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22.2 The Residue Theorem 733

Here h(z) = (z − 3π/2) has a zero of order 4 at 3π/2, and g(z) = cos (z) has a zero of order 7
4
7
at 3π/2. Therefore, f has a pole of order 3 at 3π/2.

EXAMPLE 22.8
3
9
f (z) = tan (z)/z has a pole of order 6 at 0, because the numerator has a zero of order 3 there
and the denominator has a zero of order 9.

EXAMPLE 22.9
Let
1
f (z) = .
cos (z)(z − π/2) 3
4
Then f has a pole of order 7 at π/2. f also has poles of order 4 at each (2n + 1)π/2 with n as
any nonzero integer.

SECTION 22.1 PROBLEMS

In each of Problems 1 through 12, determine all singulari- 6. z/(z + 1) 2
ties of the function and classify each singularity. In the case 2
7. (z − i)/(z + 1)
of a pole, give the order of the pole.
8. sin(z)/sinh(z)
1. cos(z)/z 2 9. z/(z − 1)
4
4sin(z + 2) 10. tan(z)
2.
2
(z + i) (z − i)
11. sec(z)
1/z
3. e (z + 2i)
12. e 1/z(z+1)
4. sin(z)/(z − π) 13. Let f be differentiable at z 0 and f (z 0 ) =0. Let g have
cos(2z) a pole of order m at z 0 . Show that fg has a pole of
5. order m at z 0 .
2
2
(z − 1) (z + 1)
22.2 The Residue Theorem
We will use singularities and a single term of the Laurent expansion to develop a powerful method
for evaluating integrals.
Suppose f is differentiable in an annulus 0 < |z − z 0 | < R and has an isolated singularity at

z 0 .Let γ be a closed path in this annulus enclosing z 0 . We want to evaluate f (z)dz.Atleast
γ
in theory, we can write the Laurent expansion
∞
n
f (z) = c n (z − z 0 ) .
n=−∞
Recall the formula for the c n ’s:
1 f (z)
c n = dz
2πi γ (z − z 0 ) n+1
for n = ··· ,−1,−2,0,1,2,···. The coefficient of 1/(z − z 0 ) in the expansion is
1
c −1 = f (z)dz.
2πi γ

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October 14, 2010 15:37 THM/NEIL Page-733 27410_22_ch22_p729-750