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

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

The equation of the circle is-

Or

Now for the lower semi-circle-

where C is |z + 3i| = 2

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-

A5)

where f(z) = cosz

           = by cauchy’s integral formula

                        =

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-

It is given that-

Now-

We know that, Taylor’s series-

So that

Hence

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

Here we have-

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

That means-

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)