1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions 599

Proof Suppose that we have

+oo

f(z) = I: Am(z - ar (8.18-14)

m=-oo

and also

+oo

f(z) = I: Bk(z - a)k (8.18-15)

k=-oo

where the expansion are valid in open rings G 1 and G 2 , respectively, con-

taining the same circle r, and the coefficients Am are given by (8.18-2).

From (8.18-15) we have, for t E r,

+oo

f(t) = I: Bk(t -a)k

k=-oo

the convergence being uniform on r by Theorem 8.43. Hence we obtain

Am= 1 J f(t) dt = 1 J L:t:-^00 Bk(t -a)k dt

27ri (t-a)m+l 27ri (t-a)m+l

since

r r

+oo

= ~ ~ Bk j(t-a)k-m-l dt = Bm

27ri ~

k=-oo r

J(t-a)k-m-1 dt = { ~7ri

r

Examples 1. To find:

if k = m

if k # m

(a) The Laurent expansion of

valid in 2 < lzl < 4.

1

f(z)= (z-2)(4-z)

(b) The Taylor expansion of the same function valid in lzl < 2.

( c) The Laurent expansion of f ( z) valid for lz I > 4.

(a) We have

f(z)= 1 [ 1 - -+-1 ]

. 2 z-2 4-z