Complex Analysis - Maths KU

5.2 Complex Integrals 251
Now in view of (4), the last integral is the same as
�
C
� 4
¯zdz=
=
−1
(2t 3 � 4
+9t) dt + i
�
1
2 t4 + 9
2 t2
� �4 �
���
−1
−1
�4
+ it 3
�
�
�
�
3t 2 dt
−1
=195+65i.
EXAMPLE 2 Evaluating a Contour Integral
�
1
Evaluate dz, where C is the circle x = cos t, y = sin t, 0≤ t ≤ 2π.
z
C
Solution In this case z(t) = cos t + i sin t = e it ,z ′ (t) =ie it , and f(z(t)) =
1
z(t) = e−it . Hence,
�
C
1
dz =
z
� 2π
0
(e −it ) ie it dt = i
� 2π
dt =2πi.
As discussed in Section 5.1, for some curves the real variable x itself
can be used as the parameter. For example, to evaluate �
C (8x2 − iy) dz
on the line segment y = 5x, 0 ≤ x ≤ 2, we write z = x +5xi, dz =
(1 + 5i)dx, �
in the usual manner:
�
(8x 2 � 2
− iy) dz =(1+5i) (8x 2 − 5ix) dx
C
C (8x2 − iy) dz = � 2
0 (8x2 − 5ix)(1 + 5i) dx, and then integrate
�
8
=(1+5i)
3 x3
�2 0
0
0
�
5
− (1 + 5i)i
2 x2
�2 =
0
214 290
+
3 3 i.
In general, if x and y are related by means of a continuous real function
y = f(x), then the corresponding curve C in the complex plane can be
parametrized by z(x) =x + if(x). Equivalently, we can let x = t so that
a set of parametric equations for C is x = t, y = f(t).
Properties The following properties of contour integrals are analogous
to the properties of real line integrals as well as the properties listed in (5)–(8).
Theorem 5.2 Properties of Contour Integrals
Suppose the functions f and g are continuous in a domain D, and C is a
smooth curve lying entirely in D. Then
(i) �
C
kf(z) dz = k� f(z) dz, k a complex constant.
C
(Theoremcontinues on page 252 )