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20.1 The Integral of a Complex Function 697

it

it
f (z(t))z (t) = e ie = i.
Therefore,
π
f (z)dz = idt = πi.
γ 0

EXAMPLE 20.5

Evaluate z dz if ϕ(t) = t + it for 0 ≤ t ≤ 1.
2
ϕ
The graph of ϕ is the straight line segment from 0 to 1 + i. On the curve, z(t) = ϕ(t) =
t + it = (1 + i)t. Then z (t) = 1 + i and

2
2
3 2
f (z(t))z (t) = ((1 + i)t) (1 + i) = (1 + i) t = (−2 + 2i)t .

Then
1 2
2
2 (−2 + 2i)t dt = (−1 + i).
z dz =
3
γ 0
EXAMPLE 20.6
2
Evaluate zRe(z)dz if γ(t) = t − it for 0 ≤ t ≤ 2.
γ
2
On this curve, z(t) = t − it ,so
f (z(t)) = z(t)Re(z(t))
2
3
2
2
2
= (t − it )Re(t − it ) = (t − it )(t) = t − it .

Furthermore, z (t) = 1 − 2it,so
3
4
3
2
2

f (z(t))z (t) = (t − it )(1 − 2it) = t − 3it − 2t .
Then

2
3
4
2
f (z)dz = (t − 3it − 2t )dt
γ 0
2 2 152
4
2
3
= (t − 2t )dt − 3i t dt =− − 12i.
15
0 0
Thus far, we can only integrate over a smooth curve. Often we want to integrate over a
piecewise smooth curve, which is a curve that is made up of a finite number of smooth arcs.
Such a curve has a continuous tangent except perhaps at finitely many points (such as corners
or “sharp points”). In this case, we think of γ as made up of curves γ 1 ,γ 2 ,···γ n , with each γ j
smooth and the terminal point of γ j equal to the initial point of γ j+1 , as in Figure 20.3. We call γ
the join of γ 1 ,··· ,γ n and write

γ = γ 1 γ 2 ··· γ n ,
as we did in Section 12.1.
Define
n

f (z)dz = f (z)dz.
γ γ j
j=1
This is the analog of
c b b

+ =
a c a
for real integrals.
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October 14, 2010 15:32 THM/NEIL Page-697 27410_20_ch20_p695-714