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698 CHAPTER 20 Complex Integration

y
4

1 5

2 x

3

FIGURE 20.3 The join of γ 1 ,··· ,γ n .

EXAMPLE 20.7

it 2
Let γ 1 (t) = 3e for 0 ≤ t ≤ π/2. Let γ 2 (t) = t + 3(t + 1)i for 0 ≤ t ≤ 1. Let γ = γ 1 γ 2 . We will

compute f (z)dz, where f (z) = Im(z).
γ
γ is shown in Figure 20.4. γ 1 is the quarter circle part from 3 counterclockwise to 3i, and γ 2
2
is part of the parabola x = (y − 3) /9 from 3i to 1 + 6i.
it
it
For the integral over γ 1 , z(t) = 3e ,so f (z(t)) = Im(3e ) = 3sin(t). Since
it

z (t) = 3ie =−3sin(t) + 3i cos(t),
we have
π/2
Im(z)dz = 3sin(t)[−3sin(t) + 3i cos(t)]dt
γ 1 0
π/2 π/2
2
=−9 sin (t)dt + 9i sin(t)cos(t)dt
0 0
9 9
=− π + i.
4 2
2

On γ 2 , z(t) = t + 3(t + 1)i,so f (z(t)) = 3(t + 1), and z (t) = 2t + 3i. Then
1
Im(z)dz = 3(t + 1)(2t + 3i)dt
γ 2 0
1 1
2
= (6t + 6t)dt + 9i (t + 1)dt
0 0
27
= 5 + i.
2
y
1 + 6i

3i

x
3

FIGURE 20.4 γ in Example 20.7.

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October 14, 2010 15:32 THM/NEIL Page-698 27410_20_ch20_p695-714