Unit-3
Complex Function
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 
Let a point 
in the complex plane and z be any positive number, then the set of points z such that-
|
|<ε
Is called ε- neighbourhood of 
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 
-
The-
Example-1: Find-
Sol. Here we have-
Divide numerator and denominator by 
, we get-
Continuity- A function w = f(z) is said to be continuous at z = 
, if
Also if w = f(z) = u(x , y) + iv(x , y) is continuous at z = 
then u(x , y), v(x , y) are also continuous at z = 
.
Differentiability-
Let f(z) be a single valued function of the variable z, then
Provided that the limit exists and has the same value for all the different ways in which 
approaches to zero.
Example-2: if f(z) is a complex function given below, then discuss 
Sol. If z→0 along radius vector y = mx
But along 
,
In different paths we get different value of 
that means 0 and –i/2, in that case the function is not differentiable at z = 0.
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
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-
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.
Example-1: Check whether the function w = sin z is analytic or not.
Sol. Here-
Now-
And
Here we see that C-R conditions are satisfied and partial derivatives are continuous.
Example-2: Prove that the real and imaginary parts of the function w = log z satisfies C-R equations.
Sol.
We put 
and 
to separate the real and imaginary parts of log z.
w = log z = log (x + iy)
Here 
On differentiating u and v, we get-
From the above two equations, we have-
Again differentiating u and v, we get-
We have form the above equation-
Hence the C-R condition is satisfied.
Conformal mapping
If the sense of the relation as well as magnitude of the angle is preserved the transformation is said to be conformal.
Example-1: Find the conformal transformation of 
.
Answer. Let 
Theorem: If W=f(z) represents a conformal transformation of a domain D in the z-plane into a domain D of the W plane then f(z) is an analytic function of z in D.
Proof: We have u+iv=u(x,y)+iv(x,y)
So that u=u(x, y) and v=v(x,y)
Let ds and 
denote elementary arc length in the z-plane and w-plane respectively Then
Now 
Hence 
Or 
Where 
Now 
is independent of direction if
Where h depends on x and y only and is not zero. Thus the conditions for an isogonal transformation
And 
The equation are satisfied if we get
Then substituting these values in 2 we get
Taking 
i.e. 
Also 
Hence 
Similarly 
i.e. 
The equation (4) are the well-known Cauchy -Reimann
Conformal mapping
Show that the mapping 
is conformal in the whole of the z plane.
Sol.
Let z=x+iy
Then 
Consider the mapping of the straight line x=a in z plane the w plane which gives 
which is a circle in the w plane in the anticlockwise direction similarly the straight line y=b is mapped into 
which is a radius vector in the w plane.
The angle between the line x=a and y=b in the z plane is a right angle. The corresponding angle in the w plane between the circle e = constant and the radius vector 
is also a right angle which establishes that the mapping 
is conformal.
Example: Show that the curve u = constant and v = constant cut orthogonally at all intersections but the transformation w = u + iv is not conformal. Where-
Sol.
Let 
…………. (1)
Differentiate (1), we get-
…………… (2)
Now-
…………….. (3)
Differentiate (3), we get-
………. (4)
As we know that for the condition for orthogonallity, from (2) and (4)
So that these two curves cut orthogonally.
Here,
And
Here the C-R equation is not satisfied so that the function u + iv is not analytic.
Hence the transformation is not conformal.
Bilinear transformation
Bilinear transformation is a correction of backwards difference method.
The bilinear transformation (also known as Tuatn’s method transformation)is defined as substitution:
Example 1:
Find the bi-linear transformation which aps points z=2,1,0 ontpo the points w=1,0,i
Sol.
Let 
Thus we have
=
Example 2:
How that the bilinear transformation w= 
transforms 
in the z-plane to 4u+3=0 in w-plane.
Sol.
Consider 
the circle in z-plane
= 0
Thus, centre of the circle is (h,k)
c(2,0) and radius r=2.
Thus in z-plane it is given as 
=2....(1)
Consider w= 
W(z-4) = 2z+3
Wz-4w=2z+3
Wz-2z=4w+3
Z(w-2) = (4w+3)
z = 
z-2 = 
- 2
In case of a complex function f(z) the path of the definite integral 
can be along any curve from z = a to z = b.
In case the initial point and final point coincide so that c is a closed curve, then this integral called contour integral and is denoted by-
If f(z) = u(x, y) + iv(x, y), then since dz = dx + i dy
We have-
It shows that the evaluation of the line integral of a complex function can be reduced to the evaluation of two-line integrals of real function.
Example: Evaluate 
along the path y = x.
Sol.
Along the line y = x,
Dy = dx that dz = dx + i dy
Dz = dx + i dx = (1 + i) dx
On putting y = x and dz = (1 + i)dx
Example: Evaluate 
where c is the circle with center a and r. What is n = -1.
Sol.
The equation of a circle C is |z - a| = r or z – a = 
Where 
varies from 0 to 2π
Dz = 
Which is the required value.
When n = -1
Cauchy’s integral theorem
If a function f(z) is analytic and its derivative f’(z) continuous at all points inside and on a closed curve c, then 
Proof: Suppose the region is R which is closed by curve c and let-
By using Green’s theorem-
Replace 
by 
and 
by 
-
So that-
Example-1: Evaluate 
where C is |z + 3i| = 2
Sol.
Here we have-
Hence the poles of f(z),
Note- put determine equal to zero to find the poles.
Here pole z = -3i lies in the given circle C.
So that-
Example 2:
where C = 
Sol.
where f(z) = cosz
= 
by cauchy’s integral formula
= 
Example 3:
Solve the following by cauchy’s integral method:
Solution:
Given,
= 
= 
= 
Cauchy’s integral formula-
Cauchy’s integral formula can be defined as-
Where f(z) is analytic function within and on closed curve C, a is any point within C.
Example-1: Evaluate 
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.
There is one pole inside the circle, z = 2,
So that-
Now by using Cauchy’s integral formula, we get-
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.
We get-
There are two poles in the circle-
Z = 0 and z = 1
So that-
Example-3: Evaluate 
if c is circle |z - 1| = 1.
Sol. Here we have-
Find its poles by equating denominator equals to zero.
The given circle encloses a simple pole at z = 1.
So that-
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
…… (1)
In some cases it may happen that the coefficient 
, then equation (1) becomes-
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-
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-
Sol.
As we know that-
So that there is a number of singularity.
is not analytic at z = a
(1/z = ∞ at z = 0)
Example: Find the singularity of 
Sol.
Here we have-
We find the poles by putting the denominator equals to zero.
That means-
Example: Determine the poles of the function-
Sol.
Here we have-
We find the poles by putting the denominator of the function equals to zero-
We get-
By De Moivre’s theorem-
If n = 0, then pole-
If n = 1, then pole-
If n = 2, then pole-
If n = 3, then pole-
Cauchy’s residue theorem-
If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then-
;DO Proof:
Suppose 
be the non-intersecting circles with centres at 
respectively.
Redii so small that they lie within the closed curve C. Then f(z) is analytic in the multipUYle connected region lying between the curves C and
Now applying the Cauchy’s theorem-
Example: Find the poles of the following functions and residue at each pole:
and hence evaluate-
where c: |z| = 3.
Sol.
The poles of the function are-
The pole at z = 1 is of second order and the pole at z = -2 is simple-
Residue of f(z) (at z = 1)
Residue of f(z) ( at z = -2)
Example: Evaluate-
Where C is the circle |z| = 4.
Sol.
Here we have,
Poles are given by-
Out of these, the poles z = -πi , 0 and πi lie inside the circle |z| = 4.
The given function 1/sinh z is of the form 
Its poles at z = a is 
Residue (at z = -πi)
Residue (at z = 0)
Residue (at z = πi)
Hence the required integral 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.
.
Sol. 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
Sol.
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.
Sol.
=
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 integrand
Show that 
Sol.
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=
- 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 
Sol.
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 
Sol.
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
Example: Evaluate
Sol.
We have-
Consider the integral-
Where,
Taken round the closed contour c consisting of the upper half of a large circle |z| = R and the real axis from -R to R
Poles of f(z) are given by-
The only pole which lies within the contour is at z = i.
The residue of f(z) at z = i
Then by Cauchy’s residue theorem, we have-
Equating real parts, we get-
Example: Evaluate-
Sol.
Let us consider the integral
Where
We find the poles by putting the denominator equal to zero,
Since there is no pole on the real axis, so that we may take the contour C consisting of the semi circle which is upper half of a large circle |z| = R and the real axis from –-R to R.
Hence by the Cauchy’s residue theorem we have-
The only pole within the contour C is z = a, and it is of order 2.
Here
Where,
Residue at the double pole (z = ai)-
And
Hence when 
the relation becomes-
