Complex Variables

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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 p c p ‘w’ ‘ ’ c 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 = with the

possible exception of z = itself

z f(z)₎ u X v Y z - Plane pPlanePl w - Plane

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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.

Derivative of f(z)

= ( ) ( ) ( ) , 2 , ’( ) ( ) ( ) ( ) ( ) 3 = + i = – 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 . / =

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Theorem

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

, are continuous functions of x and y in R.

ii) ,

, Cauchy-Riemann equations (CR Equations)

P(x, y)

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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 ___________ (i) Now let along y-axis, then x = 0 z = iy ̅ = – iy ̅ = = -1 ___________ (ii)

Hence limit do not exists.

Example f(z) = ̅ Is it differentiable? Solution z = x + iy = ( ) = . ( ) ( ) / = . ̅̅̅̅̅̅̅ ̅) / = 0 ( ) 1 = If ( ) = 1 (path 1) If ( ) = 1 (path 2) Not differentiable z Path 2 x y Path 1 z+

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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

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2. If u (x, y) is known, then to find v(x, y) we have dv = dx + dy dv = dx + dy

Integrate this equation to find v. f (z) = u(x, y) + i v(x, y)

3. 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 + c i = 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) = + c i Approach 3: dv = dx + dy = ( 6xy) dx + ( ) dy = (6xy) dx + ( ) dy

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D D C V = ∫ +∫( ) [( ) ] V = 6y. + (-3 + 3y) + c = 3 y + 3y + c f(z) = u + iv = – 3x + 3x + 1 + i (3 y - +3y) + c i = ( ) + 3x + 1 + 3yi + c i = ( ) + 3(x + iy) +1 + 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 h p h (However analytic function in simple connected domain is independent of path.)

C ch ’ theorem

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

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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, Here C : |z – 1| = 1

1

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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 ∮ = ? Solution ∮ 2π Example ∮ _{( )( )} =? If circle C is as shown as Solution ∮ _{( )( )} = ∮ = ∮ ( ) 2π , - 2π , ] = - 2π Example

∮ _{( )( )} , if the circle is as shown Solution

∮ _{( )( )}

1

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Example

∮ _{( )( )} = ? where C is circle |z| = 3 Solution

0 c _{( )( )} 1 , [ note: 3>2, 3>1] inside circle Hence ∮ _{( )( )} ∮ ( ) ∮ ( ) 2π , π(2) c π(2) ] 2π , π( ) c π( ) ] 2π ( ) - 2π ( – 1) = 4π Example ∮ _{( )} where C is circle |z| = 3 Solution f(z) = at a = 1, ∮ _{( )} ( ) (a) = ∮ ( ) ( ) Comparing n = 3 2π ( ) = (-1) = ( ₎ = π {Since f(z) = f ’( ) 2 f ”( ) f ”’( ) f ”’(-1) = } 3 1

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Example ∮ = ?

around a rectangle with vertices 2 , 2 Solution ∮ = ∮( ) c π = ∮ ∮ = ,2π c (π )- ,2π c (π ( )- = [ 2π - [ 2π -

’ h

If f(z) is continuous in a region and ∫ ( ) = 0 around every simple closed C then f(z) is analytic in that region.

T ’ S

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

(z-a) + - - -

f(z) = ∑ ( ) Where = ∫_{( )} ( ) Other form, put z = a + h f(a+h) = f(a) + h f ’( )

f ”( ) - - -

L ’ 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, = ∫ ( ) ( ) 2 1 -i i -2

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If f(z) is analytic inside the curve then L c T ’

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

I ( ) c h h h p h T ’ h f(z) = ( ) ( ) ( )

=∑ ( ) Where, = ( )

if = = = - - - = 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 ( ) L ’ , h c an essential singularity.

( ) does not exist in this case

4. Poles

If all the negative power of (z ) L ’ are missing then, the singularity at z = a is called a pole of order n.

p c “ p p ” Example

F T ’ p ( )

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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 ( ) ( )_{( )}

3. If ( ) has a pole of order n at z=a , then ( ) _{( )}2 ,( ) ( )-3

Here n =order of singularity Example

Solve using Residue Theorem ∮ _{( )( )} , if the circle is as shown Solution ∮ _{( )( )} 2π *R p w h C+ 2π R (-1) 2π ,( ) ( )- 2π ,( ) ( )( )- = 2π -1

C

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Note

C p h R h C ch ’

Basic concepts of complex number if z = x + iy

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

Its residue at this pole is always zero. x r z y r =√ |z| = a

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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

ω2_{= (} √ ₎ Point to remember 1. ω 2. ω ω2_{= 0} 3. , ω, ω2 (1, 0) ω ( √ )