S1) Here we have-

Divide numerator and denominator by , we get-

S2) 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

S3) Given that

Since      

V=2xy

Now

But                                      

Hence 

S4)

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.

S5)

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

S6)

As we know that-

So that there is a number of singularity.

is not analytic at z = a

(1/z = ∞ at z = 0)

S7) 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-

S8) Here we have-

Find its poles by equating denominator equals to zero.

The given circle encloses a simple pole at z = 1.

So that-

S9)

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)

S10)

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 =