Maximum Modulus Principle:
If f is analytic and not constant in a given domain D, then |f(z)| has no maximum
value in D.
That is, there is no z0 in the domain such that |f(z)||f(z0)| for all points z in D.
Proof: Assume that |f(z)| does have a maximum value in D.
R
z0
CR
1
f ( z0 ) 
2
1
f ( z0 ) 
2
2

f ( z0  re i )d
0
2

0
f ( z0  re i ) d
Alternatively
Theorem: If f is analytic, continuous and not constant in a closed bounded region D,
then the maximum value of |f(z)| is achieved only on the boundary of D.
Some other aspects of the maximum modulus theorem:
Assume that f(z) is not 0 in a region R. Then if f(z) is analytic in R, then so is 1/f(z).
Result: minimum of |f(z)| also occurs on the boundary.
If g ( z )  u ( x, y )  iv ( x, y ) then e g ( z )  eu iv  eu eiv  eu .
Since the max and min of |f(z)| are on the boundary, so is the max and min of u(x,y).
Same applies to v(x,y).
Indented Contour
• The complex functions f(z) = P(z)/Q(z) of
the improper integrals (2) and (3) did not
have poles on the real axis. When f(z) has
a pole at z = c, where c is a real number,
we must use the indented contour as in
Fig 19.13.
Fig 19.13
THEOREM 19.17
Behavior of Integral as r → 
Suppose f has a simple pole z = c on the real axis. If
Cr is the contour defined by z  c  rei , 0     , then
lim  f ( z ) dz   i Res( f ( z ) , c)
r 0 Cr
THEOREM 19.17 proof
Proof
Since f has a simple pole at z = c, its Laurent
series is
f(z) = a-1/(z – c) + g(z)
where a-1 = Res(f(z), c) and g is analytic at c.
Using the Laurent series and the
parameterization of Cr, we have
C
f ( z ) dz
r

 a1 
0
(12)

ire i
i
i
d  ir  g (c  re ) e d
i
0
re
 I1  I 2
THEOREM 19.17 proof
First we see

ire i
I1  a1  i
d  a1  id
0 re  9
0
 ia1  iRes( f ( z ), c)

Next, g is analytic at c and so it is
continuous at c and is bounded in a
neighborhood of the point; that is, there
exists an M > 0 for which |g(c + rei)|  M.
Hence

i
i

I 2  ir  g (c  re )e d  r  Md  rM
0
0
It follows that limr0|I2| = 0 and limr0I2 = 0.
We complete the proof.
Example 5
Evaluate the Cauchy principal value of

sin x
 x( x 2  2 x  2) dx
Solution
Since the integral is of form (3), we consider
the contour integral
eiz dz
1
C z ( z 2  2 z  2), f ( z )  z ( z 2  2 z  2)
Fig 19.14
f(z) has simple poles at z = 0 and z = 1 + i in
the upper half-plane. See Fig 19.14.
Example 5 (2)
• Now we have
C
r
  
CR
R
C r
R
   2 iRes( f ( z )eiz , 1  i ) (13)
r
Taking the limits in (13) as R   and r 
0, from Theorem 19.16 and 19.17, we
have

eix
iz
P.V.

x( x  2 x  2)
2
dx  iRes( f ( z )e ,0)
 2iRes ( f ( z )eiz ,1  i )
Example 5 (3)
1
Res( f ( z )e , 0) 
2
Now
iz
1i
e
iz
Res( f ( z )e , 1  i ) 
(1  i )
4
Therefore,
 e 1i

eix
1
P.V.
dx  i  2i 
(1  i ) 
2
  x ( x  2 x  2)
2
4



Example 5 (4)
Using e-1+i = e-1(cos 1 + i sin 1), then
cos x
 1
P.V.
dx  e (sin 1  cos 1)
  x ( x 2  2 x  2)
2

sin x

1
P.V.
dx

[
1

e
(sin 1  cos 1)]
2
  x ( x  2 x  2)
2

Indented Paths
Theorem : Suppose that
(i) a function f(z) has a simple pole at a point z  x0 on the real axis, with
a Laurent series representa tion in a punctured disk 0 | z  x0 | R2 (Fig. 96)
and with residue B0 ;
(ii) C  denotes the upper half of a circle | z  x0 |  , where   R2 and
the clockwise direction is taken.
Then, lim
 0
 f ( z )dz   B i
0
C
C

x0

sin x

lim I 2  
dx 
 0
x
2
0

  e iz 
sin x
 x dx  Im  z dz 
CR
C
1 1
  0 (Jordan' s Lemma applies)
x R


I2
I1
I1  I 2  C R  C   0

sin x
sin( u )
sin u
I1  
dx   
du  
du  I 2
x
(u )


 u
Hope the integral over C  goes to zero?
eiz
C z dz 

2
eiz
C z dz 

2
ei (cos i sin )
i
i (cos  i sin )
i

e
d


ie
d
0 ei
0
2
2
2
2
0
0
0
0
i (cos  i sin )
ie
d 

i cos
  sin
  sin

e
e
d


e
d


e


 d
CR
eiz
has a simple pole at the origin.
z
C
Laurent series representa tion exists.
2
3

eiz 1  iz iz  iz 

 1  

 
z z  1!
2!
3!

B0  1
Alternativ ely to find B0
 ( z )  eiz
B0  e  1
0
eiz
C z dz   iB0   i

I2
I1
I1  I 2  C R  C   0

eiz
eiz
eiz
 z dz  C z dz  C z dz  0

R

eiz
 z dz   i
  eiz 
sin x
dz   
0 x dx  Im lim
 0  z




Contour Integration Example
The graphical interpretation
1
Find p (0) 
2
f ( x) 
sin( x)
x

sin(  )
  / 2 d. (  2)
>> x=[-10*pi:0.1:10*pi];
>> plot(x,sin(x)./x)
>> grid on
>> axis([-10*pi 10*pi -0.4 1])
sin( x) sin( z )
f ( z) 

x
z
z  x  iy is on the Real axis
>> x=[-10*pi:0.1:10*pi];
>> plot3(x,zeros(size(x)),sin(x)./x)
>> grid on
>> axis([-10*pi 10*pi -1 1 -0.4 1])
 eiz 
f ( z )  Im  z  x  iy is on the Real axis
z
eiz
g ( z) 
z
cos( z )
Reg ( z ) 
z
Img ( z ) 
sin( z )
z
f ( z) 
sin( x) sin( z )

x
z
z  x  iy is on the Real axis
>> x=[-10*pi:0.1:10*pi];
>> y = [-3:0.1:3].';
>> z=ones(size(y))*x+i.*(y*ones(size(x)));
>> mesh(x,y,cos(z)./z)
cos( z )
Reg ( z ) 
z
>> mesh(x,y,sin(z)./z)
Img ( z ) 
sin( z )
z
Cauchy’s Inequality: If f is analytic inside and on CR and M is the maximum
value of f on CR, then
R
z0
f
CR
(n)
n! M
( z0 )  n
R
(n  1,2, )
Note : Any point on CR can be written as : z  z0  Rei
Proof:
i
f
(
z

R
e
)
n
!
n!
0
f ( n ) ( z0 ) 
dz

2i CR ( z  z0 ) n1
2i
n!
2i
n!
2
2

0
2

0
i
f ( z0  R e )
n!
i
Rie d 
n 1 i ( n 1)
R e
2
f ( z0  Rei )
R n 1ei ( n 1)

0
f ( z0  Rei )
i
Rie
d
n 1 i ( n 1)
R e
f ( z0  Rei )
2
n!
i
Rie d 
2
2

R n 1ei ( n 1)
0
2

0
Rie i d
M
n! M
R d  n
n 1
R
R
f
(n)
n! M
( z0 )  n
R
M
f ' ( z0 ) 
R
(n  1,2, )
As R goes to infinity, then f’(z) must go to zero,
everywhere. Then f(z) must be constant.
Liouville’s Theorem: If f is entire and bounded in the complex plane, then f(z) is
constant throughout the plane.
Gauss’s Mean Value Theorem: If f is analytic within and on a given circle, its value
at the center is the arithmetic mean of its values on the circle.
Proof:
1
f ( z)
1
f ( z0 ) 
dz

2i CR z  z0
2i
2

0
f ( z0  re i ) i
1
ire
d


re i
2
2

0
f ( z0  re i )d

cos 3x
 x

i

1
2
dx 
cosh 3iy
  iy 
2



1
2
dy 
cosh 3iy
  iy 
2

i
i

1
2
dy 
cosh 3z
  z 
i
2

1
2
dz
i
cosh 3z
1
e3 z  e 3 z
dz  
dz  
dz
2
2
2
2
2
2
2 i 1  z  1  z 
 z   1
i 1  z  1  z 

i
2

cosh 3z

e 3 z  e 3 R cos i 3 R sin
C1
For    , e 3 z  e 3 R cos i 3 R sin  e3 R
For    , e3 z  e3 R cos i 3 R sin  e 3 R
On the left half plane
i
e 3 z
i 1  z 2 1  z 2 dz  0
The integral does not go to zero on the circle, the integral can’t be solved this way.
  x 3eiax 
  x 3eiax 
x 3 sin ax
 x 4  4 dx  Im x 4  4dx  Im  x 4  4dx

x4  4  0

x 4  4e i 3 ,4e i ,4ei ,4ei 3
x
 2e
i 3 / 4

, 2e i / 4 , 2ei / 4 , 2ei 3 / 4


z 3eiaz
z 3eiaz
z3
z3
 2i
Res
4
4
 z 4  4dz  C z 4  4  2i z  2Res
exp( i / 4 ) z  4
z  2 exp( i 3 / 4 ) z  4
x3
R3
R 



 0
4
4
x 4 R 4
Jordan’s Lemma
y
CR
R0
x
R
Suppose that
(i) a function f ( z ) is analytic at all points z in the upper half plane y  0 that are
exterior t o a circle | z | R0 ;
(ii) CR denotes a semicircle
z  Rei (0     ) where R  R0 ;
(iii) for all points z on C R , there is a positive constant M R such that | f ( z ) | M R ,
where lim M R  0
R 
Then, for every positive constant a, lim
R 
 f ( z )e
CR
iaz
dz  0

  ei 3 z

dx  Re  
dz 
2
2
2
2
x 1
 z  1


cos 3x

ei 3 z




z
1
2

1
2

R
1
2

1
2
R 
 M R 
 0

ei 3 z
 z 2  12 dz   z  i 2 z  i 2 dz
C1
C2


ei 3 z
ei 3 z
 z 2  12 dz   z  i 2 z  i 2 dz
z
1
2

1
2

R
1
2

1
2
R 
 M R 
 0
ei 3 z
 i12e 3  4ie 3
3
Res



ie
z i  z  i 2  z  i 2
16
z i
ei 3 z
 ( z) 
z  i 2
i3ei 3 z z  i   2z  i ei 3 z
 ' ( z) 
z  i 4
2

 z

ei 3 z
2
 1
2
dz  2i (ie 3 ) 
2
e3
C2