Transcript Slide 1

COMPLEX INTEGRATION
INTRODUCTON
 DOMAIN: A set S of points in
the Argand plane is said to be
connected set if any two of its
points can be joined by a
continuous curve , all of those
points belongs to S.
Open and Closed Domain

An open connected set is called an
open domain. If the boundary
points of S are also added to an
open domain then it is called
closed domain.
CONTOUR



A continuous chain of a finite
number of regular arcs is called a
contour.
If a contour is closed and doesnot
intersect itself ,then it is called
closed contour.
Example: Boundaries of triangles
and rectangles.
Simply & Multiply connected
Domains


A domain in which every closed
curve can be shrunk to a point
without passing out of the region is
called simply connected domain.
If a domain is not simply connected
then it is called multiply
connected.
Complex line integral
Suppose f(z) is continuous at every
point of a closed curve C having a
finite length.i.e C is a rectifiable
xn
zk
curve.
Zn=b
Zn-1
x3
a
z1
O
x2
z2
z3
Zk-1
Divide C into n parts by the means of
points z0,z1,z2,…………,zn
Let a=z0, b=zn
we choose a point xn on each arc joining
zk-1 to zk.
Form the sum
Sn=∑f(xr)(zr-zr-1)
Where ‘r’ varies from 1 to n.
Suppose maximum value of (zr-zr-1) tends
to zero as n tends to zero.
Then the sum sn tends to a fixed limit which
does not depends upon the mode of
subdivision and denote this limit by

C
f ( z )dz
Which is called the complex line integral or line
integral of f(z) along C .
THEOREM
CAUCHY’S INTEGRAL FORMULA
Statement:
Let f(z) be analytic over a region R enclosed by a
closed path C. If z0 is a point within R, then

C
f ( z)
dz  2 i f ( z0 ).
z  z0
Note that while f(z) is analytic throughout R,
f(z)/(z-z0 ) is not analytic (z0 is a pole).
Im{z}
R
C
z0
Re{z}
Proof:
We add and subtract f(z0) to the numerator of the
integrand so as to split-up the integral into two
terms:
f ( z )  f ( z0 )  f ( z0 )
f ( z)
dz

dz
C z  z0 C
z  z0

C
f ( z )  f ( z0 )
dz
dz  f ( z0 ) 
.
C zz
z  z0
0
If f(z) is analytic within the region R, then it is also
continuous. So, for any e>0, we can find a d>0 such
that for
z  z0  d ,
we have
f ( z)  f ( z0 )  e .
Let us choose a r  d such that z  R, i.e., the disk
|z-z0| < r is totally within R. Let us denote the path
|z-z0| = r by the symbol C’ .
r
Im{z}
R
C
z0
C’
Re{z}
So if
z  z0  r  d ,
f ( z)  f ( z0 )  e ,
and
f ( z )  f ( z0 )
e
e


(on the pathC' ).
z  z0
z  z0 r
So for appropriate values of d and r, the integrand in

C'
f ( z )  f ( z0 )
dz
z  z0
can be made arbitrarily small. Now since the
integrand is analytic except at z = z0, we have

C'
f ( z )  f ( z0 )
f ( z )  f ( z0 )
dz  
dz.
C
z  z0
z  z0
The integral

C
dz
z  z0
is equal to 2i. So,
f ( z)
C z  z0 dz  0  f ( z0 )2 i
and the theorem is proved.
THEOREM
Cauchy’s integral formula for the
derivative of an analytic function
Statement:
if a function is analytic within and on a
closed contour c and ‘a’ is any point lying in
it then
1
f ( z)
f ' ( z0 ) 
dz.
2

2 i C z  z0 
Proof:
Let a+h be a point in the neighbourhood of a
point ‘a’ ,then by cauchy’s integral formula,

C
and
f ( z)
dz  f (a )2 i
za
f ( z)
C z  (a  h) dz  f (a  h)2i
From which we get
f ( a  h)  f ( a )
h
1
f ( z) 
1
1 


dz.



2 i C h  ( z  a  h) ( z  a) 
1




1
f ( z)
h
  1 dz.

1 

2 i C ( z  a)h 
( z  a) 


2
3


 h   h 
1
f ( z)
h
  
  ....... dz.

 


2 i C ( z  a)h  ( z  a)  ( z  a)   ( z  a) 


taking limit as h
0,we get
2


1
f ( z)
1
h
h



 ....... dz.

2
3

2 i C ( z  a)  ( z  a) ( z  a) ( z  a)


f ( a  h)  f ( a ) 1
f ( z)  1
lim

 0  0  ........ dz.


h 0
h
2 i C z  a  ( z  a)

1
f ( z)
f (a) 
dz.
2

2 i C ( z  a)
THEOREM
Cauchy’s integral formula
for higher order derivative
Statement:If a function f(z) is analutic within and on a closed
contour C and ‘a’ is any point within C then
derivatives of all orders are analytic and are given by
f
(n)
n!
f ( z)
( z0 ) 
dz
.
n

1
2 i C z  z0 
This formula is very useful in deriving the Taylor
series.
Proof:
We know that
1
f ( z)
f ' ( z0 ) 
dz.
2

2 i C z  z0 
This proves that the required result is true for
n=1. let us suppose that the required result is
true for n=m, so that
f
( m)
m!
f ( z)
( z0 ) 
dz
.
m 1

C
2 i z  z0 
Let ‘a+h’ be any point in the neighbourhood of ‘a’ .
we observe that
( a  h)  f  m  ( a )
h

m!
f ( z) 
1
1


 m 1 
 m 1  dz.

C
2 i
h  ( z  a  h)
( z  a)

f
m!

2 i

m 
C
 ( m 1)




f ( z)
h

1
 1 dz.
 m 1 

( z  a)
h 
( z  a) 


2


m!
f ( z)
h(m  1) (m  1)( m  2)  h 

  ..... dz.



m 1

C
2 i ( z  a) h  ( z  a)
2!

 ( z  a) 

m!
f ( z)

2 i C ( z  a) ( m1)
 (m  1) (m  1)( m  2

h
)
 ....... dz.
 ( z  a) 
2
2!
( z  a)


taking limit as h
0,we get
f ( m ) (a  h)  f ( m ) (a) m!
f ( z)
lim

h 0
h
2 i C ( z  a) ( m1)
 m 1

 ( z  a)  0  0  ........ dz.


f
f
( m 1)
(m  1)m!
f ( z)
(a) 
dz.
m

2

C ( z  a)
2 i
( m 1)
(m  1)!
f ( z)
(a) 
dz
.
m 2

C
2 i
( z  a)
This proves that the required result is true for n=m+1.
Thus the required result is true for any positive integral value of
m
Since f´(a), f´´(a) ,f´´´(a)………..all exist .
And hence are analytic within C
THEOREM
Cauchy’s integral formula for
multiply connected region
STATEMENT:
If f (z) is analytic in ring shaped region bounded
by two closed curves C1 and C2 and
‘a’ is a point in the region between C1 and C2.
then
f (a)
1
2 i C2
f ( z)
1
f ( z)
dz 
dz
za
2 i c1 z  a
Proof: Describe a circle γ about the point
z=a of radius ‘r’ such that the circle lies in the
ring shaped region.
C2
C1
a
γ
The function
f ( z)
za
is analytic in the region bounded
by three closed curves C1,C2 and γ. By cauchy’s theorem

C2
f ( z)
f ( z)
f ( z)
dz  
dz  
dz
C1 z  a
za
 za
Where the integral along each curve is taken in anti-clockwise
direction.using cauchy’s integral formula,
We have ,

C2
f ( z)
f ( z)
dz  
dz  2if (a)
C1 z  a
za
1
f ( z)
1
f ( z)
f (a) 
dz

dz
2 i C2 z  a 
2i C1 z  a
Which is the required result
Independence of the Path
Let z0 and z1 be points in a domain D. A contour
integral
C f ( z) d z
is said to be independent of
the path
if its value is the same for all contours C in D with
an
initial point z0 and a terminal point z1.
Fig
Note that C and C1 form a closed contour.
If f is analytic in D then
C f ( z) d z  C
f ( z ) d z  0 ……
(2)
1
Thus
C
f ( z) d z  
C1
f ( z) d z
…….
(3)
EXAMPLES ON COMPLEX
INTEGRATION
Example 1: Evaluate the integral

C
z 1
dz,
z 1
where C is a closed curve around z = 1.
Solution:
z0  1. f ( z)  z  1.
So,

C
z 1
dz  f ( z0 )2 i  (1  1)2 i  4 i
z 1
Example 2:
1
C z( z   i) dz,
Evaluate the integral
Solution:
Given integral is
Take f(z)=1/z
Therefore
I=
I
1
C z( z   i) dz,
=
f ( z)
C ( z  (  i)) dz,
=2πi f(-πi)
=2πi(1/(-πi)
=2
Example 3: Evaluate the integral
z 1
C z  12 dz,
where C is a closed curve around z = 1.
Solution:
So,
z0  1. f ( z)  z  1. n  1.
z 1
(1)
dz

f
(
z
)
2

i

(
1
)
2

i

2

i
0
2
C z  1
Example 4: Evaluate the integral
z 1
C z  13 dz,
where C is a closed curve around z = 1.
Solution:
So,
z0  1. f ( z)  z  1. n  2.
z 1
2 i
( 2)
C z  13 dz  f ( z0 ) 2!  (0) i  0.
Example: 5 Evaluate the integral
z3 1
C z  23 dz,
where C is a closed curve around z = 2.
Solution:
z0  2. f ( z)  z  1. n  2.
3
f
( 2)
( z )  6 z.
z3 1
2 i
( 2)
C z  23 dz  f ( z0 ) 2!  (6)(2) i  12 i
Example 6
5z  7
C z 2  2 z  3 dz
where C is the circle |z – 2| = 2.
Solution
5z  7
3
2


2
z  2z  3 z  1 z  3
and so
5z  7
dz
dz
C z 2  2 z  3 d z  3 C z  1  2 C z  3
Evaluate
Y
Center (2,0)
Radius=2
C
O
1
2
3
4
X
Given circle is Iz-2I=2 ,whose center is 2 and radius is
also 2 units
Since z = 1 is interior to C
and z = −3
is exterior to C, we have
5z  7
d
z

3
(
2

i
)

2
(
0
)

6

i
C z 2  2 z  3
Example 7
Evaluate
dz
C z 2  1
where C is the circle |z| = 3.
Solution
1
1 / 2i 1 / 2i


2
z 1 z  i z  i
dz
1
C z 2  1  2i
 1  1  dz
C  z  i z  i 
Y
Radius =3
Center (0,0)
O
3
X
Given circle is IZI=3 , whose center is (0,0) and radius is 3
We now surround the points z =
i and z = −i by circular contours
C1 and C2. , we have
dz
C z 2  1
1  1
1 
1
1 

  

dz   

dz


C
C
2 z  i
2i 1  z  i z  i 
z  i
1
dz 1
dz
1
dz 1
dz
 
 
 
 
C
C
C
2i 1 z  i 2i 1 z  i 2i 2 z  i 2i C2 z  i
dz
Since 
i  2i,
C1 z  i
thus (7) becomes zero.
dz
C2 z  ii  2i
Example 8
Evaluate
C 2 z d z
where C is shown in Fig
Solution
Since f(z) = 2z is entire, we choose the
path C1 to replace C (see Fig ). C1 is a
straight line segment x = − 1, 0  y 1 .
Thus z = −1 + iy, dz = idy.
C2 zdz  C2 zdz
1
1
1
0
0
 2 ydy  2i  dy  1  2i
TEST-1
NOTE: Attempt any
one from each
unit.
UNIT-1
1. Expalin cauchy’s integral formula
for the derivative of an analytic
function?
2. (a) Prove that zeros of an analytic
function are analytic?
(b) Find the singularities of
function :
F(z)=cotπz/(z-a)2
UNIT -2
1.Let f(z) be analytic inside and on a simple
closed curve C except for a finite number of
poles inside C and f(z) ≠ 0 on C .then prove that
1
f ' ( z)
dz

N

P

2 i C f ( z ) 0
Where N=no. of zeros of f(z)
P= no. of poles of f(z)
2.State and prove maximum modulous principle?
TEST - 2
NOTE: Attempt any
one from each
unit.
UNIT-1
1.State and prove taylor’s theorem?
2. (a) Prove that the poles of an analytic
function are isolated?
z exp( z )
C z  12 dz,
(b)Evaluate:
UNIT-2
1.State and prove schwarz lemma?
2.If f(z) and g(z) are analytic inside
and on a simple closed curve C
and if Ig(z)I<If(z)I
on C then f(z) and f(z)+g(z) both
have same number of zeros inside
C?