1550251515-Classical_Complex_Analysis__Gonzalez_

498 Chapter^7

for lzl > R. On the compact set lzl SR the real continuous function lg(z)I

is bounded: i.e., there is f{ > 0 such that

Jg(z)I < K

for I z I ~ R. Therefore,

lg(z)I <min(~, K)

for all z EC. By Theorem 7.33 and the fact that g(z) =/= 0 for all z, we have

g(z)=cf=O (caconstant)

so that
1
f(z) = -
c

for all z E C. This is a contradiction since f is supposed to be a polynomial

of degree n :;:::: 1 (the identity principle for polynomials would imply ao =
ai = · · · = an-1 = 0 and an = 1/c).
Other proofs of the fundamental theorem of algebra are given in Sections
9.14 and. 9.15. A good many proofs of this theorem have been published.
See the bibliography at the end of the chapter.

Note An entire transcendental (nonalgebraic) function may have no ze-

ros. For instance, ez =/= 0 for all z E C. The proof above does not apply to

this function since the inequality I ez I > M holds good only for Re z > ln M.

7.26 Riemann's Theorem

Theorem 7.36 Suppose that f is analytic in an open set A except at

a point a E A where f is locally bounded. Then there exists a unique

number b such that the extension of the domain of definition of f obtained

by letting f(a) = b yields a function that is analytic at a.

Proof Let -y: ( - a = re it, 0 S t S 211", be a circle with radius small enough

so that -y* C A (Fig. 7.27). By Theorem 7.24 we have

J(z) = ~ j !(0 d( (7.26-1)

27ri ' - z

'Y

for any z E Int -y* such that z =/= a. However, by Theorem 7 .28 the integral

on the right-hand side of (7.26-1) defines an analytic function throughout

the region Int -y*. Hence the integral has a well-defined value at a, namely,

b= _1 J J(()d(

271'i ( - a

'Y