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 is analytic in c.

Hence, by Cauchy's formula, $\int_{c} \cfrac{f(z)}{(z-z_{o})^{n}} dz = \cfrac{2 \pi i}{(n-1)!} f^{n-1}(z_{o})$

$\int_{c} \cfrac{dz/(z+4)}{z^{3}} = \cfrac{2 \pi i}{(3-1)!} f^{3-1}(z_{o})= \cfrac{2 \pi i}{2!} f^{2}(z_{o}) = \pi i f''(z_{o}) = \pi i \cfrac{d^{2}}{dz^{2}} \left ( \cfrac{1}{z+4} \right)$

$= \pi i \cfrac{d}{dz} \left [ \cfrac{(z+4)(0)-(1)(1+0)}{(z+4)^{2}} \right ]$ at $z_{o}=0$

$= \pi i \cfrac{d}{dz} \left [ \cfrac{-1}{(z+4)^{2}} \right ]$ at $z_{o}=0$

$= \pi i \left [ \cfrac{(z+4)(0)-(-1)2(z+4)(1+0)}{(z+4)^{4}} \right ]$ at $z_{o}=0$

$= \pi i \left [ \cfrac{2(z+4)}{(z+4)^{4}} \right ]$ at $z_{o}=0$

$= \pi i \left [ \cfrac{2}{(z+4)^{3}} \right ]$ at $z_{o}=0$

$= \pi i \left [ \cfrac{2}{(4)^{3}} \right ]$

$\therefore \int_{c} \cfrac{dz}{z^{3} (z+4)} = \cfrac{2 i \pi}{64} = \cfrac{i \pi}{32}$