Complex Analysis - Maths KU

250 Chapter 5 Integration in the Complex Plane
and dy by x(t), y(t), x ′ (t) dt, and y ′ (t) dt, respectively, the right side of (9)
becomes
�
C udx−vdy
�
� b
�� �
[u(x(t),y(t)) x ′ (t) − v(x(t),y(t)) y ′ (t)] dt
a
�
C
+ i
vdx+udy
�
� b
�� �
a
[v(x(t),y(t)) x ′ (t)+u(x(t),y(t)) y ′ (t)] dt.
(10)
If we use the complex-valued function (1) to describe the contour C, then
(10) is the same as � b
a f(z(t)) z′ (t) dt when the integrand
f(z(t)) z ′ (t) =[u(x(t),y(t)) + iv(x(t),y(t))] [x ′ (t)+iy ′ (t)]
is multiplied out and � b
a f (z(t)) z′ (t) dt is expressed in terms of its real and
imaginary parts. Thus we arrive at a practical means of evaluating a contour
integral.
Theorem 5.1 Evaluation of a Contour Integral
If f is continuous on a smooth curve C given by the parametrization
z(t) =x(t)+iy(t), a ≤ t ≤ b, then
�
� b
f(z) dz = f(z(t)) z ′ (t) dt. (11)
C
a
The foregoing results in (10) and (11) bear repeating—this time in somewhat
different words. Suppose z(t) =x(t) +iy(t) and z ′ (t) =x ′ (t) +iy ′ (t).
Then the integrand f (z(t)) z ′ (t) is a complex-valued function of a real variable
t. Hence the integral � b
a f(z(t)) z′ (t) dt is evaluated in the manner defined
in (4).
The next example illustrates the method.
EXAMPLE 1 Evaluating a Contour Integral
Evaluate �
C ¯zdz, where C is given by x =3t, y = t2 , −1 ≤ t ≤ 4.
Solution From(1) a parametrization of the contour C is z(t) =3t + it 2 .
Therefore, with the identification f(z) = ¯z we have f(z(t)) = 3t + it 2 =
3t − it 2 . Also, z ′ (t)=3+2it, and so by (11) the integral is
�
C
� 4
¯zdz= (3t − it 2 � 4
)(3+2it) dt =
−1
−1
� 2t 3 +9t +3t 2 i � dt.