or

(5.5.3.3)

.A 1 defined either by (5.5.3.2) or (5.5.3.3)i is called the order (of growth) of the entire function f(z) of finite order. ~ote that the defini tion (5.5. 3.3) is still Ya.lid for infinite order. For instance,

a polynomial is of order 0 (because p

AI (r) < Arn < er for anyp > O);

174

Chap. 5

Fundamental Theory

Ao

ez (k: a positive integer) is of order k; sin z and cos z are of order 1; co

JZ is of order ~ I cos vzl ""'cosh

(because, setting

(r!sin~)

rv

z

=

reiO)

er 1 sin~ ).

(5.5. 3.4)

For our clai1ned purpose: we need a preparatory result relating the inoclulus lf (z)I on a circle to the moduli of the zeros of f (z) inside the circle. It is the Jensen's formula. Let f(z) be analytic on lzl < r with zeros ai, ... , an (n1ultiple zeros being repeated) in lzl < r and f (0) -f. O. Then n

log l.f(O)I = -

~ log ,:kl + 2~ lo

2n 9

log lf(r ei )ldB.

In general, for lz l < r, r 2 - -akz loglf(z)l=- L: log ( ) k=l r z - ak 11

1

+-

1 211"

2n o

re'•(} + z · Re .8 loglf(rei8 )1dB, re' - z

(5.5. 3.5)

called the Poisson- Jensen formula.

As a matter of fact: Exercise B(l ) of Sec. 3.4.4 shows that lf(O) I < ''~f~) la 1 · · · a11 I and v~e) dp < log (Jc~~, where v(p) counts the number of zeros of f on lzl < p < r . Can you figure these out now?

J;

Proof. First, suppose f (z) -f. 0 on lzl < r. Then log lf(z)I is hannonic in lz l < rand hence (see (3.4.3.2) and (3.4.3.5)) , 1

log If (O)I =

7r

2

{27r

./o

'8

log lf(re'' ) lclB.

In general, we have Poisson 's forn1ula (see Exercise B( 4) of Sec. 3.4.2 or (6.3.2.1)) log lf( z )I = -

1

27r

1 27r Re r eiB. + z log I f(rei.9 )iclB, 0

rei8 - z

lz l < r.

(*3) re1nains valid even if f (z) has zeros on lz I = r. It is sufficient to assume that f (z) has only one zero reiBo along lzl = r . Then zJ1~=)00 has a removable singularity at re iBo and thus, is analytic on lzl < r and free of zeros there. Applying (*3) to this a new function. we obtain 1

log lf(O) I - logr = 27r

1211 log If (rei8)1d0 0

1

-')

~7r

127r log lreiB - r eiOo ldO. 0

Sec. 5.5.3

Hadamard's Order Theorem

175

All one needs to do is to show that

~

log r

= 2_ f 21r log rl ei8 -

fo

log lei9

2n ./0

2 1r

2n- 8o

1

-

eieo ldO

eiBo ldO

=0

log Il - eit/J Id'lj;

or

=0

(with 0 - Oo = 'If;)

- fJo

~

ionlog sin OdO =

- 7r

log 2.

This last integral was established in Example 4 of Sec. 4.12.l. Returning to t he general case 1 the function

F (z)

= f(z)

n

IT r k= I

2

-

- akz

r(z -

ak)

is analyt ic on izl < r, free fron1 zeros in lzl < r, and IF(z)i = lf(z)I on lzl = r. Applying ( *3) to t his F (z), the required ident ity fo llows: while, by invoking (*4) to F (z), the Poisson- Jensen forn1ula follows, too. 0 The genus (see (5.5.2.6) and Exercise A(ll ) t here) and the order (see (5.5.3.3)) are closely related in the following H ad ama r d 's order theor e m . The genus h and the order >. of an entire fun ction (of finit e order) satisfy the inequality h < >. < h + l.

(5.5.3.6)

The following proof is based on that appeared in Ref. [l], pp. 209- 211. P roof. >. < h + 1: It is hannless to suppose that h < oo. In thi. case; the entire fun ction f (z) can be expressed as

f(z)

= z m eg(z)

IT Eh(_.:._). n= l

with

E1i(w) = (1 - w)eL~=t t w',

Cln

and

E o(w)

=l - w

1

h > 1:

176

Chap. 5

Fundamental Theory

where m, > 0 and g(z) is a polyno1nial of degree< h (refer to Exercise A(l l ) of Sec. 5.5.2). By (5.5.3.4), zm is of order 0 and eg(z) has order < h. Since the order of a product cannot exceed the orders of its factors (see E xercise A( l )), it is sufficient to how that

P (z) =

J1 (:J E,,

is of order < h + 1. We need the following esti1nate for Eh(w ) in ( *6): log IEh(w) I < (2h + l)lw lh+l:

w E C.

To begin with, observe that log IE1i(w) I < log IE1i-1(w)I

+ lwlh for any

w E C

While 1 if lwl < 1, we have, by using log(l - w) = - 2::~ 1 oo

L

loglE1i(w)I
.1 and >.2, respectively. Show that the orders of f(z)g(z) and f(z) + g(z) are < max(>. 1, >. 2). In case >. 1 '/= >. 2: the order off (z) + 9(z) is just equal to max(>.1 , >.2). (2) Suppose A, B , and a are positive constants so that an entire function f(z) satisfies lf(z ) I < eAlzl"'+B for all large lzl. Show that the order of f (z) is < a . (3) Let f (z) be entire of finite order >. and p(z) be a polyno1nial. Show that the order of p(z)f(z) is still>.. (4) F ind the order of ee"' . Exercises B

(1) Let f(z) be an entire function of finite order >. and an '/= 0, n > 1, be its zeros. Then 00

1

""'"' lan l-\+e < 00 ~ for each s > 0. Try t he following steps: Choose 0 la1 I < la2 I < · · · < Ja ni< · · · ·

< a < s.

1. Set Qn(z) = (z - ai) · · · (z - an) . Then -

1

.

27ri

f

f(() d( =

} J( J=R (Qn(()

f(O) a1 ··· an

if lan l < R.

Suppose

Sec. 5.5.3

2. NI(R) < T hen

eR>.+a

if R

Hadamard 's Order Theorem

181

> 0 is large enough. Choose lzl = R = 3lan j.

If (O) I < eC31anl)-\+a

---'"~...;.....;...-

la1 ·· · an l -

--~....;...._~

2n janln

3. Note t ha.t lan fX+e 1 andL 1;., < oo. Let 8 = inf a where a > 0 are such that L - 1 °' < oo. This o is called 1 1 t he index of convergence of zeros of f( z) . Then o > 0. As a m atter of fact ) we ha.ve a ·1i

o-

1< h

(the genus of the canonical product associated with the zeros an)


0, g(z) is a polynomial of degree< A and

is the canonical product associated with t he zeros an, n > 1, of f( z ). As a beginning, note that (5.5.3.9) a nd (5.5.2.2) guarantee that f(z) = Z 111'eg(z) p (z) with g(z) an entire function only. Try t he following steps to prove that g(z) is a. polynon1ia.l of degree < A.

1. Fix any R > 1. Choose an integer iV and Ian I > R if n > fll. Let

> 1 so that Ianl < R if n < iV

f( z) = f1(z)f2(z) ,

182

Chap. 5

where

h(z)

= zm

Fundamental Theory

ft ( :J + i; :J LJ1 :J 1-

and

n=l

Fh (

h(z) =exp {g(z)

(1 -

eF•(,: )

with

1

1

h

F1i(w) =

L yw

1

•

l =l

It is tacit ly understood that Fh(w) = 0 if h = 0. 2. On lzl = 2R, JVI(2R) > lf( z) I > lh (z)I holds. Consequently, lf2(z)I < lv1 (2R) if lzl < R . 3. Rewrite f2(z) = e >.>h. Differentiate cp(z)m time to get (m) (z)

cp

=

(m) (z)

g

n=N+l

+ ~ (- l )m(m - l )!. (z - an)m

L

n = N+l

Try to show that

g (m) (z)

_ 0. According to

cp(m) (0) 4 maxlzi=r Recp(z) - 2Re cp(O) < m! rm '

=> (using step 2 and letting r

R)

-

O ( notmg g(m) (0)

m .l

. and R large.

f\ f

-

0 a R - oo.

(2R) < e( 2R)"+e for 0 < c
.6.1 will discuss in detail t he background why we choose (3) as our definition for I'(z) and then, prove its equivalent representations

184

Chap. 5

Fundamental Theor1J

shown in (5.6.1). Other representations will b e introduced as we proceed to Sec. 5.6.2. Basic and characteristic properties for r (z) will be the core of Sec. 5.6.2. Using LindelOf's residue nlethod, Sec. 5.6.3 studies the asymptotic behavior of r (z), na111ely, t he well-known Stirling's formula . 5.6 .1

D efinition and represent ations

To start with, we pose the problem: Does there exist a rncr01norphic function f (z) in C , satisfy ing the following conditions: 1. f(l) = 1, and

2. f (z + 1) = z f( z ), wherever f(z) is defined in C ?

(5.6. 1.1)

If such a funct ion does exist, it should be a m eromorphic .function with simple poles at z = 0, -1 , - 2, ... and the corresponding residues Re s(.f( z); -n)

=

(- l )n

n! ,

n

> 0.

Since .f(z) i analytic at z = 1, and li1nz-.O f (z) = limz~O f(z; l ) = oo and limz~ozf (z) = limz-+O f (z+ l ) = 1holds, 0 is a sin1ple pole ( ee (4.10.2.1 )) with residue equal t o 1. By inductive process. it is easily seen that f (z) has a simple pole at z = -n (n > 1) and

f (z + n + 1) = (z + n)(z + n - 1) · .. (z + l )z f (z), ::::?

(z

+ n)f(z) =

f (z + n + 1) z(z + 1) · · · (z + n - l ) (- 1r n!

as

z

~

z # 0, - 1, .... -n - 1

f (1) - t - - -- - -

- n( - n+l)· .. (- 1)

-n.

Even if (5.6.1.1) has a solution, it is definitely not a unique one. One nlight choose any meromorphic funct ion r.p( z) with the properties that r.p( l ) = 1 a.nd r.p(z + 1) = r.p(z ), z E C. Then r.p(z )f(z ) will be another solut ion. For instan ce, r.p(z ) = tanf~~~nz) is a possible cand idate. But, in this case, r.p(z )f(z) turns out to have additional zeros (due to t an(i + 27rz) = 0) and simple poles (due to tan(i + 27rz) = oo). The solution is also not unique if we further impose t he restriction that the olution has neither zeros nor poles other than 0, - 1, - 2, . ... This is becau e all one needs to do is to choose an entire function r.p(z) sat isfyin g cp( l ) = 1 and r.p(z + 1) = r.p(z), z E C . Then r.p(z) f (z) will work , too.

Sec. 5. 6.1

Definition and R epresentations

185

Consequently) the best we can do toward the solution of (5.6.1.1) is the following

C laim. Suppose f (z) is a meromorphic function in C satisfying: (1) f(l) = 1; (2) f(z + 1) = zf(z), wherever f(z) is defined in C . and

(3) f(z) does not have zeros and has simple poles only at nonpositive integers. Then , f( z) should be of the form

f( z) = e-g(z)

1

z ITn=l (1 + 00

•

~) e- ;;: ' n

where g(z ) i:::; an entire function having the properties:

g(l) = l + 2k1fi; g(z + 1) - g(z) = / in which k and l a.re integers) and '"'( constant.

+ 2l1fi

= limn-oo ('L~=l

t - log n) is Eule1->s (5.6.1.2)

Sket ch of proof. Note t hat f fz ) is still a m eromorphic function, no poles and having simple zeros z = 0. -1 1 -2..... According to Exa111ple 3(3) in Sec. 5.5.1 and (5.5.2.2), t he ent ire function z (1 + ~) e- fi has sin1ple 1

zeros only at z

= 0, -

rr:=

1, - 2 1 • • • • Hence, the entire function

(f (z)) - 1 z IT~= 1 ( 1 + ; ) e- -;; does not have any zero in C and thus, can be written as eg(z) (see (5.5. 1)), where g(z) is an entire function. The claimed form for f( z) then follows. R.ewri te f (z) a.s

f( z) = lin1 fn(z),

where

'/l-'00

+ L ~=l f.} z(z + 1) · .. (z + n)

e-g(z)

fn(z)=

z

n ! exp { - g(z)

rrnk=l (1 + .! ) e-~" k

T hen 1

=

z f(z) f(z + 1)

=

lim z fn(z) n-. fn(Z + 1)

= lin1 (z + n + 1) exp n-

{

1}

g(z + 1) - g(z) - "1l . -

~ k

k=l

186

Chap. 5

Fundamental Theory

1 = nl~~ (1 + z : ) e xp { g(z + ! ) - g( z) = exp{g(z + 1) =?

g(z ) - / }.

[t. ~ -logn]}

z EC

g(z + 1) - g(z) = 'Y + 2lni, l an integer. Sin1ilarly.

1=

lim f n(l) = lim exp {-g(l ) + n~oo

n-+

[L~=l f

1 + 1n

- logn J}=exp {- g(l) + 1 }

and thus, g(l) = 'Y + 2kni for integer k.

0

The simplest such function g(z) in (5.6.1.2) is the one given by g(z ) = "(Z . The corresponding f( z ) is specifically denoted as

r( Z ) = e-1z 1 - -

r (z)

= e'Yz z

noo z

n= 1

IT

n=l

1 (l

(1 +

+ -z ) e - "-n , z

E C - {O, - 1, - 2 , . . .} or

n

.:_) e-;7 n

)

z E C.

(5.6.1.3)

r (z) is called the gamma function in W eierstrass's form. Refer to Ref. [5 Vol. II, p. 30 for the graph of its modulus surfaceµ= lf (z)i or

m.

I

L

Now (1) , (2) and (4) in (5.6.1) have the following extensions to t he complex variable z . Equivalent representations of I'(z) .

(1) Gauss's representation: If z E C - {O , -1. -2 , ... }, n !n z f (z) = li1n . n -+oo z(z + 1) · · · (z + n)

(2) E uler 's integral representation: In Re z > 0, 00

f (z) =

/

Jo

e-ttz- 1 dt

(t E R ) .

The funct ional equation I' (z + 1) = z I'(z) permits f (z) to be extended to all z :/:- 0, - 1, - 2, . ... (3) P artial fractional representation: If z E C - {O, - 1, - 2, . .. }, I'(z) = f

(-l)n

n = O n !(z

+ n)

+

f' ooe-te- 1dt,

./1

where the series I:~ converges absolutely and locally uniformly in 00 C - {O, - 1. - 2, ... } and t he integral f1 represents an entire function. (5.6.1.4)

Sec.. 5. 6.1

Definition a.nd R epresentations

Proof. (1) Returning to (*i) with g(z) = I

r (z )

{ '\""n

z

n. exp 6 k = l "k - "f Z . 11111 = n-oo z(z + 1) · · · (z + n) = lim

n! exp { (~~=l

z (z

n-

= n~ lim

"fZ,

187

we have

}

f: - log n - 1) z + z logn} + 1) · · · (z + ri)

n ! exp{z logn}

.

z(z+ l ) ··· (z + n)

n !nz

= n-.ooz(z hm . +l) ·· ·(z+n)

Regarding (2) and (3), set temporarily the function 00

f .lo

f( z ) =

e -ttz- 1dt

(t E R , z E C )

rl e-tt z-ldt + roo e-tt z- l dt ,

.lo

.J1

where the path of integration is a.long the nonnegative real axis [O, oo), and t z-1 = e(z-1) logt. Since le -tt z-11 = e- t.fle z-1 and lim (e-ttRez-l)t 1 -Rez

t-0

= lin1 e-t = 1, if Rez > O; t~ O

tlim (e-ttRez - l)t2 = 0 foranyzE C , it follows , respectively, that

Ii (z) = h(z) =

f 1 e-t~- 1 dt

.J0

f

00

e- ttz-Idt

absolutely in Rez > Oi absolutely in C.

. 1

According to (4.7.4) , Exercise B (2) of Sec. 5.3.1 and Exercises B of Sec. 5.3.3 (see also Exercises A(2) and A(3)) , Ii (z) is analytic in Re z > 0 while h (z ) is entire. Suppose, for the moment , that the function .f(z) defined in (*2) is really identical with r( z), namely, (2) is proved. Note that

L -n.l )n tn 00

e-t =

(

1

absolutely and uniformly in [O)].

n= 0

00 (- 1)7l -- """"' L..J n!(z +n)'

ri=O

Rez > 0.

188

Chap. 5

Fundamental Theory

But the series in the right converges absolutely and locally uniformly m C - {O, - 1, - 2, ... } and hence, by W eierstrass's theorem (5.3.1.1), represents the analytic function I 1 (z) there; inoreover, Ii (z) has simple poles at z = -n (n > 0) with residues ( ~;r. Therefore, by ( *2) .

I'(z) - I1(z) =

1=

e-te- 1 d(

Rez > 0

h olds identically. Both functions in the left and the right are entire. They should be coincident everywhere in C owing to the interior uniqueness property. Hence, (3) holds. To prove (2), we want to show that, in (*2),

= n-+oo lim fn( z ) = I'(z),

f( z)

=

fn(z)

•n (1 - -t )ne-

.lco

1

n

Re z > 0, dt

where

for n > 1.

J;1. e-ttz- 1 dt in Rez > 0, it is sufficient to show that

Since f( z) = lim 11 _ lim

n-

rl, [ ( t) n-11 e-

.lo

e-t -

1- n

1

dt

=0

in order to justify that f (z) = limn- oo f n(z) . In passing, note that fn(z) is analytic (entire) for each n > 1 (see (4.7.4)). Now

I+: 0.

D

R e mark. It i worthy to mention t he following relations obtained during the above proof (see (1). (*s), and (*6)):

.Ion(1 -

!)

n tz-ldt = n z .fol (1 - t)ntz -ldt

z( z + 1) · .. ( z + n) '

z =J. 0,-1, . . . ,-n;

Rez > O. Recall th at e-t

(5.6.1.5)

= limn-.oo (1 - !; )n .

In what follows, we try to find some other integral representations for

r (z) . For our purpose and in1portance by itself, we need the following

Euler's functional equation.

r (z)f (l - z) = In particular,

r

(~) =

Jif.

.

1f

Sill 1fZ

;

z =J. 0. -1. - 2, .... (5. 6.1.6)

190

Chap . 5

Fimdamental Theory

-F

Fig. 5.15

This is easily seen by using (5.6.1.3) and the infinite product for sin 7rZ (see Example 1 in Sec. 5.5.2). Also) refer to Exercises A(4) and A (9). Here comes Th e complex integral representation of f(z) . Fix any£ > 0. Let 'Ye denote the contour composed of t he lower side of the cut (-oo , O] oriented \~rith the increasing t, follO\ved by the circle lzl = c in counterclockwise direction and then the upper side of the cut ( -oo: O] oriented with the decreasing t. See F ig. 5.15. Then, if z -/= 0, - 1, - 2, .. .,

r (z) = 1

l

. .1

2i sin 7rz . 'Ye 1

r (.z) = 21ri

l

• "fo

t

et

ettz-ldt;

-z

dt ,

where t- z = e-zLog t is the principal branch in C -(-oo , O]: -7r N1oreover, the integrals are independent of the choice of £.

< Arg t < 7r . (5. 6 .1. 7)

Proof. Ovving to identity (5.6.l.6)i it is sufficient t o prove the validity of the first integral for r(z) . Denote by '"'fe,n the part of 'Ye that lies in the closed disk lzl < n for n = 1, 2, ... . Along the lower side of 'Ye,n, t = ltle-ni; while, along the upper side, t = 1 tle7fi. Hence,

=

{- e ete- z(log 1t l-7ri)dt

J_ n

= (e1riz _ e -7riz)

J n

e

+ { 1f

J_1f

e- tt-zdt

eee; 9 6 - z(loge+iB)sieie dB

+i

111 eee'. . c l-ze(l - z)BidB. 9

.

-7r

(*7)

Sec. 5. 6.1

Definitfon and Representations

191

To estimate the last integral in (*7 ) 1 consider Re z < 1 and then

1:

e~c' 8 .:::1-ze(l-z}Oi dO
0,

( *8)

where -y denotes the path (-oo) O], oriented both forward and backward. vVhat remains is to show that, for any 0 < s1 < .:::2,

.lei ett- zdt

=

.l,,.2 ett-zdt

holds and hence, t his com1non function is the one represented by J'Y ett-z dt in (*8). Let r denote the contour composed of the circles lzl = .:::1, lzl = s2: and the segments [-.:2, -ci] oriented twice. See Fig. 5.16. By Cauchy's integral theorern,

and (*9) is t hus followed. Combining(* ) and (*9), we obtain, for a.ny.::: > 0, f (l - z) =

:::} f(z) =

1

.. 2i sm 1T z

1 .. 2i sm 1T z

1 1

ett- zdt,

Re (1 - z) > 0

'Yz

ettz- ldt.

Re(z) > 0.

le

Finally, t he recursive identity r (z + 1) = zr (z) permits us to extend the above integral to C - {O, - 1, -2, . .. }. 0

192

Chap. 5

Fundamental Theory

Fig. 5.16

Finally i the

Other two complex integral representations for I'( z) . (1) Let /e be as in (5. 6.1.7) (see F ig. 5.15) . Then

r(

Z

r

1

+ 1) = _..!.._ 21fi } 1e

e~l dtl

tz

z

# 0, -

11 - 2, ....

(2) (Refer to Sec. 4.14.) - 1

< Rez < Oi a> 0, t > 0. (5.6.1.8)

Recall that we had used (2) in Exa1nple 2 of Sec. 4.14. Proof. ( 1) The integral can be deco1nposecl as -

1

1-e

27ri _

1

-

et

t z+l

l

dt

(along the lower side of ( -oo, -c]) et

1

+ -27rl. . ltl =e ~ dt + -2 . tz . 7r'l

1-e

et

t z+l dt

(along the upper side of ( -00 1 -c]).

Sec. 5. 6.1

Definition and R epresentations

193

Note t hat, along t he lower side of (-oo, O], t z+l = e-(z+l)7ri jtjz+1; along the upper side, t z+l = e(z+l)7riJt jz+l. Hence, in (* 10 ), the first integral+ the third integral

=

~ (e(z+ 1 )1l'i 2rri

=

sin (z + l )rr

1

1=

e -tt- z -ldt

c

00

e

- t

t

- z-l

dt

sin (z + l )rr

--7 - - - - -

c

7f

= -

_ e - (z+ l )1l'·i )

S ll1 Z 7f I'(-z)

7f

as c

--7

l .

0

0.

(*11)

7f

=e

vVhilc, along jtj

or t

= seie , 0 < B < 2rr ,

the second integral in (* 10) 7r

1

1 1-11 e ~ei9 · eW

e:ei9

= -21T'i. -7r e(z+le)(loge: +iB) ~r.-ie ·e dB = -21T' . _ --7

1

1-

11

szei(z+1)8

dB

0 ass --7 0, if Re z < 0.

Putting these results in (*10) , we have

f

1

21T'i

et

.J,e t z+i dt =

-

sin 1T'Z 1 rr r (- z) = I'(z + l ),

if Re z < O.

Note that integral in the left still converges on Re z > 0. Hence, via r (z + 1) = zr( z), the above expression can be extended to c -

{O, - 1, - 2, . . .}. (2) Let 'Y be t he path shown in Fig. 5.17. Then

The integra.1 a.long imposed: lim

~ -

0

_1_

2 . 7ft

1

'Y~. R

can be evaluated as 1n (*11) when Re z < 0 is

_

~ sin 11z l'z+l d( 7f

"/c.R ',

loo

e

- t(

(

- z-l

• Q

(1 + z)'

t

7f

R~

r

_ sin rrz . z (- )

d( -

t > O· )

r

z

Chap. 5

194

Fundamental Theory

F ig . 5.17

if - 1 < Re z is considered , then

l

1

lim - 1 _e_t( d( = 0 and ~---o 2ni . C e (z+l

l

et(

liin - -d( = 0. R->oo 2rii . C n (z+ l

Substituting these results in (*12), we obtain the claimed integral.

D

Exercises A (1) Try to use (5.6.1.3) to derive (1) in (5.6.1.4) . following the steps: (a) Use the definition of Euler's constant 'Y to show t ha t e - -yz =

IJ n

liin n ---+OO

(

1+

k=l

1).: e- r. .

-

k

(b) Then , rewrite r (z) as

zr(z ) = liin n -+oo

IT (i + k~)

k= 1

z

n ->OO

(2) Adopt the follov,ring steps to show that I' (z) in Rez

> 0.

t

(1 + zk ) -1 = lim ~~;,(n 1):). + le= 1 Z

00

= .[0 t z- le-tdt is analytic

J:

(a) For any 0 < a < b < oo , show that e -ttz- 1 c1t is entire. (b) R.ewrite f (z) =I1 (z) + I2(z ) as in (*2). Show that 1 1 (z) =

=

f 1 e-ttz -ldt .lo

L un(z) 1

absolutely and locally uniformly in Rez > 0,

Sec. 5. 6.1

l w1ere

Un (

z) =

195

Definition and R epresentations

> _ 1;

l/n e -ttz-ldt , n Jl/(n+l)

00

=

L vn(z)

absolutely and locally uniforntly in C ,

n=l where Vn(z) = J;:+l e-ttz- 1dt, n > 1. (c) Then, use \.Veierstrass's t heorern (5.3.1.1).

(3) Use the following steps to define r (z), z =I= 0, -1 , -2, .... (a) Use Morera's theorem (3.4.2.13) or Integrals of Cauchy type (4.7.1) 00 and Exercise B(2) of Sec. 5.3.l to show t hat I'(z) = f0 e-ttz- 1 dt is analytic in Re z > 0. (b) Show that I'(z + 1) = z I' (z) holds in Rez > 0. (c) Use I'(z) = r(z: l) to extend I'(z) fr01n Rez > 0 to C - {O, -1)

- 2) ... }.

(4) Use sin 1T'Z =

00 (1 - ;) efi to define I'(z) in the following steps.

1fZ f1 ~

(a) Note that sin 1T'Z is the simplest entire function having all the integers a.s its only si1nple zeros. Then, the simplest entire function having all negative integers as its only simples zeros is 00

G( z) =

IT (1 + ~) e- ;7 . n=l

Vlhat is G(-z)? Does zG(z)G( -z) = sii~1fz hold? (b) Show that G(z - 1) = zei(z)G(z) for some entire function f'(z) . T ry to show t hat "((z) _ 0. Hence f'(z) = '"'(, a constant , and

G(z - 1) = ze1 G(z) , z E C . (c) Choose z 1. Show that f'

TI~=l (1 + ;) e-*

t-

lin1n->oo (2=~- l log n) . (cl) Let 1-I(z) = G(z)e"z · Then H (z -1) function by

r (z) Then I'(z

=

1

z H( z)

= z JI( z) . Define

00

z IT (1 +-) z n

1 = e-/Z _

-l

,

e n.

n=l

+ 1) = zI'(z) and r (z)I'(l - z)

=

.

7r

Slll 1!' Z

)

z =I= 0, ±1 , ±2, ....

the gamma

Chap. 5

196

Fundamental Theory

r(z) to express t he following products. (a) Tin=l (1 - 2~) et,; .

(5) Try to

(b)

(c)

USC

n:,1 (1 + 2nz-1) (1 n:=l (1 + (-2!{).

(6) In Re z

2zn ).

> 0, show that

roo e - t t2z - l dt = lorl (log t1 )z-1 dt.

= 2 lo

r(z) (7) Integrate

e-( ( z- l

f

2

along t he path shown in Fig. 5.1 and show that

00

lo

e- i.ttz-ldt

= I' (z)e-¥ z

1

0

< Rez < l.

(8) Integrate e-((z-l = e-(+(z-l)Log( i Log( = log l( I + i Arg( and 0 < Arg ( < 27T, along the pat h re. shown in Fig. 5.19. Show that, .if z =f. 0, - 1, - 2, ...

Fig. 5.18

\0

~

~~,=.=======~======~~~·======

_.,.,, Fig. 5.19

Sec. 5. 6. 2

Basic and Characteristic Properties

197

Fig. 5.20

Alternatively) designate Log( = log l( l+i Arg( and a< Arg ( < 21f+a and integrate e -((z-l = e -(+(z -l)Log( along the path 1~ shown in Fig. 5.20. \i\That happens to these two integral expressions? (9) Adopt Euler's integral r(z) = f(;° e - ttz- 1 dt to prove Euler's functional equation (5.6.1.6). Fill up t he gaps in t he following process:

r (z)I'(l - z) =

f

00

f

00

Jo Jo

e- (t+s)t z- l s- zdtds,

0 < Rez < 1

oo 1o=e -xy z- dxdy ( setting t + s = x .lao . o l +y oo e- xdx · l oo ~z-l dy . = l

=

l

.o

.o

and

l+ y

It is advised to use t he residue method to justify that s in7T7rz, 0 < Re z < 1.

5 .6.2

- - 1

J(;° 1f+y dy

=

B asic and ch ar acteristic p r operties

The most basic ones among a ll) in addition to Euler's equation (5.6.1.6) , are the following BaS'ic properties.

(1) r (1) = i. 2 (2) r (~) = J000 e- tt-t dt = 2 J000 e- x d X = ,;:;.

198

Chap. 5

Fundamental Theory

(3) f (z + 1) = zf (z), z =/= 0, -1 , -2 , .... Hence, 1. r (z

+ n) = z(z + 1) ... (z + n - l )f (z) if n > 1 is a n integer ; in

particular ,

r

( + 21) n

(2n) !

= 411 n !

Ji.

n = 0, 1, 2.....

2. r (n + 1) = n !.

3.

lim z~-n r

( z) =

00

and

( 1)n Res(f(z ), -n) = _ n_! _

for n

> 0.

(4) L egendre's product formula:

(z+ ~).

J7fr (2z) = 22z - 1 r(z)r (5) Gauss's product fo rmula:

(27r) ¥ r (z) = nz- ! r

(~) r

(z:

1

) ... r

( z +: -

1

).

(5.6.2.1)

Proof. Only (3)4, (4), and (5) will be proved. About (3)4: By Gauss·s representation for f (z),

. f (z + n) lnn

n~oo

nz f (n)

li z(z + 1) · · · (z + n = n~oo m nz(n - 1)!

n

= n-oo lim z +n

1)

n !nz

· ---:------,---:----"""':'" z(z + 1) · · · (z + n )

= 1.

About (4): Adopt Weierstrass 's representation for f(z). Differentiate Log f(z) logarithmically and we obtain

r'(z) 1 r (z) = - ry - ; - ;

00 (

z

1

+n

1)

- n

)

(5.6.2.2)

a bsolutely and locally tu1iformly in C -{0, -1 ,-2, ... }. This function is called Gauss ·s 7/J-function or digamma function. Differentiate (5. 6.2.2) again to obtain d dz

(r'(z) ) r (z)

1 =

] ; (z+ n )2l

Sec. 5.6.2

Basic and Charact eristic Properties

199

Now, I' (z) I'(z + ~ ) and I'(2z) obviously have the same set of poles and corresponding residues. Hence, there is an ent ire function g(z) such t hat I' (z) I' (z + t ) = eg(z)r (2z) holds. By using (*1) .

d I'' (z) d I'' (z + ~ ) dz I'( z) + dz r (z +

00

1

1

~) = .,~ (z + n) 2 + ]; ( z + n + t)2 _ -

=? g" (z)

4

t

_ 9 .!:._ f' (2z) n = O (2z + n) 2 - ~ dz f (2z) 1

=0

=? g(z) = az + b for sorne constants a and b and hence,

r (z)r ( z + ~) = eaz+ br(z ). Setting z = 1 and! , respectively, and using r (!) = J'ii, I'(l ) = 1, r( ~ ) = ~ J'ii, and f (2) = 1, we find t hat a = -2 log 2 and b = ~Jog 7r + log 2. The result follows. One can also prove (4) by setting nz for z in (5) (see (5.6.2.1)) and t hen lett ing n = 2. For other proofs, see Exercises A(2) and A(3). Yet for another proof, we use Euler·s integral for I'(z) a our start ing point. Replacing t by x 2 , t hen

f (z) = 2

f

00

2

e-x x 2=- 1 dx,

Rez > 0

Jo =? r (z + ~) = 2 .l oo e-Yy 2 zdy , 2

Re z > 0

=? (mult iplying both side by side, and then the factor 22z-l )

22z-1r(z )f ( z + 22z-1r (z)f

~) = 4 .fooo .loo e - (x2+y 2) (2 xy) 2z-lydx dy;

(z+ ~) =

4

fooo .loo e - (x2+y2) (2xy)2z-lxdxdy

(interchanging t he role of x and y in the first integTal) =?(adding up) 22z- 1r (z)f ( z+ = 4

fl

2

~)

e-(x +y2) (2xy) 2 z - 1 (x + y)dxdy,

200

Chap. 5

Fiindamental Theory

where n = { ( x, y) JO < x < oo, 0 < y < x}. Invoke the change of variables: u = x 2 + y 2 and v = 2xy with the Jacobian det.ern1inant (4(x2 - y 2 ))- 1 . Note that u - v = (J; - y) 2 > 0. Hence 2

z- r(z)I' ( z + ~) .loo v

2

1

=

= 2 =

t)Q Jo

2

z-

1

dv

.fo

J:-:

00

e-vv2 z- 1 dv

f

v du

00

Jo

e- t

2

dt

(setting u = v

+ t 2)

v '1ff(2z). 1

About (5): Use nz to replace z in (5). \'!\Te just only need to prove n- I

1

n ~-nz(27r) -rr (nz) =

nrr -l r

(

z

+ nk ) .

k=O

To see this, set n-1

f(z) =g r(z+:) This f(z) has the san1e sin1ple poles at -m-~, m > 0, and 0 < k < n-1, as the function I'(nz) does. Therefore, t here is an entire function g(z) so that

f(z) = eg(z) r(nz) ~ (by

definition (5.6.JL.3) for r (z) and the process similar to t he proof of (4)) g(z) is a polyn o1nial az + b of degree< 1 .a.nd hence

~ (replacing z by z

+ ;) f(z + ~)

=

f(z)I'(z

+ 1) =

ceaz+~r(nz

+ 1) (*3)

r (z + 1) = z I'(z)) f(z) = ceaz+f;: . n I' (nz) (comparing with (*2)) ne~ = 1 and hence, e0 · = n-n.

~ (using ~

Setting z = 0 in (*3), vve obtain

) IT r (k ~ = ce~ =

n -1

k=I

~.

Sec. 5.6.2

Basic and Characteristic Properties

201

To c01npute c in ( *4 ): Observe that

n-1

n-l 1T

-

1T

( rr~:isink?T/n) - n/2n-1 -

(21f)n-l

n

where we have used t he identity IT~:{ sin~; = 2,-:'.:_1 (see Exercise A(12) of Sec. 1.5). It follows that

c = n·

IJ r -k) ;:; = n · (2n)(n-1)/2 n = vn(21f)n-1/2_

n-l

(

vn

k =l

Substituting these values of c and ea into (*2) 1 the result follows.

D

For another proof) see Exercise A{4) . (5.6.1.5) indicates t hat, no matter \i\Teierstrass 's infinite product representation or Euler 1s integral representation being adopted 1 the gan1ma funct ion f (z) can always be represented in Gauss's form a nd hence, enjoys . n-+oo nr(z(n? zr n = l . t h e property t h at 1ll11 As a matter of fact, we have the following

Characteristic properties of I' (z) . There exists a unique analytic function F(z) in C - {O) -1, -2, ... }, satisfying: 1. F (l) = 1; 2. F (z

+ 1) = zF( z),

3 . }'lffin-+oo

F(z+n) n ~ F(n)

=

a nd l

.

Indeed, F(z) = f (z), holds for z

=J 0, - 1, -2, ...

(5.6.2.3)

This is easily seen as follows: Note that F(z + n) = z(z + 1) · · · (z + n - l)F(z), for integers n > 1 and z =J 0 , -1, ... , -n, and thus F (n + 1) = n! , n > 0. Therefore we h ave

.

1 = 1nn

n-+oo

. { z(z + 1) · · · (z + n -1)} F( ) F(z) F(z + n) = 1nn z =n ZF(n) n-+OO nZ(n - l)! r (z)

for a ny z =J 0, -1, -2, .... As two important applications of the gamn1a function, we will sketch the Beta. and hypergeometric functions in the following two sections, respectively.

Chap. 5

202

Fundamental Theory

Section (1) The J3-function The function defined by 1

B (z ) w) =

{

lo

tz- l ( 1 - t) w - l dt

(t E R )

(5.6.2.4)

is called the B eta f1inction of two con1plex variables z and w , commonly knows as Euler's second integral. Since the integral converges absolutely and locally uniformly if Re z > 0 and Rew > 0, it follows that B (z i w) is an analytic function on Re z > 0 and on Rew > 0) respectively. This can be justified by using (4.7.4) and Exercise B(2) of Sec. 5.3.1 (refer to (*2)) (*3) of Sec. 5.6.1 a nd Exercise A(2) there). l\!Ioreover) we have following Integral representation of B(z) w). If Re z > 0 a nd Re w> 0, then

_ r(z) r (w) ( ) B z '. w - r(z+w ) oo tz-1

l l

- - - - dt

(1 + t)z+w

.0

= 2

•1f

/2

.o

(sinB) 2 z - 1 (con0) 2w- 1 dw

1 - (1- e211·iz)( l - e27riw)

x {

.f~ro,1

( z- 1 (1- ()w- 1 d(

(due to L. Pochhan1n1er (1890)) .

The first formula can be used to define B (z, w) in C 2 - {( z )w )lz, w, z +w =

0, -1, -2) .. .}. In the fourth representation, (P- 1(1 - oq-\ where p = Re z > 0 and q = Rew > 0) has a single-valued analytic branch in C - ( -oo) O] - [1 ) oo) (refer to Example 5. in Sec. 2.7.3) and l'o,l is the path (see Fig. 5.21), composed of four .Jordan closed curves APA and AQA (both in counterclockwise direction)) ARA and ASA (both in clockwise direction) , where A is a fixed point in the interval (0, 1). (5 .6.2.5) Proof. Use t(l - t)- 1 to replace t in (5.6.2.4) and one obtains the second representation; substitute t = sin2 0 (or cos2 0) and the third one follows. As for the first fonnula: Set F(z) = B(z, w)r(z+ w)) Re z > 0. It is easy to check that B (l ) w) = ; , w =f:. 0, and hence F(l) = r (w) holds . In case

Sec. 5.6.2

Basic and Characteristic Properties

203

F ig. 5.21

z '/:- 0 and w B (z

+ 1, w)

'/:- 0) by using integration by parts (see (2.9.10)),

=

fo

le (l -

t) w- 1 dt

= - -1 tz( l - t) w 1+ fol ~ tz- 1 (1 - t)wdt w

. Q

z = - B(z, w

w

=?B(z+ l ,w)=

and =?

z

+ 1) = -

z z+ w

w

[B(z , w) - B (z

B (z,w),

z + w -/:-0,

0

. Q

w

+ 1, w)] Rez>- 1,

Rew>O

zw # 0

(by induction) B (z + n , w)

Rez > -n, Consequently, F(z

Rew> 0,

=

and

(z + n-l) ... z (z+wn ) B (z, w).

z +w

# 0,

zw # 0.

+ 1) = B(z + l ,w) f (z + 1 + w) = z.F(z ) and

. .F(z +n) lnn nz F (n ) n-.oo

=

. B (z + n , w) f (z + n+w) Inn = 1. n~oo nzB (n, w)r (n + w )

Con1paring to (5.6.2.3), F(z) = f (z)f (w) does hold and hence the forn1l1la follows. For other proofs see Exercises A(ll ), A(12) and A(14). As for the fourt h one: The Jordan closed curve APA can be deforn1ed continuou ly in C - (-oo, O] into a closed curve /P shown in Fig. 5.22. In this case. as the point (starts fron1 a point tin (0. 1) and winds around the small circle in the counterclockwise sense and back to the original point t . t he values assumed by t he function f (z) = (P(l - ()q-l are changed from

204

Chap. 5

Fundamental Theory

Fig. 5.22

~'

A

;..___---~~~~~~-

0

(I

Fig . 5 .23

f(t) continuously to f(t)e 2 7riq . Hence, by Cauchy's integral thcore1n,

f

JAPA

J(()d( =

j

J(()d(

= f l- -y J(t)dt +

.la

-yp

+ e2 rdq

.l :'Y f(t)dt

.l

-

1 c..,

.l l

J(()d 1,

(5) (R ef. [1], pp. 316- 317) F(a ,b, c,z) is a solution of the hypergeometric differential eqirntion

(z - z 2 )w" + [c - (a + b + l )z]w' - abw

=0

(see also Exercise B (l) of Sec. 3.3.2). (6) (Ref. [1], p. 318) If Ree> Reb > 0, then F(a , b,c, z) =

f(b)~~~)- b) fo

1 1

1

tb- (1- t) c-b- (1 - t z)-a dt ,

lzl < 1.

Also the integral in the right converges to an analytic function absolutely and locally uniformly in C - [1. oo), so it extends F (a, b,c. z) analytically fron1 lzl < 1 to C - [1. oo). (7) Suppose c # 0, -1, -2, ... (and a and b may assume the value 0 or negative integers) and Re (c - a - b) > 0. Then F(a , b, c, 1) is analytic in a or b or c, respectively.

Sec. 5.6.2

Basic and Characteristic Properties

207

(8) In case Re (c - a - b) > 0, then r (c)f(c- a - b) F(a )b) c, l ) = f(c - a) f(c - b)"

(5.6.2.9)

Proof. (1)- (4) are left as Exercise A(18). (5) : For si1nplicity, set D = zd~: a differential operator, and w = F(a , b1 c1 z) . Since (D + k )zn = z · nzn-l + kzn = (n + k)zn for constants k, direct co1nputation shows that

D (D + c - l)w =

t (a~n(b)n

D(D - c + l) zn

n .(c)n

n =O

~

(a)n(b)n n = ~ (n - l)!(c)n-1 z

f

= z

n=O

=

*

z(D

(a + n)(a)n · (b + n)(b)n zn n!(c)n

+ a)( D + b)w

[D(D + c - 1) - z(D + a)(D + b)]w = 0.

The result follows by noting that Dw = zw' , D(D- l)w = z 2 w". (6): Using f(b+ n) = (b)nf(b) and (5.6.2.5) , then

f(b + n) r (c) r(b)f(c+n)

(b)n (c)n

r(c) r(b)f(c-b)

f(b+n)f(c-b) r(c + n)

r(c) r(n) f (c-b) B (b+ n , c-b) =

r(c) r(b)f(c - b)

rl tb+n-1(1 - t)c-b-ldt Jo '

Re c > R eb

*

> 0 for n > 0. 00

F (a b c z ) = ) ) )

=

1

f (c) ~ (a)n zn · { tb+n- 1(1 - tr-b-1 dt f (b )r(c - b) 0 n! J0 f (c) f (b)f(c - b)

rl tb-l(l - t) c-b-I (t (a)n (zt) n) dt Jo n! n=O

(see t he reason below) 1

f (c) { b-1(. c-b-I )-a = f (b)f (c-b) Jo t l-t) (l-tz dt ,

lzl < 1.

( *7)

208

Chap. 5

Fundamental Theory

1 The interchange of l:~=O a.nd f0 is valid because the series 2::~ (~~) (zt)n, lz l < 1, converges uniformly to (1 -tz)-a in the t-interva.l [O)]. The right integral in (*7) converges, indeed, locally uniformly in C - (1, oo). To see this, choose (1- t.z )-a = e-aLog(l-tz>, the principal branch in C - [l,oo) . Fix any s1nall r > 0, c > 0 , and large R > 0. On the compact set]{ = {zlr < lz - l l < Rand c < Arg(z- 1) < 27r- c }, we have the estin1ate {1 tb-1(1- t)c-b-1(1 - tz)-adt

Jo

< fol itb-1(1 -t)c-b-1(1 - tz)-a ldt


0. This shows t hat the right integral in (*7) converges uniformly in the con1pact set J(. (7): F ix 5 > 0 so that Re (c-a- b) > 25 holds. Using Gauss's representation for r(z) ,

(a)n(b)n/ _ l_ n!(c)n

nl+z

(a)n+l n!na

r( c)

(b)n+l n !nb

n!nc (c)n+l

.

~ r(a)f(b) · 1·0 = 0

n(c + n) (a+n)(b+n)

1

n c- a- b- 8

asn~oo.

00 Co1npa.rison test says that t he series F (a ,, b' c ' PI• = 1 + ""' n (b ),, u n=l (a) n !(c)n converges locally unifonnly in Re (c - a - b) > 0. Since ea.eh tenn in the series is a.n analytic function in a or b or c, the claim follows by \:Veierstrass's theore1n (5.3.1.1) . (8) By (7) and Abel's li1nit theore1n (5.1.3.1), if z = x (0 < x < 1) is real, then

F(a, b, c, l) = lim F(a, b, c, x) X -+1 -

-

i·

- x2!i1-

r (c) {1 tb-1 (1 r(b)r(c - b) .lo

-

t)c-b-1(1

-

t )-ad . x

x,

Sec. 5. 6. 2

209

Basic and Characteristic Properties

under the additional assumption that Re c > Re b > 0 and then, using (6). 1 = r(c) f tb-l (1 - t)c-b-l . li1n (1 - tx )-adx f (b) f (c-a) }o x -+1 (see t he reason below) r (a) { 1 tb-l(l - t)c-b-a-ldt r (b)r( c - b) ./0

=

r(c)r(c - b- a) f (b)f(c-b) and

if Re ( c - a - b) > 0

Re c > Re b > 0. 1

To see that limx_. 1- and f 0 is interchangeable, all one needs to do 1 is to show that the integral f0 converges uniformly in the x-interval [O, 1]. Since

ltb-1(1 - t)c-b-1(1 - tx)-al

< { tReb-1(1- t)Re(c-a-b)-\ if R ea>() ~

tReb-l(l - t)Re(c-b )-l ,

if R ea< 0

(under the assu1nption that Re (c - a - b) > 0 and Re c > Re b > 0)

.fol tReb( l -t)Re(c-a-b)-ldt = B (Reb Re(c- a- b)) , and 1

{ 1 tReb-1 (1 - t)Re (c-b)-1 dt = B (Re b, Re ( c - b))

Jo

both exist. Therefore the claim follows. Finally, by (7) , we know that both sides of (*8) are analytic in the do1nain Re (c - a - b) > 0 and coincident in t he subdo1nain Re c > Re b > 0. Hence 1 they coincide throughout Re ( c - a - b) > 0. o Exercises A

(1) Prove t he following ident ities.

(a.) r (z) = I'(z ), namely) r (z) is symmetric wit h respect to the real axis. (b) f (z) f (-z) y = In1 z .

=

zsi7i71'Z ;

Consequently, lf (iy) l2 =

y si:h1ly

for

Chap. 5

210

Fundamental Theory

(cl)

1r (~ + iy) 12 = cos~ rry and h ence, limy-+± r (t + iy) = o. J01 (1 - t)1'"tz-ldt = z(z+it'· 0. for integer n > 1.

(e)

IT11=1 (-1.!L) 2n-1 e.fn = e 2 e~ ·

(f)

r ( ~ + z) r

(c)

(!- z) =

co: Jrz ·

(g) (1 - z) (1 +

~z) (1- !z) (1 + t z) · · · =

(h) r(t) = 2- i

(~)~ [r(~) ] 2 .

r(i + t zfr( i - t z) .

22 "r (z)r(z+ l2 ) . Adopt the following steps to reprove q 2; )

(2) Set f (z) = Legendre's duplication formula.

(a) f( z ) is an entire function satisfying f(z + 1) = f( z), z E C. (b) u~e f( z) = J27f(z - 1)z-! e - z+l (1+ 0 (~))for all large lzl (see Example 3 in Sec. 4.14 or Sec. 5.6.3 below) to show that f( z) is bounded in lzl > 1.

(c) Use Liouville's theorem to show that f( z) = limlzl-oo f (z) = J2;.

J2rr by noting t hat

(3) Use (5.6.2.2), the series expansion of ~l:l) to show th at 2f'(2z) f ' (z) r' (z + f (2z) - r (z) - r (z +

!)

~)

=

210 2 g

to r eprove Legendre's duplication formula. By the way, show that r' (~) = - (1'+2 log2)J1f. (4) Set

*) ...

nnzr (z) f (z + r(z + n~ l) f( z ) = nr (nz) Imitate the inethod given in Exercise (2) to reprove Gauss's product formula. Let z = ~ and show that

r ( ~) r ( ~) ... r ( n :

1)

=

(2n);

i ) ;2

(5) Basic properties of Gauss's 'if; -function (5.6.2.2). (a) 'if;(l ) = -I' (hence, Euler;s constant 'Y is the slope of the tangent to the curve y = -r(x) at x = 1).

±

(b) 'if;(z + 1) = + 'if;(z) and hence, 'if;(z + n) = I:~:J z~k integer n > 1 and 'if;(n + 1) = L~=l "f.

t-

+ 'if;(z)

for

Sec. 5.6.2

(c) 'lf;(z)

= 'lf;(l -

Basic and Characteristic Properties

211

z) - 7T cot7Tz .

(cl) Emn-. [7/;(z +n)-7/;((l+ n)] = limn_,oo['!fi(z +n) - Log((+n)] = 0 forz,(;zf 0, -1;-2, .. ..

(e) 7/;(z) = Logz + E : =o [ z!n - Log ( 1+ z! n )J z ;if 0, - L-2 ..... (f) J~ 'if;(z + t)dt = Log z, z E C - (-oo: O]. 1

(6) Characteristic properties of 'l/J-function. T here exists a unique analytic function G(z) in C - {O, - 1, -2, . . .} sat isfying: 1. G(l) = -')';

2. G(z + 1) - G(z) = ~ ; 3. limn -+oo[G(z

+ n)

- G (l

+ n)] =

0.

Indeed, G(z) = 'l/J(z) on z E C - {0 ,-1 , - 2, .. .} . (7) Integral representation of 7/;(z) . If R ez > 0, t hen (refer to Exercise (8) if necessarily)

'lj;(z) =

rx:, ( e- t -

lo

= - ')' +

e-z~ t ) dt

t

1- e

1

e-t _ e- zt

00

0

1- e-

t

dt .

T1y to u e Exercise (6) . By which, one needs to show that

roe ( e-t -

Jo

t

roo e- tc1 -

e-t ) 1 - e-t dt - ./0

t - e-t)

t( l - e-t)

- dt - I ·

( ) Integral representation of Log r (z) . If Re z > 0, t hen Log f (z)

{oo {

=lo

(z - l )e-t -

-zt } dt

e 1 ~ ee- t -t

t ._

Try t he following steps: Set r 11 (z) = z(z+If~(~~n -i)' n > 1. By Gauss's representation, then r (z ) = limn_, 00 r n(z) .

(a) Log f n (z) = (z - 1) fooo e-t ~e-nt dt - E~=l Jo"° c -H-c ~ O.

212

Chap. 5

Fundamental Theory

(9) Use Exercise ( ) to reprove t he first integral formula in Exercise (7). (10) Show that

(1 - ta)(l - tb) dt 1- t log l / t

= Loo- r (a + b + 1) 0

. f (a+ l )I'(b+ l ) '

R eb

>

R e(a + b) > - 1.

Rea> - 1 ,

- 1.

What is

(1 - ta) (l - tb)( l - tC)

.lo

l - t

dt log l / t1

if Re (a + b + c)

Re( a+ b) > - 1, Re (a + c) > - 1 and

=

> - 1,

Re a > - 1't

e

=

psin B(O < < ~ ) in f (z) f (w) 4 J0 J0 e- 0, Rew > 0. c·ive precise

(11) Set u.

pcosB, v 2

2

et u = tv in I'(z) I'(w) = f0 t7' - 1 e-tdt · f 0 uw-le-u du and t hen change the order of integration (why?) to reprove B(z, w) = rr(zz~: , Re z > 0. R ew > 0. (13) Basic properties of the B eta-fu.nction. 00

(12)

(a) B(z, 1) = ~ '

z =J= 0. (b) B(z,w )= B (w ,z).

(c) B(z, w + 1) = z~w B (z, w), z + w =/= 0, Rez > - 1, Rew> - 1 and zw =I= 0. (d) B (z + w, n )B (z, w) = B (z, w + n )B (w, n ), Rez > 0, Rew > - n , Re(z

+ w) > 0, w =/= 0 and integer

n

>

1.

(c) B(z,w )= B (z 1 1,w) I B (z,w 11 ), undcrthcnssumptionso.s(c).

(f) zB(z, w + 1) = wB (z + 1, w ), tmder t he assun1ptions as (c). B (x, y ) 1 > 0 , y > 1. (g) 1l.lTiy ~oo y:r r(x) = ,X (14) Try the following steps to prove B (z, w) R ew> O. (a) U e (d) in Exercise (13) to show that B (x, y) = __._-....................._...... x > 0, y > 0, and n > 1.

Sec. 5. 6. 2

Basic and Characteristic Properties

213

(b) Use (g) in Exercise (13) to show that (due to T . S. Nanjundiah

(1969)) .

B (x y ) = 1im 1

n_,.oo

(y+n)- xr(x) · n-Y f (y) ( ) n- x+y I' (x + y)

- r (x)f (y) r (x + y) ,

x > 0,

y

> o.

(c) Finally, use the interior uniqueness theorem.

(15) Show the following.

(a.)

J:(t - a)z-l(b - t)w- dt = (b - a)z+w- l B(z , w) , where a and b 1

are real with a < b1 R ez > 0, R ew> 0. (b) J01 (1 - xP) 11CJdx = J01 (1 - x CJ) 11Pdx, p > O, q > O. (16) Prove (5.6.2.7) in d etail. (17) Prove (5. 6.2.8) in d etail. (18) Prove (1)- (4) in (5.6.2.9) . Exercises B Recall t hat, in Exa1nple 3 of Sec. 4.9.2 1 t he coefficients Jn (z) of the Laurent 1 series e ! ((- (- ) = Z::~oo Jn (z)( 1\ 0 < lzl < oo, ar e called the B essel junctions of the .first kind. Jn (z) can be cmnputed by residues as follow. Sett ing ,... = 1?J.

.,

z '

(")nj'

Jn (z) = - 1 . ::::. 27ri 2

.2 _2 -1-eTJ- t:ij dry = ( -z ) n R es ( -1- e'l'J -4,1 · 0) . J·11l =~ l z l 1r+1 2 77n+ 1 ' ,

where R es

1 17 (1711,+-e 1 -~·o) l

= t he coefficient of 17n in t he Laurent series _2

expansion of eTJ- trj at 17 = 0. = t he coefficient of 17n in the product series

~

={:a (-l)

k

1 (k + n) !k !

(z)2k 2

·

Chap. 5

214

Fundamental Theory

Imitating this forn1 of Jn (z ) 1 we are able to define the gen eralized Bessel function J v (z) =

where gers .

LI

(~Y/ £= k!f(~ -:~ + 1) (~) k=O

2 LI'/=

k,

0, -1 , -2, ...

is any presumed complex nu1nber other than the nonpositive inte-

(1) (a) Show that) for

I/'/=

0, -1 ) -2, ... ,

Jv (z ) =

(z)

v

2

·-

1. ·

27ri

1

_2 -1- et;-4Z d(

l' v+ l /e '>

where 'Ye. is the contour shown in Fig. 5.15. (b) If -1 < Re LI < 0, then Z

Jv( z) = ( - ) 2

v· - 1 . · l a+ioo - 1 . a-ioo ': . 21fi

!' v+l

·

,,2

e 0 (refer to Sec. 4.13) . (c) In case R ez > 0 and LI i= 0, -1) -2 , ... , then Jv (z ) =

I_ . 27r

reiz

J,

sin t;-

i vt; d(

where 'Y is t he contour shown in F ig. 5.24. Note. One ina.y need to use (5.6.1.8) in (a) and (b). The integral in (c) may be written as

l oo e- zsrnht . e- vtdt+ -1 1·1f cos (zsint- L1t)dt ;

J 11 (z ) = - sin //1f . 1f

1f

. Q

/\

- tr

Il

I\

!(

\ f

Fig . 5 .24

0

Sec. 5. 6. 2

Basic and Characteristic Properties

215

in particular, if 1; is chosen to be a positive integer n, then

lo7r cos (z sin t -

Jn(z) = -1

nt)dt.

o

n

F. \\!. Be sel introduced this integral when solving 71 as a function of B from Kepler·s equation() = 77-csin 77(-l 0.

2 11i . a - ioo

T1y the following steps: (a) Integrate r(z) over a rectangle/ vvith vertices at a±iR and b± iR , where 0 < a < b and R > 0. Then r (z) = -21 . r ~(z) d(, z E 7!'t .J , "-z Int I'· Along t he upper side, the absolute value of the integral is ( b- a)r(b) O as R ~ oo. So d oes aong 1 1 1ower s1·de . < 7r{Ja2+ R!i -lzl} ~ t1e 2

Hence, after letting R f (z)

~

oo,

1 l a+ioo r (() 1 1 b+ioo r (() = -21fi . . a-ioc Z - ( d( + -21fi . d(, b-ioc Z - (

xz - l

{l

a+ioo

.

r (()x- (d(

}

a< Rez

0, aJ1d

l oo

1T .

or

Rez > 0

f(z ) = J2nzz-t e- zeJ(z)i

J (z) = -1

Re z > 01

Z 1_ log drt('r/ E R ), 2 2 o rt + z 1 - e 21f"T/

n = L (-l)k-1 (2k ~2~)2k

Rez > 0

. z2;-1+ Jn(z)

k =l

rv

Loo ( -

l) n- 1

B 2n

.

(2n- 1)2n

n =l

1

z2 n -l

asn-700

in which lirnz- oo J(z) = 0 if Re z > x o > 0 and numbers; n1oreover , J n (z) --

1·= + (· /

(- 1)71 1 ·z 2n+ l 7f o

1

772n

1

T/

z )2

satisfying limz~oc z 2nJn(z) = 0 if Rez

r (Z ) rv V~1fZ 0'""" z- l2e-

z (

B 2n

log 1 _ -27rT/ dn.,, e

are Bernoulli's

n> l

> xo > 0. Consequently,

1 + 288z2 1 + ... ). 1+12z

T he following proof is based on R ef.

(5 .6.3.4)

[l], pp. 201- 206.

Pro of. Ste p 1. To re present (5. 6.3.1) as the complex line integra l (5.6.3.2). The appearance of the series in (5 .6.3.1) suggests that we need a function having si1nple poles at integers k > 0 with the associated residues (z) kp. Since 7f cot7rz has simple pole at k with t he residue 1 (refer to (4.15.3)), 7f cot nz (() = (z + ()2

does work. Here ( is t he variable while z (Re z > 0) stands as a parameter for the in01nent . F ix an integer n > 1. Integrate (z) a.long t he path 'Yn shown in F ig. 5.25, where 'Yn. is a rectangle having vertices at ±Ri and n + ~ ± Ri

Chap. 5

218

Fundamental Theory

..

"'

)

"

I Q

Q

0

\_

ii\

>

I

,~.t------------j;>------o

Fig. 5.25

with an outward dented half circle around the point z = 0. Then , applying (4.12.3.3) to the dented half circle, t he residue theorem says that

P.V.~ j (()d( = 27f'l .,,..

(

3 n2

1 = 2z2

t

~) ~Res(((); 0) + k= 2 27ri 1

n

+ ~ (z + k )2

1 = - 2z2

R.es(; k) 1

1

n

+ {;

(z + k)2 ·

(*i)

We try to let c ---+ 0, R ---+ oo, and n ---+ oo in order to calculate the integral in the left side. Along the horizontal side ( = CJ + iR, 0 < CJ < n + ~: . e 2-;ri u e -2-nR

cot 7rZ

~

r

} upper side

= i e 27rur. e - 2

+1

R 7r

-

1

.

uniformly as R

--+ i

r

I (1') r 7f Icot 7rzl o ~ d~ < }upper l(j 2 - jzj2

ld(I

---+

(if R is large enough)

s ide

l' l n (n + !) R2 - lzl2


0, a constant)

while n is fixed.

Sec. 5.6.3

The Asymptotic Function of I'(z)

219

cot?Tz = -i uniforntly, similarly we have

Since limR-

Jim

R-+

f

0.

where Log z and Log f (.z) are real logarith1ns in case .z = .r > 0 is real. Therefore, both c1 and c2 are real constant. The integral in the right is specifically denoted as 1

J(z ) = -

i 'CX>

7f . Q

2z

'T/ +.z

1 2 log1 -e-27f'T/d771 Re z

> 0.

(*10)

In order to determine ci and c2 in (*9), we need to prove t hat lin1z_... 00 J( z ) = 0 in case Re z > xo > 0. Separate J( z) into two integrals as

J (z) On [O,

lf J, 111

2

=

lo

lzl/2

+

l oo = Ji (.z) + J2 (z) .

. lzl/2

+ z 2 1> lzl 2 - 11112 > lzl 2 4 IJ1 (z)I < 3nlzl

-

r=log

Jo

~ lzl 2 = f lzl 2 hold and thus, 1 1-

e -2r.11

d71.

\Vhile, on [ ~ 1 00), 17'J2 + z 2 = lz -i11l l.z+ i77I > xol z l (this follows obviously by the relative position of the points z , ±i77, the line Re z = xo and the set 17 > ~). Consequently, 1

Since

by compari on, it follows that f 0 log (l - e- 2;r11) - 1 dT] < oo. Hence both J 1(z) - 0 and J2 (z) - 0 hold as z ~ oo. And the claim follows. 00

To prove c1 = 0: Applying I'(z + 1) = z f (z) to (*9) and using t he fact that lilnz~oo ( z + ~ ) . Log (1 + ~) = 1 (here, one may just asstm1e that

Sec. 5.6.3

F~Lnction

The Asymptotic

of I'(z)

223

z = x > 0) , then c2 + c1(z + 1) + ( z + = Log z + =?

c1 = -

c2

(z+ ~)Log

as z

-+

~)Log (z + 1) -

z - 1 + J(z +1)

~) Log z -

+ c1z + ( z -

z + J (z)

(1+;) + 1 + J( z) + J( z + 1)--+ -1+1 = 0

oo.

To prove c2 = ! log27r: Set z = ~ + iy inI'(z)I'(l- z ) = sin7r7rz · By ( *8 )

with c1 = 0, we have

(~ + iy) -

2c2 - 1 + iy Log = log 7r

-

iy Log ( ~ - iy) + J ( ~ + iy) + J ( ~ - iy) (*11)

log cosh ny + 2k7fi,

·where k is a constant. Since Log is the principle branch in C - (oo, OJ, each term in both sides, except 2k7ri, turns out to be real if y = 0 is assumed. It follows that k = 0. On the other hand, if IYI >

!,

· L (1 ·) . L (12 -iy. )

iy og

~+ iy

-iy og

i2 - iy. = iy { Jri +Log 11. -+ r2iy1~Y }

:-1 + iy .

= iy Log

00

.

= iy

{

.

ni

+ I:

(

n

n =l

= -Jry

if

-1) n- 1

+1+ 0

(

-. 2iy

1 ) n

00

+ I: -1 n =l

-. 2iy

( 1 ) n}

n

( _; ) : Y" ,

IYI > 0, log cash ny = log

1 + e-

2

'/rY

e_ Y 2 11

= ny - log 2 + 0

(

1 )

e 27rY

.

Putting these two results in ( *11), known that k = 0, then 2c2 - 1 - ny

+ 1+ 0

(y

1 3)

+J

(

~ + iy) + J ( ~ -

iy)

1 = log7r - ny + log2 + 0 ( -e 21fY ) . Let y

-+

oo in both sides after ny is cancelled and we get c2 =

! log 2n.

224

Chap. 5

Fundamental Theory

! + iy in f (z) r (l -

A nother proof fo r c2 = ~log 211: Set z = Then

r (~ + iy ) r (~ 2

2

=> (observing t hat

r (~ -

iy )

iy) =

=

z)=

sii; -;r z ·

211e-rry

1 + e- 2 rry

r (~ + iy)

(refer to Exercise A(l )(a) of Sec. 5.6.2))

1 + iy) = log v~ 1 log ( 1 + e - 2rry) . R e Log r ( 2 27r - 1rY - 2 2

On the other hand, set z = ~

Re r

( ~ + iy) =

c.2

+ iy in (*9) -

ytan -

1

with c1 = 0 and we have

2y -

~ + Re J ( ~ + iy)

1

where tan - 1 denotes t he principal branch. Combining the last two expressions, we have

For x and y restricted to x > 0 and y > 0 1

R e J (x + iy) =

x(x2 + y 2 + rJ2)

~ f 1f

Jo


log

x

( x2

1-

e-rry

< _4_x {oo log 7ry2

100

1

Jo

17)2]

1 . cly 1 - e-2rrry

+ ~) .Jo

+lou 0

+ (y -

1 dry 1 - e-2rr71

x

x2

u 2

+ (y -

d 77) 2 TJ

1 d77 + log __1__ 2 1 - e- 7r77 1 - e-rry

=> y-oo lim Re J

(~ + iy) = ~

-!;

j

_

x

x

2

+ 172 cl17

0.

Since limy- y (tan- 1 2y - ~ ) = setting y ~ oo in (* 12); it follows t hat c2 =log ,/2;. In conclusion , we have proved t hat log I'(z) = ~ log 211 - z + (z - t ) log z + J( z), Re z > 0. vVhat remains is to estimate J (z).

Sec. 5.6.3

225

The Asymptotic Function of I'(z)

Ste p 4. The estimat e of J( z). Using t he identity

z T/2

1

+ z2

1

= ; . 1 + (TJ/ z)2

1 { 1 - 2-+ 2 =z z2

+

1n

( '2._2) 2z2

···+ (-l )n-1

( 2_2) n-1 z2

} ('112/z2)n ., .

(-) l +(TJ/z) 2 '

J (z) in ( *10) can be rewrit t en as

,n

ck

J (z) = ~ k =l

z

Zk- 1

C k = (- l )k- l !.,

+Jn (z),

where

loo TJ

1 ·log 1 -e -

11-,

Jn(z) = (- 1)

1

n

Zk- 2

0

1 laoo . ·7f . 0

z2n+l

T/ 2n

1 + (ry/ z)2

log

for 1 < k < n,

dTJ

271'77

1 1_

e-Zr.17

dry , '

and

n > 1.

To evaluate Ck : By invoking the change of the variables t = 27f'r), then Ck = (- l )k- l _1100 ( - t ) 2k-2 {-lo0'(1 -e- t) }· -dt 1 = (- l )k -. 1f

27f

0

1f

1

27f

b

loo t k- 2 log(l-e- t)dt . 2

(27r) 2k-1 . 0

It is well-known that log (1 -e-t) = uniforn1ly in (0)oo) . Therefore,

L: =oe:;:t (t > 0) converges locally

(-t ) 2k-2 e -t -dt -- - 2= - 1 loo t2k - 2e - tdt -- - 2= -m1 100 o rn ni m oo

oo

m= l

m=l

1 2= -v;, m 00

= - r (2k - 1)

1n= l

1

< k < m.

2k

' 0

226

Chap. 5

Fundcimental Theory

s

·-. 5 o f ec. or.;: .4 .1, we l1ave . Accor·d'mg to E xerc1::;e k > 1, it follows that Gk

""00 1 = L....rn =l mrr

22k-1 (B21, k) ! 7r 2k , 2

=

) 1 2 2k- 1 B 2k 2k 1 ( . ) ( k,, ( -l ) k ;1 · (27r)2kl. - l 2 - 2 .· . (2k)!1r

=

(-l)

k- 1

B 2k

k > 1.

(2k-1)(2k)'

To the last, in case Re z > xo > 0, .

l

1

1

z 2n J (.,.) I < . n ~

-

Iz

·J

loo I1 + (

172n

looo I

1

= -

'f]2n

T/ 2

7r

'fJ / z) 2

7r . 0

lzI

1og

+ z 21

lou b

J

1 _2

1- e

exactly the same as the proof that J(z)

--+

1

1 - e - 2m7

'lr'fJ

dry

--+

dn .i

0 as z

--+

oo ,

0, if Rez > xo > 0; in (*io) . D

As a supplement to (5.6.3.4), we want to introduce another form of Stirling's forn1ula and sketch as follows. Set

µ(z) = Log r (z + 1)- ( z +

~) Logz + z -

logJ2;,

z E C -(oo, OL (5.6.3.5)

called t he Binet function . µ(z) assumes real values in case z = x > 0. µ(z) 1 has s01ne other representations such as µ(z) = J0 (t - ~) 'lj;(z + t)dt (where

!

'lj; (z) is Gauss's 'lj;-function ~{:/ ) = - J0 7Z~f dt (where p(t) = t - [t] for real t) , both having t heir own interests (see Exercises A(l) and A(2)) . Differentiate now µ( z) and we obtain 00

1

µ' (z) = 'lj;(z + 1) - Log z + - . 2z Using the first integral representation of 'lj;(z) in Exercise A(7) of Sec. 5.6.2 and the facts that Logz = J000 e- t_/ - tz dt (see Exercises B of Sec. 4.12.4C) and ~ = J000 e-zt dt , it follows t hat

'() .looo (t

-1 - -1 -

µ z =

2

1 ) e - zt dt , et - 1

R e z > 0.

!-

Since It - ! - et:._ 1 < !if t > 0 and limt-.o+ (t et:._ 1 ) = 0, t he above integral converges locally unifonnly in Re z > 0. Then, integrate µ' (z) along the line seg1nent joining 1 to z (R e z > O) a.nd change t he order 1

Sec. 5.6. 3

The Asymptotic Function of f'( z)

227

of integration. As a result , we have

µ(z) - µ( 1)

=

lo

1 1 1) e-zt + et

oo (

-1

0

2

-

t

t

dt

.fooo (et~ 1 + ~ - ~) e ~t dt. Applying the known expansion (see Exercise A(4){h) of Sec. 5.4.2) 00

1 +~- ~ et - 1 2 t l

1

=> ( e t - l

= 2t ~ L

n=l

1) 1

t t
O

(*1s)

Rez > 0.

the above as the

Stirling's form'Ula in Binet form .

r( z + 1) = (~) z Jz;;:;e~i(z),

z E C - (oo, OJ,

where the Binet function µ(z) (see (5.6.3.5)) satisfies Jµ( z)J Re z > 0. In particular, for real x > 0 ,

f (x) with 0 < B(x) < 1.

=

~

.

J.


e2 1Ttz - 1 2

'Yn

+ ~ f (k) . ~

k=l

(*16)

232

Chav. 5

Fundamental Theory

Finally, t he restricted condition will impose that t he second and the t hird integrals in the left approach 0 as n - oo. D

(1) Choose J(z +t) = (z ~t)2 , Re z > 0, and use (5.6.3.7) to derive (5.6.3.2). (2) Apply (5.6.3.7) to J(z) = (z; lF and show that

l oo 6y +- y2)3 2y + o . 3

1

" -=1 ~ n3

.

(1

1 e21fy -

1

dy

.

(3) Let / = liinn-+oo {1 + ~ + · · · + ~ - logn} be Euler 's const ant. Show that, by applying (*1 6) to f( z) = z~l, 1

'Y

5 .7

loo 1 +Y

= 2+ 2· O

1 y

2

e 21fy -

1 dy.

The Riem ann zet a Function ( (z)

Let z = x + iy a nd n > 1 be integers. In this section , log n in n - z = e - z log n is always understood as the real logarithm. Since ln-zl = n-x and I:~ ;"' < L~ ~ for x > 8 > 1, hence 2::: 1 7~" converges unifonnly in Re z = x > 8 > 1. It follows that (refer to Example 3 in Sec. 5.1.1)

The Riemann zeta function . The series 1

=" L......t nz 00

( (z)

n=l

converges a bsolutely in Re z > 1 and uniformly in every clo::;ed half-plane Re z > o > 1 (and, hence, locally uniformly in Re z > 1) to an analytic function ( (z), called the Riemann zeta function , simply denoted as the

(-series.

(5.7.1)

This function originated from number theory . It plays an important role in the study of distribut ion of prime nun1bers . Let 2, 3, 5, 7: 11 , 13 , 17, 19, 23, ... , Pn, .. . be the sequence of all t he prime ntunbers ( uppose there are infinitely many of t hem a nd a re arranged in increasing magni tude). A bsolute convergence of t he ( -series in H ez > 1 permits us to write

Sec. 5. 7

where

ni2

The Riemann Zeta Function ((z)

> 1. Since

runs through all the odd positive integers

L

m2~l

1 1nz


1 and unifonnly in Re z > 8 > 1, and approaches 1 in Re z > 1 as n --t oo . Therefore 1 setting n --t oo in ( *1 ). we obtain n

((z) lim

n -->oo

II(l -p; z) = 1

(*2)

k= l

absolutely in Re z > 1 and uniformly in Re z summarize the above as

> 8 >

1. In conclusion,

Th e infinite product representation of the (-function. Let p1 = 2, p3 = 3, ... , Pn , . .. , be t he sequence of all pri1ne numbers in increasing magnitude. Then, 00

((z)

=II (1 - p:;; z )-\

Rez > 1 or

n= l

1 - -

((z)

absolutely in Re z

=

II (1 -

n =l

_.!_) .' rn

Rez > 1

> 1 and unifonnly in each closed half-plane Re z > 8 > 1. (5.7.2)

From (*2), it is easily seen that ((z) =I= 0 in Re z > 1. Note that. in the above process lending to ( *2), it is preswned that there are infinitely inany priines. In case it is not and PN is the largest prime, then we would stop at

Chap. 5

234

Fundamental Theory

(*1) which turns out to be ((z)(l - 2-z) · · · (1 - p·;/) = 1. This will i1nply that ((z) has a finite lin1it as z - 1) contradicting to 2:: 1 ~ = oo. T his argument, due to Euler; obtains purely algebraic result through analytic method. The following is divided into two subsect ions. Section ( 1) Complex line integral r e presentation of the ( -function and it s a n a lytic ext e n sion Use nt to replace t in f (z) = 000 e- ttz- 1dt; Re z > 0, n > 1, and we have

J

n-zr (z )

/ oo t z- le- ntdt , =Jo

Re z > 0,

When restricted to Re z > 1 + xo ( xo > 0):

r(z )( (z)

2. n

= r (z) ;~~ = lim

=

oo

t z- l

k- z

= ri~'!!?oo .la

e- t _

n-+oo .l0 oo t z-1 dt .l 0 et - 1

oo

> 1. n

e-

1

{;

e-ktdt

e - (n +l )t

dt

1 - e- t

. hm

n

l oo

n ~oo . 0

tz-l

e-(n+l)t 1 - e -t

dt,

o- I

where fa ~z _ 1 dt converges absolutely and uniformly in Re z > 1 (why?). \i\Thile, on using 1 +t 0,

l

oo

.o

tz-1 e

-(n-Ll)t

1 - e-t

dt
1 0

converges absolutely and locally uniformly in Rez > 1. (5.7.3) Note that , in tz-l = e 0 along the upper side of (0, oo) and Log t = log t + 27ri a.long t he lower side of (0, oo). Integrate F(t) along 'Ye. a.ncl we have

1

/e

t z-1 - -dt = et - 1

le:

•

00

t z-1 dt et - 1

= (e27ri z

-1)

+

l

.

IZ 1 ~ -~

tz-1 dt et - 1

+

l oo tz-l e27ri(z -1) dt

•

et - 1

e.

roo . : z-1. dt + r :z-1. dt. Jc: e J lz l=e. e 1

(*5)

1

o,;\ring to Cauchy's integral theorem, fit l=e.i = fitl=c: 2 holds for any 0 < s 1 < c2 < 27r. This indicates that the integral fi tl =e: in (*5) is a constant, independent of the choice of such £ . It can b e evaluated as follm,.rs: Along ltl = £, if z = x + iy,

!tz-11

= e(x - l)loge. -yArg t

let - 1I > Nlc

=?

r :z-1 dt .flt l=c e - 1

(NI > 0 , a constant)

r

sx- le211lyl

- .fltl=e.

.fl!k


1.

On letting £ --t 0 in the right side in ( *5) and using the above result, we have, by (5.7.3).

l

. "!
1.

236

Chap. 5

Fundamental Theory

Summarize as

The complex line integral representation of the ( -function. Let /, (0 < (:; < 27r) be as in Fig. 5.27. Then, if Re z > 1: ((z)=

.

(e21Tiz

1

1 -

l )f (z)

l

_ _ f (l - z) 2 . 7rt .

"'fe

tz - l

et - 1

dt =

e - 1Tizf

( l - z)

27ri

1 "Y~

t z- 1

et - 1

dt

(-ty - 1 t 1 dt, e -

"'f,,

where t acts a::> a complex variable, and p-l = e(z- l ) Log t and Log t is the principal branch in C - [O, oo) (and hence, designate Log (- t) = log I - t l + iArg ( - t) with - 7r < Arg ( -t) = Arg t - 7r < 7r on the third integral) . (5.7.4) The integral in (5. 7.4)

1 'Ye

t z -1

et - 1 dt.'

ZE C

defines an entire function. This is so according to (4.7.4), or , more accurately, according to Exercise B (2) of Sec. 5.3.3. Therefore, the meron1orphic function e-1Tiz I' ( l. -

z)

27ri

l

.

''fo

t"t"'- 1

dt,

e - 1

z E C - {1. 2.... }

has possible simple poles at z = 1, 2, .. .. Since ((z) is already known to be analytic in Rez > 1, the for111ula in (5.7.4) shows that t he poles of (* 7 ) at z = 2, 3, .. . must cancel out the zeros of (*6). So, the fun ction in (* 1) has only one simple pole at z = 1 with t he associated residue given by e - 7Tiz (

r

t z- 1

7ri .J'Y~ et _ 2 =

e - 1Ti Re

1

s(

)

dt

evaluated at z = 1 · Re s(f (l - z );1)

1 ; e - 1 t

o) Re s (f (

1 - z ); 1)

= (-

1) · 1 · ( - 1)

= 1.

In summary, the importance of (5. 7.4) lies in the following fact.

The analytic continuation of the ( -function f orm Re z > 1 to C - {1}. Define

zE C - {l}.

The Riemann Zeta Function ((z)

Sec. 5. 7

237

Then (( z) is meromorphic in C - {1} with only one si1nple pole at z = 1 and Res(((z); 1) = 1.

""oo

1 - 2t + D n =l (2n)! B2n t2n K nown that. et t.'._ 1 , Jtl < 2n (see Example 5 of Sec. 4.8) , it follows immediately that

((0) =

1

-2;

((- 2n) = 0 ,

n > - 1· )

-B2n

n , n > 1. 2 The points -2n(n > 1) are called the trii1ial zeros of the ( -function.

((1 - 2n) =

(5.7.5)

Sorne b ool 0. Choose a path '"'In shown in Fig. 5.28, composed of the following curves: the upper side of t he cut [(2n + l )7r, oo), descending from oo to (2n + 1)7r ; then followed by a square wit h vertices at (2n + l )7r(±l ± i), in counterclockwise direction; and finally, a.long t he lower side of [(2n+ 1)7r, oo) form (2n + l)7r to oo. Let rc.,n be the circle ltl = ~,where 0 < $ < 271'. Note that '"'(n - '"'(,,n approaches ~le. in Fig. 5.27 a.s n --+ oo. Now, for 1 < k < n, Res(F (t): 2kni) + Re s(F (t); -2k7ri) = (2k7ry-1 eni(z - 1) { e'lri(z-1) / 2

+ e-1l'i(z-1 )/2}

. 1 = -2(2kn)z- 1 e111 z sin 211z .

... 0 ].,,,,,

I\

0 \ I

'.:;1

l(11 l1:r -r ~'I '~ •• - ..~------------.....- - - - - : : - ' i1 '----::-:. -----'

I\

~11~l1:n

Fig. 5.28

Sec. 5. 7

239

The Riemann Zeta Functi on ((z)

By the residue theorem, t z -1

1

t

e - 1

fn - "{""'

t z -1

l

=?

n

.

• "{o,n

et - 1

dt

= 27ri

L

{Res(F(t ); 2k7ri) +Re s(F(t); -2k7ri )}

k=l

dt =

1

tz- 1

, et - 1

dt

n

+ 47rie71'iz sin ~ ~ (2k7ry- 1 . 2 L

k= l

In

Wha t re1nains is to evaluate the integral

J'

a.s

'"'f1i

oo.

ri -

Divide !n into 11\.1) + ry~1 ) where /~l) is t h e part on the square and 1~2 ) that outside the sq0_)re, namely; the upper and lower side of the cut [(2n+ l)7r, oo) . Along 'Yn :

jet - 1I > inin let - 1 I = rn > 0 ....,(1} ,n

ltz - 1 1 = e(Rez- l)log Jtl- (Im z)Argt


1) of sin rr2z can cel against t he simple poles 2n of f (l - z) , and t he simple zeros 2n + l (n > 1) of ((1 - z) cancel against the simple poles 2n + 1 off (1 - z L t oo. Consequently, t his function is indeed analytic in C - {1} and agrees \vith ( (z) , by (*9) , on the open set Re z < 0. Hence, ( *9) is true for all z # 1. By t he \Vay, t o compute ( '(0): Differentiate logarit hmically the equivalent equation obtained by repla,cing z by 1 - z in (*9) and we obtain

( '(1 - z) 7r 1fZ f' (z) ( ' (z) - ((1- z ) = -log 27f- 2 tan 2 + r (z) + ((z ) .

(*10)

The mero1norphic functions in the right have t he Laurent expansions at z = 1 as: 7f 1rZ 1 - tan - = +O( Jz - 11) 2 2 z- 1 (see Exercise A(5) of Sec. 4.8 and Exercise A(2) of Sec. 4.9.2);

r' (z ) r' (l ) r (z) - r(l ) + .. . = -1 + analytic t enns in (z - 1) (see (5.6.2.2));

( ' (z) _ _ _(__ z _ l __ ) -_2_+_c_+_·· _·_ ( (z) - (z - 1)- 1 + r + c(z - 1) + · .. 1 + ')' + · · · (by using (5.1.1.4)), z- 1 where / is Euler's constant. Letting z -+ 1 in ( *10) a nd not ing that ((0) = (see (5.7.5)) , it comes ( ' (O) = - ~log 2n. D

-t

z to can cel out t he simple pole of ((z) at z = 1 in (1 - z)((z) , and the simple As for u,n othcr form of the function u,l equa tion , we use t h e fu,ct or 1

Sec. 5. 7

The Riemann Zeta Function ( ( z)

241

poles of r (~) to cancel against the simple zeros - 2n (n > 1) of ( (z) in r (~) (( z) . And one more factor z is needed in order to remove t he simple pole of r ( ~) at O. Consequently, we formulate

Another form of the (-function. The function

~(z) = ~z(l 2

z)7r- z/2r

(:_) ((z) 2

is entire of order 1 and satisfies the relation: '(z) = ~( l -z);

e(~ + ·iz)

zE C -{0,1} ,

='(~ -iz) , z

EC - {

or

±~i} ,

·which i::; equivalent to Legendre's duplication formu la (see (4) in (5.6.2.1)):

J7[r(2z) = 22z -l r (z) r ( z + ~)

.

(5.7.7)

Sketch of p r oof. In case z =/= 0: 1, ~(z) =

{:::>(by definition

of~(z)and

(~) ((z)

7r -z/2r

cos

~z r(z)r ( 1 ;

{:::> ( on using r (

1

;

z ) = 2z- 1 7r 1! 2 r

z) r ( 1 ~ z) = ~

cos

~z

(~) 7r

COS7l' Z

/'> ) J;r(z) ~

Set 2z in place of z in this last identity and we get Legendre s formula. As for the order of '(z), it is sufficient to estimate the growth of l((z)I for Rez > because of c;(z) = , (1 - z) . Owing to (5.5.3.4) and Exercise A(l ) there, it is thus sufficient to estimate the growth of r ( ~) and ((z) , respectively.

!

Chap. 5

242

Fundamental Theory

Th e growth of r (~): By Stirling's forrnula. (5.6.3.4) ,

r ( ~) =? log

·\ /'2; (~) (z-1) / 2 e-zf2eJ(z/2)

=

jr (~) \ = ~log 211 +Re ( z ; - Im ( z ;

=?log

1

)

jr (~) 1 < J\1lz l log lzl

and Rez

1

) log

I~ j

Arg~ - ~Re z + Re J ( ~) for s01ne consta.nt 111

> 0 and large lzl

> 0.

(*n)

Moreover , by Stirling's fonnula in Binet fonn (5.6.3.6) , this estimate is precise for real z = x .

The growth of ((z): Denote by [xJ the Gaussian integer for real x . It can be shown that , if Re z > 1,

((z) = z

f,

. 1

00

x~J 1 dx

N 1 Nl-z =~- + -z L..t kz z- 1 k=l

loo . N

X -

[x]

. dx

xz+l

for any positive integer N > 1. The adva.ntage of the second expression in ( *12) over the first one lies on the fact that the integral .fi~ :~41 1 dx converges for Rez > 0. Therefore, (*12) is still valid for Rez > 0. Vilhen restricted to Re z >

!,

l

oo x - [x] dx < loo

·N

xz+l

.JV

1 x Re z+l

.2 < Nflzl3,

lzl is

N-Rez

dx =

Rez

< 21v-!

where NI is a constant )

(*13)

')

large and JV is chosen as the integer closest to lz l ~. Co1nbining ( *n) and ( *13)) and the very definition (5.5.3.3) of the order, it follows that ~ (z) is of order 1. D

if

The infinite product representation (5.7.2) and the fact that ((z) is analytic in {Re z > 1} - { 1} so that (z - 1)( (z) is ana.lytic and zero-free in Re z > 1 can be used to prove the following ren1arkable

Sec. 5. 7

The Riemann Zeta Function ((z)

243

Prime numbe r t h eore m . Let 7r(1'l) denote the number of the primes not larger than the positive integer N. Then,

7r(N) log N _ . 1nn N - 1, N~oo namely, 7r(

)

rv

Jl,T/ log J -.

(5.7. )

One may refer to Ref. [71] for a proof. R e mark (A mille nnium prize proble m ). Let C be an elliptic curve over Q (the rational field). An affine inodel for the curve in \Veierstrass form it> given by y 2 --

x 3 +ax+ b,

a, b E Z (the set of integers).

Set

JVp

=

the discriminant of the cubic; the cardinal number of the set {the solutions of y 2 n1od p}, and

Clp

=

p-

l::i=

=

x3

+ ax + b

p

The incomplete L-series of C is defined as

L (C, s)

=

IT (1 - app-s + Pl-2s)-1, pj2c.

considered as a function of the complex variables s (compare to Exercise A(lO) of Sec. 5.5.2). It has been proved in 1995, 2001, that L(C, s) can be analytically continued fro1n Res > ~ to the whole con1plex plane. Hence comes the

Conjectures (B. Birch and H. Swinnerton~Dyer, 1965). The Taylor series expansion of L( C, s) at s = 1 has the form L( C, s)

= c(s -

l)r + higher order terms

with c :f. 0 and r = rank(C(Q)) .

For details, read Andrew vViles: The Birch and Swinnerton- Dyer Conjecture, Clay NIath Institilte Nfillenniitm Problems, www.claymath.org/ millennium.

244

5.8

Chap. 5

Fimdamental Theory

Normal F amilies of Analytic (Me romorphic) Functions

The concept of normal family was initiated by P . l'vlontel in 1907. Let n be a. domain in C * (see Remark 2 in Sec. 2.4) and ~be a family of analytic or n1ero1norphic function in n. If every sequence in ~ contains a subsequence that 1. converges locally uniformly in

n to an

analytic or a mer01norphic func-

tion, or 2. converges locally uniformly in n to oo (namely, for each compact set K inn and each positive number R , there is an integer N = lV(K, R) > 1 so that lf n(z)I > R for all n > N and z EK), (5.8.1) then ~ is said to be normal in n. In case n is a domain in the finite complex plane C and J contains only analytic functions and Case 2 does not happen, the convergence is in the Euclidean metric; otherwise, in spherical chord metric (see Sec. 1.6). This is the classical definition for the nonnal fa1nily and is the version for functions of Bolzano-Weierstrass's theore1n for bounded infinite point set or sequf)nce (see Sec. 1.9) . Section 5.8.1 will establish criteria for normality. Here, in Exercises B, we will fonnulate the concept of nonnal fan1ily in tenns of metric spacf) with Ascoli- Arzela's theoren1 as our ma.in concern. Lots of concrete examples are listed in Sec. 5.8.2. Important ones among a.11 are fan1ilies of uniforn1ly bounded analytic functions, of univalent functions in the unit disk, etc. The rnain the1ne in Sf)c. 5.83 is to prove J\!lontel's normality criterion, obtained by l\/Iontel in 1912, v,rhich relates to the values taken by inembers of the fan1ily. There are several proofs for it. The one we used is an application of the elliptic inodula.r function (see (5.8.3.5)) and the n1onodromy theorem (see (5.2.2.1) and (5.2.2.2)). This criterion will, in turn, be adopted to prove Picard 's first and second theorems as we have mentioned in (4.10.3.3) and even n1ore such as .Julia direction, covering problems, Landau and Schottky theorems, etc., all beyond the scope of this book. Also, SchwarzAhlfors 's Lemn1a in Sec. 5.8.4 can be used to prove Picard 's and Montel 's theore1ns through the method of Gauss 's curvature. See Refs . {2, 51] for details. As a n1eaningful application of the concept of the normal family and Schwarz- Pick's theorem , and as a continuation of Sec. 5.4.3, Sec. 5.8.5 presents some results in con1plex dynamical systen1s which are of current research .

Sec. 5. 8.1

5.8.1

Criteria for Normality

245

Criteria for normality

Section (1 ) Family of analytic functions vVe had hown in (5. 3.3.10) t hat the local uniforn1 boundedne s is a sufficient condition for a sequence of analytic functions to be normal in a domain. It is indeed a necessary con dition , t oo. To extend this result to a family J of analytic functions defined in a don1ain n, some terminologies about ~ are introduced as follows: 1. J is locally uniformly bounded in n if, for each compact set sup sup lf(z) < oo /E& zEK

J(

in n ,

(5.8.1.1)

J

holds. This is equivalent to say that J is uniformly bounded in a neighborhood of each point in n. 2. J. is locally ( un'iformly) equicontinuous if, for each con1pact set J( in n and c > 0, there is a = (I 0 such that

o o

sup

sup

lf(z)-f(z')l (2): This is obvious. (2) => (3): Suppose, on the contrary, t hat J fails to be locally unifonnly bounded in n. Then , there is a. compact subset K of n such that. for each integer n > 1, there a.re a function fn E J and a point Zn E K so that If n(zn)I > n. \ \Te inay suppose the sequence Z n itself converging to a point

zo E ](. By assumption , J is normal in a neighborhood Ozo of zo. vVe may, as well , assume that the sequence fn itself converges locally uniformly to an analytic fun ction fin Ozo . In particular, fn(zn) _... f( zo) as n _... oo (why? see Exercises B(4) and (5) of Sec. 5.3.1). This obviously contradicts to the fact that lfn(z.1JI > n fo r n > 1. (3) => (4): Only the local equicontinuity is needed to be proved. Fix any point (o E n. Choose a closed disk lz - (o l < r , contained inn. For any two points z and z' in I( - (o I < r, v,re have by Cauchy's integral theorem,

f( z ) - f (z') =

_ 2 1. 7rt

r

11( - (ol=r

= z - z'

((

1

-

- ( 1 ,) Z

-

J(()d(

Z

r

!(() d( 21fi 1 1( - (ol =r (( - z)(( - z')

=> IJ(z ) - f (z' )I

lz - z' I
l

is locally bounded either in Imz > 0 or in I1n z < 0, but not so on domains with nonempty intersection with the real axis. (3) i cannot be uniformly bounded in any don1ain in C and hence, is not nonnal there in the Euclidean sense. vVhile, the family 2l al 1 + laz

+ bl2 '

a, b E C and a

f. 0

is not bounded at z = 0. This shows t hat J is not normal in C in the spherical sense. (4) Set fn(z) = e( 2n+ ~)zsin(2n + ~) z, n > 1. Thenli1nn- oo fn(1T) = oo while limn->oo fn.(27r) = 0. Consequently, f n(z) cannot have a subsequence converging locally uniformly to either an analytic function or oo in Iz - 7r I < 1T + 1. Then J is not nonnal in lz - 7r I < 7r + 1 in the spherical metric. As a prelude to the next exa1nple, consider the sequence f n (z) = ;(enz - 1) for n > 1. Observe that lin1n-oo fn(O) = 0 and li1nn-oo fn(x) = oo, where x is a fixed real number between 0 and 1 (excluded) . Hence, f,.(z ), n

> 1, contains no subsequence locally uniformly convergent in lz I < 1 to

an analytic function or to oo. And then , t he family of analytic functions f( z) on lzl < 1 that satisfy f(O) = 0 and f'(O) = 1 is not nonnal. Refer to Exercise A(3) of Sec. 5.8.3. Yet we have the following in1porta.nt

Normality of the family of normalized univalent (analytic) functions on the open disk. The family .& = {f(z) is univalent on lz l < 1. lf(O) = 0 and f'(O) = l}

256

Chap. 5

Fundamental Theory

is normal in lzl < 1 and is compact in the sense that the limit function of any locally uniformly convergent sequence from -8 is still in -8. "tvioreover , (1) (Koeb ·s t- theorem, 1907; Bieberbach and Faber. 1916) For each f E -8 , the in1age f ( {lzl < 1} ) always contains the open disk lwl < ~. (2) {!' If E -8} is norn1al in lzl < 1. (3) { If E -8} is nonnal in lzl < 1.

J,

Consequently, letting -8 (11; zo) = {f : n

n be a ---+

domain in

C is univalent.

c

and z o En a fixed point, then

IIf(zo)I < J..;J

(a constant ) and

lf'(zo)I < NI} is both norma.l and co1npact in D. Note. A one-to-one analytic function will always be si1nply called an univalent function. (5.8.2.3) Recall that (1) is the one mentioned in Exercise B(4) of Sec. 4. and the const ant is sharp as the muvalent function (l~z )'i , lzl < 1, h aving its image the domain C - (oo , ~] (see Example 1 in Sec. 3.5.7). For a proof, we adopt the one appeared in Ref. [63], pp. 215- 217.

t

Proof. Let f be univalent in

f ( =?

t:

(~)

lzl < 1. Then

is tmivalent and analytic in

lzl < 1 for

each fixed

(Kl < 1.

The normalized

::fz)-

J( J(() g(z) = f'(()(l -1(12) =?

The second coefficient of t he T aylor series expansion of g( z ) a t z = 0 is

g"(O) _ 2! =?

E -8 .

! { !"(()(1 -1(1 2 )

- 2

!'(()

(by using Exercise B (3) of Sec. 4.8 that says ( by z)

zf"(z) f' (z)

_

-}

2 (

.

l.i.}Qll < 2) and

replacing

2 2lzl 4lzl 2 . - -""""' 2 < 1 - lzl - 1 - lzl

Here. we h ould ren1ind the readers that J' (z) =/= 0 throughout lz l < 1 owing to the univalence (see (3.5.1.9)) . Consequently, log J' (z) has single-valued

Sec. 5.8.2

Examples

257

branches in lzl < 1 (see (3.3.1.8) or (4.4.2)) . Fix such a branch. According to Exercise A (6) of Sec. 3.2.2,

d

I

dz log J (z)

a

z f"( z)

=?

r(8Br log If (z) I + i Bra arg f (z)

= r 8r

Re f'(z)

I

=~

log If

/

(z)I,

I

lzl = r

)

.

lz I = r < 1

< 1.

Then, ( *i) reduces to 2r - 4 8 , 4 + 2r 1 - r2 < log If (z)I < 1 - r2 ,

or

=?

lzl

= r < i.

(integrating these inequalities from 0 to

lzl and

assuming that

f' (0) = 1, a nd t hen by taking exponentials) 1 - lzl , 1 + lzl (1 + lzl)3 < If (z)I < (1 - lzl)3' lzl < 1.

(5.8.2.4)

This is a local distortion t heorem for If' (z) I of the univa lent mapping w = J(z) with J'(O) = 1. Both the lower a nd the upper bounds in (5 .. 2.4) are sharp a the function cl-=-.:)3 shows. On integrating the second inequality in (5 .. 2.4) along the segment connecting 0 to z and by assuming that f(O) = 0, we obtain

r

lzl

= r

!or

L

l +r r {z J'(z) dz < lf'(z)ldr < ( ) dr = ( ')2, 3 o o l - r l - 1 .J0

lf(z)I =

< 1,

(*2)

an upper bound for lf(z)I . To obtain a luwer bounded for lf(z) I, consider first t he case t hat lf(z) I < By (1) , Koebe's quarter theorem, t he line segment connecting 0 to f (z) in t he w-plane lies entirely in t he ra nge J({lzl < 1}). Let 'Y be the arc in lzl < 1, which is mapped by w = f( z ) onto this line segment. Then dw = f' (z) dz > 0 along 'Y and then, we have

i·

If (z)I =

f J'(z)dz = f lf'(z)ldr >

.J'Y

.J'Y

1 r· ( - :; clr = ( r ·) .Jo 1 +1 1+ 1 3

2

.

(* 3 )

lf(z)I > i · Combining (* 2) and (*3), we have a local distortion theorem for lf(z)j if f E -8:

Since (l~-r)2 < ~if 0 < r < 1,

lzl

(*3)

< lf(z)I
1, vVeierstrass·s theorem (5.3.1.1) saysthat/(O) = Oandf'(O) = 1 hold. too. That/'(O) =1 shows that f is not a constant ftmction. Hurwitz·s theorem (5.3.2.2), in ttun, implies that f is univalent in izl < 1. Hence f E -8 and -8 is compact. (2) is a consequence of (5.8.1.3). As for (3): Suppose on the contrary that { If E -8} is not normal in lz l < l. Then there are a compact set J( in lz l < 1, and functions fn E -8 a nd points Zn E J( for n > 1, such t hat

J,

1

f/i (Zn )

==>

> n,

n> l

-

lf~(zn)I < .!., n

n > 1.

After adj usting to subsequences, we may assume that z 11 converges to a point zo E J( and t he sequence f n itself converges in izl < 1 locally uniformly to an analytic ftmction f E J. Since f is univalent; so J'( zo) :/= 0. On the other hand, since f n converges uniformly to f in J( , therefore (refer to Exercises B(4) and (5)(b) of Sec. 5.3.1) limnJ;1(z11) = J'(zo) = 0 holds, a contradiction. Thus, the claim follows. To prove the last statement , we 1na.y just assmne that n is an open disk lz - zo l < Rowing to (2) in (5.8. 1.3). Set

r) = f (zo + R () - f (zo) ( g "' Rf1 (zo) '

1(1

< 1 and f

E -S(n;zo).

Then {g(()l.f E J(!l; zo)} is t he family J on 1( 1 < 1 and is thus compact. Let 9n(() be a sequence converging locally uniformly tog(() in 1( 1 < 1. Note that fn( zo +R() = RJ;1 (zo)gn(() + f n(zo), n > 1, 1(1 < 1. Since lf n(zo) I < JVJ and lf~(zo) I < lvf for n > 1, t here are convergent subsequenccs fn; (zo) and 1 J(zo) for j > 1. Under t his circumstance, f n(z) has a subsequence f ni(z) converging locally lmiformly to an analytic function in iz - zo I < R and this limit function is still in J (!l; zo) . O

J:

S ke t ch o f a n ot h er proof for (5 .8.2.3) . Since each f E -8 satisfies f(O) = 0 a nd /'(O) = 1, it is certain that 0 is an interior point of the image f(D ), where D denotes t he unit disk lzl < 1 in what follows . If we are able to show that each f E J , after some adjustment, will not assume points in the

Sec. 5.8.2

Examples

259

same open disk, then (5 .. 2.2) says that -8 is thus normal. Ideas here will be helpful in proving t he R iemann mapping t heorem (sec (6.1.1)). First. note that; for each f E -8 , CXJ = dist(O . C - f( D )) sat isfies 0

< CtJ
0 is obvious owing to t he openness off. The disk lwl < C\'f is contained in f(D). The analytic function J- 1 (cx1() form 1(1 < 1 to lzl < 1 maps 0 to 0 and hence, by Schwarz's lemma, IJ- 1 (a1() I < Kl on 1(1 < 1. Setting a1

1- 1 (01() = z

1

we obtain a1(

lzl

!

nc C 'lj;(Dr) C 'lj;(Ds) C nc5 if 8 > 0 is sufficiently small.

Sec. 5.8.4.

R emarks on Schottky's Theorems and Schwarz- Ahlfors' Lemma

287

....

.. ...

• •

•

1/

.'. ... .

J

••

''

, ''

'

,,'

''

.' •.

•

,•'

· ······-··-·· Fig. 5.34

Finally 1 let such f (z) come into play. et f (0) = ao for a moment. Recall that ao =f. 0) 1. There is a zo, lzol < 1 so that 'lj;(zo) = ao . Pick up any fixed single-valued branch g(w) of ( = 'lf;- 1 (w) in an open neighborhood Vo of ao in C (0.1) so that g(ao) = zo . Choose an open neighborhood Uo of 0 in lzl < 1 so that f(Uo) C Vo. Just as in the proof of :Niontel's criterion, the function elen1ent (g o f 1 Uo) can be analytically continued into a single-valued an alytic function in lzl < 1, owing to t he monodromy theorem (5.2.2.2). Still denote t his extended function by

h(z ) = go f( z):

{lzl
0. 1

288

Chap. 5

Fundamental Theory

In conclui:>ion, we have another

Schottky's theorem (II). Suppose f (z) is analytic in JzJ < R and omits the valttes 0 and 1. Then, for any 0 < ::; < ~ and !II > 0, there is a 6 = c5(s, l\I) > 0 satisfying the property: for each z, JzJ < R, and f (z) E Sle: (see (*6) and Fig. 5.34): then

f (() whenever

1(1< R with the hyperbolic

E

D8

distance d( ~ , ~)

< NI.

(5 .. 4.4)

Fix a point zo, Izol < R , note t hat the hyperbolic open disk: d(71 , 7/0) < M, . . Rl z-zo l {:} the Euclidean open disk: IR2 _ zoz l 0 throughout n, dsp and the Euclidean metric Jdz J are conformal. The p-distance between two distinct points Z1 and z2 in n is defined by

(*11)

dp(z 1 .z2) = inflp(z)Jdzj . I

I

·where inf is taken over all rectifiable curves/ inn, connecting z 1 to z 2 . An extremal curve is called a geodesic connecting z1 and z 2 . The curvat'ure of p( z) at a point z E r2 is defined by ~

logp(z)

(5.8.4.12)

p(z)2 .,2

,,2

2

2

l

2

.

·

.

0 8 8 0 where ~ = c)x .u 2 + . 2 = 4. fJ _ = 8y ()z{)z 8r 2 + -r -8r + J,,. r - "§'§'1 (z = x + ?/Y = reie) is the L aplacian operator. This is the Gaussian curvature in the Riemannian space (n , dsp)· So far in this book, we have mentioned the following metrics, except the usual Euclidean (or parabolic) metric Jdz I (see Appendix B):

1. Th e spherical or elliptic metric (see Exerci e B(2) of Sec. 1.6 or

Appendix B)

ds

= 1 2Jdzj + Jzl 2 .

zE

C

. h K( ) wit z, a

=1

where a(z) =

2

I .

1 + z 12

Sec. 5.8.4

R emarks on Schottky's Theorems and Schwarz- Ahlfors ' Lemma

291

2. The non-Euclidean or hyperbolic metric (see Exercise B(5) of Sec. 3.4.5 or Appendix B) 2jdzl ds = l - lz l2'

ds = ldzl Imz '

izl < 1 with J( (z, >-.) = -1 wheTe >-.(z) =

2 - lzl 2 ; or 1

1 Iin z > 0 with K (z, >-.) = -1 ·where >-.(z) = - Iin z

They a.re also called Poincare1s metrics on lz l < 1 and 1111 z > 0, respectively. 3. Poincare's metric on the punctured sphere C*(O, 1,oo) (see (5.8.3.10))

ds = l(-.) (z)

= >-.(j(z))IJ '(z) I = (def. )

21 f'(z) J 2

1 - lf (z) I

or (f *>-.)(z) ldz l,

z En, (*13)

Poincare's metric on n induced by f (z) . It is invariant under conformal self mappings of n. Quite generally, an induced inetric, such as (* 13 ), can be loosely defined as follows. Let f(z) : D1 (a d01na.in) --7 f22 (a. domain) be a nonconstant analytic function (not necessarily univalent) and p( w) be a metric on D2. Then

(f* p)(z) = p(f(z)) lf'(z)I , (clef.)

z E n1,

(5.8.4.13)

is a. metric on f2 1, induced fro1n p by f ( z) . A vivid example is provided by (5.$.1.7) . J'vforeover, J{ (z, f *p) = J (in terms of (* 13)) (f* >.)(z) < >.(z) = dsr>.

1

lz2I < 1

2 _ lzl 2 ,

lzl < 1 or

< cls>.

(*14)

with equality only if f (z) is a bilinear tra.nsfonnation of lzl < 1 onto itself. Via the concept of curvature int roduced in (5.8.4. 12), we have

>.(z) 2

=-

J((z, >.)- 1 Ll log >.(z)

= Ll log >.(z);

(j*>.. )(z) 2 = - J((z, j *>.)- 1 Ll log(j*>.)(z) = Lllog(f* >.)(z), and hence, ( *14 ) is equivalent to the validity of the inequality Ll[log>.(z)-log(j* >.)(z)] > 0,

lzl < 1.

Owing to the fact that lin1 lzl - l log >.(z) = , if it i po sible to keep log(f*>.)(z) remain bounded in lzl < 1. then lim1z1-i[log >.(z) log (!* >.) ( z) l = 00 will i1nply that the C 2 function log >. (z) - log (f* >.)( z) would have a minin1un1 at some point zo in lzl < 1. This, in turn , implies that Ll [log >.(zo) - log(f* >.) (zo)] > 0 holds (•Yhy? the trace of the positive semidefinjte Hessian matrix at zo) . Then (*15) holds throughout lz l < 1. This ob ervation see1ns reasonably lead to the fo llowing Sch warz- Ahlfors's Lemma (1938). Suppose j(z): {lz l < l} ~ n (a domain) is analytic and there is a metric p(w) on n s·uch that its curvature I< (w, p) < - 1 holds throughout n. Then the induced m etric

(f*p)(z) holds everywhere in

= p(f(z))lf'(z) I
.(z)

=

2 l - lz l2 ,

lzl < 1

lz I < 1; in short, dsrP

Hence, an analytic function on lzl

< ds>..

< 1 will

decrease Poincare 1s 1netric. (5 . .4.14)

The cla ical Schwarz·s Le1nma (3.4.5.1) and Schwarz- P icks Le1nma (3.4.5.2) are indeed special cases of this one. See Exerci e A (9). S01ne historical com1nents: This theoren1 appea.red in Ref. [5] which initiated the

Sec. 5.8.4

Remarks on Schottky's Theorems and S chwarz- A hlfors' Lemma

293

study of geometric beha.viors of analytic functions by mean of differential geometry. Based on t his point of view, R. JVL Robinson [69] proved Picard's t heorems wi thout recourse to the tradit ional method using t he elliptic modula r function. Some mathematicians, like H. Grauert and H. Reckziegel [37], Z. Kobayashi [49], D. ~linda and G. Schober [59], made contributions in this direction. And theorems of this type appeared at first t ime in t he books by Ahlfors [2] a nd S. G. Krantz [51]. It is highly recommended to read t he article [59] and the books [2, 51] for detailed account a bout materials within the present Section (2). Sket c h of proof (Ref. [2]). Fix 0 < r < 1. Construct Poincar e's metric

2r

Ar(z) = r 2 - Iz 12,

iz l < r

on izl < r induced by h(z ) = ~ frorn izl < r onto lwl < 1 (see (*13)). The process leading to ( *15) reads now as (with f *p replacing f *>.) ~ [log(f* p)(z) - log>.r(z)]

> (f*p)(z)2 - >.r(z) 2 ,

(*16)

owing to J((z, f *p) = I< (f (z), p) < -1. The set E - {ziizl < rand log (f* p)(z) > log >.r(z) } is open in lz l < r and does not cont ain

the isolated singularit ies of f *p. F (z) = log(f *p)(z) - log Ar(z) is posit ive in E and (*16) shows that it is subharmonic there (see (6.4.1)) . Hence F(z) does not assume local maximum except it is a constant. In ea e E =fi 0. then t he boundary 8E =fi (/J . 8E has empty intersection wit h t he circle iz l = r, because F (z) = 0 along 8E by the continuity of F (z), while lim1z1-r log >..,.(z) = +oo and liin1z1-n· log(f *p)(z) > -oo. Consequently, F (z ) a ttains a local maxi111um on the compact set E which is contained in lzl < r; and hence F( z ) - 0 on lzl < r, a contradiction. Thus E = 0 holds and log (f*p)(z ) < log Ar(z) ,

::;. (f *p) (z ) < >.r(z),

lzl

lz l < r

.(z ),

lzl < 1.

T his finishes the proof. While, lVIinda and Schober [59] improved the above argument as follows. Consider - (! * p)(z) ( ) v z - Ar(z) ' lzl < r .

v(z) is cont inuous on lzl < r and> O; v(z) ~ 0 as lzl ~ r because (f *p)(z) is bounded in izl < r a nd >..,.(z) ~ oo as lzl ~ r . Hence v(z) assumes its

294

Chap. 5

Fundamental Theory

maximum at some zo , lzol < 1. It is t hen sufficient to show that v(zo) < 1. In case f *(p)(zo) = 0 1 it follows that f *(p)(z) - 0 in lzl < r . If f*(p)(zo) > 0, then logv(z ) assmnes its inaximum at zo . Hence 0 > ~ log v(zo) =~ log( /* p)(zo) - ~ log Ar(zo) =

- J( (zo .f*p)(f *p)(zo) 2 + I (f *p)(zo) 2 - Ar(zo) 2 ::::;. v ( zo) < 1. What remains is t he same as near the end of t he last paragraph.

D

A slight generalization of (5.8.4.14) is at hand . Observe tha.t t he metric c-\(z) , \;i,rhere c > 0 is a constant , has t he curvature - 1;r. As a consequence 1 for a fixed constant A > O, ,\A (z) R

-

2R VA(R 2 - lzi2) '

II z


. ~) = - A. On applying (5.. 4.14) t o Poincan~'s metric ,\(() = JAR,\~(R(), 1(1< 1, and the metric /Bp(w) on n, where B > 0 is a cons tant , we have

./Bp( f (R()) R lf' (R 0 is a constant. Th en

(f*p)(z ) = p(f (z)) if'(z) I < where >.~(z) is given by (5.8.4.15) .

.~ ,\~(z),

lz l < R , (5.8.4.16)

R em a rk (Ult r a h y p erbolic metrics and Sch ott ky's theorem) . A careful examination of Ahlfors's proof for (5.8.4.14) shows t hat the central idea. behind it lies on t he fact t hat the continuous function log >.r(z) log(! *p)(z) at tains its (local) ininirnum in izl < 1. In Ref. [2], Ahlfors formally introduced the notion of an ultrahyperbolic metric in a domain n: It is a metric PIdzl in n, satisfying t he following two propertie : 1. p is upper emicont inuous (nainely 1 lim z~ zo P (z) < p(zo) for each zo E f2). 2. At every point zo E 0 wit h p(zo) > 0, t here is a C 2 metric po in a ncighborhood Dzo of zo inn such that I po in Ozo

and p(zo) = po(zo) .

(5 . .4.17)

Sec. 5.8.4

Remarks on Schottky's Theorems and Schwarz- Ahlfo1·s' Lemma

295

In this case, log ).(z) - logp(z) is lower se1nicontinuous inn and the local minimum of log >..(z) - logp0 (z) is still assured. Localize the proof for (5 .. 4.14) at each point zo ·where p(zo) > 0 and we obtain

A gen er a lization of S ch warz- Ahlfors's Le mma . L et f(z) be analytic from lzl < 1 into a domain n on which there is given an ultrahyperbolic metric p(z) . Then (f*p)(z) =

p(f (z))IJ'(z)I < >..(z)

=

2

l - lzJ 2 1 lzl < l.

(5.8.4.18)

This result remains valid even if n is a. Riemann surface. The notion of an ultra.hyperbolic metric has recently found many new applications in the theory of several complex variables. There is no ult ra.hyperbolic metric either jn the ·whole plane C or in a punctured plane C - {zo}, where zo E C is a, fixed point. ( *i2) shows that we do have one in the punctured plane C (0, 1) via the elliptic modular fun ction. Indeed, we do have Th e existence and uniqueness of an ultrahyperbolic metric. Let n be a domain in C whose con1plement (in C ) contains at least two distinct points.

( 1) There is a metric p( z) in n whose curvature K (z, p)

< - A, where A > 0

is a constant. (2) There exists a unique maximal ultrahyperbolic metric .An in that

J( (z, An) = -1 ,

z

E

n , such

n.

Here "m aximal" ineans that , for any ultra.hyperbolic metric p(z) in n, p(z ) < >..n(z) always holds. Such a >..n(z) is called Poincar e's m etric in n. (5.8.4.19)

(1) is left as Exercise A(ll) . T he existence proof in (2) is concerned with the uniformization of the Riemann surface (see Chap. 7; in particular , the last paragraph on the last page) and one may refer to Ref. [2], p. 151 or Ref. [51], Chap. 3. Ana lytic expression such as t he one appeared in ( *l 2) is not of great value. vVhat we need at the present are the following Bounds for Poincare's metric (Ahlfors, 1938).

(1) In ea e f21 C f22. t hen A.r22 < >.n1 in f21. (2) For each z E n,

296

Chap. 5

Fundamental Theory

always holds . In case z is the center of an open disk n, the equality hold. (3) Let >.o, 1(z) denote Poincare)s metric of the three-point. puncttu-ed sphere C*(0.1,oo) = C (0 ,1). Then

Ao,1(z)
0 so that a(w) p(w) < lvh if w E K. Consequently,

a(w) p(w) < 111 = inax{A1li .1112} ,

w E C (O, 1) .

C01nbini11g (* 1 ) and (*19), we have

(f*a)(z) = a(f(z)) lf'(z) I < Nlp(f(z))lf'(z)I 21\llr , A(r2 - lz - zo l2)

< JA

-

lz - zol < r

and

f

E

This shows that {f*alf E J} is unifonnly bound ed on lz - zol hence is locally so in D. The normality of J is thus proved.

J.

< r and D

Exercises A

(1) (Landau) Suppose f (z) is analytic in Jz l < R, f'(O) 1 throughout. Then R < R(ao,a1)

~

0, and f(z) =/= 0,

1 - Jg(ao) I

= Jg'(ao)llail' ao = f(O) and a1 = f'(O),

where g(w) is any fixed single-valued branch at ao of the inverse function ( = 'lji- 1 (w) of the elliptic modular function w = '!/J(() : {lz l < 1} - C (O, 1) and R(ao , a1) is a constant. dependent only on ao and

Sec. 5.8.4.

Remarks on Schottky's Theorems and Schwarz- Ahlfors ' Lemma

299

a1 , independent of the choice of g(w). The equality R = R(ao ,a1 ) holds if and only if f(z) = 'lj;(h(z)), lzl < R , where ( = h(z) : { lz l < R} ~

{I(I < 1} is a linear transfonna.tion. .\tioreover for any fixed 1

ao # 0 1 1, and ai # 0, there exists s1ich a function J( z), satisfying f(O) =ao, f' (O) = ai, so that R = R (ao,a 1 ) holds. T ry the following steps. (a) Apply Schwarz- Pick's Lennna to the function h(z) in (* ) and con1pute h(O), h'(O). {b) In case g(w) is another such a branch, then g = T o g where T : {1(1 < 1} ~ {1(1 < 1} is a inodular transformation (see -· . , th . t IT'(zo)I 1 _ 1 ( l ?) (5'· 8"'· 3 9)) · Obsene a 1- JT(zo)l - 1-lzol· wiy. · (c) Set R = 1;- (ao) for given ao a.nd a1 _/.. 0. Then h(z ) = z +Rq(a-o) lg a.o !!ail

r

R+g(ao)z

maps lzl < R onto 1( 1 < 1. Let fo( z) = 1/J(h(z)), lzl < R. Choose B so that eiB f'(O) = ai. Does f(z) = fo(eiOz) work? (2) Use Exercise (1) to prove Pica.rd's little theorem. (3) Suppose f( z) is analytic in lzl < Rand /(z) :/= 0, 1 there. For any fixed 0 < J( < oo and 0 < JV.l < oo, then there is an JV(O < N < oo) so that lf(()I < /'-.l whenever lzl 1, be a sequence of analytic fun ctions in lz I < R so that fn(z) # 0, 1 t here for each n > 1.

ft,

(a) If there is a point zo, lzol < R, so that limn~oo fn (zo) = 0 or 1 or oo, then fn(z) converges locally uniformly to 0 or 1 or oo in lzl < R, respectively. Try to use (5.8.4.5). (b) In case liinn-.oo fn( z) = f(z) exists for each z, lzl < R, then fn(z) converges to f( z) locally uniformly in lzJ < R. Try to use (5.8.4.4), (5.8.3.1), and Exercise A(3) of Sec. 5.3.3. (c) Use (a) and (b) to give l\/Iontel's normality criterion another proof. (5) Let f( z ) be an entire function that omits the value 0, and that f(k)(z) does not assu1ne a nonzero finite con1plex number for some nonnegative integer k. Then f(z) is a nonzero constant function. Use this fact to prove Picard's theorems. T ry to apply Niiranda's theorem to f( z) = .rc:,._z) for any fixed R > 1, under the assumption that j (J.-) (z) :/= 1, z E C . (6) (P6lya- Saxer) Let J (z) be an entire function such that f (z), f'(z) and f" (z) never vanish in C. Then f (z) = eaz+ b for some constants a and b. Tiy to use Exercise (5) .

Chap. 5

300

Fundamental Theory

(7) Suppose f(z) is an entire function, not of the fonnz +b (b , a constant). Then f of (z) inust have a fixed point. Hint: Suppose on the contrary and consider f(f (z ) )- z and use Picard's little theorem. J( z )-z (8) Show that the curvature defined in (5.8.4. 13) is indeed a conformal invariant. (9) Prove tha.t Schwarz's Lem1na (3.4.5.1) and Schwarz- Pick's Le1nma, (3.4.5.2) are special cases of Schwarz- Ahlfors's Le1nma (5.8.4.14). (10) Show t hat there ]s no ultra.hyperbolic inetric in C or in C - {zo}, where zo E C is a fixed point. (11) Let D be as in (5.8.4.19). Let a, b E C -D wh ere a =f. b. After invoking a transforn1ation t =~ , we may suppose that a = 0 and b = 1, and n = C (O, 1). Set

(1

p(z ) =

+ lzll /3)1/ 2 (1 + lz - 111/3)1/ 2 lzl5/ 6

lz - 11 5/ 6

z E

C (Oi 1).

Compute J((z, p), z E C (O, 1). Observing that J((z , p) < 0 for z E C(0, 1) and the limits of ]{ (z, p) as z ~ 0, 1, oo , respectively, show that K(z ,p) 0. Exercises B Read Ref. (2], pp. 12- 21 for a, c01nplete understanding of the Ren1ark. *5.8.5

An application: Some results in complex dynamical s ystem

As an in1portant application of both Montel's normality criterion (5.8.3.1) and Schwarz- Pick's Le1n1na (3.4.5.2), or its generalized form (5.8.5.6) below, this short section is going to prove a re1narka.ble result in inodern complex dyna1nica.l syste1n due to Sullivan in early 1980s (see (5.8.5.11)). The method of argun1ent is quite si1nilar to that given in Sec. 5.4.3; in particular, in Sections (3) and (4) t here. Let n be a d01nain in C * and f: n ____. n be an analytic (self-) ina.pping. The iterate sequence Jn, n > 1, is called a cornplex dynarnical system on n. Point sequences fn(z) , n > 1, detern1ined by a point z, is called the orbit of z. The questions are: 1. \\That is t he behavior of f 11 (z) if n is large enough? 2. How does the behavior depend on the initial point z and does the limit exist as n --+ oo?

Sec. 5.8.5

An Application

301

Fa.tou and Julia studied iterate sequences syste1natically before World W ar I. The usage of computer work draws out wonderful and complicated figures of iterate functions 1 which provide fruit ful exan1ples for fractal geometry. This revives the study of c01nplex dynan1ical system after 1980. Recall that (see Exercise A(5) of Sec. 5.3.4) : Fatou set F (j) = {z E Dlf n, n > 1, is normal in a neighborhood of z }i Julia set ~l(j) = n - F(j) = {z E Dlfn(z) , n > 1, is not normal}. (5.8.5.1) Note t hat F (f) is open, while ~(j) is closed. BothFatou and Julia sets are invariant under conjugation (see (5.3.4.2)) . J\!Iore precisely, if : n --t n is a conjugation between f and g , t hen

F(f) = (F(g))

and

8'(!) = (8'(.g)).

(5.8.5.2)

T ake, for instance, Exercise A(5) of Sec. 5.3.4 a.s an exa1nple: Let j(z ) = z 2 - 2: C * --t C*. Then (() = ( + 1(1> 1 conjugates g to f , na1nely f = o go - 1 . It is easy to find that F(g) = {lzl < l} U {lzl > l} and ~S'(g) = {lzl = 1}; as a consequence, F(f) = C * - [-2, 2] and ~'f(f) = [-2, - 2]. Readers are asked to do Exercises A(6) and A (7) t here for more practice. In case the initial point z lies in t he F a.tou set, then its orbit is sta.ble in the follow ing sense: It might be convergent, or periodic a fter n is large enough, or has a convergent subsequence in its worst situation. If ~~(j) = 1, is uniforn1ly bounded in n and hence , by (5.8.1.3) , it is normal on n. In t his case, F(f) = n and f.2~(.f) = -I < 1; superattractive if ,\

2. repellent or rep'ulsive if 3. neutral if l.\I = 1.

= 0;

l.\I > l ; (5.8.5.5)

An attractive or s uperattractive point belongs to the Fatou set (see (5.3.4.7)), while a repulsive point is in the .Julia set. A neutral point could be in the F atou set or in the .Julia set. Suppose g: n (a domain in C*) ----; { lw I < 1} is a univalent, onto (conformal ) mapping. Recall that (see ( *i3) and (5.8.4.13) in Sec. 5 . .4) ds

= >.(z) ldz l,

where >.(z)

=

29 l ' (z)I 1 - lg(z)l2

the Poincare ·s metric on n induced by g from the hyperbolic metric ,\(w) = on lwl < 1. Then Schwarz- P ick's Lemma (3.4.5.2) can be easily extended to

1-fwF

Generalized Schwarz- Pick's Le mma. Suppose domains f21and02 have Poincare's m etrics ds 1 = .\1(z)ldz l a nd ds2 = .\2(w)ldw l, respectively. In case f : f21 -+ f22 'iS analytic, then

nl with equality if and only if f is univalently from nl onto n2. >.2(! (z)) lf' (z) I < A1(z),

zE

(5.8.5.6)

Compare to Schwarz- Ahlfor's Lemma (5.8.4. 14) .

Proof. The composite map F = 92 o f o 91 1 : { 1(1 < 1} analytic. and, by the classic Schwarz- Pick 's Lemma,

-+ {

IF'(()I 1 1 - IF(()l 2 < 1 -1(1 2 =?

(by setting z = 9} 1 (())

lg2(f (z) )II!' (z )I < l - l92(f(z))l

2

-

lg} (z) I l - lg1(z)l 2 '

1771
1. Suppose G is a component of the Fatou set F(f) so that fk ( G) = G holds and the restriction f IG : G ---+ G is not a homeornorphism (i. e., fails to be univalent or onto). Then: (1)

ff fk

has a fixed point and, as n --+ oo,

zo ·i n G) then zo is attractive or superattractive

locally 'llniformly on G. (2) If fk does not have a fixed point in G , then as n---+ oo

f kn ( z) ---+ (

E [JG (a fixed bo1mdary point).

locally uniformly on G.

z E G,

(5 .. 5. )

Chap. 5

304

Fundamental Theory

The component G in (1) is called an attractive (stability) domain or a sitperattractive (stability) domain if z 0 is attractive or superattractive, respectively. vVhile G in (2) is usually called a parabolic domain. Readers should refer to (5.3.4.1 1) and (5.3.4.12) and their proofs. Before we start to prove, note that

(5 .. 5.9) holds. Since Jk(zo) = z 0 , this follows easily by invoking (5.3.4.6) to Jk. But the generalized Schwa.rz- Pick's Lemma (5.8.5.6) provide::; a quick and simpler proof for this inequality. To see t his, observe that G ~ C* - ~s( f) and ~' (f) is a nonempty perfect set. Hence C * - G contains at least two distinct points and then G has a P oincare's metric, say ,\(z)ldz l. Recalling !la: G -+ G is a self-mapping of G , it follows that

>.(fk(z))l(fk)'(z)I < >.(z), ::::} (setting z

z E G

= zo and noting that

f k (zo)

= zo)

>.(zo)l(fk)' (zo)I < >.(zo) ::::} (since >.(zo)

> O) I(fk)'(zo)I < 1.

This proves (5 ..5.9).

Proof of (5 .8 .5.8). (1) By assumption that !la: G -+ G fails to be univalent or onto, the multiplier,\ 1nust satisfy l>-1 < 1 (see (5.3.4.11)). Therefore zo is attractive or superattractive. Since l>-1 < 1, there is an open neighborhood 0 of zo and a constant 0 < a < 1 so t hat (see ( *s) in Sec. 5.3.4)

lfk(z) -

zo l < a lz- zol,

::::} lfkn(z) - zol < anlz - zol,

z E 0

z

E 0.

Hence Jk 11 (z ) converges to zo uniformly on 0 , as n -+ oo. We claim that f kn (z) will then converge to zo locally uniformly on the whole set G. Geometrically this means that for any fixed point z E G , the orbit of z is always attracted to t hat of zo: more precisely, fkn+i(z) _, f i (zo), a n -+ , for all z E G and 0 < j < k - 1. To prove the claim, we adopt contradictory argument (a we did in Sec. 5.3.4). Suppose that there is a point (in G and a subsequence fkni(z), j > 1, so that J1mi (() does not converge to zo . Since fm, ni > 11 is normal, a subsequence of fkn j(z), say itself, will converge locally uniformly on G to

Sec. 5.8.5

An Application

305

an analytic function g (see (5.3.1.1)). Consequently, g(z) _ zo on 0 and, in turn, on the whole set G. This is contradictory and the claim follows. (2) As we stated in (1) of (5.3.4.12), we first try to prove that any convergent subsequence fkni (z), j > 1, will always converge locally unifonnly on G to a constant function h . Rewrite

fknj+t (z) = fk (nj+t -nj ) (z)

0

Jni

In case h is not a constan t , it must be univalent in a neighborhood of some point; in t his case, g should be an identity on t hat neighborhood and hence on the whole set G. Consequently

fk (nj+i -nj) ~the identity inap on G , locally unifonnly on G , contra.dieting to t he assumption t hat Jk la is not homeomorphic. Compare to (*13) and ( *14) in Sec. 5.3.4. Now, suppose

fknj (z ) ~ ( (a point in G)) locally uniforrnly on G . =?

fk(() = h rn fk o fkn j (z) = hm fkn j o fk(z) = ( . ]-+ 00

J-+00

Hence ( is a fixed point of f k . Since f k is assumed to have no fixed point in G, it follows that ( E 8G should hold. F inally, we have to show that the sequence

f 1m(z)

~(

locally uniformly on G.

As we had shown that any convergent subsequence of f kn, n > 1, always converg~s locally uniforn1ly on G to a fixed point of f k, all we need to do is to prove that f k has only one fixed point. Suppose on the contrary that fk has fixed points (i , ... , (L where l > 1 (why does fk have only finitely many fixed point?). Choose disjoint open neighborhoods o.i of (j, 1 < j < l, so that

1k(oj) nLJop= c1>)

l 0. Section 6.2 also contains lots of concrete examples, especially if n is a triangle or a rectangle. The application to the elliptic function is only sketched in Exercise B of Sec. 6.2.2. The next main theme of this chapter is Dirichlel 's problem, of finding a harmonic function in a domain with a presumed boundary values. Section 6.3 olves this problem for a disk which is usually called Schwarz's theorem, by using Poisson 's integral for harmonic functions. Basic properties of Green's function of a domain with respect to a point (pole) are also 307

308

Chap. 6

Conformal Mapping and Dirichlet's Pr-oblem,s

studied for lat,er usage in Sec. 6.6. Included are the symmelnJ principle and Harnack's principle. The extension of hannonic functions to the subharmonic ones is in Sec. 6.4. Basic properties are proved for the need of Sec. 6.5 .

Section 6.5 adopts Perron's method, concerning a family of subharmonic functions wit,h specified boundary properties, to solve Dirichlet's problem for a class of general domains n. The best result that we obtained is the following: Dirichlet 's problem is solvable for any domain whose complement in C is such that no component reducp,s to a point(see (6.5.7)); in particular, a simply connected domain whose boundary contains more than one point and a domajn bounded by a finite number of.Jordan curves are in this class. Th ese res~1,lts are used in Sec. 6.6. The last section (Sec. 6.6) relies mainly on the concepts of harmonic measv,re (see Sec. 6.6.1) and Green's function (see Sec. 6.3.1) to determine the canonical mapping that maps a finitely-connected domain univalently onto a canonical domain. The canonical dmnains introduced are: 1. Annulus with concentric circular slits ((6.6.2.1); Exercise B(5) of

Sec. 6.6.2). 2. Domain wit,h spiral slits (Exercise B (l) of Sec. 6.6.2).

3. 4. 5. 6. 7. 8.

Disk with concentric circular slits (Exercise B(3) of Sec. 6.6.2). Domain with vertical slits (6.6.3.3). Domain with horizontal slits (6.6.3.5). Domain with parallel slits (6.6.3.6) . Horizontal strip with parallel segment slits (6.6.4.2). n-Sheeted open strip (6.6.4.3).

The ei'tremal problem method (as the one in proving the Riemann mapping theorem) is also touched in the exercises in finding the canonical domains and mappings mentioned above. 6.1

The Riemann Mapping Theorem

In Sees. 2.5 (about elementary rational functions) and Sec. 2.6 (about elementary transcendental functions), we experienced many examples showing how some particularly-shaped simply connected domains, whose boundaries compose mostly of line segments and circular arcs, are mapped univalently and analytically onto another simply connected domains. The highlights in this direction cuhninate in Sec. 3.5. 7, including Sees. 2.7.2- 2. 7.4 wherein Riemann sw·faces of elementary 1nttltiple-valued functions are constructed for descriptive purpose.

Sec. 6.1

The Riemann Mapping Theorem

309

The general setting is the problem: Does there always exist a univalent analytic function mapping a given domain n1 onto another domain r2 2? If it does, n 1 and n2 are said to be conformally equivalent. Riemann asserted that the answer is in the affirmative if both are simply connected domains, each of whose boundaries contains at least two distinct points. Refer to Remark 3 in Sec. 2.4 for related explanation. As a preparatory work, recall that a nonconstant analytic function maps

a domain onto a domain (see (3.4.4.4) and (3.5.1.12)); if, in addition, it is univalent, then it maps a simply connected domain onto a simply connected one (see (2.4.13), (3.5.1.9), and (3.5.5.1)). Conformal equivalence among do1nains is an equivalent relation (namely, reflexive, symmetric, and transitive). Hence, we may choose the open unit disk as one of the two simply connected domains preassigned; Liouville's theorem suggests that the other one cannot be the whole plane C or the plane C with a point removed. Riemann formulated

The Rie m a nn m a ppin g theorem. Let n be a simply connected domain in C* where boundary (in C *) contains at least two distinct points. Fix a point z 0 E n and a real number() (0 < () < 27r). Then there exists a unique univalent analytic function f (z) mapping n onto the open unit disk lwI < 1, satisfying the conditions:

f( zo) = O; 2. Arg f' (z o) = (). 1.

(6.1.1)

This is an existence and uniqueness theorem. The explicit analytic expression for such an f( z) cannot be found, in general, except in very restricted cases (see Sec. 6.2). However, the geometric properties of the domains that are being mapped lead to analytic properties of the mapping function (see, for instance, Sec. 6. 1.2). T his theorem, initiated the study of the geometric function theory, does not have its counterpart in the theory of function of several complex variables (Poincare 's theorem: There is no biholomorphic map between the open unit ball { z = (z1, ... , Zn) E C 11 II zil 2 + . . . + lzn 12 < 1} and the open polycy linder {z E C 11 llz1I < 1, ... , lznl < 1} , where n > 2). For pedagogical (not for logical) reason, we have cited this theorem previously in quite a few places, such as in Sees. 3.4.5, 3.5.7, 5.8.2, 5.8.3, and 5.8.4, etc.

310

Chap. 6

Conformal Mapping and Dirichlet's Problems

Note that only condition 1 cannot single out such an f (z) uniquely because eiO f( z) is also such a function for each real e if f( z) merely satisfies J(z0 ) = 0. It is condition 2 that does so (see the beginning of Sec. 6.1.1). Condition 2 means geo1netrically that w = f (z) maps a segment, passing z0 and parallel to the positive real axis, onto a curve passing 0 and having an inclination angle eto the positive real axis (refer to (3.5.1.1)); consequently, the tangents to all cw·ves passing through the point z 0 experience a rotation of angle when subjected to the transformation w = J( z) mapping them into curves passing through the center 0 of the unit disk lwl < 1. Two sections are divided. Sectiton 6.1.1 is devoted fully to the proof of the theorem. Sectiton 6.1.2 extends the function f(z) homeomorphically to the closure n, mapping onto lwl < 1, in case n is a J ordan domain. By a Jordan domain n in C *, we mean the one whose boundary is a closed Jordan curve 'Y in C *, namely, n = Int "I (see (2.4.11) and Remark 2 after it). We formally state as

e

The boundary corr espondence theorem. Let J(z) be a univalent analytic j1J,nction mapping a Jordan domain 0 1 onto a Jordan domain n 2 . Then f (z) can be continuously extended as a homeomorphism from nl onto n2 so that w = f (z) sets up a homeomorphism between the boundaries 801 and 8fh. (6.1.2)

For detailed account, refer to Ref. [58], Vol. III, Chap. 2; for a little up-to-date treatment, refer to Ref. [66], Sec. 4 of Chap. IX, and Ref. [9], Sec. 8 of Chap. 4. For simplicity, a univalent function will always mean an analytic function that is univalent. Recall that a univalent function is conformal (see (3.5.1.9)) but not conversely. 6.1.1

Proof

P. Koebe gave the theorem its first successful proof. Later proofs of it are variants of Koebe 1s original proof. Unique n ess: In case g : n -+ {lwl < 1} is another such function , then the composite map g o 1- 1 : {lwl < 1} --+ {lwl < 1} is univalent, onto and maps 0 into 0. According to Example 1 in Sec. 3.4.5, there is a real a so that g(J - 1 (w)) = ei0 w, lwl < 1. Since J(z0 ) = 0, then g'(z0 )(j- 1 ) 1 (0) = g'(z0 )(f'(z0 ))- 1 = eicr. and Argf'(zo) = Argg'(zo) =(} together show that a= 0 should hold. Hence g(z) = J(z) throughout n.

Sec. 6.1.1

Proof

311

Ex istence: If f (z) is known to be the required mapping, then the function g(z) = e-ill f(z) will map n univalently onto lwl < 1, satisfying the properties that g(z0 ) = 0 and g'(z 0 ) > 0. Hence, we may prove the theorem in the case that e = o. Consider the family J of all the functions g(z) satisfying the conditions: 1. g(z) is univalent (and analytic) inn;

2. lg(z)I < 1 inn; 3. g(zo) = 0 and g' (zo)

> 0.

Then we try to show that .J is nonem,pty and hence is a normal fainily. Fina.lly, try to solve the e:r;tremal problem, sup 9 Ei g' (z0 ) and show that the extremal function j(z) is the required one (refer to the Remark in Sec. 5.8.2 and (5.8.2.7) there).

Step 1. The nonemptyness of J. There is a point a =/= oo not in D, since, otherwise, D = C or C * will contradict to our assumption that contains at least two distinct points. Since z - a =!= O on n, j z - a has two distinct single-valued branches (see (3 .3.1.5) or (4.4.2) , etc.). Fix one of them as h(z). h is 11nivalent on n for, if h(z1) = h(z2 ), then

on

h(z1 ) 2 = h(z2 ) 2 , namely .z1

-

a= z 2

-

a or z 1 = Zz.

:Moreover, h(z1) =I= - h(zz) for any z1, Zz En for, if so, then h(z1) = - h(zz) implies that h(z1 ) 2 = h(z2 ) 2 , i.e., 2 1 = z 2 which, in turn, implies that h(z1) = 0, contradicting to the fact that h(z) =I= 0 throughout n. w = h(z) is open and hence the i1nage h(D) contains an open disk lw - h(zo)I < p for some p > o. Since - h(z) tt h(D) for all z En, it follows that 1- h(z) - h(z0 )1 = lh(z) + h(z0 )1 > p comes true on D; in particular, 2lh(z0 )1 > p holds. Set

90 (z)

= kh(z) - h(zo)

h(z)

+ h(zo),

n

z EH,

where k is a constant to be detern1ined later on so that this g0 will be in J. First 1

l9o(z)I = lkl lh(zo)I h(zo) - h(z) 4lkl lh(zo) I _ p

- 1)

+2 h(zo)
0 if 00

= ·l/J0

-

r.p 0

is chosen.

In conclusion, rewrite (*2 ) in its final form as

lh'(zo)I h(zo) h(z) - h(zo) go(z) = 4 h'(z0 ) lh(z0 )12 h(z) + h(z0 )' p

z E 0.

This go (z) is indeed in J.

Step 2. An extremal problem. According to (3) in (5.8.1.3), J is normal (compare to (5.8.2.3)) . Let J'\lf =

sup g' (zo) . .QE'J

At present, !VI could be +oo . However, J has a sequence 9n(z), n > 1, satisfying g~t(zo ) --. !VI as n --. oo. The sequence 9n(z) has a subsequence, say 9n(z) itself, converging to an analytic function f(z) locally uniformly inn. Observe that 9n(zo) = 0, g~(zo) > 0, and l9n(z)I < 1 for each n > 1. T hese facts lead to f( z0 ) = 0, f'(z 0 ) = M > 0 (also Alf< oo ), and If (z )I < 1 on n. Hw·witz's theorem (5.3.2.2) guarantees that f( z) is univalent in n because f (z) is not a constant. We conclude that f( z) is in~ and is the extremal function solving (*4 ). Step 3. The extremal function is required. What remains is to show that f(z) maps n onto the whole disk lwl < 1.

Sec. 6.1.1

313

Proof

lwl < 1 which function F in .J

Suppose on the opposite that there is a point w 0 in

does

not lie in the don1ain f (D,). We try to construct a with F'(z0 ) >M in order to contradict to ( *4 ). Once this is done, as a do1na.in, f(D,) = {lwl < 1} should hold because f (z) cannot assume values lying on the circle lw I = 1. Consider the following three auxiliary inaps: ( =

f1(w) = w -~o I - wow

rJ= !2(() =

: {lwl < l} - {1(1 < 1};

V(: {1(1

O} -

{l 0: A direct computation shows that

F '( zo ) = J'3 (G( z o))G'(z o) = G' (zo) = -

1

2'fJo

On choosing 80 so that G'(z0) = and 'fJo = /=Wo, we have

F1(zo) = IG'(zo)I = 1 1-

lrJol

2

1-

i Oo

e

· f'(zo)(l -

e-iOo

2

G'(zo)

1 - IrJo 12 ,

v"here "

lwol 2 ).

IC' (z 0 ) I and recalling that f'(z 0)

lwo l . M lwol 2lwol 1 l 2

= 1+

lwol M > M . 2lwol 1l 2

contradicting to ( *4). This finishes the proof of the Riemann mapping theorem.

=

M

314

Chap. 6

Conformal Mapping and Dirichlet's Problems

Remark 1 (On variants of the proof). n is always assumed to be a simply connected domain whose boundary in C * contains more than one point . About Step 1: In case n is bounded, say, n is contained in the disk lzl < R. For any fixed point Zo E f2, the function 9(z) = 2 (z - zo) is univalent in n , 9(zo) = 0,

k

91(zo) > 0, and l9(z) I < 1 on n. While the function .Qq~=~) = h(z) satisfies h(z0 ) = 0, h' (z0 ) = 1, and h(z) remains bounded on n . In case f2 is unbounded but has an exterior point, say (o E C. Suppose the disk lz - ( 0 1 < r is contained in the exterior of n. Then g(z) = z~(o is bounded by ; on n a nd univalent; moreover, 9(z) maps n onto a bounded sirnply connected domain g(D). In case n is unbounded with empty exterior, then the boundary on is a continuum (na1nely, a compact connect set in C *, containing infinitely many points) . Fix two distinct points a and b along on. Then ;-=:.g- is univalent

-/!ii

in n and never assumes the values 0 and oo there; consequently, has two distinct single-valued univalent branches, say 91 (z) and 9 2 (z) , on n (see (J.J.1.5)) . Note that, g2(z) = - g1(z) for all z E n. Therefore gl(H) and 92 (!1) are pairwise disjoint simply connected domains. Either of g1 (D) and 92 (n) is contained in the exterior of the other. According to the last paragraph , there are bounded univalent functions on 9 1 (!1) or 92 (!1) , and so are on the original n . In conclusion, there are (in fact, infinitely many) bounded univalent functions on n . If g(z) is SUCh a function and Zo is a fixed point inn, then h(z) = g(z) - 9(z0 ) is bounded, univalent and h(z0 ) = 0 and h'(z0 ) =/= O; and eie h(z) has a positive derivative at z for son1e real B; and h~{;~) has its derivative equal to 1 at zo .

0

About Step 2: It is harmless, in (* 1 ), to drop the condition 3 and, instead, consider the family J = {9(z) is univalent in n , 9(zo) = 0, and l9(z) < 1 (even if lg( z) I < 1) there}. In this case, (*4 ) is changed to lvl = sup9 E J lg' (z 0 ) I· That lvl < oo can be proved in advance. To see this, choose any 0 < r < 1 such that the disk lz - z0 1< r is contained inn. Via Cauchy's inequa.lity, lg' (z0 ) 1 < ~ for any 9 E J; it follows that M < ~· In case j(z) is the corresponding extremal function, after multiplying j(z) by a constant of modulus 1, the resulted f(z) will then satisfy f'(z 0 ) = M > 0. Of course, there are some other choices of the family J.

Sec. 6.1.1

Proof

315

In the formulation of g0 (z) in ( *3 ), p > 0 can be chosen s1naller so that the closed disk lw-h(z0 )1 p holds; consequently, l9o( z) I < 1 holds on D.. Suppose this is the case in what follows. Hence 9o(z) maps n univalently onto a domain 9o(D.) in 1(1 < 1. Note that 9o(zo) = 0 E 9o (D.). Consider the family

J = { 1, are then uniformly bounded on D., and hence is normal as a family of functions. A subsequence of it, say itself, converges to an analytic function f(z) locally uniformly in D.. f(zo) = 0 and f'( zo) = 1, in turn, ilnply that J(z) is univalent on n by Hurwitz's theorem. For any fixed (E n , l9n(()I < supzEn l9n(z)I for n > 1 indicate, after letting n--+ oo, that lf(() I < R zo for all (ED.. Hence J E J. This implies that supzEn l/(z)I > Rz0 . Hence, supzEn l/(z)I = R zo holds, namely, l/(z)I < Rzo throughout n. That l/(z)I = Rzo for some z E n is not possible (why?) shows that I/ (z) I < Rzo for all z E n. Step 3 is still

316

Chap. 6

Conformal Mapping and Dirichlet's Problems

applicable here to guarantee that f(z) indeed maps n onto the whole open disk lwl < Rzo. This f(z) is the required one. 0 Remark 2 (The conformal radius of the domain n relative to the point Zo E:: D). Let w = f (z) be the unique univalent function inapping n onto the open unit disk lwl < 1 and satisfying the extra conditions:

f'(zo) > 0 ,

f(zo) = 0 and

(6.1.1.1)

where n denotes a silnply connected domain with more than one boundary point and Zo E n is a preassigned point. Then the function

f (z) F(z) = f'(zo) inaps

n univalently onto lwl

~ < IJ(r2) -

1

82

72

f(r1)I

=

f'(z)dz =

Tl

('

.•

{

101

,y

W-}

Fig. 6.1

C=ji.

f'(( +rei8 )riei8 de

Sec. 6.1 .2

~

The Boundary Correspondence

321

(integrating both sides with respect tor from p 1 to p 2 , where 0 0 , let 0 = o(c) > 0 be such that, for any two points

ZI)

Zz En,

lf(z1) - f(z2)I
1 and Zn - (as n - 00. Then t he sequence Zn, n > 1 1 is Cauchy. There is an integer N = N(o) > 0 so that IZn - Zm I < o if n > m, > N. According to (*3 ), this means that lf(zn) - f(zm) I < € if n > m > N. That f(zn), n > 1, is Cauchy shows that f(zn) does converge in C. Suppose z;P n > 1, is another sequence of points inn, converging to the sa1ne boundary point(. Then f(z~), n > 1, also converges and, as a inatter of fact, converges to t he same limit as f(zn) does. To see this, choose an integer N > 0 so that lzn - ( I < ~and lz:i - ( I < ~whenever n > N. Hence lzn - z~ I < oif n > JV and, then, (*3 ) indicates that lf(zrJ - f(z;J I JV. The claimed result thus follows. In conclusion , we define 1

f (() = lim. f(z), z E !1

(Eon.

z- (

This very definition indicates t hat dmna.in n.

f (()

is continuous from within t he

322

Chap. 6

Gonfor·mal Mapping and Dirichlet's Problems

Let £> 0 and o > 0 be as in (*3 ). For any two points ( 1 , 1(1 - (21 2 (see Section Two of Sec. 3.5.1; in particular, the geometric meaning of (3.5.1.6)). This is clearly impossible. In Uxe'Y D (x), g(z) =/= 0 always holds and hence, logg(z) has singleva1ued branches. Fix such a branch, still denoted by g( z), and set 'u(z) = log lg(z) I and v(z) = Argg(z) . In this case, if x E "f,

ou (x) = lim u (x + h) - u(x) = lim 0 - 0 = 0 = ov (x);

OX

h -+O

h

h->O

h

oy

"lies in I m z > O) . -au() x = l'im n(x+ iy) - u(x) < O (reca111 t 1at ~,

dy

y

y -.0-1

-

&u

:::} - (x) &y If

g; (xo) = 0 for some x

0 E

d logg(z)lz=xo dz

&v

= - - (x) < 0 OX

along T

'Y, then

= ~u(xo) +i ~v (xo) = 0 = g'((xo; vx

g Xo

vx

::::} g'(xo) = 0, a contradiction. Consequently,

8v Bx(x)

=

8 ox Argg(x) > 0

along 'Y·

This means Arg g(x) is strictly increasing along 'Y (and hence moves constantly in the same direction). Thus w = g(z) maps 'Y one-to-one and onto an open a rc on the circle lwl = 1. This proves (2) . Return to the general case that 'Y is a regular , free one-sided, analytic Jordan curve. Recall (see Remark 1 in Sec. 3.4.6) that there is a function cp(t) : (a, b) -+ C satisfying:

(a, b), there is an open interval N(to) = (to - p, to+ p) ~ (a, b) such that cp(t) = L~=oan(t - t 0 )n converges in N(t 0 ), i.e. , cp(t) is real, analytic. 2. cp'(t) # 0 for all t E (a,b), i.e., cp(t) is regular. 3. cp(t 1) = cp(t2) if and only if t 1 = t 2 E (a, b), i.e., cp(t) is Jordanian or simple. 1. For each to E

Sec. 6.1.2

The Bo'undary Correspondence

327

Indeed, 00

(*9)

Hence oo is a removable singularity of g(z) (see (4.10.1.2)) . Finally, consider the entire function n

G(z) = g(z) - ~ CYk - 1. L z - ak k=l

Since G(oo) = 0 (see (* 9 )), G(z) is bounded and hence, is the constant zero function. Consequently, n

q(z) = ~ L CYk k= l

1>

z - ak

z E C * - { a1, .. . , an }

f"(z) ~ :::} (see (*6)) f ' ( ) = L z

CYk-

k= I z -

l

ak

'vvhich is analytic everywhere in Im z > O.

(*1 o)

Fix any point zo in lm z > 0 and integrate the above expression along any rectifiable curve in Im z > 0, connecting z 0 to a point z (Im z > 0). We obtain log f'( z) =

L (ak -

1) log(z - ak) +a,

k= I

where a is a constant, and the log(z - ak), 1 < k < n, are some fixed branches. Therefore, n

J'(z) = c1

IT (z- ak)°'k- 1,

c1 =ea

k=l

and another integration yields the required representation (6.2.1.1 ) below.

334

Chap. 6

Conformal Mapping an d Dirichlel 's Problems

In conclusion , we obtain

The Schwarz- Christoffe l formula for a bounded polygon with finite vertices. Let n be a bounded polygon with finite vertices W1, ... ) Wn which, in t his ordering, induce t he positive orientation of t he boundary 8!1. Let o:.k1f (0 < o:. < 2) be t he interior angle at t he vertex Wk for 1 < k < n so t hat f3k1f = (1 - O:.k)1f (- 1 < fJk < 1) is t he corresponding exterior angle (see Fig . 6.2). Then the functions w = J(z) which map Im z > 0 univalently onto n are of the form Im z > 0, ·where zo is any fixed point in Im z > 0, c1 =!= 0, and c2 a re complex constants, a nd a1, ... , an are points on the real axis , wit h increasing ordering -oo < a 1 < ... < an < + oo, and wk = J(ak), 1 < k < n . Recall that the sum of the interior angles is (6.2.1.1 )

The constants c 1 and c2 are detennined by t he size and position of t he polygon n in t he plane C (subject to a similarity t ransforma tion). Of t he n points a1, ... , an, t hree can be chosen arbit ra rily to correspond to t hree of t he given vertices w 1 , . . . , Wn provided t hat t hey have t he same ordering as t he corresponding vertices. \i\'hen t his is t he case, such a mapping is uniquely deterrnined according to (6.1.2.2): t he other vertices are then uniquely determined by n; in particular, by t he following (n - 2) independent conditions O.

(*12)

Integrating t his expression twice, we obtain

The Schwarz- Christoffel formula for an unbounded polygon with finite vertices and having oo as an interior point. Let n be such a polygon 1 and w1, ... ,wn and a1,. .. ,an be as in (6.2.1.1) . T hen any functionw = f(z) mapping l m z > 0 univalently onto n is of t he fo rm W = .f (z)

=

rzrr~=l

C1

}zo

(( - ak)Ctk - 1

(( - b)2 (( - b)2 d(

+ C2,

where zo in Im z > 0 is fixed point and b = 1- 1 (oo) is the inverse image in I mz > 0 of the point oo under w = .f(z). In this case, the sum of the interior a ngles is (6.2. 1.3)

The last identity follows from t he fact t hat the sum of all the residues is equal zero (see (4. 11.1.7)) . In this case, by (*s), (*9), (*n ), and (* 11 )', \Ve have

"n ' Res

6

(J" ) + Res (1"-!''·oo )+ Res (1"- ·b) + Res (1"-!' '·b) -

·ak

f' '

f' '

n

=

L(

CXk -

k= l

1) - ( - 2) - 2 - 2 =

o.

Chap. 6

336

Conformal Mapping and Dirichlel 's Problems

Section (3) A polygon with oo corresponding to one of its ver t ices Now, suppose an.= oo in (6.2.1.1) . Choose any fixed point a< a 1. The transformation 1 ( = a -z

maps Im z > 0 univalently onto Im( > 0, the real axis one-to-one and onto the real axis, and the points ai, ... , an - Ii oo into the points bi, ... , bn- 1,bn = 0, resp ectively , where - oo < b1 < · · · < bn- 1 < bn = 0. Hence, on applying (6.2. 1.1) to 77 = F(() = f(a - t) , we have

1II

( n-1

77 = F(() =

C1

(T -

bk)l' 0 small enough such that, if lz -

ak l

(z -

ak)

}

+ ·· · .

< r, then

(k)

b1

ak

- - (z - a1r)+ .. · < 1 (X.k + 1 b(k) . 0

:::;:. (since c

> 0 and

ak

> 0)

arg(w - wk) - argb 0(k) -

>

+ akarg(z -

ak)

b(k)

1 + a rg { 1 + O'.kCt'.k+ 1 -b(k) . (z -

ak)

+ · ··

}

0

This indicates the fact that, if z approaches ak along a ray starting at ak and pointing toward the upper half-plane and with the inclination angle

Sec. 6. 2.1

The Schwarz- Ghrisloffel Formulas for Polygons

343

e with

the positive real axis, then w = w(z) will approach wk along a curve whose tangent at wk has the inclination angle Arg b&k) + o.ke with the positive real axis. As a consequence, when z starts form ak + r and winds along the upper half circle / : lz - ak l = r, Im z > 0, in the anticlockwise direction and terminates at ak - r, the variation of Arg ( z - ak) is given by t::,.,_,Arg (z - ak) = 7r - 0 :::::} 6.7 Arg (w-wk)

= (Argb~k) +ak1T)- (Argb~k) _ O) = Cl'k7r,

1 < k < n.

Owing lo the univalence, the image curve w(1) is a Jordan curve in the polygon n, starting at a point on the side wkwk+i and ending at point on the side 'Wx:-1 Wk; it never winds around the vertex Wk. Hence, O.k7r as the interior angle of the polygon at wk, O'.k 7r < 27r does holds; namely, ak < 2 for 1 < k < n. What is Lhe interior angle at w 0 = Wn+l? It is

-{t (o:l -

l)7r - (- 7r) } =

(n -

l )7r-

l=l

t ak?T.

(*24)

l= l

See Fig. 6.8 (also, Fig. 6.6). Since n has (n + 1) vertices, the SlUil of its interior angle is [(n + 1) - 2]1f = (n - l )7r. Hence, (* 24 ) is exactly what we want; moreover , n

L a17r < 27r

(n - l )7r -

l=l

:::::} n - 3
OL and w3 = (1 + i)a. In this case, (*1 ) becomes w = f(z) =

C1

1z

an

(-i(l - ()-id(,,

':vhere

aJ'I,; [r (~) ]2.

The tessellation of the isosceles right triangle net is shown in Fig. 6.12(b). 1. The extended w = j (z) is one-to-infinite; it maps either Im z > 0 or Im z < 0 univalently onto the closure of a si1nple isosceles right triangle.

Sec. 6.2.2

LJ. --- - -: ', 0 '

'

'

''

'

''

,'

, , '

co

0

l

o.. '

00

···-- · ·--- , ·-····- --o

:~~h!