Complex Analysis - Maths KU

270 Chapter 5 Integration in the Complex Plane
log e 2−log e 3 = log e 2
3
☞
and so,
� 2i
3
1
z dz = log 2
e
3
π
+ i ≈−0.4055 + 1.5708i.
2
EXAMPLE 5 Using an Antiderivative of z −1/2
�
1
Evaluate
z1/2 dz, where C is the line segment between z0 = i and z1 =9.
C
Solution Throughout we take f1(z) =z 1/2 to be the principal branch of
the square root function. In the domain |z| > 0, −π < arg(z) < π, the
function f1(z) =1/z 1/2 = z −1/2 is analytic and possesses the antiderivative
F (z) =2z 1/2 (see (9) in Section 4.2). Hence,
� 9
i
1
z
�
�
�
�
9
dz =2z1/2
1/2
i
=(6− √ 2) − i √ 2.
See Problem 5 in Exercise 1.4
� �√
� �� �
√ ��
2 2
=2 3 − + i
2 2
Remarks Comparison with Real Analysis
(i) In the study of techniques of integration in calculus you learned that
indefinite integrals of certain kinds of products could be evaluated
by integration by parts:
�
f(x)g ′ �
(x)dx = f(x)g(x) − g(x)f ′ (x)dx. (11)
You � undoubtedly � have used (11) in the more compact form
udv = uv − vdu. Formula (11) carries over to complex analysis.
Suppose f and g are analytic in a simply connected domain D. Then
�
f(z)g ′ �
(z)dz = f(z)g(z) − g(z)f ′ (z)dz. (12)
In addition, if z0 and z1 are the initial and terminal points of a
contour C lying entirely in D, then
� z1
f(z)g ′ �
�
�
(z) dz = f(z)g(z) �
�
z0
z1
z0
� z1
−
z0
g(z)f ′ (z) dz. (13)
These results can be proved in a straightforward manner using Theorem5.7
on the function d
fg. See Problems 21–24 in Exercises
dz
5.4.