Section 4: Complex Integration I
Examples and Problems
Example 1
[integration in the complex plane]
Integrate the function f ( z )  Re z from 0 to 1  j along the straight-line joining these endpoints.
Solution
The line can be represented by the function z (t )  t  jt ( 0  t  1). Thus dz / dt  1  j and
f [ z (t )]  Re( t  jt )  t . Thus

C
1
t2
f ( z )dz   t (1  j )dt  (1  j )
2
0
1
 (1  j ) / 2
0
Example 2
[integration in the complex plane]
Integrate the function f ( z )  ( z  z 0 ) m where m is an integer and z 0 is a constant. Integrate in
the positive direction around the circle z  z 0  re j ( 0    2 ).
Solution
The path of integration is z (t )  z o  re jt ( 0  t  2 ). Thus dz / dt  jre jt and

f z (t )  z 0  re jt  z 0

m
 r m e mjt
Thus

2
f ( z )dz  jr m 1  e ( m 1) jt dt
C
0
 jr
m 1
m 1
e ( m 1) jt
(m  1) j

2
0

r
e 2 ( m 1) j  1
m 1
r m 1
cos2 (m  1)  j sin 2 (m  1)  1

m 1
0

Complex Variables Section 4 Examples & Problems - 1
This follows because cos2 (m  1)  1 and sin 2 (m  1)  0 for any integer m. This is true
for m  1 ; this formula is not valid for m  1 because we can't divide by zero ( 1 /( m  1) ). In
the case of m  1 we have

2
f ( z )dz  jr 0  e ( 0 ) jt dt
C
0
2
 j  dt
0
 2j
Thus our result is
 (z  z
C
0
 2j
) m dz  
0
(m  1)
(m  1)
Example 3
[Cauchy's Integral Theorem]
Determine the domain of analyticity of the function f ( z )  z 2 /( z  3) and apply the Cauchy
Integral Theorem to show that
 f ( z)dz  0
C
where the closed contour C is the unit circle (that is, of radius 1, centre 0).
Solution
This function is analytic everywhere in the complex plane except at the point z  3 (the
numerator, z 2 , is a polynomial which is entire, i.e. analytic everywhere). The closed contour C
is the circle of radius 1 with centre z  0 . The only singular point of this function, z  3 , lies
outside the circle/contour, and the function is analytic everywhere on and inside the
circle/contour, so by the Cauchy Integral Theorem, the integral is zero.
Complex Variables Section 4 Examples & Problems - 2
Problem 1
[Line Integration]
(i)
(ii)
(iii)
(iv)
What curves are represented by the following functions? [sketch in the complex plane]
(a) (1  2 j)t ( 0  t  3 )
(b) 2  j  2e jt ( 0  t  2 )
(c) j  e  j ( 0    2 )
Evaluate the integral of z taken counterclockwise over the triangle with vertices
j ,  1, 1  j . Does your result agree with what you expect from the Cauchy Integral
Theorem? [Draw the triangle in the complex plane and introduce a parameter t to
represent the three straight-line paths (edges) in the form z  f (t ) ].
Evaluate the integral of the function z 2 taken counterclockwise over the circle with centre
(1, j ) and radius 3 . Does your result agree with what you expect from the Cauchy
Integral Theorem? [use the exponential polar form to represent the path of integration]
Integrate the function f ( z )  Im z counterclockwise about the circle z  r .
[ANSWER:  r 2 ]
Problem 2
[Cauchy's Integral Theorem]
Determine the domain of analyticity of the following functions (that is, the domains in which the
functions are analytic) and apply the Cauchy Integral Theorem to show that
 f ( z)dz  0
C
where the closed contour C is the unit circle.
1
f ( z)  2
(i)
z  2z  2
f ( z )  tan z
(ii)
Problem 3
[Cauchy's Integral Theorem]
(i) For what contours C will it follow from the Cauchy Integral Theorem that
e1 / z
 2 dz  0
C z 9
(in other words, can you find any closed contours for which the Cauchy Integral Theorem
cannot be applied?)
(ii) Use Cauchy's Integral Theorem to show that the integrals of an analytic function around the
boundaries of a doubly connected region are equal, and hence evaluate the integral
cos z
C z dz
where C consists of z  e j (counterclockwise) and z  3e j (clockwise), 0    2 .
[ANSWER: 0]
Complex Variables Section 4 Examples & Problems - 3