Chapter 9 Complex Variable Theory
9-1 Introduction
Complex z plane can be expressed as
z = (x, y) = x + iy
i
= re
(Cartesian notation)
(Polar notation)
where x and y respectively denote the x and y axes. θ is the argument of z
1
y
x
and is defined as θ = arg(z) = tan ( ) .
The value of θ satisfying of 0 ≦ θ < 2π is the principal argument of z
and is written as
θ = arg(z) + 2kπ,
k = 0,  1,2 ,…
Ex.:
arg(1 + i) =π/4 + 2kπ,
k = 0,  1,2 ,…
Ex.:
1i  [cos( 2k)  i sin( 2k)]i  (e 2 ki )i  e 2 k ,
for k = 0,  1,2 ,…
The absolute value or modulus of z is written as r = z
The complex conjugate of z is defined as z  x  iy
Euler’s formula is e z  e x iy  e x (cos y  i sin y)
← Prove it!
z
z
Hyperbolic cosine function cosh(z) = (e  e ) / 2 .
z
z
Hyperbolic sine function sinh(z) = (e  e ) / 2
1
9-2 Differential calculus
f(z) = 1/z is not differentiable at z = 0. However, f ( z )  z is not
differentiable anywhere. The main reason is
x  iy
z 0 x  iy
f (z)  lim [f (z  z)  f (z)] / z  lim
z 0
 1 for y  0
Thus f (z)  
 1 for x  0
Assume f(z) = u(x, y) + iv(x, y), then
[u ( x  x, y)  iv ( x  x, y)]  [u ( x, y)  iv ( x, y)]
x 0
x
f (z)  lim
 f (z) 
u
v
i
x
x
and
[u ( x, y  y)  iv ( x, y  y)]  [u ( x, y)  iv ( x, y)]
y 0
iy
f (z)  lim
 f (z) 
v u
i
y y
Thus we can obtain Cauchy-Riemann equations as
u v

x y
and
v
u

x
y
The similar equations can also be obtained from the following relations.



Assume that the velocity V  u i  v j , then the continuity equation is
2
 u v
V 

0
x y
u v

x y
i.e.,
The irrotational equation is

v u
V  

0
x y
v
u

x
y
i.e.,
If f(z) is differentiable, then f ( z ) satisfies
f (z) 

x u
v
(u  iv )

i
x
z x
x
f (z) 

y 1 u
v
v
u
(u  iv )
 (
i
)
i
y
z i y
y
y
y
and
Ex.:
Prove whether f (z)  zz  x 2  y 2 is differentiable.
Sol.
∵ u  x 2  y2 ,
∴
u
 2x ,
x
v
 0,
x
v=0
u
 2y
y
v
0
y
Obviously, Cauchy-Riemann equations are not satisfied except for z
= 0. This implies that f(z) is analytic nowhere.
A function that is analytic everywhere is said to be entire.
3
Ex.:
f(z) = sin(z) is analytic everywhere.
Sol.:
∵ sin(iy) = (iy ) 
1
1
(iy ) 3  (iy ) 5      = isinh(y)
3!
5!
f(z) = sin(z) = sin(x + iy) = sin(x)con(iy) + sin(iy)cos(x)
= sin(x)cosh(y) +i sinh(y)cos(x)
∴ u = sin(x) cosh(y),
∴
∴
v = sinh(y)cos(x)
u
 c o sx() c o s h
y)(,
x
u
 s i n hy() s i nx()
y
v
  sinh( y) sin( x ) ,
x
v
 cosh( y) cos( x )
y
u v

x y
(1)
v
u

x
y
(2)
Obviously, Cauchy-Riemann equations are satisfied. This implies that f(z) is
analytic everywhere.
Ex.:
f(z) = 1/z is analytic everywhere in the cut plane.
Ex.:
The principal value of log(z) can be written as
log(z) = ln(r) + i 
Assume f (z)  u(r, )  iv(r, ) . Thus we have
f (z) 

r
u
v i
(u  iv )
(
i
)e
r
z
r
r
4
(3)
and
f (z) 

 e i u
v
e i v
u
(u  iv )

(
i
)
(
i
)

z
ri


r 

(4)
Obviously, we can obtained the following relationships from Eqs. (3) and (4).
u 1 v

r
r 
and
v
1 u

r
r 
The other proof is as follows.
u u x u y


r x r y r
∴
u u
u

cos() 
sin( )
r x
y
v v x v y


 x  y 
v
v
v


r
sin(

)

r
cos(

)
∴

x
y

v
u
u u
 r sin( )
 r cos()


y
x r
Similarly, we can determine
v
1 u

.
r
r 
Definition: A function is called the harmonic function when it satisfies
5
Laplace’s equation and is C 2 .
Theorem:
If f(z) = u(x, y) + iv(x, y) is analytic in a domain D, then u and v
are harmonic in D.
Proof:
∵ f(z) is analytic. Thus we have
 2u
2v

x 2 xy
2v
 2u
 2
xy
y
 2u  2u

0
∴
x 2 y 2
Similarly, we can also obtain
2v 2v

0
x 2 y 2
Ex.:
Given: 1. f(z) = u(x, y) + iv(x, y) is harmonic in the entire z plane.
2.
u ( x, y)  3xy 2  x 3
Find:
The conjugate harmonic function v(x, y)
Sol.:
u
v
 3y 2  3x 2 
x
y
v( x, y)  y3  3x 2 y  c
9-3 Complex integral calculus
Introduce the complex integral
 f (z)dz
c
, where f may or may not be
analytic, and c may be either a closed curve or an arc. But, c is piecewise smooth
6
and simple. The curve c is called the path of integration or contour.
Suppose that c has initial and final points z = A and z = B, respectively. Then
B2 A 2
Cdz = B – A and Czdz  2  2 .
Ex.:
Determine I =
 z dz , where C:
2
C
x  4  y 2 is the parabolic arc as shown
in Fig. 1.
Sol.:
Method 1:
I   [( x 2  y 2 )dx  2xydy ]  i  [2xydx  (x 2  y 2 )dy]
C
C
2
I   [( 4  y 2 ) 2  y 2 ]( 2 y)dy  2(4  y 2 ) ydy 
2
2
i  {2(4  y 2 ) y(2 y)dy  [( 4  y 2 ) 2  y 2 ]dy}
2
I  0
Method 2:
Ex.:
z3
I
3
16 16
i i
3
3
2 i
2i
2z 3

3
0
2i
2(2i) 3 16

 i
3
3
n
Evaluate I  C (z  a ) dz where C is a circle of radius R, centered at z =
a and oriented counterclockwise as shown in Fig. 2. a is a given complex
number. n is any integer.
Sol.:
It is obvious that the contour C is closed.
z  a  Re i
2
I   (Re i ) n Rie id
0
2
R n 1 i ( n 1) 
I
e
0
n 1
0
for n ≠ -1
7
2
I  i  d  2i
0
for n = -1
Thus
2i for n  1
for n  1
 (z  a ) dz   0
n
C
Ex.:
Find the upper bound of the integral
 sin( z)dz ,
C
where C: x = 3 - 3y/2 is a straight line as shown in Fig. 3.
sin( z)  sin( x  iy )  [sin 2 ( x ) cosh 2 ( y)  sinh 2 ( y) cos 2 ( x )]1/ 2
Sol.:
 [cosh 2 ( y)  sinh 2 ( y)]1/ 2  [cosh 2 (2)  sinh 2 (2)]1/ 2
Thus
 sin( z)dz   sin( z)dz   sin( z) dz
 [cosh (2)  sinh (2)]  dz  13[cosh
C
C
2
C
2
1/ 2
C
Ex.:
2
(2)  sinh 2 (2)]1/ 2
Find the upper bound of the integral
dz
dz , where C is a circular contour as shown in Fig. 4.
C z( z  2) 3

Sol.:
1
1

3
z(z  2)
5  33
Thus
dz
dz
2  5 2
dz  
dz 

3
3
C z( z  2)
C z( z  2)
135 27

8
9-4 Cauchy’s theorem
Due to
 f (z)dz   (u  iv )(dx  idy )

 f (z)dz   (udx  vdy)  i (vdx  udy )
C
C
C
C
C
In accordance with Green’s theorem, we have

C
f (z)dz    (
S
u v
u v
 )dxdy   (  )dxdy
S x
y x
y
The substitution of Cauchy-Riemann’ equations into the above equation yields
 f (z)dz  0
C
Theorem: Cauchy’s theorem
If f(z) is analytic in a simply connected domain D, then
 f (z)dz  0
C
for
every piecewise-smooth simple closed curve C in D.
Ex.: I  C
1
dz  0 , where the curve C is shown in Fig. 5.
(z  2)( z  3)
Theorem: Path independence
If f(z) is analytic in a simply connected domain D, then
 f (z)dz
C
is
independent of path in D.
Ex.:
Determine I  C
1

dz   i , where the curve C is
z (z  2)( z  4)
16
2
shown in Fig. 6.
9
f (z) 
Sol.:
1
1
1
A(2) 1
A(4) 1
 ' (0)  2 (0) 

z (z  2)( z  4) z
z
B' (2) z  2 B' (4) z  4
2
where A(z) =1, B(z)  (z  2)( z  4) , (z) 
' (z)  
1
1

2
(z  2) (z  4) (z  2)( z  4) 2
Let f(z) be analytic in a simply connected domain D, and let z 0 be
Theorem:
any fixed point in D. Then
z
(i)
G(z)   f ()d is analytic in D and G(z)  f (z)
z
0
(ii) If G(z)  f (z) , then
Ex.:
Ex.:
1
and
(z  2)( z  4)

3
2i

z
z0
f ()d  G(z) - G(z 0 )
sin( z)dz  cosh( 2)  cos(3)
 ln 2 5
 2  i 4 
i 1

dz

1i z 
 ln 2 3
 2  i 4 

Sol.: ln(-i) = ln( e

 i
2
3
i

3
)   i  ln( e 2 )  i
2
2

i
4
1

ln(1+i) = ln( 2e )  ln 2  i
2
4
1
3
ln(-i) - ln(1+i) =  ln 2  i
2
4
for Fig. 7(a)
10
and
1
5
ln(-i) - ln(1+i) =  ln 2  i
2
4
Theorem:
Cauchy integral theorem
for Fig. 7(b)

Cauchy’s theorem +
 (z  a )
C
n
dz
If f(z) is analytic in a simply connected domain D, C is a piecewise-smooth
closed curve C in D oriented counted counterclockwise. Then
f (z)
dz  2if (a )
C za

Proof:
Method 1:
f (z)
f (z)
dz  
dz  0
C za
C z  a

f (z)
f (z)
1
dz  
dz  f (a ) 
dz  2if (a )
C za
C z  a
C z  a

Method 2:
f (z)
f (a )
f ( z )  f (a )
dz  
dz  
dz
C z  a
C z  a
C
za

f ( z )  f (a )
dz
C
za
 2if (a )  
f ( z )  f (a )
dz 
C
za

f ( z )  f (a )
M
dz  2  2M
C
za


∵ M → 0 as ρ → 0
∴
f (z)
dz  2if (a )
C za

Ex: Determine I j  C
sin( z)
dz where C j is shown in Fig. 8.
2
j z
 iz  2
11
sin( z)
dz
C j ( z  i)( z  2i)
Ij  
Sol.:
I1 = 0
I2 
2i
sinh( 2)
3
I3  2i[
sin( 2i) sin( i) 2i

]
[sinh( 2)  sinh( 1)]
3i
 3i
3
f (z  z)  f (z)
1
1
1
 lim
[

]f ()d
z 0
z 0 2iz C   ( z  z)
z
z
f (z)  lim
1
1
f ()d
z 0 2i C [  ( z  z)](   z )
1
1

f ()d

2i C (  z) 2
 lim
f (z  z)  f (z)
1 f (z  2z)  f (z  z) f (z  z)  f (z)
 lim
[

]
z 0
z 0 z
z
z
z
1
1
2
1
 lim
[


]f ()d
2
z 0 2i( z ) C   ( z  2z )
  (z  z)   z
f (z)  lim
1
∴ f (z)  lim
z 0 2i(z) 2

f (z) 
2(z) 2 f ()
C [  (z  2z)][  (z  z)](  z) d
2!
f ( )
d

2i C (  z)3
Similarly, we can obtain
f ( n ) (z) 
n!
f ( )
d

2i C (  z) n 1
 Generalized Cauchy integral theorem
12
ez
I   3 dz , where C is the counterclockwise unit circle
C z
Ex: Determine
z  1.
Sol.:
2i d 2 z
I
(e )
 i
2! dz 2
z 0
sin( 2z)
dz , where C is the clockwise rectangular
C ( z  3)( z  1) 2
I
Ex: Determine
contour with vertices at 3+i, -2+i, -2-i and 3-i.
Sol.:
I
2i d sin( z)
i
(
)
  [4 cos( 2)  sin( 2)]
1! dz z  3 z 1
2
9-5 Taylor series
Theorem:
Taylor series
If f(z) is analytic in a domain D including a circle with the radius R and

f ( n ) (a )
(z  a ) n
z  a  R , then f (z)  
n!
n 0
Proof: Let C be the circle   a   , taken counterclockwise.
Due to f (z) 
1
f ()
d and

2i C   z
1
1
1

  z   a 1  (z  a ) /(   a )
1
za za 2

[1 
(
)  ...]
a
a a
13
f (z) 
1
f ()
f ( )
f ()
[
d  (z  a ) 
d ...  (z  a ) n 1 
d] 
2
C
C
C
2i   a
(  a )
(  a ) n
R n (z)  R n 1 (z)  ...
(5)
where the remainder term R n (z) is
(z  a ) n
R n ( z) 
2i
f ()
C (  a)n 1 d
Since f(z) is analytic on C so that it must be bounded on C. On the other
hand, f (z) < m, where m is a positive value.
Thus we have
(z  a ) n
R n (z) 
2

C
f ( )
R nm
d 
(  a ) n 1
2n 1
It is obvious that R n (z)  R n (z)  0
R nm
C d  n
for R < ρ and n →  .
The application of the Cauchy integral formular to Eq. (5) yields
f (z)  f (a )  f (a )( z  a )     
f ( n 1) (a )
(z  a ) n 1    
(n  1)!
or
za
f ( n ) (a )
f ( z)  
(z  a ) n in z  a  R or
1
n
!
R
n 0

(6)
The above result shows that the Taylor series converges to f(z). This also
implies that every analytic function is representable by a convergent power
series.
14
Ex.:
1
 1 z  
1 z
is valid in z < 1
1
 1 z2  z4   
2
1 z
is valid in z < 1
9-6 Laurent series
Given: 1. D is a closed region between and including concentric circles C i
and Co with their common center at z = a.
2. f(z) is analytic in D.
Find:
f (z) 

c
n  
n
(z  a ) n
in D
where C is any piecewise-smooth simple closed counterclockwise
path lying entirely in D and c n is defined as
cn 
1
f ()
d
2i C (  a ) n 1
Proof: The Cauchy’s integral formula gives
f (z) 
1
f ()
d

C

C

C

C
1
2
3
4
2i
z
where C1 = Co and - C 3 = C i .
f (z) 
1
f ()
1
f ()
d


d
2i C1   z
2i C3   z
1
1  za n

( )
  z   a n 0   a
15
for the C1 integral

1
f ( )
1
f ( )  z  a n
d 
(
) d   c n (z  a ) n



2i C1   z
2i C1   a n 0   a
n 0
where c n 
1
f ()
d
2i C1 (  a ) n 1
1
1  a n

( )
z
z  a n 0 z  a
for the C 3 integral
1
f ()
1
f ()    a n
d  
 ( ) d
2i C3   z
2i C3 z  a n 0 z  a

 1

  
f ()(  a ) n d (z  a ) ( n 1)

2i C3

n 0 

 1

 ( n 1)
 
f
(

)(


a
)
d

(z  a ) n


C3

n  1 2i


c
n  1
∴ f (z) 

c
n  
n
n
(z  a ) n
(z  a ) n
Remark: Explain the difference between Taylor series and Laurent series.
Ex.:
1
 1 z  z2  
1 z

in z < 1
1 1
1
1
1
1 1
  [1   ( ) 2    ]    2    
z 1 1/ z
z
z
z
z z
16
in 1 < z < ∞
Ex.:
Expand f (z) 
Sol.:
Let   z  i
1
z(z - 2)
about z = i in
5  z i  
Method 1:
1
1
1
1
1

 2
z(z - 2) (  i)(  i  2)  1  1 /  1  (i  2) / 
1
 1  i /   i 2 / 2     
1 i / 
i
1
for

i2
1
1
 1  (i  2) /   (i  2) 2 /  2      for

1  (i  2) / 
Thus
1
1
2  2i
1  6i



 
2
3
z(z - 2) (z  i)
( z  i)
( z  i) 4
for
5  z i  
Method 2:
1
1
1 1
1

 (

)
z(z - 2) (  i)(  i  2)
2 i i2
1
1
1

(

)
2 1  1 /  1  (i  2) / 
1
 1  i /   i 2 / 2     
1 i / 
i
1
for

i2
1
1
 1  (i  2) /   (i  2) 2 /  2      for

1  (i  2) / 
17

1
1
1
1 2  2i
(

) 2 
 
2 1  1 /  1  (i  2) / 

3
Thus
1
1
2  2i


 
2
z(z - 2) (z  i)
( z  i) 3
Ex.:
for
Find all the Laurent series for f ( z) 
5  z i  
1
:
z-2
(a) with the center at z = 0 (b) with the center at z = 2
Sol.:
1
1 1
1
z
z 2




[
1


(
)    ] for z < 2
(a) z  2
2 1 z
2
2
2
2
This expression is the Taylor series.
1
1
1
2
2
1 2
 [1   ( ) 2    ]   2    
z 1 2 / z z
z
z
z z
(b) The Laurent series of f(z) is
Ex.:
for 2 < z
1
for the center at z = 2.
z-2
Find all the Laurent series for f (z) 
1
:
(z - 1)(z - 2)
(a) with the center at z = 0 (b) with the center at z = 1
Sol.:
(a) f (z) 
1
1
1


(z - 1)(z - 2) z - 2 z - 1
The function has singularities at z = 1 and 2, thus the Laurent
expansions are valid about z = 0 for 0 < z < 1, 1 < z < 2, and 2
< z .
18
(A) 0 < z < 1
1
1

 (1  z  z 2    )
z 1
1 z
1
1 1
1
z
z

  [1   ( ) 2    ]
z2
2 1 z
2
2
2
2
This expression is the Taylor series.
1
1 3z 7z 2
 

    for 0 < z < 1
(z  1)( z  2) 2 4
8
(B)
1< z <2
1
1 1
1
1
1

 [1   ( ) 2    ]
z 1 z 1 1/ z z
z
z
1
1 1
1
z
z

  [1   ( ) 2    ]
z2
2 1 z
2
2
2
2
The Laurent series of this function is

1
zn
1
  ( n 1  n 1 ) for 1 < z < 2
(z  1)( z  2)
z
n 0 2
(C)
z >2
1
1 1
1
1
1

 [1   ( ) 2    ]
z 1 z 1 1/ z z
z
z
1
1 1
1
2
2

 [1   ( ) 2    ]
z  2 z 1 2 z
z
z
z
19
The Laurent series of this function is

1
2n  1

(z  1)( z  2) n 1 z n 1
for
z >2
(b) There are two representations for the center at z = 1. They are
valid for 0 < z  1 < 1 and z  1 > 1.
(A) For 0 < z  1 < 1
1
1

 [1  (z  1)  (z  1) 2    ]
z  2 (z  1)  1
Thus the Laurent series of this function is

1
1

  (z  1) n for 0 < z  1 < 1
(z  1)( z  2)
z  1 n 0
(B) For
z 1 > 1
1
1
1
1
1
1
1 2



[1 
(
)    ]
z  2 (z  1)  1 z  1 1  1 /( z  1) z  1
z 1 z 1
Thus the Laurent series of this function is

1
1
1 n 1

 (
)
(z  1)( z  2)
z  1 n 0 z  1
9-7
Singularities
f(z) has a kth-order zero at z = a if

f ( n ) (a )
f ( z)  
(z  a ) n
n!
n k
(k)
where f (a )  0 .
20
for
z 1 > 1
f(z) is not analytic at z = a provided that z = a is a singular point.
f(z) is singular at z = a but analytic in an annular 0  z  a   . Then z
= a is said to be an isolated singular point of f(z).
If f (z) 

c
n  N
n
(z  a ) n , where N is a positive integer, then the
singularity of f(z) at z = a can be regarded as an Nth-order pole. A first-order pole
is sometimes called a simple pole. If there are an infinite number of negative
powers of (z – a), then this singularity can be treated as an essential singularity.
The coefficient c1 is defined as the residue of f(z).
If f(z) has a simple pole at z = a, then it can be written as

c 1
f (z) 
  c n (z  a ) n
z  a n 0
Further, we have

(z  a )f (z)  c 1   c n (z  a ) n 1
n 0
This implies that the coefficient c1 is
c 1  lim [( z  a )f (z)]
z a
Assume that f(z) has an Nth-order pole at z = a, then
(z  a ) N f (z)  c  N 

c
n   N 1
At this time, the coefficient c1 is
21
n
(z  a ) n
1
d N 1
c 1 
lim { N 1 [( z  a ) N f (z)]}
( N  1)! za dz
Ex.:
Find the singularities and the residues of
f (z) 
Sol.:
1
z 3 (2  z)
f(z) is singular at z = 0 and z = 2.
Method 1:
1
d2 3
1
Re s f (z)  lim { 2 [z 3
]} = 1/8
z 0
2! z0 dz
z ( 2  z)
Re s f (z)  lim [( z  2)
z 2
z 2
1
] = -1/8
z 3 (2  z)
Method 2:
1
1
1
 3
 3
3
z (2  z) 2z (1  z / 2) 2z

z
 ( 2)
n
n 0
for 0  z  2
Re s f ( z ) = 1/8
z 0
1
1
1


3
3
z (2  z)
[( z  2)  2] (z  2)
(z  2)[( z  2)  6(z  2) 2  12(z  2)  8]
3

1
8(z  2)[( z  2)  6(z  2) 2  12(z  2)] / 8  1

1
{1  [( z  2)3  6(z  2) 2  12(z  2)] / 8    }
8(z  2)
3
Thus
Re s f ( z ) = -1/8
z 2
22
Theorem:
If p(z) and q(z) respectively have zeros of order P and Q at z = a,
then p(z)/q(z) has a pole of order N = Q - P if Q > P and is analytic
there if Q ≦ P.
Ex.: Determine the residues of f(z) = z/sin(z)
s f (z)  lim
Sol.: Re
z n
z  n
(z  n)z
2z  n
 lim
 (1) n n
z n cos( z )
sin( z)
sin(z) has 1st-order zeros at z = nπ.
z has 1st-order zero at z = 0 and zeroth-order zeros at z =  π,  2π, …。
Thus f(z) has 1st-order poles at z =  π,  2π, …。
9-8
Residue theorem
Consider the closed contour integral
I   f (z)dz
C
where f(z) has a finite number of singularities within C. Each of them is
isolated.
As shown in Fig. 9, we have
I   f (z)dz = C f ( z)dz  C f (z )dz
C
1
2
Expanding f(z) in the Laurent series in annuli 0  z  z1  1 and
0  z  z 2  2 yields
f (z) 

c
n  
(1)
n
(z  z1 ) n
in 0  z  z1  1
23
and
f (z) 

c
n  
( 2)
n
(z  z 2 ) n
in 0  z  z 2  2
Thus
I
C1

 c(n1) (z  z1 ) n dz  
C2
n  
 2i(c
(1)
1

c
n  
( 2)
n
(z  z 2 ) n dz
c )
( 2)
1
where c1 ' s are called the “residues”.
Theorem: Residue theorem
Let C be a piecewise-smooth simple closed curve oriented counterclockwise
and let f(z) be analytic inside and on C except at finitely many isolated singular
( j)
points z1 , z 2 ,  z k within C. If c 1 denotes the residue of f(z) at z = z j , then
k
 f (z)dz  2i c
C
j1
Ex.:
Determine I =
Sol.:
f (z) 
( j)
n
cos( z)
C z (z 2  4)dz
3
cos( z)
z 3 (z 2  4)
1
d 2 3 cos( z)
3
Re s f (z)  lim 2 [z 3 2
]
z 0
2! z0 dz
z (z  4)
16
d2
cos( z)
cosh( 2)
Re s f (z)  lim 2 [( z  2i) 3 2
]
z 2i dz
z  2i
z (z  4)
32
24
Thus
I = 2i(
Ex.:
Evaluate I =
Sol.:
f (z) 
cosh( 2) 3
 )
32
16
sin( z )
dz , where C is the unit circle z  1 .
C
z4

sin( z )
z4
Method I:
The Laurent series for f(z) about z = 0 is
1
1
z


  
z 3 3! z 5!
Thus
I= 
i
3
Method II:
1
d 3 4 sin( z)
1
Re s f (z)  lim 3 [z
]


z 0
3! z0 dz
z4
3!
Thus
I= 
Ex: Determine
i
3
ez
I   3 dz , where C is the counterclockwise unit circle
C z
z  1.
25
Sol.:
2i
d 2 3 ez
I
lim
(z 3 )  i
2! z0 dz 2
z
 Compare with the generalized
Cauchy integral theorem
Ex.:
Evaluate I =

C
cot( z)dz , where C is the unit circle z  1 traversed in
a clockwise sense.
Sol.:
f (z)  cot( z)
Method I:
The Laurent series for f(z) about z = 0 is
z2 z4
1

  
1
2
!
4
!
cot( z) 

 
z3 z5
z
z

  
3! 5!
Thus
I =  2i
Method II:
Re s f (z)  lim [z cot( z)]  1
z 0
z 0
Thus
I =  2i
Consider I =
Let z =

2
0
F( cos , sin )d
e i
26

i
dz = ie d  izd 
ei  e i z 2  1

 cos  
2
2z
ei  e i z 2  1

 sin  
2i
2iz
z2  1 z2 1 1
,
) dz
 I  C F(
2z
2iz iz
Ex.:
Evaluate I =
Sol.:
I

2
0
d
2  sin 
1
1
dz
2
C
z  1 iz
2
2iz
 I  2C
1
dz
z 2  4iz  1
1
dz
C
[ z  (2  3 )i][ z  ( 2  3 )i]

I  2 

I  2i

I
Remark: I =
lim
{[ z  (2  3 )i]
z ( 2  3 ) i
2
3

2
0
d
a  sin 
27
2
}
[z  (2  3 )i][ z  (2  3 )i]
 I  2C
1
dz
z 2  2aiz  1
2
 z  (a  a  1)i

I can not be performed for a  1 .

Ex.: I =
dx
 1  x 2
Sol::
R
1
1
1
1
dz

dz

s( 2
)
C z 2  1 CR z 2  1 R z 2  1 dz  2i Re
z i
z 1
1
1
1
dz

dz

CR z 2  1
CR z 2  1
CR (z  i)( z  i)dz
1



R

(R  1)( R 2  1)1/ 2
R

Thus, the value of

CR
e iz
dz approaches zero as R approaches infinity.
z

dx

 1  x 2

Theorem: Simple pole on the real axis
If f(z) has a simple pole at z = a on the real axis, then
lim

 0 C 
Proof:
f (z)dz  i Re s f (z)
z a
By the definition of a simple pole, f(z) can be expressed as
f (z) 
c 1
 g (z)
za
28
where g(z) is analytic on the semicircle of integration.

C
f (z)dz  

0
c 1
dz   g (z)dz  ic 1
C
z
where c  is a unit circle z   traversed in a clockwise sense.
Remark:

C
f (z)dz   f (z)dz   f (z)dz   f (z)dz  2i Re sf (z)
R
R
CR
We can prove Rlim


CR
C
f ( z)dz  0 .
Thus

pr.v. f (x)dx  i Re sf (z)  2i Re sf (z)

Ex.: Show that I =
Sol::


0
sin x

dx 
x
2
iz
iz
 e
R e
eiz
eiz
eiz
C z dz CR z dz  R z dz  C z dz   z dz  0


CR
eiz
dz 
z

CR
eiz
eiz
1
dz  
dz  R  
CR z
z
R
It is obvious that the upper bound of
prove its actual value.
29

CR
e iz
dz is
z
 . However, we cannot

CR
e iz
dz 
z
 2
/ 2
0

CR
iz
 e

e iz
dz  
Rd    e  R sin  d
0
0
z
z

e  R sin  d  2(  e  R sin  d  

0
 2  2 
/ 2

/ 2
e  R sin  d)
e  R sin  d  2  e  R sin 
If  is sufficiently small, then the value of

CR
e iz
dz approaches zero as R
z
approaches infinity.
eiz
1
z
dz

(

i

   )dz  i
C z
C z
2
Thus
e iz
 z dz  i

The above result implies that


0
sin x

dx 
x
2

Ex.: Find the principle value pr.v.
Sol::
1
dx
( x 2  3x  2)( x 2  1)
2
2
Assume f(z) is f (z)  (z  3z  2)( z  1) Simple poles have z = 1,
2 and i.
Re s f (z)  
z 1
1
2
30
Re s f (z) 
1
5
Re s f ( z ) 
3i
20
z2
and
z i
Thus

1
dx
  ( x  3x  2)( x 2  1)
3i
1 1

 2i(
)  i(  ) 
20
2 5
10
pr.v.
Ex.: Show that I =
2


0
 cos x
sin x

dx

dx

1/ 2
1
/
2
0 x
x
2
31