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-
5.7 Evaluation of real integrals of the type 
and 
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
