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(1)

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)

(2)

Learning Out comes

Objectives:

Introduction

Introduction

Discussion Points

Discussion Points

Concept of Line Integral

Simple & Multiple Connected Region

Simple & Multiple Connected Region

(3)

Learning Outcomes

Be familiar with Simple & Multiple

Connected Region

(4)

Discussion Points

Line Integrals

Simple & Multiple Connected Region

Jordan Curve

(5)
(6)

Fig-1

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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-1

to 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

(8)

( ) ( )... ...(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

.

(9)

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

(10)

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|

(11)

Home Work

Ch-4 / SL/98-99

Solved Problem: 1,2,3

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Simply and multiply connected region

׀ z

2

=

׀

׀z2= ׀

׀ z1= ׀

1<׀z

2

<

׀

(13)

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.

(14)

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

(15)

Show that a continuous, closed, non-intersecting

curve which lies in a bounded region R but which

has an infinite length.

(16)

Consider

equilateral triangle

ABC [Fig. 4-21] with

sides of

unit length

.

By

trisecting each side

, constructing equilateral

triangles DEF,GHJ and KLM.

(17)

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

(18)
(19)

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

(20)

GREEN’S THEOREM

Proof:

Firstly suppose that a line parallel to either coordinate axis meets the boundary C in at most two points.

(21)

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

(22)
(23)

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

(24)

c c h g g h

ii

dy

y

x

dy

y

x

dy

y

y

g

dy

y

y

g

)

...(

...

)

,

(

)

,

(

}

),

(

{

}

),

(

{

2 1

Subtracting (i) from (ii), we have



c R

dxdy

y

x

dy

dx

)

(

)

(25)

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

(26)

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)

(27)

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 x

dx

dy

x

y

Along the curve y2=x, the line integral becomes

0

(28)
(29)

60

1

15

14

12

11

}

)

{(

2 2

(30)
(31)

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

(32)

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

(33)

Similarly

let the equations of curves GEH and GFH be x=X1(y) and x=X2(y)

respectively

(34)

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

(35)
(36)

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

(37)

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.

(38)

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.

(39)

Home Work

Ch-4 / SL/99-117

References

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