1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications

and let

f(z) = zn, g(z) = a1zn-l +···+an

Choose a circle C: lzl = R such that

R >. max(l, ia1 I+···+ iani)

Then on iz I = R we have

lf(z)i = lzln =Rn···-

747

lg(z)i::; ia1IRn-l + ... + ianl < Rn-l(la1i + "' + iani) <Rn

so that

lg(z)I < lf(z)I

on lzl = R. It follows that f + g has inside C as many roots as f(z) = zn,

namely, n. Clearly, f + g has no roots for izl 2: R.

3. To show that the equation ez - 3z = 0 has just one zero inside lzl = 1.
   Let f(z) = -3z and g(z) = ez. On lzl = 1 we have

lf(z)I = 3lzl = 3, lg(z)I = e"'::; e < 3

Hence lg(z)I < lf(z)I for lzl = 1, so that ez - 3z has as many zeros as

f(z) = -3z inside izl = 1, namely, one zero.

Theorem 9.22 (Hurwitz's Theorem). Let Un(z)}~ be a sequence of
analytic functions in some open set A, and suppose that:

1. fn(z) ~ f(z) on every compact subset of A.


2. J(z) is not identically zero in A.

Then a point z 0 E A, z 0 -=/=-oo, is a zero of f(z) if either (1) z 0 is a zero
of each fn(z) for n large enough, or (2) z 0 is an accumulation point of the
set of zeros or the fn(z).

Proof By Theorem 8.2 the function f(z) is analytic in A. Let 'Y be a circle
around zo contained in A and such that f(z) does not vanish on or inside

"f, except possibly at zo. Since f is continuous on 'Y and f(z)-=/=-0 for z E /,

there exists c > 0 such that

lf(z)I 2: c (9.15-8)

for all z E 'Y· Also, since fn(z) ~ f(z) on compact subsets, we have

lfn(z) - f(z)I < c (9.15-9)