I have to evaluate the following integral:
$$I=\oint_C\frac{s^3+2s}{(s-z)^3}\;ds$$
Case 1: $C$ is an arbitrary closed curve oriented counter-clockwise and $z$ is within $C$.
I'll rewrite as follows:
$$I=\oint_C\frac{f(s)}{(s-z)^3}\;ds$$
I immediately notice the following:
$$I=f''(z)=\frac{2!}{2\pi i}\oint_C\frac{s^3+2s}{s-z}\;ds$$
Using Cauchy Integral Formula:
$$I=\pi i\cdot 6z=6\pi i z$$
However, I'm also to consider a second case where $z$ lies without $C$. I know the answer is supposed to be $0$, but I can't find anything in my book on when this happens and why it happens. Any insight would be much appreciated.