Mathematics of Physics and Engineering

Laurent Series 217

where

Ck = :ti(f '» '-^fc+idC, fc = 0,±l,±2,..., (4.4.3)

27riJCo(C-z 0 )fe+1'

and Cp is a circle with center at ZQ and radius p so that Ri < p < \z — zo\;
the circle is oriented counterclockwise. Representation (4-4-%) *s unique:
if f(z) = EfeL-oo ck(z ~ zo)k = Efcl-oo ak(z ~ zo)k in G, then ak = ck for
all k.

Proof. We present the main steps of the proof. The details are in the
exercise below.
Step 1. Fix the point z. Let Cr be a circle with center ZQ and radius
r so that p < r < R and \z — ZQ\ < r (draw a picture). Combining the
proof of Theorem 4.2.4, page 198, with the Integral Formula of Cauchy, we
conclude that

with both Cr and Cp oriented counterclockwise.
Step 2. We already know that

h i ^-C - J>(. - »)>, * - JL £ j^JC (4.4.5)

see the proof of Theorem 4.3.4, page 210.
Step 3. By Exercise 4.2.17, page 199, we conclude that

2™ Jr

/(C) ,r i / /(C) ,, r

Tcr (C - zo)k+1 * 2ni JCp (C - zo)fc+1

Step 4- Let us show that

h I {&« - t (7^- — Si £'<«« - *>"«•

(4.4.6)

fc=i

Once again, in (4.4.6) we have positive integer k.
To establish (4.4.6), we write
1 1
C - z z-zQ-{Q-z 0 ) (2 ZQ) ^ c^

_ y (C-o)fc _f>(C-o)fc-^1