Prove that if a function has an isolated singularity at $z_0$, then its derivative also has an isolated singularity at $z_0$

Question

Prove that if a function has an isolated singularity at $z_0$, then its derivative also has an isolated singularity at $z_0$. Find $Res(f',z_0)$.

My approach: Suppose $f$ has an isolated singularity at $z_0$. Then $f(z)=\sum_{k \in \mathbb{Z}} c_k (z-z_0)^k$ where $0<|z-z_0|<R$ for some $R>0$. Hence, $f'(z)=\sum_{k\in \mathbb{Z}} kc_k (z-z_0)^{k-1}=\sum_{k\in \mathbb{Z}} d_k (z-z_0)^{k}$. Hence, $f'(k)$ has an isolated singularity at $z_0$. Also, $Res(f',z_0)=d_{-1}=0 \cdot c_0 =0$.

My concern: How to prove $f'(z)$ has an isolated singularity at $z_0$. I can only write the Laurent series for $f'$ and don't know how to prove it.

Answer 1

You're over-complicating matters. $f$ is analytic in a deleted neighbourhood of $z_0$, therefore so is $f′$.