. Here we have-

Divide numerator and denominator by , we get-

Q2: 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.

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.

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

Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Given that

Since      

V=2xy

Now

But                                      

Hence 

Let z=x+iy

Then       

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.

Now equating real and imaginary parts-

Again

Hence

Let

Thus we have

=

w= transforms

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