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728 CHAPTER 21 Series Representations of Functions

2

z

z 1 L 2
L 2
z 0
z 0 z 0
r
1 2
L
R r 1 1
r 2
L 1

FIGURE 21.3 1 and 2 in the proof of The-
orem 21.5 for Problem 21.11, Section 21.2.

FIGURE 21.2 Circles in the proof of Theorem 21.5 for
Problem 21.11, Section 21.2.

line segments L 1 and L 2 between these circles (Figure Use these to show that
21.2), forming two closed paths 1 and 2 in the
∞
annulus (shown separately in Figure 21.3). Show that 1 f (w) n
f (z) = dw (z − z 0 )
1 f (w) n=0 2πi γ 2 (w − z 0 ) n+1
f (z) = dw
2πi w − z ∞
1
1 n 1
and + 2πi γ 1 f (w)(w − z 0 ) dw (z − z 0 ) n+1 .
1 f (w) n=0
dw = 0.
2πi w − z Finally, replace n =−m −1 in the last summation, and
2
Add these to obtain then use the deformation theorem to replace γ 1 and γ 2
1 f (w) 1 f (w) by any closed path about z 0 and in the annulus.
f (w) = dw + dw 12. Fill in the details of the following proof of the unique-
2πi w − z 2πi w − z
1 2
ness of Laurent expansions. Suppose, in the annulus,
with counterclockwise orientation on both paths. By
noting that in these integrals, the parts of the inte- ∞ ∞

n n
grals over the line segments vanish (these segments f (z) = c n (z − z 0 ) = b n (z − z 0 ) ,
are traversed in both directions), show that n=−∞ n=−∞
1 f (w) 1 f (w)
f (z) = dz − dw where the c n ’s are the Laurent coefficients. Then
2πi w − z 2πi w − z
γ 2 γ 1
f (w)
with both integrations counterclockwise over the
2πic k = dw
closed path. γ (w − z 0 ) k+1
On γ 2 , show that |(z − z 0 )/(w − z 0 )| < 1, and use ∞
1
the geometric series to write = b n (z − z 0 ) n dw
(w − z 0 ) k+1
γ
∞ n=−∞
1 1
n
= (z − z 0 ) . ∞
w − z (w − z 0 ) n+1 1
n=0 = b n dw.
γ (w − z 0 ) k−n+1
On γ 1 , show that |(w − z 0 )/(z − z 0 )| < 1toshowthat n=−∞
∞
1 1 Choose γ as a circle, and evaluate the integral in the
=− (w − z 0 ) n .
w − z (z − z 0 ) n+1 last summation to complete the proof.
n=0
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October 14, 2010 15:35 THM/NEIL Page-728 27410_21_ch21_p715-728