Unit 4 | unit 4 complex variable differentiation

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Unit - 4

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

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) Find the conformal transformation of .

A14)

Let

Q15) Show that the mapping is conformal in the whole of the z plane.

A15)

Let z=x+iy

Then

Consider the mapping of the straight line x=a in z plane the w plane which gives which is a circle in the w plane in the anticlockwise direction similarly the straight line y=b is mapped into which is a radius vector in the w plane.

The angle between the line x=a and y=b in the z plane is a right angle. The corresponding angle in the w plane between the circle e = constant and the radius vector is also a right angle which establishes that the mapping is conformal.

Q160 Show that the curve u = constant and v = constant cut orthogonally at all intersections but the transformation w = u + iv is not conformal. Where-

A16)

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.

Q16) Find the bi-linear transformation which aps points z=2,1,0ontpo the points w=1,0,i

A17)

Let

Thus we have

Q17) How that the bilinear transformation w= transforms in the z-plane to 4u+3=0 in w-plane.

A18)

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

Q18) Theorem: If W=f(z) represents a conformal transformation of a domain D in the z-plane into a domain D of the W plane then f(z) is an analytic function of z in D. Prove this theorem.

A19)

Proof: We have u+iv=u(x,y)+iv(x,y)

So that u=u(x, y) and v=v(x,y)

Let ds and denote elementary arc length in the z-plane and w-plane respectively Then

Now

Hence

Where

Now is independent of direction if

Where h depends on x and y only and is not zero. Thus the conditions for an Isogonal transformation

And

The equation are satisfied if we get

Then substituting these values in 2 we get

Taking i.e.

Also

Hence

Similarly i.e.

The equation (4) are the well-known Cauchy -Reimann

Unit - 4

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

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) Find the conformal transformation of .

A14)

Let

Q15) Show that the mapping is conformal in the whole of the z plane.

A15)

Let z=x+iy

Then

Q160 Show that the curve u = constant and v = constant cut orthogonally at all intersections but the transformation w = u + iv is not conformal. Where-

A16)

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.

Q16) Find the bi-linear transformation which aps points z=2,1,0ontpo the points w=1,0,i

A17)

Let

Thus we have

Q17) How that the bilinear transformation w= transforms in the z-plane to 4u+3=0 in w-plane.

A18)

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

Q18) Theorem: If W=f(z) represents a conformal transformation of a domain D in the z-plane into a domain D of the W plane then f(z) is an analytic function of z in D. Prove this theorem.

A19)

Proof: We have u+iv=u(x,y)+iv(x,y)

So that u=u(x, y) and v=v(x,y)

Let ds and denote elementary arc length in the z-plane and w-plane respectively Then

Now

Hence

Where

Now is independent of direction if

Where h depends on x and y only and is not zero. Thus the conditions for an Isogonal transformation

And

The equation are satisfied if we get

Then substituting these values in 2 we get

Taking i.e.

Also

Hence

Similarly i.e.

The equation (4) are the well-known Cauchy -Reimann

Unit - 4

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

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) Find the conformal transformation of .

A14)

Let

Q15) Show that the mapping is conformal in the whole of the z plane.

A15)

Let z=x+iy

Then

Q160 Show that the curve u = constant and v = constant cut orthogonally at all intersections but the transformation w = u + iv is not conformal. Where-

A16)

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.

Q16) Find the bi-linear transformation which aps points z=2,1,0ontpo the points w=1,0,i

A17)

Let

Thus we have

Q17) How that the bilinear transformation w= transforms in the z-plane to 4u+3=0 in w-plane.

A18)

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

Q18) Theorem: If W=f(z) represents a conformal transformation of a domain D in the z-plane into a domain D of the W plane then f(z) is an analytic function of z in D. Prove this theorem.

A19)

Proof: We have u+iv=u(x,y)+iv(x,y)

So that u=u(x, y) and v=v(x,y)

Let ds and denote elementary arc length in the z-plane and w-plane respectively Then

Now

Hence

Where

Now is independent of direction if

Where h depends on x and y only and is not zero. Thus the conditions for an Isogonal transformation

And

The equation are satisfied if we get

Then substituting these values in 2 we get

Taking i.e.

Also

Hence

Similarly i.e.

The equation (4) are the well-known Cauchy -Reimann

Unit - 4

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

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) Find the conformal transformation of .

A14)

Let

Q15) Show that the mapping is conformal in the whole of the z plane.

A15)

Let z=x+iy

Then

Q160 Show that the curve u = constant and v = constant cut orthogonally at all intersections but the transformation w = u + iv is not conformal. Where-

A16)

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.

Q16) Find the bi-linear transformation which aps points z=2,1,0ontpo the points w=1,0,i

A17)

Let

Thus we have

Q17) How that the bilinear transformation w= transforms in the z-plane to 4u+3=0 in w-plane.

A18)

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

Q18) Theorem: If W=f(z) represents a conformal transformation of a domain D in the z-plane into a domain D of the W plane then f(z) is an analytic function of z in D. Prove this theorem.

A19)

Proof: We have u+iv=u(x,y)+iv(x,y)

So that u=u(x, y) and v=v(x,y)

Let ds and denote elementary arc length in the z-plane and w-plane respectively Then