If $f'/f$ has a pole of order 1 at $z=0$, then $f$ either has a pole or zero at $z=0$

Question

Problem: Suppose that a function $f$ is analytic in $\{z\in\mathbb{C}\ :\ 0 <|z|<r\}$ for some $r>0$, and that $f'/f$ has a pole of order one at $z=0$. Prove that then $f$ has a pole or a zero at $z=0$.

To arrive at a contradiction, suppose that $f$ neither has a zero nor a pole at $z=0$.

If $f$ has a removable singularity at $z=0$. Then $f'$ also has a removable singularity at $z=0$ and since $f(0)\ne0$, we get that $1/f$ is analytic near $z=0$. Hence $f'/f$ cannot have a pole of order one at $z=0$.

If $f$ has an essential singularity at $z=0$. Then $f'$ also has an essential singularity at $z=0$.

I am not sure how to proceed here and finish this argument. Any suggestions?

Answer 1

Here is a small hint:
Choose a small enough $s\lt r/2$ such that $f$ has no zeros or poles in the domain $\{z\in C|0\lt |z|\lt 2s\}$. Use Rouche's theorem and Residue theorem to compute the integral $\int_{|z|=s} \frac{f'(z)}{f(z)}dz$.