Not an essential singularity

Question

The following is an old qualifying exam that I cannot solve:

Suppose $f$ is holomorphic on the punctured unit disk and $ \frac{f '}{f} $ has a simple pole at $0$. Show that $f$ does not have an essential singularity at $0$.

Answer 1

By considering a smaller punctured disk if necessary, we may assume that $f$ has no zeros.

The crucial point is to recognise that

$$\operatorname{Res}\left(\frac{f'}{f};0\right) = k \in \mathbb{Z}.$$

Because

$$\operatorname{Res}\left(\frac{f'}{f};0\right) = \frac{1}{2\pi i}\int_{\lvert \zeta\rvert = \rho} \frac{f'(\zeta)}{f(\zeta)}\,d\zeta$$

is the winding number of the closed path $t \mapsto f\left(\rho e^{it}\right),\: t\in [0,2\pi]$ around $0$.

Once you know that, consider $g(z) = z^{-k}f(z)$. You have

$$\frac{g'(z)}{g(z)} = \frac{-kz^{-k-1}f(z) + z^{-k}f'(z)}{z^{-k}f(z)} = -\frac{k}{z} + \frac{f'(z)}{f(z)},$$

hence $\frac{g'}{g}$ has a removable singularity in $0$, and thus there is a holomorphic $h$ in the disk with $h' = \frac{g'}{g}$. Thus $g(z)\cdot e^{-h(z)}$ is constant, and by adding a suitable constant to $h$, we can assume

$$g(z) = e^{h(z)}.$$

But the right hand side is holomorphic on the entire disk, so $g$ has a removable singularity in $0$, and $f(z) = z^kg(z) = z^ke^{h(z)}$ has either a removable singularity ($k > 0$) or a pole ($k < 0$) in $0$, and not an essential singularity.