Page 755 - Advanced_Engineering_Mathematics o'neil

22.2 The Residue Theorem 735

EXAMPLE 22.10
2
f (z) = sin(z)/z has a simple pole at 0, and
sin(z)
Res( f,0) = lim zf (z) = lim = 1.
z→0 z→0 z
Because 0 is the only singularity of f ,if γ is any closed path enclosing the origin, then

sin(z)
dz = 2πiRes( f,0) = 2πi.
z 2
γ
Here is an alternative version of Theorem 22.5.

COROLLARY 22.1

Let f (z) = h(z)/g(z) where h is continuous at z 0 and h(z 0 ) = 0. Suppose g is differentiable at z 0
and has a simple zero there. Then f has a simple pole at z 0 , and
h(z 0 )
Res( f, z 0 ) = .
g (z 0 )

The fact that f has a simple pole at z 0 follows from Theorem 22.2. By Theorem 22.5,
because g(z 0 ) = 0, we can write

Res( f, z 0 ) = lim(z − z 0 ) f (z)
z→z 0
h(z)
= lim(z − z 0 )
z→z 0 g(z)
h(z) h(z 0 )
= lim = .
z→z 0 (g(z) − g(z 0 ))/(z − z 0 ) g (z 0 )

EXAMPLE 22.11
Let
4iz − 1
f (z) = .
sin(z)
Then f has a simple pole at π, and by the corollary,
4iπ − 1
Res( f,π) = = 1 − 4πi.
cos(π)
In fact, f has a simple pole at nπ for each integer n, and

4inπ − 1
n
Res( f,nπ) = = (−1) (4inπ − 1).
cos(nπ)
EXAMPLE 22.12

Evaluate f (z)dz where f (z) is the function of Example 22.11 and γ is the closed path shown
γ
in Figure 22.1.

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October 14, 2010 15:37 THM/NEIL Page-735 27410_22_ch22_p729-750