Unit-3

Complex Function

Question Bank

Question-1: Find-

Sol. Here we have-

Divide numerator and denominator by , we get-

Question-2: 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

Question-3: Prove that

Sol. Given that

Since      

V=2xy

Now

But                                      

Hence 

Question-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-

Question-5: 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.

Question-6: 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.

Question-7: 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.

Question-8: 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

Question-9: 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

Question-10: 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-

Question-11: Solve the following by cauchy’s integral method:

Solution:

Given,

=

=

=

Question-12: 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-

Question-13: Find the singularity of

Sol.

Here we have-

We find the poles by putting the denominator equals to zero.

That means-

Question-14: 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)

Question-15: 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=

Question-16: 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

Question-17: 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-