Suppose $f$ is analytic and not constant on the domain $D \subseteq \mathbb{C}$.
If $z_0 \in D$ is a zero of $f$ of order $k$, show that the
function $\frac{f'(z)}{f(z)}$ has a simple pole at $z_0$ with residue
$k$.
I am not entirely sure how to manipulate the definition of a residue of orders greater than 1 in order to show this.
Best Answer
Hint: $f(z) = g(z) (z - z_0)^k$ where $g(z_0) \ne 0$ is analytic on $D$. What does this say about $f'(z)/f(z)$?