COMPLEX INTEGRATION
Introduction: The advantage of complex integration is that certain complicated real
integrals can be evaluated and properties of analytic functions can be established.
Line Integrals in Complex plane:
Introduction:
Curve: Let
and
equation
be continuous functions of the real variable , then the
represents a curve in the complex plane.
Example:
plane.
represents a circle in the complex
Smooth curve: Suppose set of points
defined by
with the
parameter in the interval
. If and g are differentiable then the curve is said
to be smooth.
Simple curve: A curve C is said to be simple if it doesnot intersect itself.
Example: Semi-circle above x-axis,
Simple Closed curve: A curve is said to be a simple closed curve (Jordan curve) if
it is a simple curve and its endpoints coincide.
Example: Circle.
Contour: It is a continuous chain of a finite number of smooth curves.
Simply connected region: It is a region such that every simple closed curve in R
contains only points of R..
Example: Interior of a circle, rectangle, triangle and ellipse.
Multiply connected region: A region which is not simply connected is called
multiply connected.
Example: Open annulus.
Simply Connected
Doubly Connected
Triply Connected
Line integral in the complex plane:
Consider a function
defined at all
points of a curve C having end points
and . Divide C into parts at the
points
. Let
and
called the line integral of
Properties:
1)
2)
curve
. Then
taken along the path C. ie
be any point on the
is
3)
4)
where C consists of the curve
where M is a constant
length of the curve C.
for all
Problems:
1) Evaluate
along
i)
The line
ii)
The real axis to 2 and then vertically to
Soln:
i)
Along the line
ii)
Along x-axis
Along y-axis,
.
and
on C and
is the
2) Integrate
where
is an integer,
is a constant, in the
counter clockwise sense around the circle with radius
Soln:
and centre
.
if
If
3) If C is a circle
4) Evaluate
, evaluate
where C is
(Ans:
)
(Ans: 30)
5) Evaluate
where C is
a) The path from
to
b) Along the straight line from (1,1) to (3,1) and then from (3,1) to (3,2)
6) Evaluate
a)
where C is
b)
Square with vertices at
Theorem: If
is analytic on simple curve C, then the line integral
independent of the path chosen, it depends only on the end points.
7) Evaluate
where C is the straight line from I to
8) Evaluate
from
then the line from
to
to
9) Evaluate
a) Along the straight line from
b) Along the real axis from
along the line from
is
.
to
.
to
to
and then vertically to
CAUCHY’S INTEGRAL THEOREM
f (z)
If
is analytic in a simply connected domain D then
closed curve C lying entirely within D
Proof:
and
for simple
.
Consider
, u and v have continued partial
derivatives in D because
is analytic and
is continuous. Applying Green’s
theorem in the plane i.e.) if
and
be continuous in a domain R of the
xy-plane bounded by a closed a curve C, then
=
=0.
Note: Cauchy’s Integral Theorem is also known as Cauchy’s theorem.
Ex.1:
function.
Ex.2:
for any simple closed curve C because
but
is an analytic
is not analytic at z=0
doesnot implies that f(z) is analytic.
Ex.3:
where
. Hence the condition that the domain D is
simply connected is essential. But
is analytic in
.
Independence of path:
Let
be analytic in a simply connected domain D. Let
and
be any
two paths in D joining any two points and
in D and having no further points in common
then
both
and
traversed in the same direction ie) the integral of
of the path joining them.
from
and is independent
Proof:
The two curves
and
together form a simple closed curve C in D.
By Cauchy’s theorem,
(Or)
.
Cauchy’s theorem for multiply Connected Domains:
If
is analytic in the domain D between two simple closed curves C and
.
Proof:
By Cauchy’s integral theorem
i.e.
.
Where
Note: If
and
are traversed in same direction.
be any number of closed curves within C then
Examples: for Cauchy’s Integral Theorem
then
1)
Since z=3 lies outside
2)
Since
.
lies outside
.
Derivative of analytic functions:
If a function
derivative at any point
is analytic in a simply connected domain D, then its
of D is also analytic and is given by
Proof:
By Cauchy’s Integral formula
neighborhood of a.
.Let (a+h) be any point in the
.
=
.
In general
.
Problem:
1)
, C is a rectangle with vertices at
.
2) Evaluate
3)
4)
5)
6)
where C is the rectangle with vertices
.
where C is the circles (i)
where C:
Where (i)
(ii)
Where C:
.
.
7)
8)
Where
.
9)
Where
10)
Where C is (i)
11)
where C is (i)
Cauchy’s Integral Formula: If
and if a is any point within C then
.
(ii)
(iii)
(ii)
(iv)
(iii)
.
.
is analytic inside and on simple closed curve C
.
Note: Suppose
is not analytic at z=a and z=b. Then write
then apply Cauchy’s theorem.
Cauchy’s Integral Formula:
Let f (x) be analytic in a simply connected domain D . Let C be any simple closed
D
a
curve in enclosing any point in
in counter clockwise direction.
D.
Then
where
C
is traversed
Proof:
Consider the function
which is
analytic at all points within C except at
z=a. With a as centre and radius r , draw
a small circle C 1 lying entirely within C .
Now
being analytic in the region
enclosed by C and C 1 . We have by
Cauchy’s Theorem for multiply connected domains,
Consider
where
Therefore
As circle C 1shrinks to a point a , i.e.
From (2),
.
From (1),
.
1. Evaluate
(iii)
over each of the following contours C: (i)
.
Solution:
is not analytic at
.
(i)
lies within the circle
.
Therefore
(ii)
lies outside
.
Therefore
(iii)
lies outside C.
Therefore
2. Evaluate
(iv)
where C: (i)
.
(ii)
(iii)
(ii)
Solution:
(i)
Since
lies onside C. We have
(ii)
lies outside C. We have
(iii)
(iv)
3. Evaluate
Solution:
4. Evaluate
Solution:
where C:
.
lies outside C, we have
where C:
and
.
.
.
5. Evaluate
where C:
and
Solution:
6. Evaluate
where C:
Solution:
7. Evaluate
Solution:
where C:
. Therefore
lies outside C.
.
8. Evaluate
where C:
Solution: C:
lies outside
9. Evaluate
Therfore
where C:
Solution:
lies inside C.
Therefore
10.
where C:
Solution:
11.
where C:
Solution:
.
.
Therefore
12.
.
where C: (i)the circle
(ii)
(iii)
Solution:
(i)
.
has singularities at -1-2i and -1+2i i.e.
is not analytic.
(ii)
Centre at (-1,1) and radius 2.
Let
Distance is greater than radius.
Therefore (-1,-2) lies outside the circle.
lies inside
.
iii)
Centre at
and radius 2. Let
lies inside
.
lies outside
.
Let
.
13)
where
Soln:
14) Evaluate
around
i) the circle
ii) the circle
iii) the rectangle with vertices at
iv) triangle with vertices at
Soln: i) since
is not analytic at
We have
ii)
iii) Along
,
.
.
.
Along
,
Along
,
Along
,
iv) By Cauchy’s Integral Theorem,
Since, along
Along
,
,
, we have
we have
.
15) Evaluate
where i)
ii)
iii)
Soln: i) By Cauchy’s Theorem,
ii)
iii) By Cauchy’s Theorem,
16) Verify Cauchy’s Theorem for the function
rectangle with vertices
.
Soln: Along
Along
,
Along
,
Along
,
,
By Cauchy’s theorem,
.
with
as the boundary of the
Extra Problems:
1) Evaluate
where
.
Soln:
2) Evaluate
where
is the circle i)
Soln: i) Consider
ii) Consider
.
Where
and
iii) Consider
Since
.
both lie inside
, we have
ii)
iii)
3) Evaluate
where
Soln:
4) Determine
ellipse
if
where
.
Soln: Given
i.e,
i) Consider
but
is not analytic at
. It lies inside
By Cauchy’s Integral Formula,
where
.
is the
ii)
lies outside C. Hence by Cauchy’s integral theorem
iii)
lies inside C. Hence by Cauchy’s integral formula,
iv) For any
lying inside C,
is not analytic at
By Cauchy’s formula,
Where
and
and
5) Determine
,
if
where
C:
Soln : i) Consider
By Cauchy’s integral theorem
ii) Consider
By Cauchy’s integral formula,
iii) For any lying inside C,
By Cauchy’s integral formula,
6) Determine
C:
Sol:
By Cauchy’s integral theorem,
if
is not analytic at
where
7)
Where C:
Sol: Since
Upon solving we get,
lies inside C, we have