Unit - 5

Complex Variable - Integration

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

Fourier Integral: Improper Integral Involving Trigonometric Functions

The integrals

Can be evaluated by integrating

By residue theorem

Where the summation extends to all poles of in the upper half=plane (since | it follows that

Equating the real and imaginary

Parts of (3), we get

Example: Evaluate-

Sol:

We know that atits poles in the upper half plane. So consider

(z) has simple poles at z = ±ai, z = ±bi out of which z = ai, bi lies in the upper half plane.

Similarly,

Example: Evaluate:

Sol:

We know that atits poles in the upper half plane.

So consider which has singular points at z = for k = 0, 1, 2, 3. Out of these four, only lies in the

Upper half plane.

Which is 0/0 form, Applying L’ Hostpital’s rule

Real part of k1 = ¼ e-m/2 . Sin m/2

Similarly

Real part of k2 = ¼ e- m/2 . Sin m/2

Then

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 - Integration

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

Fourier Integral: Improper Integral Involving Trigonometric Functions

The integrals

Can be evaluated by integrating

By residue theorem

Where the summation extends to all poles of in the upper half=plane (since | it follows that

Equating the real and imaginary

Parts of (3), we get

Example: Evaluate-

Sol:

We know that atits poles in the upper half plane. So consider

(z) has simple poles at z = ±ai, z = ±bi out of which z = ai, bi lies in the upper half plane.

Similarly,

Example: Evaluate:

Sol:

We know that atits poles in the upper half plane.

So consider which has singular points at z = for k = 0, 1, 2, 3. Out of these four, only lies in the

Upper half plane.

Which is 0/0 form, Applying L’ Hostpital’s rule

Real part of k1 = ¼ e-m/2 . Sin m/2

Similarly

Real part of k2 = ¼ e- m/2 . Sin m/2

Then

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 - Integration

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

Fourier Integral: Improper Integral Involving Trigonometric Functions

The integrals

Can be evaluated by integrating

By residue theorem

Where the summation extends to all poles of in the upper half=plane (since | it follows that

Equating the real and imaginary

Parts of (3), we get

Example: Evaluate-

Sol:

We know that atits poles in the upper half plane. So consider

(z) has simple poles at z = ±ai, z = ±bi out of which z = ai, bi lies in the upper half plane.

Similarly,

Example: Evaluate:

Sol:

We know that atits poles in the upper half plane.

So consider which has singular points at z = for k = 0, 1, 2, 3. Out of these four, only lies in the

Upper half plane.

Which is 0/0 form, Applying L’ Hostpital’s rule

Real part of k1 = ¼ e-m/2 . Sin m/2

Similarly

Real part of k2 = ¼ e- m/2 . Sin m/2

Then

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