1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions

which in view of the hypothesis, reduces to

where

1

f(z) = j<m~~zo) (z - zo)m + ~:+~c;;i (z - zor+l + ...

= (z - zorg(z)

f(m)(zo) f(m+l)(zo)

g(z) = + (z - zo) + · · ·

m! (m + 1)!

is analytic in D and g(z 0 ) -:/:-0.

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Theorem 8.21 The only analytic function in a region G having in G a

zero of infinite order is the constant function 0.

Proof It follows from Corollary 8.10 that at a zero z 0 E G off of infinite

order we have

f(zo) = 0 and

for every n 2". 1. Otherwise, zo would be a zero of finite order. Hence

(8.9-2) shows that f(z) = 0 for all z such that lz-zol < R, where R is the

radius of convergence of the series. But R = oo in this case since all the

coefficients of the series are equal to zero. Therefore, f = 0 throughout G,

in fact, f = 0 throughout the complex plane.

Theorem 8.22 The zeros of an analytic function on a region G are

isolated, unless f = 0 in G.

Proof It follows at once from Theorems 8.17 and 8.21.

Theorem 8.23 If f is analytic in a region G, then f has at most a finite

number of zeros in every compact subset KC G, unless f = 0 in G.

Proof Suppose that f =/:. 0 has infinitely many zeros in K. Then the set

J of the distinct zeros in K is also infinite. By Theorem 2.21 the set J

must have at least an accumulation point c EK. Choose a sequence {en}

of zeros in J such that Cn ---+ c. By the continuity of f at c we have

limn_,. 00 f(cn) = f(c). But f(cn) = 0 for every n, so that f(c) = 0. This
is impossible since c is not isolated.

Example The function f(z) = sin[l/(2 - z)] has in the region lzl < 2

infinitely many zeros Zn= 2 -1/mr (n = 1, 2, ... ). However, any compact

subset of lzl < 2 contains at most a finite number of such zeros. The point

z = 2 is an accumulation point of the set of zeros, but it neither belongs

to the region Jzl < 2 nor is f analytic at z = 2.