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20.3 Consequences of Cauchy’s Theorem 705

The theorem states that the integral of f has the same value over both paths under the
conditions stated. This means that we can replace the integral over one curve with the integral
over the other, allowing us great flexibility in choice of paths in evaluating an integral.

EXAMPLE 20.11
Evaluate

1
dz
z − a
where is any closed path enclosing the number a (Figure 20.8(a)).
We do not know , so it might appear that we cannot evaluate this integral. Because f (z) =
1/(z − a) is not differentiable in this region, and Cauchy’s theorem does not apply. However, a
is the only point at which f is not differentiable. Place a circle γ about a of sufficiently small
radius r so that the two curves do not intersect (Figure 20.8(b)). Now f is differentiable on both
curves and on the region between them, so

f (z)dz = f (z)dz.
γ
The point is that we can easily evaluate the integral over γ . Using polar coordinates centered at
it
a, write γ(t) = a +re for 0 ≤ t ≤ 2π. Then

f (z)dz = f (z)dz
γ
1
2π 2π
it
= ire dt = i dt = 2πi.
re it
0 0
A proof of the deformation theorem is reminiscent of the argument used for the extended
Green’s theorem in Chapter 12. Figure 20.9(a) shows typical curves and γ . Insert line segments
L 1 and L 2 between these paths (Figure 20.9(b)), and use these to form two closed paths, and ,
as in Figure 20.10. Both and are oriented positively (counterclockwise), which is consistent
with positive orientations on and γ . Because f is differentiable on and γ and all points in
between, f is differentiable on and and all points they enclose, so by Cauchy’s theorem,

f (z)dz = f (z)dz = 0.

y y

a a

x x

(a) (b)
FIGURE 20.8 Enclosing a in a circle γ interior to
in Example 20.11.

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October 14, 2010 15:32 THM/NEIL Page-705 27410_20_ch20_p695-714