Evaluate $\int_{c} \cfrac{dz}{z^{3} (z+4)}$ where c is the circle $|z|=2$.

Solution:

$|z|=2$ is a circle with center at the origin and radius 2.

There are two poles at z=0 and z=-4.

When pole z=0 lies inside the circle and pole z=-4 lies outside the circle.

$\therefore z^{3}$ gives the highest order of pole, i.e., n=3.

$\therefore$ We take, $f(z)=\cfrac{1}{z+4}$ which …