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702 CHAPTER 20 Complex Integration

In Theorem 20.1, the notation is used. The oval on the integral is a reminder that the path
γ
is closed. This notation was also used in connection with line integrals in Chapter 12. It is not
required to include this oval for integrals over closed paths.

EXAMPLE 20.9
If γ is any closed path in the plane,
2
z
e dz = 0,
γ
because e is differentiable on the entire plane, which is a simply connected domain.
2
z

EXAMPLE 20.10
We will evaluate

2z + 1
dz,
2
γ z + 3iz
where γ is the circle |z + 3i|= 2 of radius 2 about −3i.
Observe that f (z) is differentiable at all points except 0 and −3i, where the denominator
vanishes. Use a partial fractions decomposition to write
1 1 6 + i 1
f (z) = + .
3i z 3 z + 3i
Then
2z + 1 1 1 6 + i 1

dz = dz + dz.
γ z + 3iz 3i γ z 3 γ z + 3i
2
Because γ does not enclose 0, 1/z is differentiable on and within the simply connected
domain enclosed by γ . By Cauchy’s theorem,

1
dz = 0.
γ z
However, 1/(z + 3i) is not differentiable in the region enclosed by γ , so Cauchy’s theorem
does not apply to the integral of this function over γ . We will evaluate this integral directly by
it
parametrizing γ(t) = z(t) =−3i + 2e (polar coordinates centered at −3i)for 0 ≤ t ≤ 2π.Now
1 1
2π

dz = z (t)dt
γ z + 3i 0 z(t) + 3i
1 it
2π 2π
= 2ie dt = idt = 2πi.
0 2e it 0
Therefore,
2z + 1 6 + i 2

dz = (2πi) = − + 4i π.
γ z + 3iz 3 3
2
A proof of Cauchy’s theorem requires a delicate argument we will not engage here. However,
a less general version can be established easily. Write f = u + iv and assume that u and v are
continuous with continuous first and second partial derivatives on G. Now we can apply Green’s

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October 14, 2010 15:32 THM/NEIL Page-702 27410_20_ch20_p695-714