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 at
then
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 except
as 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 at
then
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 except
as 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
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 at
then
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 except
as 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
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 at
then
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 except
as 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
