Unit-4
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 
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 = 
.
Example: Show that f(z) = Re z = x is continuous but not differentiable.
Sol.
Continuity:
Not differentiability-
While
Hence limit does not exist which means function is not differentiable.
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.
Key takeaways-
|
|<ε
2. 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-
3. A function w = f(z) is said to be continuous at z = 
, if
4. 
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-
Key takeaways-
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-
(ii) 
2. C-R equations in polar form are-
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
Here it satisfies Laplace equation so that u (x, y) is harmonic.
Now-
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 
Milne-Thomson method is used to construct an analytic function without finding v,
The method is given below-
Since,
So that-
The relation above can be regarded as a formal identity in two independent variables z and 
Replacing 
by z we get-
Condition-1 if u is given-
We have-
If we write-
On integrating-
Condition-2: If v is given-
f(z) = u + iv
Example: If
Then find f(z)
Sol.
Here-
Which is the required answer.
Example: Verify if 
is analytic or not?
Sol.
Here we have-
And
At origin-
Since
C-R equations are satisfied at the origin-
But-
Let 
along the real axis y = 0, then
Now let
along the curve 
, then-
Which shows that f'(0) does not exist since the limit is not unique along two different paths. Hence f(z) is not analytic at origin but C-R equations are satisfied.
Key takeaways-
2. If v is given-
If the sense of the relation as well as magnitude of the angle is preserved the transformation is said to be conformal.
Note-An analytic function f (z) is conformal everywhere except at its critical points where f (z) = 0.
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
Example: 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.
Key takeaways-
Bilinear transformation (Mobius 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=2,1,0ontpo 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
Properties-
References
