Welcome to the presentation
On
Line integral of a complex
function
Presented by :
Mohammed Nasir Uddin
Assistant Professor
Dept. Of ICT
Faculty of Science and Technology(FST)
Learning Out comes
Objectives:
Introduction
Introduction
Discussion Points
Discussion Points
Concept of Line Integral
Simple & Multiple Connected Region
Simple & Multiple Connected Region
Learning Outcomes
Be familiar with Simple & Multiple
Connected Region
Discussion Points
Line Integrals
Simple & Multiple Connected Region
Jordan Curve
Fig-1
Complex Line Integrals
Let f(z) be continuous at all points of a curve C (fig.-1)
which we shall assume has a finite length, i.e. C is a
rectifiable curve
.
Subdivide C into n parts by means of Points
z
1,z
2,...,z
n-1, chosen arbitrarily, and call a=z
0,b=z
n.
On
each arc
joining z
k-1to z
k[where k goes from 1 to
n] choose a point
k.From the sum
Sn = f(1) (z1 - a) + f( 2) (z2 – z1) + ... + f( n) (b – zn-1)... …(1)
On writing z
k– z
k-1= ∆z
k ,this becomes
( ) ( )... ...(3)
b
a c
f z dz or f z
Called the
complex line integral
of briefly
line integral
of f(z) along curve C, or the definite integral of f(z) from
a to b along curve C.
In such case f(z) is said to be
integrable along C
.
Let the number of subdivisions
n
increase in such a
way that the
largest of the chord lengths |∆z
k|
approaches zero
.
Application of Line Integral
: )
3 ( )
2 (
) 1 , 2 (
) 3 , 0 (
2 dx x y dyalong
x y
Evaluate
(a) The parabola z=2t, y=t2+3
Application of CRE
Problem – SL/99/3
Prove that if f(z) is integrable along a curve C having finite length L and if there exists a positive number M such that , |f(z)|≤ M on C, then
|∫c f(z)dz| ≤ ML
Solution : By the definition of complex line,
Where we have used the facts that |f(z)| ≤ M for all points z on C
and that represents the sum of all the chord lengths joining
points zk-1 and zk ,
where k=1,2,……..n, and that this sum is not greater than the length of C .
Taking the limit of both sides of (2),using (1), the required result follows; It is possible to show, more generally ,that
|∫c f(z)dz| ≤ ∫c |f(z)| |dz|
Home Work
Ch-4 / SL/98-99
Solved Problem: 1,2,3
Simply and multiply connected region
׀ z
2
=
׀
׀z2= ׀
׀ z1= ׀
1<׀z
2
<
׀
Simply and multiply connected region:
A region R is called simply- connected if any simple closed
curve which lies in R can be shrank to a point without leaving R .
A region R which is not simply connected is called multiply-connected.
For example, suppose R is the region defined by | z| < 2 shown in Fig. 4-2. If Г is any simple closed curve lying in R, [whose
points are in R] we see that it can be shrank to a point which lies in R, and does not leave R, so that R is simply connected.
Jordan Curve Theorem
Any continuous, closed curve which
does not intersect
itself and which may or may not have a finite length is
called a
Jordan curve
.
A Jordan curve divides the plane into
two regions
having the curve as common boundary.
That region which is
bounded
(i.e. is such that all
points of it satisfy ׀z ׀< M, Where M is some positive
constant ) is called the
interior or inside
of the
Show that a continuous, closed, non-intersecting
curve which lies in a bounded region R but which
has an infinite length.
Consider
equilateral triangle
ABC [Fig. 4-21] with
sides of
unit length
.
By
trisecting each side
, constructing equilateral
triangles DEF,GHJ and KLM.
The process how can be continued by trisecting sides
DE, EF, BG, GH, etc., and constructing equilateral
triangle as before.
By repeating the process
indefinitely
, we obtain a
continuous closed non-intersecting curve which is the
boundary of a region with finite area equal to
(1/4)√3 + (3)(1/3)2√3/4 +(9)(1/9)2√3/4 +(27)(1/27)2√3/4+ ...
= √3/4 (1+1/3+1/9+... )
= (√3/4) 1/(1-1/3)
= 3√3/8
GREEN’S THEOREM
Statement: If R is a closed region in the xy- plane bounded by a simple closed curve C and if φ(x,y) and (x,y) are
continuous functions having continuous partial derivatives in R, then
where C is a traversed in the positive (anti- clockwise) direction.
Theorem is valid for both simply and multiply connected regions.
R
c dx dy x y dxdy
GREEN’S THEOREM
Proof:
Firstly suppose that a line parallel to either coordinate axis meets the boundary C in at most two points.
Let the equation to the curves AEB and BFA, denoted by c1 and c2 , be y=f1(x) and y=f2(x ) respectively. If R is the region bounded by C, Then we have
R
b x
a x
x f y
x f y
dxdy
y
dxdy
y
) (
) (
2
1
Again let the equations to the curves EAF and FBE, denoted be x = g1(y) and x = g2(y) respectively. Then we have
R h y g y y g x y g x dxdy x dxdy x ) ( ) ( 2 1
c c h g g hii
dy
y
x
dy
y
x
dy
y
y
g
dy
y
y
g
)
...(
...
)
,
(
)
,
(
}
),
(
{
}
),
(
{
2 1
Subtracting (i) from (ii), we have
c Rdxdy
y
x
dy
dx
)
(
)
PROBLEM: Verify the Green’s theorem in the plane for
Where C is the closed curve of the region bounded by y = x and y2 = x
c
{(
xy
x
)
dx
xy
dy
}
2 2
PROBLEM: Verify the Green’s theorem in the plane for
Where C is the closed curve of the region bounded by y = x and y = x2
c
{(
xy
y
)
dx
x
dy
}
2 2
PROBLEM: Verify the Green’s theorem in the plane for
Where C is the closed curve of the region bounded by y = x and y2 = x
c
{(
xy
x
)
dx
xy
dy
}
2 2
SOLUTION: y = x and y2 = x intersect at (0,0) and (1,1). The
positive direction in traversing c is as shown in the figure.
Y=x
Y2=x
y
x
(1,1)Along the curve y = x , the line integral becomes 12 11 4 1 3 2 4 4 3 3 2 2 ) {( 1 0 1 0 1 0 1 0 3 2 1 0 3 2 2
x x dx x dx x dx x dx x xdx
dy
x
y
Along the curve y2=x, the line integral becomes
0
60
1
15
14
12
11
}
)
{(
2 2
60
1
3
1
2
1
15
4
3
4
3
1
2
/
5
3
2
1 0 3 4 2 / 5
x
x
x
Application of Green’s Theorem(GT)
Problem – CV/SL/99/4
Prove Green’s theorem in the plane if C is a simple closed curve
which has the property that any straight line parallel to the coordinate axes cuts C in at most two points.
Proof:
Let the equations of the curves EGF and EHF (fig-1) be y=Y1(x) and y=Y2(x)
respectively . If R is the region bounded by C , we have
Similarly
let the equations of curves GEH and GFH be x=X1(y) and x=X2(y)
respectively
Extend the proof of Green’s theorem in the plane to curves C for which
lines parallel to the coordinate axes may cut C in more than two points.
Proof :
Consider a simple closed curve C in which lines parallel to the axes may meet
C in more than two points.
By constructing line ST the region is divided into two regions R1 and R2 and
for which Green’s theorem applies , i.e,
Application of GT
Problem – CV/SL/100/6
Prove that a necessary and sufficient condition that
be independent of the path in R joining points A and B is that identically in R.
Application of CRE
Problem – CV/SL/102/9
Cauchy’s Theorem
Let f(z) be analytic in a region R and on its boundary C.
Then
c
z
f ( ) 0
This is fundamental theorem, often called Cauchy’s integral
Theorem or Cauchy’s theorem, is
valid
for both simply-and
multiple-connected regions.
It was first proved by use of Green’s theorem with the added restriction that f/(z) be continuous in R.
However, Goursat gave a proof which removed this restriction.
Prove Cauchy’s theorem if
f(z)
is analytic with
derivative
f
/(z)
which is continuous at all points
inside and on a simple closed curve
C.
Prove the Cauchy–Goursat theorem for
multiply-connected regions.
If f(z) is analytic in a simply-connected region R
1,
prove that is independent of the path in R
joining any two points a and b in R.
Let f(z) be analytic in a region R bounded by simple closed curves C1 and C2 and also on C1 and C2.