If $C$ is a closed path oriented in the positive direction and $$g(z_0)=\int_C \frac{z^3+2z}{(z-z_0)^3}$$ show that $g(z_0)=6\pi iz_0$ when $z_0$ is in interior of $C$ and $g(z_0)=0$ when $z_0$ is out of $C$
I could show the first part, using it $$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}}$$
anyone can help in the second part?
If $z_0$ lies outside $C$, then the integrand is analytic inside and on $C$. Hence, by Cauchy's theorem, the integral is $0$.