Complex Integral result proof

Question

Question: Let $f \colon D \to \mathbb C$ be a holomorphic function defined on an open subset $D$ of $\mathbb C$, and suppose that $z_0 \in D$ is a double zero of $f$. Show with justification that for sufficiently small $\epsilon>0$ we have $$ \int_{|z-z_0|= \epsilon} \frac{z-z_0}{f(z)}dz = \frac{4\pi i}{f''(z_0)}. $$

I´ve tried defining $f(z)$ as $f(z)=g(z)(z-z_{0})^2$ and then applying Cauchy's formula for the second derivative, replacing $f(z)$ as stated, but i couldn't reach the intended result.

If someone could guide me or enlight me on the process i would appreciate it.

Answer 1

To elaborate on @J.G.'s hint, you were almost there when you suggested writing $f$ as $(z-z_0)^2g(z)$ -- since $f$ has a zero of order $2$ at $z_0$, we have $f(z) = \frac{1}{2!}f''(z_0).(z-z_0)^2.g(z)$, where $g(z_0) = 1$ (and $f''(z_0) \neq 0$). Consider what you get if you just use this slightly finer description of $f$ near $z_0$ in the integral you are trying to evaluate.