Let $f:S\to \mathbb{C}$ be analytic, $z_0 \in S$ be of order $k$ of $f$. Prove that there exists an analytic function $g$ satisfying $$\frac{f'(z)}{f(z)}=\frac{k}{z-z_0}+g(z)$$ $\forall z$ in $\varepsilon$-neighbourhood of $z_0$.
My proof:
Since $f$ is analytic at $z_0$, $\exists$ an analytic function $h(z)$ satisfying $f(z)=(z-z_0)^k h(z)$ on $N_\varepsilon(z_0)$. Then $f'(z)=k(z-z_0)^{k-1}h(z) + (z-z_0)^{k}h'(z)$ on $N'_\varepsilon(z_0)$. Hence, by division, $\frac{f'(z)}{f(z)}=\frac{k}{z-z_0}+\frac{h'(z)}{h(z)}$ on $N_\varepsilon(z_0)$. Set $g(z):=\frac{h'(z)}{h(z)}$, where $g(z)$ is analytic since $\frac{h'(z)}{h(z)}$ is analytic.
Please let me know if you think this proof is good enough?