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736 CHAPTER 22 Singularities and the Residue Theorem

y

x
–π π 2π 3π

FIGURE 22.1 γ in Example 22.12.

The singularities of f enclosed by γ are 0,π,2π,3π and −π. By the residue theorem and
the conclusion of Example 22.11,
3

4iz − 1
dz = 2πi Res( f,nπ)
γ sin(z) n=−1
= 2πi[−(−1 − 4πi) − 1 − (−1 + 4πi) + (−1 + 8πi) − (−1 + 12πi)]
= 8 + 2πi.

Next we treat poles of multiplicity greater than one.

THEOREM 22.6 Residue at a Pole of Order m

Let f have a pole of order m at z 0 . Then

1 d m−1 m
Res( f, z 0 ) = lim [(z − z 0 ) f (z)].
(m − 1)! z→z 0 dz m−1

The theorem can be proved by manipulating the Laurent expansion of f (z) about z 0 , which
in the case of a pole of order m is

c −m c −m+1
f (z) = + + ··· ,
(z − z 0 ) m (z − z 0 ) m−1
so that

m
(z − z 0 ) f (z) = c −m + c −m+1 (z − z 0 ) + ··· + c −1 (z − z 0 ) m−1 + ··· .
Differentiation of this equation m − 1 times isolates c −1 , yielding the residue of f at z 0 .
Theorem 22.6 reduces to Corollary 22.1 if m = 1 with the conventions that 0!= 1 and the
zero-order derivative of a function is the function itself.
By this result, to obtain the residue of f at a pole of order m,
m
1. Multiply f (z) by (z − z 0 ) .
2. Differentiate m − 1 times.
3. Take the limit as z approaches z 0 .
4. Divide by (m − 1)!.

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October 14, 2010 15:37 THM/NEIL Page-736 27410_22_ch22_p729-750