1550251515-Classical_Complex_Analysis__Gonzalez_

672 Chapter^9

that g(O) = g'(O) = 1, h(O) = h'(O) = h"'(O) = 0, h"(O) = 1. Hence by
using (9.8-4) we obtain

ez

Res =2

z=O 1-COSZ
( c) Residue at DO. Residues at DO are computed in general by apply-
ing either (9.7-4) or (9.7-5). However, in some simple cases the following
formulas are useful.

1. f is regular at DO but f(DO) f 0. Then f(z) has in some deleted

neighborhood of DO the representation

Bo -=f. 0

so that

Hence

Resf(z) = -B-1 = lim z^2 f'(z)

z=O z-+oo

(9.8-6)

2. f is regular at DO and has a simple zero at DO. In this case we have
   the representation

Hence

Res f(z) = -B-1 = - lim zf(z)

z=oc z--+oo

(9.8-7)

3. f is regular at DO and has a zero of order m ;::: 2 at DO. For this
   case we have

so that B-1 = 0 and Resz=oo f(z) = 0.

4. f has a simple pole at DO. In this case we have

so that

Hence

f(z) = Blz +Bo+ B1z-^1 + B 2 z-^2 + · · ·,

J'(z) =Bi - B_1z-^2 - 2B-2z-^3 - • • •

J"(z) = 2B-1z-^3 + 6B_ 2 z-^4 + · · ·

z=oo Res f(z) = -B-1 = -1/^2 z--+oo lim z^3 J"(z) (9.8-8)