9.
Complex Integrals
9.1. The Complex Integral
We define the complex line integral of f (z) along a curve C in the complex plane as:

C
f ( z )dz   (u  jv)( dx  jdy )
C
  udx  vdy  j vdx  udy
C
C
which may be evaluated as a combination of two real line integrals.
Example

Evaluate
C
zdz from z = 0 to z =1+j along each of the paths shown.
Im z
1+ j
C1
C2
C2
Re z
1
Sol:

C
zdz   xdx  ydy  j  ydx  xdy
C
C
(i)
let y = x
1

zdz  
x 0
C1
0  j  2 xdx  j
x 0
(ii)

C2
1
zdz  
x 0
1
1
x 0
y 0
xdx  j  0  
1
 ydy  j  1dy
y 0
 j
Note that both paths give the same result of j which is also the value of
1 j
z2 
 
 2 0
1
Example
Evaluate

C
zdz along each of the paths of the previous example.
Sol:

C
zdz   xdx  ydy  j   ydx  xdy
C
C
(i) Let y  x

C1
1
1
x 0
x 0
zdz   2 xdx  j  0
1
(ii)

C2
1
1
1
x 0
x 0
y 0
zdz   xdx  j  0  
1
ydy  j  1dy
y 0
 1 j
Note that in this case the result depends on the path. Some integrals which will be useful later are:
Example
Evaluate

C
1
dz where C is the Unit circle. (  denotes integration around a closed curve in an
z
anticlockwise (positive) direction).
Sol:
On the unit circle z  e j and hence
dz
 je j giving
d
2
1
- j
j
dz

C z
 0 e  je d  2j
Example
0
dz

n
C (z  z )
2j
0
Proof:

n 1
where C is the circle with centre z 0 and radius r.
n 1
The path is given by z  z 0  re j for 0    2 .
Hence
j
2 jre
2
dz
e (1-n ) j

d


j
 0 r n e jn
 0 r n1 d
C ( z  z )n
0
I 
2
2
If n  1
j  e (1n ) j 
j
I  n 1 
 n 1

r  (1  n ) j  0
r
 e (1n ) j2
1 
 (1  n ) j  (1  n ) j 


0
If n  1
I 
2
0
jd  2πj
Complex integrals have the following properties:
(i)

C
f ( z )dz   f ( z )dz where –C denotes the path C but the reverse direction.
C
(ii)
If C  C1  C2 , then

C
f ( z )dz   f ( z )dz   f ( z )dz
C1
C2
(iii)
 [k
f ( z )  k2 f 2 ( z )]dz  k1  f1 ( z )dz  k 2  f 2 ( z )dz where k1 and k 2 are complex constants.
1 1
C
C
C
(iv)

C
f ( z )dz  ML where L is the length of the contour C and M is such that
f ( z)  M
everywhere on C.
8.2. Cauchy’s Integral Theorem
A curve C in the complex plane is:
(i)
simple if it does not cross itself;
(ii) simple closed if z(a) = z(b) (C is anticlockwise);
(iii) smooth if it has a unique tangent at each point;
(iv) piecewise smooth if it consists of a finite number of smooth curves joined end to end.
A region R is connected if every two points of R can be connected by a curve in R,
A connected region R in the complex plane is simply-connected if every simple closed curve in R
encloses only points of R. Otherwise the region is multiply-connected.
3
R
R
Multiply-connected
Simply-connected
Cauchy’s Theorem
If f (z) is analytic in a simply-connected region D, then for every simple closed curve C in D,

C
f ( z)dz  0
Proof:

C
f ( z )dz   u d x v d y j  v d x u d y
C
C

Now apply Green’s Theorem in the Plane  F1dx  F2 dy  
C

 F2
  x
R


F1 
dxdy to the real and
y 

imaginary parts. We shall also assume that f ' ( z ) is continuous and hence that u x , u y , v x , v y
are continuous.
 v
u 
 udx  vdy      x  y dxdy  0
C
R
using the C  R equation u y  v x
and
 u
v 
 vdx  udy     x  y dxdy  0
C
R
using the C  R equation u x  v y
where R is the region enclosed by C.
Hence

C
f ( z)dz  0
In fact it is not necessary that f ' ( z ) be continuous for the above result to be true. In this case the
theorem is known as the Cauchy-Goursat Theorem.
Example
The function f ( z )  1, f ( z )  z, f ( z )  z 2 etc., and f ( z )  e z are analytic everywhere in the
complex plane. Hence in each ease

C
f ( z)dz  0 for all closed contours C.
For example, it is easy to verify explicitly, that
z
C
4
2
dz  0 where C is a circle of radius r, centre
the origin. On C, z  re j , 0    2 , hence
2
 z dz  
2
C
0
2
2 2 j
r e
 e 3 j 
r 3 6j
rje d  r j 

e 1  0

3
 3 j 0
j

3

Example
we have that

C
dz
 0 , n  1 where C is the circle with centre z 0 and radius r.
( z  z0 ) n
However these results do not follow from Cauchy’s Theorem since f (z) has a singularity at z  z 0
which lies within C and hence f (z) is not analytic within C. This example shows that the condition
that
If
f(z) be analytic is sufficient but not necessary for

C

C
f ( z)dz  0 to be true.
f ( z)dz  0 for every closed curve C in a simply-connected region R where f(z) is continuous,
then f(z) is analytic. This converse of Cauchy’s Theorem is known as Morera’s theorem.
Some Consequences of Cauchy’s Theorem
Let f (z) be analytic in a simply-connected region R. Then the following results hold.
(a)
Independence of Path
If z1 and z 2 are any two points in R, then

z2
z1
f ( z )dz
is independent of the path joining z1 and z 2 .
Proof:
By Cauchy’s Theorem

ADBEA
ADB

ADB
and
E
C1
f ( z )dz  0
A
z1
Hence

Im z
f ( z )dz  
BEA
f ( z )dz  
AEB

C1
B
z2
D
f ( z )dz  0 , i.e.
C2
f ( z )dz  0
Re z
f ( z )dz   f ( z )dz
C2
This result explains the “coincidence” in the first example in Section 8.1
(b) Indefinite integration
z
If z 0 and z are any two points in R then F ( z )   f ( s )ds is analytic in R and F ' ( z )  f ( z ) .
z0
5
Proof:
In order to prove this we consider


z
F ( z  z )  F ( z )
1 z z
 f ( z) 
f ( s )ds   f ( s )ds  f ( z )

z
z
0
z
z 0
1 z z
1 z z
 
f ( s )ds  f ( z )   ( f ( s )  f ( z )) ds
z
z 0
z z0
since z is fixed and

 1 z z f ( z )ds  f ( z ) z z ds  f ( z )
 z z

z z
Now f (x) is continuous. Hence for any   0,   0 such that f (s)  f ( z)   whenever
sz 
The last integral

z z
z
f ( z )ds above is independent of the path from z to z  z by Cauchy’s
Theorem and hence we take the straight line segment. We may choose z such that z   . Then
F ( z  z )  F ( z )
1
 f ( z) 
z
z

z z
z0
( f ( s )  f ( z )) ds 
z z
1

z z
ds 
1
 z  
z
Thus,
F ' ( z )  lim
z 0
F ( z  z )  F ( z )
 f ( z ) and F is analytic in R.
z
This enables us to evaluate complex line integrals by indefinite integration since we have,

z2
z1
f ( z )dz  F ( z 2 )  F ( z1 ) where F ' ( z )  f ( z ) for any path joining z1 to z 2 .
Example
1 j

0
1 j
 z2 
(1  j ) 2
zdz    
 j as we had in the first example of Section 8.1
2
 2 0
Example
z2
d
z
(sin z )  cos z . Hence,  cos zdz  sin z z12  sin z 2  sin z1
z
1
dz
Example
6
d
1
1
(ln z )  and
is analytic in the region D  {z : Re z  0} for any path in D.
dz
z
z
Hence,

j
-j
1


j
dz  ln z - j  ln j-ln (- j )  j-( - j )  j
z
2
2
(c) Deformation of Path
If f (z) is analytic in a region R bounded by two simple closed curves C1 and C2 , where C2 lies
within C1 as shown in the figure, and is analytic on C1 and C2 , then
Im z

C1
f ( z )dz   f ( z )dz
C2
C1
C2
R
Re z
Proof:
Im z
Consider cutting the region
R by a line AB. Then by
Cauchy’s Theorem

AEFABGHBA
C1
AB
H
C2
f ( z)dz  0
G
and

E
F
B
R
A
Re z
f ( z )dz   f ( z)dz
BA
Hence,

AEFABGHBA
f ( z )dz  
AEFA

C1

AB

BGHB

BA

C2
Therefore

C1
f ( z )dz   f ( z )dz
C2
This result shows that if we wish to integrate f (z) around any closed curve C1 , we may replace C1
by any other curve C2 as long as f (z) is analytic in the region between C1 and C2 .
7
Often we wish to evaluate integrals where f (z) has a singularity at some point within C (for
otherwise the integral is zero). This result may be used to replace C by a circle around the
singularity.
Example
Evaluate
1
dz
C z

where C is any closed curve enclosing the origin.
Sol:
Im z
C
1
1
Re z
We may replace C by the unit circle and use a previous example to give 
C
1
dz  2j
z
The result we proved above easily generalizes to:
Let f (z) be analytic in a region R bounded by the non-overlapping simple closed curves
C , C1 , C 2 ,..., C n (where C1 , C2 ,..., Cn are inside C), and on these curves. Then

C
f ( z )dz   f ( z )dz   f ( z )dz  ....   f ( z )dz
C1
C2
Cn
Im z
C
C2
Cn
R
C1
Re z
8
Example
Evaluate

C
z
dz around
( z  1)( z  2 j )
(i)
C1 : z  1 / 2
(ii)
C2 : z  3 / 2
(iii)
C3 : the rectangle with vertices 2+j, -1+j, -1-3j, 2-3j.
Sol:
Im z
1
C2
C1
-1
1
-1
2
Re z
C3
-2
-3
The integrand has singularities at z = 1 and z = - 2j and, using partial fractions, may be written as
z
1 2 j 1
42j
1




( z  1)( z  2 j )
5
z 1
5
z2j
Hence

C
(i)
f ( z )dz 
1 2 j
1
42j
1
1 2 j
42j
dz 
dz 
I1 
I2


C z2j
5 C z 1
5
5
5
C1 contains no singularities, i.e f(z) is analytic within and on C1 , hence

C1
f ( z )dz  0
(ii) C2 contains the singularity at z = 1 but not the singularity at z = 2j, hence I 2  0 and
I1  2πj . For I 1 we have replace C2 by a small circle with centre z = 1 and used the result
1
C z  z0 dz  2πj where C is a circle centre z0 , radius r. Thus

C2
f ( z )dz 
1 2 j
2
 2j 
(2  j )
5
5
(iii) C 3 contains both singularities and hence I1  2j and I 2  2j
9
where C 3 has been
replaced by two small circles centered on z = 1 and z = -2j respectively. Hence,

C3
f ( z )dz 
1 2 j
42j
 2j 
 2πj  2j
5
5
+
10