Unit-6

Complex Variables

Q1) Find-

A1) Here we have-

Divide numerator and denominator by , we get-

Q2) If w = log z, then find . Also determine where w is non-analytic.

A2) 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

Q3) Prove that

A3) Given that

Since      

V=2xy

Now

But                                      

Hence 

Q4) Show that polar form of C-R equations are-

A4)             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-

Q5) Prove that the real and imaginary parts of the function w = log z satisfies C-R equations.

A5)

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.

Q6) Show that the curve u = constant and v = constant cut orthogonally at all intersections but the transformation w = u + iv is not conformal. Where-

A6)

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.

Q7) How that the bilinear transformation w= transforms in the z-plane to 4u+3=0 in w-plane.

A7)

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

Q8) Evaluate where C is |z + 3i| = 2

A8)

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-

Q9) Evaluate the integral given below by using Cauchy’s integral formula-

A9) 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-

Q10) Evaluate  where c;|z|=4

A10)

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

Q11) Evaluate 

A11)

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