Unit-5

Complex Variable –Integration

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.

Properties of line integral-

2. Sense reversal-

3. Partitioning of path-

4. ML – inequality-

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

Example: Evaluate where c is the upper half of the circle |z – 2| = 3.

Find the value of the integral if c is the lower half of the above circle.

Sol.

The equation of the circle is-

Or

Now for the lower semi circle-

Key takeaways-

3. Sense reversal-

4. Partitioning of path-

5. ML – inequality-

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-

Key takeaways-

Cauchy’s integral formula-

Taylor’s series-

If f(z) is analytic inside a circle C with centre ‘a’ then for z inside C,

Laurent’s series-

If f(z) is analytic in the ring shaped region R bounded by two concentric circles C and of radii ‘r’ and where r is greater and with centre at’a’, then for all z in R

Where

And

Example: Expand sin z in a Taylor’s series about z = 0.

Sol.

It is given that-

Now-

We know that, Taylor’s series-

So that

Hence

Example: Expand f(z) = 1/ [(z - 1) (z - 2)] in the region |z| < 1.

Sol.

By using partial fractions-

Now for |z|<1, both |z/2| and |z| are < 1,

Hence we get from second equation-

Which is a Taylor’s series.

Example: Find the Laurent’s expansion of-

In the region 1 < z + 1< 3.

Sol.

Let z + 1 = u, we get-

Here since 1 < u < 3 or 1/u < 1 and u/3 < 1,

Now expanding by Binomial theorem-

Hence

Which is valid in the region 1 < z + 1 < 3

Example: Expand about the singularity z = 1 in Laurent’s series.

Sol.

Here to expand about z = 1, which means in powers of z -1, we put z – 1 = t or z = t + 1

So that-

Key takeaways-

2. Laurent’s series-

Where

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-

Zero of an analytic function-

The value of z is said to be zero of the analytic function f(z) when f(z) = 0.

If f(z) is analytic in the neighbourhood of z = a, then by Taylor’s theorem-

If , then f(z) is said to have a zero of order n at z = a.

The zero is said to be simple if n = 1.

for a zero of order m at z = a,

Thus in the neighbourhood of the zero at z = a of order n

Where is analytic and non-zero at and in the neighbourhood of z = a.

Example: Find out the zero of the following-

Sol. Zeroes of the function-

Key takeaways-

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

Residue at a pole-

If z = a is an isolated singularity of f(z) then f(x) can be expressed expanded in Laurent’s series about z = a

So that-

Note- the coefficient of which is is called the residue of f(z) at z = a and it is written as-

Since-

So that-

Method of finding residues-

Example: Find the residue of f(z) = z cos (1/z) at z = 0.

Sol.

Which is the Laurent’s series expansion about z = 0

So that-

By the definition of residue-

Residue of f(z) at z = 0 is = -1/2.

Example: Find the residue at z = 0 of the function-

Sol. Z = 0 is a pole of order 1.

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

Sol.

Poles are given by-

Z = 1 is a simple pole while z = 2 is a double pole.

Now

Residue of f(z) at simple pole (z = 1) is-

Residue of f(z) at double pole (z = 2) is-

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 multipUYle 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 =

Key takeaways-

2. Cauchy’s residue theorem-

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

Integral of the type -

Example: Evaluate .

Sol.

Here put

Then-

Where C is the circle |z| = 1

The pole of the integrand are the roots of which are-

Out of the two poles, here z lies inside the circle C.

Residue at z is-

By residue theorem-

So that-

Example: Evaluate where a > |b|.

Sol.

As we know that-

So that-

Or

Now on putting , we have-

Where c is the circle |z| = 1.

Residue at z = is-

By residue theorem-

Hence-

Example: Evaluate

Sol.

Over the closed contour C consisting of the real axis from –R to R and the semi-circle of radius R in the upper half of plane.

Now-

……(1)

Poles of

Are given by

Which means-

Residue of

Residue of

Now by Cauchy’s residue theorem-

2.

3. Also

So that-

……(4)

From the above four equations-

Hence-

References