Unit - 5

Complex Variable - Integration

Q1) Solve where

A1)

Where

Poles of the inter are given by putting the denominator equal to zero.

Z(z-1)(2z+5)=0

Z=0,1,-5/2

The integrand has three simple poles at

Z=0,1,-5/2

The given circle |z|=3/2 with centre at z=0 and radius =3/2 encloses two poles z=0 and 1

Q2) Solve

A2)

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Q3) Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

A3)

Given,

f(n)(a) = n!/2πi

f(k + 1)(a) = d/da f(k)(a)

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Q4) Evaluate dz by using Cauchy’s integral formula.

Here c is the circle |z - 2| = 1/2

A4)

It is given that-

Find its poles by equating denominator equals to zero.

z2 – 3z + 2 = 0

(z – 1)(z – 2) = 0

z = 1, 2

There is one pole inside the circle, z = 2,

So that-

Now by using Cauchy’s integral formula, we get-

= 2πi [ z / z-1]z = 2 = 2πi(2/2 -1) = 4πi

Q5) Evaluate the integral given below by using Cauchy’s integral formula-

A5)

Here we have-

Find its poles by equating denominator equals to zero.

z(z – 1)(z – 2) = 0

We get-

z = 0,1, 2

There are two poles in the circle-

Z = 0 and z = 1

So that-

= 2 πi [ (4 – 3z)/(z – 1)(z – 2)]z=0 + 2πi [(4 – 3z)/z(z-2)]z = 1

= 2πi . 4/(-1)(-2) + 2πi (4 -3)/1(1 -2) = 2πi(2 – 1) = 2πi

Q6) Evaluate dz if c is circle |z - 1| = 1.

A6)

Here we have-

Find its poles by equating denominator equals to zero.

z2 – 1 = 0 or z2 = 1 or z = ± 1

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

So that-

= 2πi [(3z2 + z)/(z+1)]z = 1 = 2πi((3+1)/(1+1))

= 4πi

Q7) Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

A7)

Zeroes of the function-

f(z) = 0

(z – 2)/z2 sin 1/(z – 1) = 0

(z – 2)/z2 = 0, sin 1/(z – 1) = 0

z = 2, 1/(z – 1) = nπ ( n = 0, ±1, ±2 ….)

z = 2, 1 + 1/nπ (n = 0, ±1, ±2 ….)

Q8) Find the singularity of the function-

f(z) = sin 1/z

A8)

As we know that-f(z) = a0 +a1(z – a) + a2(z – a)2 + . . . + 1/(z – a)m[ b1(z – a)m – 1 + b2(z – a)m – 2 + b3(z – a)m – 3 + . . . + bm}

So that there is a number of singularity.

Sin 1/z is not analytic at z = a

(1/z = ∞ at z = 0)

Q9) Find the singularity of 1/(1 – ez) at z = 2πi

A9)

Here we have-

f(z) = 1/(1 – ez)

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

That means-

1 – ez = 0

ez = 0 = ( cos 2n π + i sin 2n π) = e2nπi

z = 2nπi(n = 0, ±1, ±2, ….)

z = 2n π i is a simple pole

Q10) Determine the poles of the function-

f(z) = 1/(z4 +1)

A10)

Here we have-

f(z) = 1/(z4+1)

We find the poles by putting the denominator of the function equals to zero-

We get-

z4 + 1 = 0 and z4 = - 1

f(z) = a0 + a1(z – a) + a2(z – a)2 + . . . + b1/z - a

By De Moivre’s theorem-

Sin 1/z = 1/z – 1/3!z3 + 1/5!z5 + . ….+ (-1)n 1/(2n+1)! z2n+1

z = (-1)1/4 = (cos π + i sin π)1/4

If n = 0, then pole-

= [ cos (2n + 1)π + i sin (2n + 1)π]1/4

If n = 1, then pole-

= [ cos (2n + 1)π/4 + i sin (2n+1)π/4]

If n = 2, then pole-

z = [ cos π/4 + i sin π/4] = (1/2 + i 1/2)

If n = 3, then pole-

z = [ cos 3π/4 + i sin 3π/4] = (- 1/2 + i 1/2)

Q11) Expand the function

In a Taylor's series about the point

A11)

Using partial fraction method

(τ-1)(τ-3)

Both series converge when |4|<1

Therefore The series converges in the circle centred at with radius of 1.

Taylor’s series expansion is

Q12) Show that when 0<|z|<4

A12)

When |z|<4 we have

Q13) Expand for the regions

- 0<|z|<1
- 1<|z|<2
- |z|>2

A13)

Let

Hence resolving into partial functions we get

1) For 0<|z|<1 we have

2) For 1<|z|<2 we have

3) For |z|>2 we have

Q14) Obtain the Taylor’s and Laurent’s series which represents the function in the regions

1) |z|<2

2) 2<|z|<3

3) |z|>3

A14)

We have

1) For |z|<2 we have

Which is Taylor’s series valid for |z|<2

2) For 2<|z|<3 we have

3) For |z|<3

Q15) Find residue of the function

A15)

Let

The singularities of f(z) are given by

Which is of the form

Q16) Find the residue of at z=1

A16)

Let f(z)=

The poles of f(z) are determined by putting the denominator equal to zero

(z-1)(z-2)(z-3)=0

Z=1,2,3

Residue of f(z) at z=1=

=1/2

Q17) Find the residue of

A17)

f(z)=

Poles are determined by putting sinz=0=

Hence the residue of the given function at pole is

Q18) Evaluate the following integral using residue theorem

Where c is the circle..

A18)

The poles of the integral are given by putting the denominator equal to zero

The integral is analytic on and all points inside exceptas a pole at is inside the circle

Hence by residue theorem

Q19) Evaluate where c;|z|=4

A19)

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

Q20) Prove that

A20)

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 )

Q21) Evaluate

A21)

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

Q22) Prove that

A22)

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

Q23) Evaluate-

A23)

We know that atits poles in the upper half plane. So consider

(z) has simple poles at z = ±ai, z = ±bi out of which z = ai, bi lies in the upper half plane.

Similarly,

Q24) Evaluate:

A24)

We know that atits poles in the upper half plane.

So consider which has singular points at z = for k = 0, 1, 2, 3. Out of these four, only lies in the

Upper half plane.

Which is 0/0 form, Applying L’ Hostpital’s rule

Real part of k1 = ¼ e-m/2 . Sin m/2

Similarly

Real part of k2 = ¼ e- m/2 . Sin m/2

Then