Unit - 5

Complex Variable - Differentiation

In the narrow sense of the term, the theory of function of a complex variable is the theory of analytic functions (cf. Analytic function) of one or several complex variables. As an independent discipline, the theory of functions of a complex variable took shape in about the middle of the 19th century as the theory of analytic functions.

Complex function-

x + iy is a complex variable which is denoted by z

If for each value of the complex variable z = x + iy in a region R, we have one or more than one values of w = u + iv, then w is called a complex function of z.

And it is denoted as-

w = u(x , y) + iv(x , y) = f(z)

Neighbourhood of z0

Let a point z0 in the complex plane and z be any positive number, then the set of points z such that-

|z – z0|<ε

Is called ε- neighbourhood of z0

Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighbourhood of point z0-

The-

Example-1: Find-

Sol. Here we have-

Divide numerator and denominator by z3, we get-

Continuity- A function w = f(z) is said to be continuous at z = z0, if

Also if w = f(z) = u (x, y) + iv (x, y) is continuous at z = z0 then u (x, y), v (x, y) are also continuous at z = z0.

Differentiability-

Let f(z) be a single valued function of the variable z, then

f’(z) =

Provided that the limit exists and has the same value for all the different ways in which δz approaches to zero.

Example-2: if f(z) is a complex function given below, then discuss dz/dz at z = 0

Sol. If z→0 along radius vector y = mx

=

=

But along v = x3,

In different paths we get different value of df/dz that means 0 and –i/2, in that case the function is not differentiable at z = 0.

Key takeaways-

1. Neighborhood of z0

|z – z0|<ε

2. Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighborhood of point z0-

The-

3. A function w = f(z) is said to be continuous at z = z0, if

f’(z) = 4

In Cartesian form-

Theorem; The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

Proof:

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

C-R equation in polar from-

C-R equations in polar form are-

Proof:

As we know that-

x = r cos and u is the function of x and y

z = x + iy = r ( cos

Differentiate (1) partially with respect to r, we get-

Now differentiate (1) with respect to , we get-

Substitute the value of , we get-

Equating real and imaginary parts, we get-

Proved

Key takeaways-

- The necessary condition for a function to be analytic at all the points in a region R are

(ii)

2. C-R equations in polar form are-

A function is said to be analytic at a point if f is differentiable not only at but every point of some neighborhood at .

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

3. An entire function is always analytic, differentiable and continuous function. (converse is not true)

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Important note-

1. If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.

2. C-R conditions are necessary but not sufficient for analytic function.

3. C-R conditions are sufficient if the partial derivative are continuous.

State and prove sufficient condition for analytic functions

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Proof:- Let f(z) be a simple valued function having at each point in the region R. Then Cauchy-Reimann equation are satisfied by Taylor’s Theorem

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q. Show that is analytic at

Ans. The function f(z) is analytic at if the function is analytic at z=0

Since

Now is differentiable at z=0 and at all points in its neighbourhood Hence the function is analytic at z=0 and in turn f(z) is analytic at

Example-1: If w = log z, then find . Also determine where w is non-analytic.

Sol. Here we have

Therefore-

and

Again-

Hence the C-R conditions are satisfied also the partial derivatives are continuous except at (0 , 0).

So that w is analytic everywhere but not at z = 0

Example-2: Prove that the function is an analytical function.

Sol. Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Example-3: Prove that

Sol. Given that

Since

V=2xy

Now

But

Hence

Example-4: Show that polar form of C-R equations are-

Sol. z = x + iy =

U and v are expressed in terms of r and θ.

Differentiate it partially w.r.t. r and θ, we get-

By equating real and imaginary parts, we get-

Key takeaways-

- A function is said to be analytic at a point if f is differentiable not only at but a every point of some neighborhood at .
- A point at which the function is not differentiable is called singular point.
- A function which is analytic everywhere is called an entire function.
- If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.
- C-R conditions are necessary but not sufficient for analytic function.
- C-R conditions are sufficient if the partial derivative are continuous

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Example: Prove that and are harmonic functions of (x, y).

Sol.

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Example: Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

Sol.

We have,

U(x, y) = 2x (1 – y)

Let V is the harmonic conjugate of U.

So that by total differentiation,

Hence the harmonic conjugate of U is

If y/x ≠ v/y and y/y ≠ - v/x

Then function is harmonic conjugate.

Contour integration is the process of calculating the values of a contour integral around a given contour in the complex plane.

Example: Solve where

Answer

Where

Poles of the inter are given by putting the denominator equal to zero.

Z(z-1)(2z+5)=0

Z=0,1,-5/2

The integrand has three simple poles at

Z=0,1,-5/2

The given circle |z|=3/2 with centre at z=0 and radius =3/2 encloses two poles z=0 and 1

Example: Solve

Answer

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Derivation of Cauchy Integral theorem:

Example 1:

where C = | z – 3| = 2

where f(z) = cosz

= ½ (2 πi) f(5/2) by cauchy’s integral formula

= πi. Cos (5/2)

Example 2:

Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

Solution:

Given,

f(n)(a) = n!/2πi

f(k + 1)(a) = d/da f(k)(a)

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Cauchy’s integral formula-

Cauchy’s integral formula can be defined as-

f(a) = 1/2πi

Where f(z) is analytic function within and on closed curve C, a is any point within C.

Example-1: Evaluate dz by using Cauchy’s integral formula.

Here c is the circle |z - 2| = 1/2

Sol. It is given that-

Find its poles by equating denominator equals to zero.

z2 – 3z + 2 = 0

(z – 1)(z – 2) = 0

z = 1, 2

There is one pole inside the circle, z = 2,

So that-

Now by using Cauchy’s integral formula, we get-

= 2πi [ z / z-1]z = 2 = 2πi(2/2 -1) = 4πi

Example-2: Evaluate the integral given below by using Cauchy’s integral formula-

Sol. Here we have-

Find its poles by equating denominator equals to zero.

z(z – 1)(z – 2) = 0

We get-

z = 0,1, 2

There are two poles in the circle-

Z = 0 and z = 1

So that-

= 2 πi [ (4 – 3z)/(z – 1)(z – 2)]z=0 + 2πi [(4 – 3z)/z(z-2)]z = 1

= 2πi . 4/(-1)(-2) + 2πi (4 -3)/1(1 -2) = 2πi(2 – 1) = 2πi

Example-3: Evaluate dz if c is circle |z - 1| = 1.

Sol. Here we have-

Find its poles by equating denominator equals to zero.

z2 – 1 = 0 or z2 = 1 or z = ± 1

The given circle encloses a simple pole at z = 1.

So that-

= 2πi [(3z2 + z)/(z+1)]z = 1 = 2πi((3+1)/(1+1))

= 4πi

The value of z is said to be zero of the analytic function f(z) when f(z) = 0.

If f(z) is analytic in the neighbourhood of z = a, then by Taylor’s theorem-

f(z) = a0 + a1(z – a) + a2(z – a)2 + a3(z – a)3 + . . . + an(z – a)n+ . . .

If a0 = a1 = a2 = . . . = an – 1 = 0 but an is non- zero, then f(z) is said to have a zero of order n at z = a.

The zero is said to be simple if n = 1.

an = fn(a)/n!

For a zero of order m at z = a,

f(a) = f’(a) = f”(a) = fn – 1(a) = 0 but fn(a) ≠ 0

Thus in the neighbourhood of the zero at z = a of order n

f(z) = an(z – a)n + an+1 (z – a)n + 1 = (z – a)n [ an + an+1(z –a) + …] = (z – a)n ∅(z)

Where ∅(z) = an + an+1 (z – a) + . . . Is analytic and non-zero at and in the neighbourhood of z = a.

Example: Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

Sol. Zeroes of the function-

f(z) = 0

(z – 2)/z2 sin 1/(z – 1) = 0

(z – 2)/z2 = 0, sin 1/(z – 1) = 0

z = 2, 1/(z – 1) = nπ ( n = 0, ±1, ±2 ….)

z = 2, 1 + 1/nπ (n = 0, ±1, ±2 ….)

Key takeaways-

- The pole is said to be simple pole when m = 1.
- Cauchy’s residue theorem-

If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then-

z = [ cos 5π/4 + i sin 5π/4] = ( - 1/2 – i 1/2)

A point at which a function f(z) is not analytic is known as a singular point or singularity of the function.

Isolated singular point- If z = a is a singularity of f(z) and if there is no other singularity within a small circle surrounding the point z = a, then z = a is said to be an isolated singularity of the function f (z); otherwise it is called non-isolated.

Pole of order m- Suppose a function f(z) have an isolated singular point z = a, f(z) can be expanded in a Laurent’s series around z = a, giving

= 2πi [ (4 – 3z)/(z – 1)(z – 2)]z = 0 + 2πi [(4 – 3z)/z(z -2)]z = 1

= 2πi. 4/(-1)(-2) + 2πi . (4 – 3)/1(1 – 2) = 2πi(2 – 1) = 2πi …….(1)

In some cases it may happen that the coefficient bm + 1 = bm + 2 = bm + 3 = 0 , then equation (1) becomes-

= 2πi[(3z2+z)/(z+1)]z = 1 = 2πi(3+1)/(1+1)

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/(z – a) + b2/(z – a)2 + . . .+ bm/(z – a)m+ bm+1/(z – a)m +1 + bm+2/(z – a)m+2 + . . .

Then z = a is said to be a pole of order m of the function f(z).

Note- The pole is said to be simple pole when m = 1.

In this case-

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/(z – a) + b2/(z – a)2 + . . .+ bm/(z – a)m

Working steps to find singularity-

Step-1: If exists and it is finite then z = a is a removable singular point.

Step-2: If does not exists then z = a is an essential singular point.

Step-3: If is infinite then f(z) has a pole at z = a. The order of the pole is same as the number of negative power terms in the series expansion of f(z).

Example: Find the singularity of the function-

f(z) = sin 1/z

Sol.

As we know that-f(z) = a0 +a1(z – a) + a2(z – a)2 + . . . + 1/(z – a)m[ b1(z – a)m – 1 + b2(z – a)m – 2 + b3(z – a)m – 3 + . . . + bm}

So that there is a number of singularity.

Sin 1/z is not analytic at z = a

(1/z = ∞ at z = 0)

Example: Find the singularity of 1/(1 – ez) at z = 2πi

Sol.

Here we have-

f(z) = 1/(1 – ez)

We find the poles by putting the denominator equals to zero.

That means-

1 – ez = 0

ez = 0 = ( cos 2n π + i sin 2n π) = e2nπi

z = 2nπi(n = 0, ±1, ±2, ….)

z = 2n π i is a simple pole

Example: Determine the poles of the function-

f(z) = 1/(z4 +1)

Sol.

Here we have-

f(z) = 1/(z4+1)

We find the poles by putting the denominator of the function equals to zero-

We get-

z4 + 1 = 0 and z4 = - 1

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/z - a

By De Moivre’s theorem-

Sin 1/z = 1/z – 1/3!z3 + 1/5!z5 + . ….+ (-1)n 1/(2n+1)! z2n+1

z = (-1)1/4 = (cos π + i sin π)1/4

If n = 0, then pole-

= [ cos (2n + 1)π + i sin (2n + 1)π]1/4

If n = 1, then pole-

= [ cos (2n + 1)π/4 + i sin (2n+1)π/4]

If n = 2, then pole-

z = [ cos π/4 + i sin π/4] = (1/2 + i 1/2)

If n = 3, then pole-

z = [ cos 3π/4 + i sin 3π/4] = (- 1/2 + i 1/2)

Let be analytic at all points within a circle with centre and radius. Then

Example: Expand the function

In a Taylor's series about the point

Ans. Using partial fraction method

(τ-1)(τ-3)

Both series converge when |4|<1

Therefore The series converges in the circle centred at with radius of 1.

Taylor’s series expansion is

Example: Show that when 0<|z|<4

Solution When |z|<4 we have

Example: Expand for the regions

- 0<|z|<1
- 1<|z|<2
- |z|>2

Solution Let

Hence resolving into partial functions we get

1) For 0<|z|<1 we have

2) For 1<|z|<2 we have

3) For |z|>2 we have

Let be analytic in the ring shaped region D bounded by two concentric circles with centre and radii and let be any point of D. Then

Example: using Taylor's series

Example: Obtain the Taylor’s and Laurent’s series which represents the function in the regions

1) |z|<2

2) 2<|z|<3

3) |z|>3

Solution We have

1) For |z|<2 we have

Which is Taylor’s series valid for |z|<2

2) For 2<|z|<3 we have

3) For |z|<3

Let be a pole of order m of a function and circle of radius r with centre at which does not contain any other singularities except atthen is analytic within the can be expanded by Laurent’s series

The coefficient of is called residue of at the pole then

Method of finding residue

(1) If has a simple pole at then

(2) If is of the form

(3) If has a pole of order n at then

(4) Residue at a pole of any order

(5) Residue of at

Example: Find residue of the function

Answer

Let

The singularities of f(z) are given by

Which is of the form

Example: Find the residue of at z=1

Answer

Let f(z)=

The poles of f(z) are determined by putting the denominator equal to zero

(z-1)(z-2)(z-3)=0

Z=1,2,3

Residue of f(z) at z=1=

=1/2

Example: Find the residue of

Answer

f(z)=

Poles are determined by putting sinz=0=

Hence the residue of the given function at pole is

If is analytic in a closed curve c except at a finite number of poles within c then

[Sum of residue at the pole within c]

Example: Evaluate the following integral using residue theorem

Where c is the circle..

Ans.

The poles of the integral are given by putting the denominator equal to zero

The integral is analytic on and all points inside exceptas a pole at is inside the circle

Hence by residue theorem

Example: Evaluate where c;|z|=4

Answer

Here f(z)=

Poles are

Sin iz=0

Poles

Lie inside the circle |z|=4

The given function is of the form

Its pole at z=a is

Residue (at

Residue at z=0=

Residue at =

Residue at are

Respectively -1,1 and -1

Hence the required integrand

Example: Evaluate :c is the unit circle about the origin

Answer

=

This shows that z=0 is a pole of order 2 for the function and the residue of the poles is zero(coefficient of 1/z)

Now the pole at z=0 lies within c

Example: Evaluation of definite integral

Show that

Solution

I=

Real part of

Now I= =

Putting z= where c is the unit circle |z|=1

I=

Now f(z) has simple poles at and z=-2 of which only lies inside c.

Residue at is

=

=

Now equating real parts on both sides we get

I=

Example: Prove that

Solution

Let

Putting where c is the unit circle |z|=1

2ai

Poles of f(z) are given by the roots of

Or

Let

Clearly and since we have Hence the only pole inside c is at z=

Residue (at )

Example: Evaluate

Answer

Consider

Where c is the closed contour consisting of

1) Real axis from

2) Large semicircle in the upper half plane given by |z|=R

3) The real axis -R to and

4) Small semicircle given by |z|=

Now f(z) has simple poles at z=0 of which only z=is avoided by indentation

Hence by Cauchy’s Residue theorem

Since and

Hence by Jordan’s Lemma

Also since

Hence

Hence as

Equating imaginary parts we get

Example: Prove that

Solution

Consider

Where c is the contour consisting of a large semicircle in the upper half plane indented at the origin as shown in the figure

Here we have avoided the branch point o, of by indenting the origin

Then only simple of f(z) within c is at z=i

The residue(at z=i) =

Hence by residue theorem

Since on -ve real axis.

Now

Similarly

Hence when

Equating real parts we get

References:

- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
- T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010

Unit - 5

Complex Variable - Differentiation

In the narrow sense of the term, the theory of function of a complex variable is the theory of analytic functions (cf. Analytic function) of one or several complex variables. As an independent discipline, the theory of functions of a complex variable took shape in about the middle of the 19th century as the theory of analytic functions.

Complex function-

x + iy is a complex variable which is denoted by z

If for each value of the complex variable z = x + iy in a region R, we have one or more than one values of w = u + iv, then w is called a complex function of z.

And it is denoted as-

w = u(x , y) + iv(x , y) = f(z)

Neighbourhood of z0

Let a point z0 in the complex plane and z be any positive number, then the set of points z such that-

|z – z0|<ε

Is called ε- neighbourhood of z0

Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighbourhood of point z0-

The-

Example-1: Find-

Sol. Here we have-

Divide numerator and denominator by z3, we get-

Continuity- A function w = f(z) is said to be continuous at z = z0, if

Also if w = f(z) = u (x, y) + iv (x, y) is continuous at z = z0 then u (x, y), v (x, y) are also continuous at z = z0.

Differentiability-

Let f(z) be a single valued function of the variable z, then

f’(z) =

Provided that the limit exists and has the same value for all the different ways in which δz approaches to zero.

Example-2: if f(z) is a complex function given below, then discuss dz/dz at z = 0

Sol. If z→0 along radius vector y = mx

=

=

But along v = x3,

In different paths we get different value of df/dz that means 0 and –i/2, in that case the function is not differentiable at z = 0.

Key takeaways-

1. Neighborhood of z0

|z – z0|<ε

2. Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighborhood of point z0-

The-

3. A function w = f(z) is said to be continuous at z = z0, if

f’(z) = 4

In Cartesian form-

Theorem; The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

Proof:

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

C-R equation in polar from-

C-R equations in polar form are-

Proof:

As we know that-

x = r cos and u is the function of x and y

z = x + iy = r ( cos

Differentiate (1) partially with respect to r, we get-

Now differentiate (1) with respect to , we get-

Substitute the value of , we get-

Equating real and imaginary parts, we get-

Proved

Key takeaways-

- The necessary condition for a function to be analytic at all the points in a region R are

(ii)

2. C-R equations in polar form are-

A function is said to be analytic at a point if f is differentiable not only at but every point of some neighborhood at .

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

3. An entire function is always analytic, differentiable and continuous function. (converse is not true)

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Important note-

1. If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.

2. C-R conditions are necessary but not sufficient for analytic function.

3. C-R conditions are sufficient if the partial derivative are continuous.

State and prove sufficient condition for analytic functions

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Proof:- Let f(z) be a simple valued function having at each point in the region R. Then Cauchy-Reimann equation are satisfied by Taylor’s Theorem

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q. Show that is analytic at

Ans. The function f(z) is analytic at if the function is analytic at z=0

Since

Now is differentiable at z=0 and at all points in its neighbourhood Hence the function is analytic at z=0 and in turn f(z) is analytic at

Example-1: If w = log z, then find . Also determine where w is non-analytic.

Sol. Here we have

Therefore-

and

Again-

Hence the C-R conditions are satisfied also the partial derivatives are continuous except at (0 , 0).

So that w is analytic everywhere but not at z = 0

Example-2: Prove that the function is an analytical function.

Sol. Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Example-3: Prove that

Sol. Given that

Since

V=2xy

Now

But

Hence

Example-4: Show that polar form of C-R equations are-

Sol. z = x + iy =

U and v are expressed in terms of r and θ.

Differentiate it partially w.r.t. r and θ, we get-

By equating real and imaginary parts, we get-

Key takeaways-

- A function is said to be analytic at a point if f is differentiable not only at but a every point of some neighborhood at .
- A point at which the function is not differentiable is called singular point.
- A function which is analytic everywhere is called an entire function.
- If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.
- C-R conditions are necessary but not sufficient for analytic function.
- C-R conditions are sufficient if the partial derivative are continuous

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Example: Prove that and are harmonic functions of (x, y).

Sol.

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Example: Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

Sol.

We have,

U(x, y) = 2x (1 – y)

Let V is the harmonic conjugate of U.

So that by total differentiation,

Hence the harmonic conjugate of U is

If y/x ≠ v/y and y/y ≠ - v/x

Then function is harmonic conjugate.

Contour integration is the process of calculating the values of a contour integral around a given contour in the complex plane.

Example: Solve where

Answer

Where

Poles of the inter are given by putting the denominator equal to zero.

Z(z-1)(2z+5)=0

Z=0,1,-5/2

The integrand has three simple poles at

Z=0,1,-5/2

The given circle |z|=3/2 with centre at z=0 and radius =3/2 encloses two poles z=0 and 1

Example: Solve

Answer

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Derivation of Cauchy Integral theorem:

Example 1:

where C = | z – 3| = 2

where f(z) = cosz

= ½ (2 πi) f(5/2) by cauchy’s integral formula

= πi. Cos (5/2)

Example 2:

Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

Solution:

Given,

f(n)(a) = n!/2πi

f(k + 1)(a) = d/da f(k)(a)

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Cauchy’s integral formula-

Cauchy’s integral formula can be defined as-

f(a) = 1/2πi

Where f(z) is analytic function within and on closed curve C, a is any point within C.

Example-1: Evaluate dz by using Cauchy’s integral formula.

Here c is the circle |z - 2| = 1/2

Sol. It is given that-

Find its poles by equating denominator equals to zero.

z2 – 3z + 2 = 0

(z – 1)(z – 2) = 0

z = 1, 2

There is one pole inside the circle, z = 2,

So that-

Now by using Cauchy’s integral formula, we get-

= 2πi [ z / z-1]z = 2 = 2πi(2/2 -1) = 4πi

Example-2: Evaluate the integral given below by using Cauchy’s integral formula-

Sol. Here we have-

Find its poles by equating denominator equals to zero.

z(z – 1)(z – 2) = 0

We get-

z = 0,1, 2

There are two poles in the circle-

Z = 0 and z = 1

So that-

= 2 πi [ (4 – 3z)/(z – 1)(z – 2)]z=0 + 2πi [(4 – 3z)/z(z-2)]z = 1

= 2πi . 4/(-1)(-2) + 2πi (4 -3)/1(1 -2) = 2πi(2 – 1) = 2πi

Example-3: Evaluate dz if c is circle |z - 1| = 1.

Sol. Here we have-

Find its poles by equating denominator equals to zero.

z2 – 1 = 0 or z2 = 1 or z = ± 1

The given circle encloses a simple pole at z = 1.

So that-

= 2πi [(3z2 + z)/(z+1)]z = 1 = 2πi((3+1)/(1+1))

= 4πi

The value of z is said to be zero of the analytic function f(z) when f(z) = 0.

If f(z) is analytic in the neighbourhood of z = a, then by Taylor’s theorem-

f(z) = a0 + a1(z – a) + a2(z – a)2 + a3(z – a)3 + . . . + an(z – a)n+ . . .

If a0 = a1 = a2 = . . . = an – 1 = 0 but an is non- zero, then f(z) is said to have a zero of order n at z = a.

The zero is said to be simple if n = 1.

an = fn(a)/n!

For a zero of order m at z = a,

f(a) = f’(a) = f”(a) = fn – 1(a) = 0 but fn(a) ≠ 0

Thus in the neighbourhood of the zero at z = a of order n

f(z) = an(z – a)n + an+1 (z – a)n + 1 = (z – a)n [ an + an+1(z –a) + …] = (z – a)n ∅(z)

Where ∅(z) = an + an+1 (z – a) + . . . Is analytic and non-zero at and in the neighbourhood of z = a.

Example: Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

Sol. Zeroes of the function-

f(z) = 0

(z – 2)/z2 sin 1/(z – 1) = 0

(z – 2)/z2 = 0, sin 1/(z – 1) = 0

z = 2, 1/(z – 1) = nπ ( n = 0, ±1, ±2 ….)

z = 2, 1 + 1/nπ (n = 0, ±1, ±2 ….)

Key takeaways-

- The pole is said to be simple pole when m = 1.
- Cauchy’s residue theorem-

If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then-

z = [ cos 5π/4 + i sin 5π/4] = ( - 1/2 – i 1/2)

A point at which a function f(z) is not analytic is known as a singular point or singularity of the function.

Isolated singular point- If z = a is a singularity of f(z) and if there is no other singularity within a small circle surrounding the point z = a, then z = a is said to be an isolated singularity of the function f (z); otherwise it is called non-isolated.

Pole of order m- Suppose a function f(z) have an isolated singular point z = a, f(z) can be expanded in a Laurent’s series around z = a, giving

= 2πi [ (4 – 3z)/(z – 1)(z – 2)]z = 0 + 2πi [(4 – 3z)/z(z -2)]z = 1

= 2πi. 4/(-1)(-2) + 2πi . (4 – 3)/1(1 – 2) = 2πi(2 – 1) = 2πi …….(1)

In some cases it may happen that the coefficient bm + 1 = bm + 2 = bm + 3 = 0 , then equation (1) becomes-

= 2πi[(3z2+z)/(z+1)]z = 1 = 2πi(3+1)/(1+1)

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/(z – a) + b2/(z – a)2 + . . .+ bm/(z – a)m+ bm+1/(z – a)m +1 + bm+2/(z – a)m+2 + . . .

Then z = a is said to be a pole of order m of the function f(z).

Note- The pole is said to be simple pole when m = 1.

In this case-

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/(z – a) + b2/(z – a)2 + . . .+ bm/(z – a)m

Working steps to find singularity-

Step-1: If exists and it is finite then z = a is a removable singular point.

Step-2: If does not exists then z = a is an essential singular point.

Step-3: If is infinite then f(z) has a pole at z = a. The order of the pole is same as the number of negative power terms in the series expansion of f(z).

Example: Find the singularity of the function-

f(z) = sin 1/z

Sol.

As we know that-f(z) = a0 +a1(z – a) + a2(z – a)2 + . . . + 1/(z – a)m[ b1(z – a)m – 1 + b2(z – a)m – 2 + b3(z – a)m – 3 + . . . + bm}

So that there is a number of singularity.

Sin 1/z is not analytic at z = a

(1/z = ∞ at z = 0)

Example: Find the singularity of 1/(1 – ez) at z = 2πi

Sol.

Here we have-

f(z) = 1/(1 – ez)

We find the poles by putting the denominator equals to zero.

That means-

1 – ez = 0

ez = 0 = ( cos 2n π + i sin 2n π) = e2nπi

z = 2nπi(n = 0, ±1, ±2, ….)

z = 2n π i is a simple pole

Example: Determine the poles of the function-

f(z) = 1/(z4 +1)

Sol.

Here we have-

f(z) = 1/(z4+1)

We find the poles by putting the denominator of the function equals to zero-

We get-

z4 + 1 = 0 and z4 = - 1

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/z - a

By De Moivre’s theorem-

Sin 1/z = 1/z – 1/3!z3 + 1/5!z5 + . ….+ (-1)n 1/(2n+1)! z2n+1

z = (-1)1/4 = (cos π + i sin π)1/4

If n = 0, then pole-

= [ cos (2n + 1)π + i sin (2n + 1)π]1/4

If n = 1, then pole-

= [ cos (2n + 1)π/4 + i sin (2n+1)π/4]

If n = 2, then pole-

z = [ cos π/4 + i sin π/4] = (1/2 + i 1/2)

If n = 3, then pole-

z = [ cos 3π/4 + i sin 3π/4] = (- 1/2 + i 1/2)

Let be analytic at all points within a circle with centre and radius. Then

Example: Expand the function

In a Taylor's series about the point

Ans. Using partial fraction method

(τ-1)(τ-3)

Both series converge when |4|<1

Therefore The series converges in the circle centred at with radius of 1.

Taylor’s series expansion is

Example: Show that when 0<|z|<4

Solution When |z|<4 we have

Example: Expand for the regions

- 0<|z|<1
- 1<|z|<2
- |z|>2

Solution Let

Hence resolving into partial functions we get

1) For 0<|z|<1 we have

2) For 1<|z|<2 we have

3) For |z|>2 we have

Let be analytic in the ring shaped region D bounded by two concentric circles with centre and radii and let be any point of D. Then

Example: using Taylor's series

Example: Obtain the Taylor’s and Laurent’s series which represents the function in the regions

1) |z|<2

2) 2<|z|<3

3) |z|>3

Solution We have

1) For |z|<2 we have

Which is Taylor’s series valid for |z|<2

2) For 2<|z|<3 we have

3) For |z|<3

Let be a pole of order m of a function and circle of radius r with centre at which does not contain any other singularities except atthen is analytic within the can be expanded by Laurent’s series

The coefficient of is called residue of at the pole then

Method of finding residue

(1) If has a simple pole at then

(2) If is of the form

(3) If has a pole of order n at then

(4) Residue at a pole of any order

(5) Residue of at

Example: Find residue of the function

Answer

Let

The singularities of f(z) are given by

Which is of the form

Example: Find the residue of at z=1

Answer

Let f(z)=

The poles of f(z) are determined by putting the denominator equal to zero

(z-1)(z-2)(z-3)=0

Z=1,2,3

Residue of f(z) at z=1=

=1/2

Example: Find the residue of

Answer

f(z)=

Poles are determined by putting sinz=0=

Hence the residue of the given function at pole is

If is analytic in a closed curve c except at a finite number of poles within c then

[Sum of residue at the pole within c]

Example: Evaluate the following integral using residue theorem

Where c is the circle..

Ans.

The poles of the integral are given by putting the denominator equal to zero

The integral is analytic on and all points inside exceptas a pole at is inside the circle

Hence by residue theorem

Example: Evaluate where c;|z|=4

Answer

Here f(z)=

Poles are

Sin iz=0

Poles

Lie inside the circle |z|=4

The given function is of the form

Its pole at z=a is

Residue (at

Residue at z=0=

Residue at =

Residue at are

Respectively -1,1 and -1

Hence the required integrand

Example: Evaluate :c is the unit circle about the origin

Answer

=

This shows that z=0 is a pole of order 2 for the function and the residue of the poles is zero(coefficient of 1/z)

Now the pole at z=0 lies within c

Example: Evaluation of definite integral

Show that

Solution

I=

Real part of

Now I= =

Putting z= where c is the unit circle |z|=1

I=

Now f(z) has simple poles at and z=-2 of which only lies inside c.

Residue at is

=

=

Now equating real parts on both sides we get

I=

Example: Prove that

Solution

Let

Putting where c is the unit circle |z|=1

2ai

Poles of f(z) are given by the roots of

Or

Let

Clearly and since we have Hence the only pole inside c is at z=

Residue (at )

Example: Evaluate

Answer

Consider

Where c is the closed contour consisting of

1) Real axis from

2) Large semicircle in the upper half plane given by |z|=R

3) The real axis -R to and

4) Small semicircle given by |z|=

Now f(z) has simple poles at z=0 of which only z=is avoided by indentation

Hence by Cauchy’s Residue theorem

Since and

Hence by Jordan’s Lemma

Also since

Hence

Hence as

Equating imaginary parts we get

Example: Prove that

Solution

Consider

Where c is the contour consisting of a large semicircle in the upper half plane indented at the origin as shown in the figure

Here we have avoided the branch point o, of by indenting the origin

Then only simple of f(z) within c is at z=i

The residue(at z=i) =

Hence by residue theorem

Since on -ve real axis.

Now

Similarly

Hence when

Equating real parts we get

References:

- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
- T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010

Unit - 5

Complex Variable - Differentiation

In the narrow sense of the term, the theory of function of a complex variable is the theory of analytic functions (cf. Analytic function) of one or several complex variables. As an independent discipline, the theory of functions of a complex variable took shape in about the middle of the 19th century as the theory of analytic functions.

Complex function-

x + iy is a complex variable which is denoted by z

If for each value of the complex variable z = x + iy in a region R, we have one or more than one values of w = u + iv, then w is called a complex function of z.

And it is denoted as-

w = u(x , y) + iv(x , y) = f(z)

Neighbourhood of z0

Let a point z0 in the complex plane and z be any positive number, then the set of points z such that-

|z – z0|<ε

Is called ε- neighbourhood of z0

Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighbourhood of point z0-

The-

Example-1: Find-

Sol. Here we have-

Divide numerator and denominator by z3, we get-

Continuity- A function w = f(z) is said to be continuous at z = z0, if

Also if w = f(z) = u (x, y) + iv (x, y) is continuous at z = z0 then u (x, y), v (x, y) are also continuous at z = z0.

Differentiability-

Let f(z) be a single valued function of the variable z, then

f’(z) =

Provided that the limit exists and has the same value for all the different ways in which δz approaches to zero.

Example-2: if f(z) is a complex function given below, then discuss dz/dz at z = 0

Sol. If z→0 along radius vector y = mx

=

=

But along v = x3,

In different paths we get different value of df/dz that means 0 and –i/2, in that case the function is not differentiable at z = 0.

Key takeaways-

1. Neighborhood of z0

|z – z0|<ε

2. Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighborhood of point z0-

The-

3. A function w = f(z) is said to be continuous at z = z0, if

f’(z) = 4

In Cartesian form-

Theorem; The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

Proof:

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

C-R equation in polar from-

C-R equations in polar form are-

Proof:

As we know that-

x = r cos and u is the function of x and y

z = x + iy = r ( cos

Differentiate (1) partially with respect to r, we get-

Now differentiate (1) with respect to , we get-

Substitute the value of , we get-

Equating real and imaginary parts, we get-

Proved

Key takeaways-

- The necessary condition for a function to be analytic at all the points in a region R are

(ii)

2. C-R equations in polar form are-

A function is said to be analytic at a point if f is differentiable not only at but every point of some neighborhood at .

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

3. An entire function is always analytic, differentiable and continuous function. (converse is not true)

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Important note-

1. If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.

2. C-R conditions are necessary but not sufficient for analytic function.

3. C-R conditions are sufficient if the partial derivative are continuous.

State and prove sufficient condition for analytic functions

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Proof:- Let f(z) be a simple valued function having at each point in the region R. Then Cauchy-Reimann equation are satisfied by Taylor’s Theorem

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q. Show that is analytic at

Ans. The function f(z) is analytic at if the function is analytic at z=0

Since

Now is differentiable at z=0 and at all points in its neighbourhood Hence the function is analytic at z=0 and in turn f(z) is analytic at

Example-1: If w = log z, then find . Also determine where w is non-analytic.

Sol. Here we have

Therefore-

and

Again-

Hence the C-R conditions are satisfied also the partial derivatives are continuous except at (0 , 0).

So that w is analytic everywhere but not at z = 0

Example-2: Prove that the function is an analytical function.

Sol. Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Example-3: Prove that

Sol. Given that

Since

V=2xy

Now

But

Hence

Example-4: Show that polar form of C-R equations are-

Sol. z = x + iy =

U and v are expressed in terms of r and θ.

Differentiate it partially w.r.t. r and θ, we get-

By equating real and imaginary parts, we get-

Key takeaways-

- A function is said to be analytic at a point if f is differentiable not only at but a every point of some neighborhood at .
- A point at which the function is not differentiable is called singular point.
- A function which is analytic everywhere is called an entire function.
- If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.
- C-R conditions are necessary but not sufficient for analytic function.
- C-R conditions are sufficient if the partial derivative are continuous

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Example: Prove that and are harmonic functions of (x, y).

Sol.

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Example: Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

Sol.

We have,

U(x, y) = 2x (1 – y)

Let V is the harmonic conjugate of U.

So that by total differentiation,

Hence the harmonic conjugate of U is

If y/x ≠ v/y and y/y ≠ - v/x

Then function is harmonic conjugate.

Unit - 5

Complex Variable - Differentiation

Unit - 5

Complex Variable - Differentiation

Complex function-

x + iy is a complex variable which is denoted by z

And it is denoted as-

w = u(x , y) + iv(x , y) = f(z)

Neighbourhood of z0

|z – z0|<ε

Is called ε- neighbourhood of z0

Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighbourhood of point z0-

The-

Example-1: Find-

Sol. Here we have-

Divide numerator and denominator by z3, we get-

Continuity- A function w = f(z) is said to be continuous at z = z0, if

Differentiability-

Let f(z) be a single valued function of the variable z, then

f’(z) =

Example-2: if f(z) is a complex function given below, then discuss dz/dz at z = 0

Sol. If z→0 along radius vector y = mx

=

=

But along v = x3,

Key takeaways-

1. Neighborhood of z0

|z – z0|<ε

2. Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighborhood of point z0-

The-

3. A function w = f(z) is said to be continuous at z = z0, if

f’(z) = 4

Unit - 5

Complex Variable - Differentiation

Unit - 5

Complex Variable - Differentiation

Unit - 5

Complex Variable - Differentiation

Complex function-

x + iy is a complex variable which is denoted by z

And it is denoted as-

w = u(x , y) + iv(x , y) = f(z)

Neighbourhood of z0

|z – z0|<ε

Is called ε- neighbourhood of z0

Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighbourhood of point z0-

The-

Example-1: Find-

Sol. Here we have-

Divide numerator and denominator by z3, we get-

Continuity- A function w = f(z) is said to be continuous at z = z0, if

Differentiability-

Let f(z) be a single valued function of the variable z, then

f’(z) =

Example-2: if f(z) is a complex function given below, then discuss dz/dz at z = 0

Sol. If z→0 along radius vector y = mx

=

=

But along v = x3,

Key takeaways-

1. Neighborhood of z0

|z – z0|<ε

2. Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighborhood of point z0-

The-

3. A function w = f(z) is said to be continuous at z = z0, if

f’(z) = 4

In Cartesian form-

Theorem; The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

Proof:

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

C-R equation in polar from-

C-R equations in polar form are-

Proof:

As we know that-

x = r cos and u is the function of x and y

z = x + iy = r ( cos

Differentiate (1) partially with respect to r, we get-

Now differentiate (1) with respect to , we get-

Substitute the value of , we get-

Equating real and imaginary parts, we get-

Proved

Key takeaways-

- The necessary condition for a function to be analytic at all the points in a region R are

(ii)

2. C-R equations in polar form are-

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Important note-

1. If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.

2. C-R conditions are necessary but not sufficient for analytic function.

3. C-R conditions are sufficient if the partial derivative are continuous.

State and prove sufficient condition for analytic functions

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q. Show that is analytic at

Ans. The function f(z) is analytic at if the function is analytic at z=0

Since

Example-1: If w = log z, then find . Also determine where w is non-analytic.

Sol. Here we have

Therefore-

and

Again-

So that w is analytic everywhere but not at z = 0

Example-2: Prove that the function is an analytical function.

Sol. Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Example-3: Prove that

Sol. Given that

Since

V=2xy

Now

But

Hence

Example-4: Show that polar form of C-R equations are-

Sol. z = x + iy =

U and v are expressed in terms of r and θ.

Differentiate it partially w.r.t. r and θ, we get-

By equating real and imaginary parts, we get-

Key takeaways-

- A point at which the function is not differentiable is called singular point.
- A function which is analytic everywhere is called an entire function.
- If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.
- C-R conditions are necessary but not sufficient for analytic function.
- C-R conditions are sufficient if the partial derivative are continuous

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Example: Prove that and are harmonic functions of (x, y).

Sol.

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Example: Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

Sol.

We have,

U(x, y) = 2x (1 – y)

Let V is the harmonic conjugate of U.

So that by total differentiation,

Hence the harmonic conjugate of U is

If y/x ≠ v/y and y/y ≠ - v/x

Then function is harmonic conjugate.

Contour integration is the process of calculating the values of a contour integral around a given contour in the complex plane.

Example: Solve where

Answer

Where

Poles of the inter are given by putting the denominator equal to zero.

Z(z-1)(2z+5)=0

Z=0,1,-5/2

The integrand has three simple poles at

Z=0,1,-5/2

The given circle |z|=3/2 with centre at z=0 and radius =3/2 encloses two poles z=0 and 1

Example: Solve

Answer

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Derivation of Cauchy Integral theorem:

Example 1:

where C = | z – 3| = 2

where f(z) = cosz

= ½ (2 πi) f(5/2) by cauchy’s integral formula

= πi. Cos (5/2)

Example 2:

Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

Solution:

Given,

f(n)(a) = n!/2πi

f(k + 1)(a) = d/da f(k)(a)

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Cauchy’s integral formula-

Cauchy’s integral formula can be defined as-

f(a) = 1/2πi

Where f(z) is analytic function within and on closed curve C, a is any point within C.

Example-1: Evaluate dz by using Cauchy’s integral formula.

Here c is the circle |z - 2| = 1/2

Sol. It is given that-

Find its poles by equating denominator equals to zero.

z2 – 3z + 2 = 0

(z – 1)(z – 2) = 0

z = 1, 2

There is one pole inside the circle, z = 2,

So that-

Now by using Cauchy’s integral formula, we get-

= 2πi [ z / z-1]z = 2 = 2πi(2/2 -1) = 4πi

Example-2: Evaluate the integral given below by using Cauchy’s integral formula-

Sol. Here we have-

Find its poles by equating denominator equals to zero.

z(z – 1)(z – 2) = 0

We get-

z = 0,1, 2

There are two poles in the circle-

Z = 0 and z = 1

So that-

= 2 πi [ (4 – 3z)/(z – 1)(z – 2)]z=0 + 2πi [(4 – 3z)/z(z-2)]z = 1

= 2πi . 4/(-1)(-2) + 2πi (4 -3)/1(1 -2) = 2πi(2 – 1) = 2πi

Example-3: Evaluate dz if c is circle |z - 1| = 1.

Sol. Here we have-

Find its poles by equating denominator equals to zero.

z2 – 1 = 0 or z2 = 1 or z = ± 1

The given circle encloses a simple pole at z = 1.

So that-

= 2πi [(3z2 + z)/(z+1)]z = 1 = 2πi((3+1)/(1+1))

= 4πi

The value of z is said to be zero of the analytic function f(z) when f(z) = 0.

If f(z) is analytic in the neighbourhood of z = a, then by Taylor’s theorem-

f(z) = a0 + a1(z – a) + a2(z – a)2 + a3(z – a)3 + . . . + an(z – a)n+ . . .

If a0 = a1 = a2 = . . . = an – 1 = 0 but an is non- zero, then f(z) is said to have a zero of order n at z = a.

The zero is said to be simple if n = 1.

an = fn(a)/n!

For a zero of order m at z = a,

f(a) = f’(a) = f”(a) = fn – 1(a) = 0 but fn(a) ≠ 0

Thus in the neighbourhood of the zero at z = a of order n

f(z) = an(z – a)n + an+1 (z – a)n + 1 = (z – a)n [ an + an+1(z –a) + …] = (z – a)n ∅(z)

Where ∅(z) = an + an+1 (z – a) + . . . Is analytic and non-zero at and in the neighbourhood of z = a.

Example: Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

Sol. Zeroes of the function-

f(z) = 0

(z – 2)/z2 sin 1/(z – 1) = 0

(z – 2)/z2 = 0, sin 1/(z – 1) = 0

z = 2, 1/(z – 1) = nπ ( n = 0, ±1, ±2 ….)

z = 2, 1 + 1/nπ (n = 0, ±1, ±2 ….)

Key takeaways-

- The pole is said to be simple pole when m = 1.
- Cauchy’s residue theorem-

If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then-

z = [ cos 5π/4 + i sin 5π/4] = ( - 1/2 – i 1/2)

A point at which a function f(z) is not analytic is known as a singular point or singularity of the function.

Isolated singular point- If z = a is a singularity of f(z) and if there is no other singularity within a small circle surrounding the point z = a, then z = a is said to be an isolated singularity of the function f (z); otherwise it is called non-isolated.

Pole of order m- Suppose a function f(z) have an isolated singular point z = a, f(z) can be expanded in a Laurent’s series around z = a, giving

= 2πi [ (4 – 3z)/(z – 1)(z – 2)]z = 0 + 2πi [(4 – 3z)/z(z -2)]z = 1

= 2πi. 4/(-1)(-2) + 2πi . (4 – 3)/1(1 – 2) = 2πi(2 – 1) = 2πi …….(1)

In some cases it may happen that the coefficient bm + 1 = bm + 2 = bm + 3 = 0 , then equation (1) becomes-

= 2πi[(3z2+z)/(z+1)]z = 1 = 2πi(3+1)/(1+1)

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/(z – a) + b2/(z – a)2 + . . .+ bm/(z – a)m+ bm+1/(z – a)m +1 + bm+2/(z – a)m+2 + . . .

Then z = a is said to be a pole of order m of the function f(z).

Note- The pole is said to be simple pole when m = 1.

In this case-

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/(z – a) + b2/(z – a)2 + . . .+ bm/(z – a)m

Working steps to find singularity-

Step-1: If exists and it is finite then z = a is a removable singular point.

Step-2: If does not exists then z = a is an essential singular point.

Step-3: If is infinite then f(z) has a pole at z = a. The order of the pole is same as the number of negative power terms in the series expansion of f(z).

Example: Find the singularity of the function-

f(z) = sin 1/z

Sol.

As we know that-f(z) = a0 +a1(z – a) + a2(z – a)2 + . . . + 1/(z – a)m[ b1(z – a)m – 1 + b2(z – a)m – 2 + b3(z – a)m – 3 + . . . + bm}

So that there is a number of singularity.

Sin 1/z is not analytic at z = a

(1/z = ∞ at z = 0)

Example: Find the singularity of 1/(1 – ez) at z = 2πi

Sol.

Here we have-

f(z) = 1/(1 – ez)

We find the poles by putting the denominator equals to zero.

That means-

1 – ez = 0

ez = 0 = ( cos 2n π + i sin 2n π) = e2nπi

z = 2nπi(n = 0, ±1, ±2, ….)

z = 2n π i is a simple pole

Example: Determine the poles of the function-

f(z) = 1/(z4 +1)

Sol.

Here we have-

f(z) = 1/(z4+1)

We find the poles by putting the denominator of the function equals to zero-

We get-

z4 + 1 = 0 and z4 = - 1

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/z - a

By De Moivre’s theorem-

Sin 1/z = 1/z – 1/3!z3 + 1/5!z5 + . ….+ (-1)n 1/(2n+1)! z2n+1

z = (-1)1/4 = (cos π + i sin π)1/4

If n = 0, then pole-

= [ cos (2n + 1)π + i sin (2n + 1)π]1/4

If n = 1, then pole-

= [ cos (2n + 1)π/4 + i sin (2n+1)π/4]

If n = 2, then pole-

z = [ cos π/4 + i sin π/4] = (1/2 + i 1/2)

If n = 3, then pole-

z = [ cos 3π/4 + i sin 3π/4] = (- 1/2 + i 1/2)

Let be analytic at all points within a circle with centre and radius. Then

Example: Expand the function

In a Taylor's series about the point

Ans. Using partial fraction method

(τ-1)(τ-3)

Both series converge when |4|<1

Therefore The series converges in the circle centred at with radius of 1.

Taylor’s series expansion is

Example: Show that when 0<|z|<4

Solution When |z|<4 we have

Example: Expand for the regions

- 0<|z|<1
- 1<|z|<2
- |z|>2

Solution Let

Hence resolving into partial functions we get

1) For 0<|z|<1 we have

2) For 1<|z|<2 we have

3) For |z|>2 we have

Let be analytic in the ring shaped region D bounded by two concentric circles with centre and radii and let be any point of D. Then

Example: using Taylor's series

Example: Obtain the Taylor’s and Laurent’s series which represents the function in the regions

1) |z|<2

2) 2<|z|<3

3) |z|>3

Solution We have

1) For |z|<2 we have

Which is Taylor’s series valid for |z|<2

2) For 2<|z|<3 we have

3) For |z|<3

Let be a pole of order m of a function and circle of radius r with centre at which does not contain any other singularities except atthen is analytic within the can be expanded by Laurent’s series

The coefficient of is called residue of at the pole then

Method of finding residue

(1) If has a simple pole at then

(2) If is of the form

(3) If has a pole of order n at then

(4) Residue at a pole of any order

(5) Residue of at

Example: Find residue of the function

Answer

Let

The singularities of f(z) are given by

Which is of the form

Example: Find the residue of at z=1

Answer

Let f(z)=

The poles of f(z) are determined by putting the denominator equal to zero

(z-1)(z-2)(z-3)=0

Z=1,2,3

Residue of f(z) at z=1=

=1/2

Example: Find the residue of

Answer

f(z)=

Poles are determined by putting sinz=0=

Hence the residue of the given function at pole is

If is analytic in a closed curve c except at a finite number of poles within c then

[Sum of residue at the pole within c]

Example: Evaluate the following integral using residue theorem

Where c is the circle..

Ans.

The poles of the integral are given by putting the denominator equal to zero

The integral is analytic on and all points inside exceptas a pole at is inside the circle

Hence by residue theorem

Example: Evaluate where c;|z|=4

Answer

Here f(z)=

Poles are

Sin iz=0

Poles

Lie inside the circle |z|=4

The given function is of the form

Its pole at z=a is

Residue (at

Residue at z=0=

Residue at =

Residue at are

Respectively -1,1 and -1

Hence the required integrand

Example: Evaluate :c is the unit circle about the origin

Answer

=

This shows that z=0 is a pole of order 2 for the function and the residue of the poles is zero(coefficient of 1/z)

Now the pole at z=0 lies within c

Example: Evaluation of definite integral

Show that

Solution

I=

Real part of

Now I= =

Putting z= where c is the unit circle |z|=1

I=

Now f(z) has simple poles at and z=-2 of which only lies inside c.

Residue at is

=

=

Now equating real parts on both sides we get

I=

Example: Prove that

Solution

Let

Putting where c is the unit circle |z|=1

2ai

Poles of f(z) are given by the roots of

Or

Let

Clearly and since we have Hence the only pole inside c is at z=

Residue (at )

Example: Evaluate

Answer

Consider

Where c is the closed contour consisting of

1) Real axis from

2) Large semicircle in the upper half plane given by |z|=R

3) The real axis -R to and

4) Small semicircle given by |z|=

Now f(z) has simple poles at z=0 of which only z=is avoided by indentation

Hence by Cauchy’s Residue theorem

Since and

Hence by Jordan’s Lemma

Also since

Hence

Hence as

Equating imaginary parts we get

Example: Prove that

Solution

Consider

Where c is the contour consisting of a large semicircle in the upper half plane indented at the origin as shown in the figure

Here we have avoided the branch point o, of by indenting the origin

Then only simple of f(z) within c is at z=i

The residue(at z=i) =

Hence by residue theorem

Since on -ve real axis.

Now

Similarly

Hence when

Equating real parts we get

References:

- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
- T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010