Unit – 6
Complex Variables
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.
Example 1:
Solve the following complex function:
Solution:
Given,
Example 2:
Using the residue theorem to evaluate:
Solution:
The singularity z=0,is in our region and we will add the following residue
The singularity z = -3/2 will be skipped because the singularity is not in our region.
The singularity z= -2/3 is in our region and we will add the following residue
Our sum is,
Therefore the solution is,
A function 
is said to be analytic at a point 
if f is differentiable not only at 
but an every point of some neighborhood at 
.
Example, Prove that the function 
is an analytical function.
Solution. Let 
=u+iv
Let 
=u and 
=v
Hence e-R-Equation satisfied.
Prove that 
Answer Given that
Since 
V=2xy
Now 
But 
Hence 
State and prove sufficient condition for analytic functions
Answer 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 
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
Along imaginary axis
From equation (2) and (3)
Equating real and imaginary parts
This is called Cauchy Riemann Equation
If the sense of the relation as well as magnitude of the angle is preserved the transformation is said to be conformal.
Example; 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.
Ans 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.
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
Solution:
Let 
Thus we have
= 
Eample 2:
How that the bi-liear transformation w= 
transforms 
in the z-plane to 4u+3=0 in w-plane.
Solution:
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
Derivation of Cauchy Integraltheorem:
dz
f(z).dz
Example 1:
where C = 
where f(z) = cosz
= 
by cauchy’s integral formula
= 
Example 2:
Solve the following by cauchy’s integral method:
Solution:
Given,
= 
= 
= 
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]
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
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
2. 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
3. Evaluation of definite integrand
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=
4. 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 
)
5. 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
6. 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
Reference Books:
1. Erwin Kreyszig, “Advanced Engineering Mathematics”, Wiley India,10th Edition.
2. M.D. Greenberg, “Advanced Engineering Mathematics”, Pearson Education, 2 nd Edition.
3. Peter. V and O‟Neil, “Advanced Engineering Mathematics”, Cengage Learning,7th Edition.
4. S.L. Ross, “Differential Equations”, Wiley India, 3rd Edition.
5. S. C. Chapra and R. P. Canale, “Numerical Methods for Engineers”, McGraw-Hill, 7th Edition.
6. J. W. Brown and R. V. Churchill, “Complex Variables and Applications”, McGraw-Hill Inc, 8th Edition.
