I am working on this problem :
Suppose $z_0$ is any constant complex number interior to any simply closed curve contour C. Show that for all $n>1$, $∫ \frac{dz}{(z-z₀)^n} = 0$.
I know that that if $f$ is holomorphic on the simply connected domain $U$ where $U$ is a subset of $\mathbb{C}$ then $f$ is path-independent. I'm not exactly sure for cases like
$∫_{|z|=2} \frac{dz}{(z-1)^2}$. Here clearly $f(z) = \frac{1}{z-1}$ is analytic everywhere on $\mathbb{C} - \{1\}$ but there's a singularity inside the disc $|z|=2$. From a prior problem, I know that $∫_{C} \frac{dz}{z} = 2\pi i$ and from my understanding this was because $f(z) = \frac{1}{z}$ in this case had a singularity at $z=0$.