Unit - 5

Complex Variable - Differentiation

Q1) Find-

A1)

Here we have-

Divide numerator and denominator by z3, we get-

Q2) If f(z) is a complex function given below, then discuss dz/dz at z = 0

A2)

If z→0 along radius vector y = mx

=

=

But along v = x3,

In different paths we get different value of df/dz that means 0 and –i/2, in that case the function is not differentiable at z = 0.

Q3) Prove: The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

A3)

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

Q4) What is Analytic function and write is necessary conditions?

A4)

A function is said to be analytic at a point if f is differentiable not only at but every point of some neighborhood at .

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

3. An entire function is always analytic, differentiable and continuous function. (converse is not true)

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Q5) State and prove sufficient condition for analytic functions

A5)

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Proof:- Let f(z) be a simple valued function having at each point in the region R. Then Cauchy-Reimann equation are satisfied by Taylor’s Theorem

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q6) Show that is analytic at

A6)

The function f(z) is analytic at if the function is analytic at z=0

Since

Now is differentiable at z=0 and at all points in its neighbourhood Hence the function is analytic at z=0 and in turn f(z) is analytic at

Q7) If w = log z, then find . Also determine where w is non-analytic.

A7)

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

Q8) Prove that the function is an analytical function.

A8)

Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Q9) Prove that

A9)

Given that

Since

V=2xy

Now

But

Hence

Q10) Show that polar form of C-R equations are-

A10)

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-

Q11) Prove Harmonic function theorem.

A11)

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Q12) Prove that and are harmonic functions of (x, y).

A12)

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Q13) Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

A13)

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

Q14) Solve where

A14)

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

Q15) Solve

A15)

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Q16) Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

A16)

Given,

f(n)(a) = n!/2πi

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

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Q17) Evaluate dz by using Cauchy’s integral formula.

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

A17)

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

Q18) Evaluate the integral given below by using Cauchy’s integral formula-

A18)

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

Q19) Evaluate dz if c is circle |z - 1| = 1.

A19)

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

Q20) Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

A20)

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 ….)

Q21) Find the singularity of the function-

f(z) = sin 1/z

A21)

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)

Q22) Show that when 0<|z|<4

A22)

When |z|<4 we have

Q23) Find the residue of at z=1

A23)

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

Unit - 5

Complex Variable - Differentiation

Q1) Find-

A1)

Here we have-

Divide numerator and denominator by z3, we get-

Q2) If f(z) is a complex function given below, then discuss dz/dz at z = 0

A2)

If z→0 along radius vector y = mx

=

=

But along v = x3,

In different paths we get different value of df/dz that means 0 and –i/2, in that case the function is not differentiable at z = 0.

Q3) Prove: The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

A3)

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

Q4) What is Analytic function and write is necessary conditions?

A4)

A function is said to be analytic at a point if f is differentiable not only at but every point of some neighborhood at .

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

3. An entire function is always analytic, differentiable and continuous function. (converse is not true)

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Q5) State and prove sufficient condition for analytic functions

A5)

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Proof:- Let f(z) be a simple valued function having at each point in the region R. Then Cauchy-Reimann equation are satisfied by Taylor’s Theorem

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q6) Show that is analytic at

A6)

The function f(z) is analytic at if the function is analytic at z=0

Since

Now is differentiable at z=0 and at all points in its neighbourhood Hence the function is analytic at z=0 and in turn f(z) is analytic at

Q7) If w = log z, then find . Also determine where w is non-analytic.

A7)

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

Q8) Prove that the function is an analytical function.

A8)

Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Q9) Prove that

A9)

Given that

Since

V=2xy

Now

But

Hence

Q10) Show that polar form of C-R equations are-

A10)

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-

Q11) Prove Harmonic function theorem.

A11)

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Q12) Prove that and are harmonic functions of (x, y).

A12)

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Q13) Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

A13)

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

Q14) Solve where

A14)

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

Q15) Solve

A15)

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Q16) Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

A16)

Given,

f(n)(a) = n!/2πi

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

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Q17) Evaluate dz by using Cauchy’s integral formula.

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

A17)

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

Q18) Evaluate the integral given below by using Cauchy’s integral formula-

A18)

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

Q19) Evaluate dz if c is circle |z - 1| = 1.

A19)

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

Q20) Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

A20)

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 ….)

Q21) Find the singularity of the function-

f(z) = sin 1/z

A21)

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)

Q22) Show that when 0<|z|<4

A22)

When |z|<4 we have

Q23) Find the residue of at z=1

A23)

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

Unit - 5

Complex Variable - Differentiation

Q1) Find-

A1)

Here we have-

Divide numerator and denominator by z3, we get-

Q2) If f(z) is a complex function given below, then discuss dz/dz at z = 0

A2)

If z→0 along radius vector y = mx

=

=

But along v = x3,

In different paths we get different value of df/dz that means 0 and –i/2, in that case the function is not differentiable at z = 0.

Q3) Prove: The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

A3)

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

Q4) What is Analytic function and write is necessary conditions?

A4)

A function is said to be analytic at a point if f is differentiable not only at but every point of some neighborhood at .

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

3. An entire function is always analytic, differentiable and continuous function. (converse is not true)

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Q5) State and prove sufficient condition for analytic functions

A5)

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Proof:- Let f(z) be a simple valued function having at each point in the region R. Then Cauchy-Reimann equation are satisfied by Taylor’s Theorem

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q6) Show that is analytic at

A6)

The function f(z) is analytic at if the function is analytic at z=0

Since

Now is differentiable at z=0 and at all points in its neighbourhood Hence the function is analytic at z=0 and in turn f(z) is analytic at

Q7) If w = log z, then find . Also determine where w is non-analytic.

A7)

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

Q8) Prove that the function is an analytical function.

A8)

Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Q9) Prove that

A9)

Given that

Since

V=2xy

Now

But

Hence

Q10) Show that polar form of C-R equations are-

A10)

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-

Q11) Prove Harmonic function theorem.

A11)

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Q12) Prove that and are harmonic functions of (x, y).

A12)

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Q13) Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

A13)

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

Q14) Solve where

A14)

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

Q15) Solve

A15)

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Q16) Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

A16)

Given,

f(n)(a) = n!/2πi

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

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Q17) Evaluate dz by using Cauchy’s integral formula.

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

A17)

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

Q18) Evaluate the integral given below by using Cauchy’s integral formula-

A18)

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

Q19) Evaluate dz if c is circle |z - 1| = 1.

A19)

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

Q20) Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

A20)

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 ….)

Q21) Find the singularity of the function-

f(z) = sin 1/z

A21)

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)

Q22) Show that when 0<|z|<4

A22)

When |z|<4 we have

Q23) Find the residue of at z=1

A23)

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

Unit - 5

Complex Variable - Differentiation

Q1) Find-

A1)

Here we have-

Divide numerator and denominator by z3, we get-

Q2) If f(z) is a complex function given below, then discuss dz/dz at z = 0

A2)

If z→0 along radius vector y = mx

=

=

But along v = x3,

Q3) Prove: The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

A3)

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

Q4) What is Analytic function and write is necessary conditions?

A4)

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Q5) State and prove sufficient condition for analytic functions

A5)

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q6) Show that is analytic at

A6)

The function f(z) is analytic at if the function is analytic at z=0

Since

Q7) If w = log z, then find . Also determine where w is non-analytic.

A7)

Here we have

Therefore-

and

Again-

So that w is analytic everywhere but not at z = 0

Q8) Prove that the function is an analytical function.

A8)

Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Q9) Prove that

A9)

Given that

Since

V=2xy

Now

But

Hence

Q10) Show that polar form of C-R equations are-

A10)

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-

Q11) Prove Harmonic function theorem.

A11)

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Q12) Prove that and are harmonic functions of (x, y).

A12)

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Q13) Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

A13)

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

Q14) Solve where

A14)

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

Q15) Solve

A15)

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Q16) Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

A16)

Given,

f(n)(a) = n!/2πi

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

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Q17) Evaluate dz by using Cauchy’s integral formula.

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

A17)

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

Q18) Evaluate the integral given below by using Cauchy’s integral formula-

A18)

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

Q19) Evaluate dz if c is circle |z - 1| = 1.

A19)

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

Q20) Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

A20)

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 ….)

Q21) Find the singularity of the function-

f(z) = sin 1/z

A21)

So that there is a number of singularity.

Sin 1/z is not analytic at z = a

(1/z = ∞ at z = 0)

Q22) Show that when 0<|z|<4

A22)

When |z|<4 we have

Q23) Find the residue of at z=1

A23)

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

Unit - 5

Unit - 5

Complex Variable - Differentiation

Q1) Find-

A1)

Here we have-

Divide numerator and denominator by z3, we get-

Q2) If f(z) is a complex function given below, then discuss dz/dz at z = 0

A2)

If z→0 along radius vector y = mx

=

=

But along v = x3,

Q3) Prove: The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

A3)

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

Q4) What is Analytic function and write is necessary conditions?

A4)

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Q5) State and prove sufficient condition for analytic functions

A5)

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q6) Show that is analytic at

A6)

The function f(z) is analytic at if the function is analytic at z=0

Since

Q7) If w = log z, then find . Also determine where w is non-analytic.

A7)

Here we have

Therefore-

and

Again-

So that w is analytic everywhere but not at z = 0

Q8) Prove that the function is an analytical function.

A8)

Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Q9) Prove that

A9)

Given that

Since

V=2xy

Now

But

Hence

Q10) Show that polar form of C-R equations are-

A10)

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-

Q11) Prove Harmonic function theorem.

A11)

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Q12) Prove that and are harmonic functions of (x, y).

A12)

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Q13) Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

A13)

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

Q14) Solve where

A14)

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

Q15) Solve

A15)

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Q16) Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

A16)

Given,

f(n)(a) = n!/2πi

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

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Q17) Evaluate dz by using Cauchy’s integral formula.

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

A17)

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

Q18) Evaluate the integral given below by using Cauchy’s integral formula-

A18)

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

Q19) Evaluate dz if c is circle |z - 1| = 1.

A19)

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

Q20) Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

A20)

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 ….)

Q21) Find the singularity of the function-

f(z) = sin 1/z

A21)

So that there is a number of singularity.

Sin 1/z is not analytic at z = a

(1/z = ∞ at z = 0)

Q22) Show that when 0<|z|<4

A22)

When |z|<4 we have

Q23) Find the residue of at z=1

A23)

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

Unit - 5

Unit - 5

Unit - 5

Complex Variable - Differentiation

Q1) Find-

A1)

Here we have-

Divide numerator and denominator by z3, we get-

Q2) If f(z) is a complex function given below, then discuss dz/dz at z = 0

A2)

If z→0 along radius vector y = mx

=

=

But along v = x3,

Q3) Prove: The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

A3)

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

………… (1)

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

Q4) What is Analytic function and write is necessary conditions?

A4)

Note-

1. A point at which the function is not differentiable is called singular point.

2. A function which is analytic everywhere is called an entire function.

4. Analytic function is always differentiable and continuous but converse is not true.

5. A differentiable function is always continuous but converse is not true.

The necessary condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

Provided exists

Equation (1) and (2) are known as Cauchy-Riemann equations.

The sufficient condition for f(z) to be analytic-

f(z) = u + i(v) is to be analytic at all the points in a region R are-

1. …………. (1)

2. ……...…. (2)

are continuous function of x and y in region R.

Q5) State and prove sufficient condition for analytic functions

A5)

Statement – The sufficient condition for a function to be analytic at all points in a region R are

1.

2. are continuous function of x and y in region R.

Ignoring the terms of second power and higher power

We know C-R equation

Replacing

Respectively in (1) we get

Q6) Show that is analytic at

A6)

The function f(z) is analytic at if the function is analytic at z=0

Since

Q7) If w = log z, then find . Also determine where w is non-analytic.

A7)

Here we have

Therefore-

and

Again-

So that w is analytic everywhere but not at z = 0

Q8) Prove that the function is an analytical function.

A8)

Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Q9) Prove that

A9)

Given that

Since

V=2xy

Now

But

Hence

Q10) Show that polar form of C-R equations are-

A10)

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-

Q11) Prove Harmonic function theorem.

A11)

A function which satisfies the Laplace equation is known as a harmonic function.

Theorem- if f(z) = u + iv is an analytic function, then u and v are both harmonic functions.

Proof:

Suppose f(z) = u + iv, be an analytic function, then we have

Differentiate (1) with respect to x, we get

Differentiate (2) with respect to y, we get

Add 3 and 4-

Similarly-

So that u and v are harmonic functions.

Q12) Prove that and are harmonic functions of (x, y).

A12)

We have

Now

u/x = 2x, 2u/x2 = 2, u/y = -2y, 2u/y2 = -2

2u/x2 + 2u/y2 = 2 – 2 = 0

Here it satisfies Laplace equation so that u (x, y) is harmonic.

Now-

v = y/(x2 + y2), v/x = - 2xy/(x2 + y2)2

2v/x2 =

On adding the above results-

We get-

So that v(x, y) is also a harmonic function.

Q13) Find the harmonic conjugate function of the function U (x, y) = 2x (1 – y).

A13)

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

Q14) Solve where

A14)

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

Q15) Solve

A15)

Poles are

|z-0|=2

Poles 1 and -1 inside the circle

Q16) Solve the following by cauchy’s integral method:

f(n)(a) = n!/2πi

A16)

Given,

f(n)(a) = n!/2πi

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

= k!/2πi

= k!/2πi

= (k+1)!/2πi

Q17) Evaluate dz by using Cauchy’s integral formula.

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

A17)

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

Q18) Evaluate the integral given below by using Cauchy’s integral formula-

A18)

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

Q19) Evaluate dz if c is circle |z - 1| = 1.

A19)

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

Q20) Find out the zero of the following-

f(z) = (z – 2)/z2 sin 1/(z – 1)

A20)

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 ….)

Q21) Find the singularity of the function-

f(z) = sin 1/z

A21)

So that there is a number of singularity.

Sin 1/z is not analytic at z = a

(1/z = ∞ at z = 0)

Q22) Show that when 0<|z|<4

A22)

When |z|<4 we have

Q23) Find the residue of at z=1

A23)

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