Complex-Variables.pdf

Chapter 6

Mathematics

CHAPTER 6 Complex Variables Complex Variable , where x & y are real numbers and they are called real and imaginary part of Z. | |

√

( )

. /

Function of a Complex Variable I

wh ch ‘ ’

c

p

c

p

‘w’

D,

then , w = f (z) Where, z = x + iy, w = f (z) = u(x, y) + i v(x, y) Limits and Continuity Let w = f(z) be any function of z defined in a bounded or closed domain D, then, ( ) =

, if for every real

we can find real

Such that | ( ) | |< for | Basically it means: Single value for all values of z in the neighbourhood of z = possible exception of z = itself

Y

with the

v

f(z) )

z

z - Plane

pPlanePl ane PVT.LTD. THE GATE ACADEMY

X

w - Plane

u

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Chapter 6

Mathematics

Note: In real calculus x approaches only along a line, whereas in complex calculus, z approaches from any direction in the z- plane [i.e. limit is independent of the manner in which ] Continuity of f (z)  

( ) A function = f (z) is said to be continuous at if ( ) Further f (z) is said to be continuous in any region R of the z-plane, if it is continuous at every point of that region. Also is w = f (z) = u(x, y) + i v(x, y) is continuous at , then u(x, y) and v(x, y) are also continuous at x = xo & y = yo.



Q (x +

,y+

)

( )

( )

P(x, y)

Derivative of f(z) =

(

( ) (

, ’( )

2 =

+i

=

) )

( ) ( )

, 3

–i

Provided the limit exists and has the same value for all the different ways in which approaches zero. Differential rule are same as real calculus . / =

( )

Theorem The necessary and sufficient conditions for the derivative of the function f( ) to exist for all values of in a region R. i) ii)

,

,

, ,

, are continuous functions of x and y in R. ,

Cauchy-Riemann equations (CR Equations)

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Chapter 6

Mathematics

Example ̅

Evaluate

, if it exists.

Solution If limit exists, it is independent of the manner in which Now let

along x-axis, then y = 0

z = x + iy =x ̅ = x – iy = x ̅

= =1

Now let

___________ (i)

along y-axis, then x = 0

z = iy ̅ = – iy ̅

=

= -1

___________ (ii)

Hence limit do not exists.

Example f(z) = ̅

Is it differentiable?

Path 2

y

z+

z Solution

Path 1

z = x + iy

x

=

( )=

. =

(

)

( )

̅̅̅̅̅̅̅

/=.

0

(

̅)

/ )

1

= If

( ) = 1 (path 1)

If

( ) = 1 (path 2) Not differentiable


Chapter 6

Mathematics

Analytic Functions (or Regular Functions) or Holomorphic Functions A single valued function which is defined and differentiable at each point of a domain D is said to be analytic in that domain. A point at which an analytic function causes to posses a derivative is called Singular point. Thus if u and v are real Single – valued functions of x and y such that

,

,

,

are

continuous throughout a region R , then C–R equations = h“

c

= c

”c

h

c

( )

c

R

Real and imaginary part i.e. u, v of the function is called conjugate function. Analytic function posses derivatives of all order and these are themselves analytic.

Harmonic functions If f(z) = u + iv be an analytic function in some region of the z – plane then the C –R equations are satisfied. =

,

Differentiating with respect to x and y respectively, =

, =0

(Laplace Equation)

Methods of constructing Analytic functions 1.

If real part of a function is given then, f ’( )

-i

Integrate with points at (z, 0) f (z) = ∫ . /

( , )

dz

i ∫. /

dz + c

( , )

Similarly in case v(x, y) is known, then ’ (z) =

+i

f (z) = ∫ . /

( , )

dz + i ∫ . /

( , )

dz + c


Chapter 6

2.

Mathematics

If u (x, y) is known, then to find v(x, y) we have dv =

dx +

dv =

dy

dx +

dy

Integrate this equation to find v. 3.

f (z) = u(x, y) + i v(x, y) If a real part of the analytic function f(z) is a given harmonic function u (x, y), then f(z) = 2u . , / u(0, 0)

Example Find analytic function of u = Solution Approach 1 = = - 6xy ’( )

.

/ ( , )

(

=

)

= f(z) = =

z+ci i

Approach 2: f(z) = u + iv where, u = u . , / = . / – 3 . / . / + 3. / + 1 =

+

+

=

+

+1

+1

u (0, 0) = 1 f(z) = 2u . , /

u (0, 0)

f(z) =

+ci

Approach 3: dv = =

dx +

dy

( 6xy) dx + (

= (6xy) dx + (

) dy ) dy


Chapter 6

V=∫

+ ∫(

V = 6y. =3

+ (-3

y

)

[

(

)

]

+ 3y) + c

+ 3y + c f(z) = u + iv =

– 3x

=(

) + 3x + 1 + 3yi + c i

= (

) + 3(x + iy) +1 + c i

=

Mathematics

+ 3x + 1 + i (3

y-

+3y) + c i

+ 3z + 1 +c i

Complex integration Line integral = ∫ ( )

, C need not be closed path Here, f(z) = integrand curve C = path of integration

Contour integral = ∮ ( )

, if C is closed path

If f(z) = u(x, y) + i v(x, y) and dz = dx + i dy ∫ ( )

=∫(

)

∫(

)

Theorem f(z) is analytic in a simple connected domain then ∫

( )

= f( )

( )

Integration is independent of the path Dependence on path I “C p ” p h p (However analytic function in simple connected domain is independent of path.)

C

ch ’

h p h

theorem

If f(z) is analytic in a simple connected domain D, then for every simple closed path C in D,

∮

( )

C

=0 -------(A)

D


D

Chapter 6

Mathematics

Note I h w , C ch ’ h (z) is analytic on a simple closed path C and everywhere inside C (with no exception, not even a single point) then ∫ ( ) C

ch ’ I

If f(z) is analytic within and on a closed curve and if a is any point within C, then f(a) =

∫

( )

( )

f ’( )

∫

f ”( )

∫

. f n(a) =

∫

(

) ( )

(

) ( )

(

)

Note complex analytic function

has derivative of all order.

in real calculus if a real function is differentiated once, nothing follows, about the existence of second or higher derivative Example Find the complex integral of ∮

(C : circle of radius 3 )

Solution 2π

∮

, because

is analytic over entire region

Example = ? ( C : circle of radius 3 )

∮

Solution =0

∮

Example If g(z) = ∮ ( )

=?

where, C is circle shown as,

0

1

2

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Chapter 6

Mathematics

Solution g(z) =

=

(

)(

)

g(z) is not analytic at z = +1 & z = 1 (Hence watch these points ) Hence circle C, |z – 1| = 1 encloses

= 1,

Where g(z) is not analytic

Hence we write g(z) = (∮ ( )

= ( )

) =∮

=

( )

2π ( )

2π

Example In the above problem, if the circle shown is 1

=?

∮ Solution

2π

∮

Example ∮

(

)(

)

=? If circle C is as shown as

Solution ∮

(

= ∮

)(

)

( )

=∮ 2π ,

-

2π ,

] = - 2π

Example ∮

(

)(

)

, if the circle is as shown

Solution ∮

-1 (

)(

1

)


Chapter 6

Mathematics

Example ∮

(

)(

=?

)

where C is circle |z| = 3

Solution c

0

(

)(

Hence ∮

(

2π ,

)(

π(2)

2π (

1 , [ note: 3>2, 3>1] inside circle

)

∮

)

c

π(2) ]

(

∮

)

2π ,

π( )

(

c

)

π( ) ]

) - 2π ( – 1) = 4π

Example ∮

(

)

where C is circle |z| = 3 Solution f(z) = ∮

(

at a = 1,

( ) )

(a) =

∮

(

( ) )

Comparing n = 3 ( )

2π =

(

= )

(-1) 1

3

= π {Since f(z) = f ’( )

2

f ”( ) f ”’( ) f ”’(-1) =

}


Chapter 6

Mathematics

Example

i =?

∮

around a rectangle with vertices 2

1

-2

, 2

Solution

2

-i =

∮

∮(

)c

π

=

∮

=

,2π c (π )-

∮

= [ 2π -

,2π c (π (

)-

[ 2π -

’ h If f(z) is continuous in a region and ∫ ( ) analytic in that region. T

= 0 around every simple closed C then f(z) is

’ S

If f(z) is analytic inside a circle C with centre at a then for z inside C f(z) = f(a) + f ’( ) ( f(z) = ∑ Where

( =

a) +

( )

(z-a) + - - - - - - -

) ∫

() )

(

Other form, put z = a + h f(a+h) = f(a) + h f ’( )

L

f ”( )

-------

’ S

If f(z) in analytic in the ring shaped region R bounded by two concentric circles c and c of radii and ( ) and with centre at a then for all z in R (

f(z) = where,

=

) ∫

(

(

)

(

)

(

)

() )


Chapter 6

If f(z) is analytic inside the curve then

Mathematics

L

c

T

’

’ of Analytic Function The value of z for which f(z) = 0 I ( )

c (

f(z) = =∑

(

Where, if

=

h )

h

h

(

p

h

)

(

T

’ h

)

) =

=

( )

=------

= 0, then f(z) is said to have a zero of order n at z =a.

Singularities Of An Analytic Function “

p

”

c

h p

wh ch h

c

c

c

1.

Isolated Singularity If z =a is a singularity of f(z) such that f(z) is analytic at each point in its neighborhood (i.e. there exists a circle with centre a which has no other singularity 1, then z =a is called an isolated singularity).

2.

Removable Singularity If all the negative powers of (z-a) in Laurent series are zero then f(z) = ∑ ( ) Singularity can be removed by defining f(z) at z = a is such a way that it becomes analytic at z =a ( ) exists finitely, then z = a is a removable singularity. Example: f(z) = , then z = 0 is a removable singularity.

3.

Essential singularity If the numbers of negative power of ( an essential singularity. ( ) does not exist in this case

4.

)

L

’

Poles If all the negative power of (z ) L ’ singularity at z = a is called a pole of order n. p c “ p p ”

, h

c

are missing then, the

Example F

T

’

p

( )

(

)

about the point z = i


Chapter 6

Mathematics

Solution To expand about z = i, i.e. power of z +i, put t =z + i, ( )

=(

)

.

=(

)

.

∑

= .

(

)

(

) 0

1

/ (

(

)

(

( (

)

) )

)

(

/

)

/

Residue Theorem If f(z) is analytic in and on a closed curve C except at a finite number of singular point within C then ∫ ( )

2π (

h

h

p

w h

C)

Calculation of Residues 1. If f(x) has a simple pole at z=a , then Res f(a) =

,( ( ) ( )

2. If ( )

) ( )-

where ( )

(

) ( ), ( )

( ) ( )

Res ( )

C

3. If ( ) has a pole of order n at z=a , then ( )

(

)

2

,(

( )-3

)

Here n =order of singularity Example Solve using Residue Theorem ∮

(

)(

w h

C+

, if the circle is as shown

)

Solution ∮

(

)(

)

2π *R 2π R 2π

p

-1

(-1) ,(

) ( )-

2π

,(

)

(

)(

)

- = 2π


Chapter 6

Mathematics

Note C

p

h

R

h

C

ch ’

Basic concepts of complex number if z = x + iy r =√

z r

y x

In trigonometric from C

c

S c (c

I

)

Modulus of complex number |z| = r .√ Argument of complex Number = tan

1

=. /

In exponential form ,

c

+

=1 [ |

] |

√c

=1 |z| = r =√ 

If the any pole is outside the closed contour

|z| = a

|z| = a Its residue at this pole is always zero. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 185

Chapter 6

Mathematics

We always find the residue at the poles, where poles are inside the closed contour, and for any outside point residue is zero. Cube root of unity ω

√

(

)

(1, 0)

√

ω2 = (

)

Point to remember 1. ω 2. 3.

ω

ω2 = 0 , ω, ω2


Complex-Variables.pdf

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