Module 4

Functions of complex variable

Functions of a Complex variable-

In the narrow sense of the term, the theory of function of a complex variable is the theory of analytic functions (cf. Analytic function) of one or several complex variables. As an independent discipline, the theory of functions of a complex variable took shape in about the middle of the 19th century as the theory of analytic functions.

Complex function-

x + iy is a complex variable which is denoted by z

If for each value of the complex variable z = x + iy in a region R, we have one or more than one values of w = u + iv, then w is called a complex function of z.

And it is denoted as-

w = u(x , y) + iv(x , y) = f(z)

Neighbourhood of

Let a point in the complex plane and z be any positive number, then the set of points z such that-

||<ε

Is called ε- neighbourhood of

Limit of a function of a complex variable-

Suppose f(z) is a single valued function defined at all points in some neighbourhood of point -

The-

Example-1: Find-

Sol. Here we have-

Divide numerator and denominator by , we get-

Continuity- A function w = f(z) is said to be continuous at z = , if

Also if w = f(z) = u(x , y) + iv(x , y) is continuous at z = then u(x , y), v(x , y) are also continuous at z = .

Differentiability-

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.

Example-2: if f(z) is a complex function given below, then discuss

Sol. 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.

Analytic functions-

A function is said to be analytic at a point if f is differentiable not only at but an 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.

Important note-

1. If a function is analytic in a domain D, then u and v will satisfy Cauchy-Riemann conditions.

2. C-R conditions are necessary but not sufficient for analytic function.

3. C-R conditions are sufficient if the partial derivative are continuous.

State and prove sufficient condition for analytic functions

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

Show that is analytic at

Ans 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

Example-1: If w = log z, then find . Also determine where w is non-analytic.

Sol. 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

Example-2: Prove that the function is an analytical function.

Sol. Let =u+iv

Let =u and =v

Hence C-R-Equation satisfied.

Example-3: Prove that

Sol. Given that

Since

V=2xy

Now

But

Hence

Example-4: Show that polar form of C-R equations are-

Sol.

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-

Theorem; The necessary condition for a function to be analytic at all the points in a region R are

(ii)

Provided,

Proof:

Let be an analytic function in region R.

Along real axis

Then f’(z), becomes-

Along imaginary axis

From equation (1) and (2)

Equating real and imaginary parts

Therefore-

and

These are called Cauchy Riemann Equations.

Example-1: Check whether the function w = sin z is analytic or not.

Sol. Here-

Now-

And

Here we see that C-R conditions are satisfied and partial derivatives are continuous.

Example-2: Prove that the real and imaginary parts of the function w = log z satisfies C-R equations.

Sol.

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.

In case of a complex function f(z) the path of the definite integral can be along any curve from z = a to z = b.

In case the initial point and final point coincide so that c is a closed curve, then this integral called contour integral and is denoted by-

If f(z) = u(x, y) + iv(x, y), then since dz = dx + i dy

We have-

It shows that the evaluation of the line integral of a complex function can be reduced to the evaluation of two line integrals of real function.

Example: Evaluate along the path y = x.

Sol.

Along the line y = x,

Dy = dx that dz = dx + i dy

Dz = dx + i dx = (1 + i) dx

On putting y = x and dz = (1 + i)dx

Example: Evaluate where c is the circle with center a and r. What is n = -1.

Sol.

The equation of a circle C is |z - a| = r or z – a =

Where varies from 0 to 2π

Dz =

Which is the required value.

When n = -1

If a function f(z) is analytic and its derivative f’(z) continuous at all points inside and on a closed curve c, then

Proof: Suppose the region is R which is closed by curve c and let-

By using Green’s theorem-

Replace by and by -

So that-

Example-1: Evaluate where C is |z + 3i| = 2

Sol.

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-

Example 2:

where C =

Sol.

where f(z) = cosz

= by cauchy’s integral formula

=

Example 3:

Solve the following by cauchy’s integral method:

Solution:

Given,

=

=

=

Cauchy’s integral formula-

Cauchy’s integral formula can be defined as-

Where f(z) is analytic function within and on closed curve C, a is any point within C.

Example-1: Evaluate by using Cauchy’s integral formula.

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

Sol. It is given that-

Find its poles by equating denominator equals to zero.

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

So that-

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

Example-2: Evaluate the integral given below by using Cauchy’s integral formula-

Sol. 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-

Example-3: Evaluate if c is circle |z - 1| = 1.

Sol. Here we have-

Find its poles by equating denominator equals to zero.

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

So that-

A point at which a function f(z) is not analytic is known as a singular point or singularity of the function.

Isolated singular point- If z = a is a singularity of f (z) and if there is no other singularity within a small circle surrounding the point z = a, then z = a is said to be an isolated singularity of the function f (z); otherwise it is called non-isolated.

Pole of order m- Suppose a function f(z) have an isolated singular point z = a, f(z) can be expanded in a Laurent’s series around z = a, giving

…… (1)

In some cases it may happen that the coefficient , then equation (1) becomes-

Then z = a is said to be a pole of order m of the function f(z).

Note- The pole is said to be simple pole when m = 1.

In this case-

Working steps to find singularity-

Step-1: If exists and it is finite then z = a is a removable singular point.

Step-2: If does not exists then z = a is an essential singular point.

Step-3: If is infinite then f(z) has a pole at z = a. The order of the pole is same as the number of negative power terms in the series expansion of f(z).

Example: Find the singularity of the function-

Sol.

As we know that-

So that there is a number of singularity.

is not analytic at z = a

(1/z = ∞ at z = 0)

Example: Find the singularity of

Sol.

Here we have-

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

That means-

Example: Determine the poles of the function-

Sol.

Here we have-

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

We get-

By De Moivre’s theorem-

If n = 0, then pole-

If n = 1, then pole-

If n = 2, then pole-

If n = 3, then pole-

Cauchy’s residue theorem-

If f(z) is analytic in a closed curve C, except at a finite number of poles within C, then-

Proof:

Suppose be the non-intersecting circles with centres at respectively.

Redii so small that they lie within the closed curve C. Then f(z) is analytic in the multiple connected region lying between the curves C and

Now applying the Cauchy’s theorem-

Example: Find the poles of the following functions and residue at each pole:

and hence evaluate-

where c: |z| = 3.

Sol.

The poles of the function are-

The pole at z = 1 is of second order and the pole at z = -2 is simple-

Residue of f(z) (at z = 1)

Residue of f(z) ( at z = -2)

Example: Evaluate-

Where C is the circle |z| = 4.

Sol.

Here we have,

Poles are given by-

Out of these, the poles z = -πi , 0 and πi lie inside the circle |z| = 4.

The given function 1/sinh z is of the form

Its poles at z = a is

Residue (at z = -πi)

Residue (at z = 0)

Residue (at z = πi)

Hence the required integral is =

If is analytic in a closed curve c except at a finite number of poles within c then

[Sum of residue at the pole within c]

Example: Evaluate the following integral using residue theorem

Where c is the circle..

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

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

Hence by residue theorem

Example: Evaluate where c;|z|=4

Sol.

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

Example: Evaluate : c is the unit circle about the origin.

Sol.

=

This shows that z=0 is a pole of order 2 for the function and the residue of the poles is zero(coefficient of 1/z)

Now the pole at z=0 lies within c

Example: Evaluation of definite integrand

Show that

Sol.

I=

Real part of

Now I= =

Putting z= where c is the unit circle |z|=1

I=

Now f(z) has simple poles at and z=-2 of which only lies inside c.

Residue at is

=

=

Now equating real parts on both sides we get

I=

Prove that

Solution 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 )

Example: Evaluate

Sol.

Consider

Where c is the closed contour consisting of

Real axis from

Large semicircle in the upper half plane given by |z|=R

The real axis -R to and

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

Example: Prove that

Sol.

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

If a function f(z) is analytic and its derivative f’(z) continuous at all points inside and on a closed curve c, then

Proof: Suppose the region is R which is closed by curve c and let-

By using Green’s theorem-

Replace by and by -

So that-

Example-1: Evaluate where C is |z + 3i| = 2

Sol.

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-

Example 2:

where C =

Sol.

where f(z) = cosz

= by cauchy’s integral formula

=

Example 3:

Solve the following by cauchy’s integral method:

Solution:

Given,

=

=

=