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 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
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 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
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
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
Unit - 5
Complex Variable - Differentiation
Unit - 5
Complex Variable - Differentiation
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 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
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
