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22.2 The Residue Theorem 737

EXAMPLE 22.13
Let
cos(z)
f (z) = .
(z + i) 3
f has a pole of order 3 at −i, and
1 d 2 cos(z)
Res( f,−i) = lim (z + i) 3
2! z→−i dz 2 (z + i) 3
1 d 2 1 1
= lim cos(z) =− cos(−i) =− cos(i).
2 z→−i dz 2 2 2

The following example emphasizes that the value of f (z)dz depends on the residues of
γ
f at singularities enclosed by γ . Any other singularities (outside of γ ) are irrelevant for this
integral.

EXAMPLE 22.14

Evaluate f (z)dz where
γ
2iz − cos(z)
f (z) =
z + z
3
if γ is any closed path not passing through a singularity of f .
The singularities of f are simple poles at 0,i,−i. We will need the residues:
−cos(0)
Res( f,0) = =−1,
1
2
2i − cos(i) −2 − cos(i) 1
Res( f,i) = = = 1 + cos(i),
3i + 1 −2 2
2
2i(−i) − cos(−i) 1
Res( f,−i) = =−1 + cos(i).
2
3(−i) + 1 2
The following cases occur.

Case 1 If γ does not enclose any of the singularities of f , then f (z)dz = 0 by Cauchy’s
γ
theorem.
Case 2 If γ encloses 0 but not ±i,

f (z)dz = 2πiRes( f,0) =−2πi.
γ
Case 3 If γ encloses i but not 0 or −i,
1

f (z)dz = 2πi 1 + cos(i) .
2
γ
Case 4 If γ encloses −i but not 0 or i,

1
f (z)dz = 2πi −1 + cos(i) .
2
γ

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October 14, 2010 15:37 THM/NEIL Page-737 27410_22_ch22_p729-750