# Determining type of singularity

Let $f$ be holomorphic in $\mathbb{C} \setminus \{0\}$.

Prove or disprove: If $|f(\frac{i}{n})| \ge n$ for all $n \in \mathbb{N}$, then $f$ has a pole in $0$.

The equation yields that $f$ is not bounded near $0$, from which I know it is not a removable singularity. So it's either a pole or an essential singularity.

I have tried to use Casorati-Weierstraß theroem, but I couldn't make any sense of it.

Define $f(z)=e^{\frac{i}{z}}$. Then for every $n\in\mathbb{N}$ you get $f(\frac{i}{n})=e^n\geq n$, just like you need. On the other hand for every $n\in\mathbb{N}$ we have $f(\frac{-i}{n})=e^{-n}$, so the limit of $f(\frac{-i}{n})$ when $n\to {\infty}$ is $0$. Hence the singularity of $f$ at zero is essential.

• Is the singularity in this case essential because the two limits from the two different directions do not even have the same value? – Friedrich Sep 19 '18 at 16:17
• @Clip Yes. If it had a pole, then the value of the limit must be $\infty$. – b00n heT Sep 19 '18 at 16:19
• This argument is good. Alternatively, we could show that in the above (counter-)example $f(z)=exp(\frac{i}{z}) = \sum_{k=0}^{\infty} \left( \frac{i^k}{k!}\right)z^{-k}$. Since the principal part of the Laurent series has an infinite number of summands, the singularity in $0$ is essential. – Friedrich Sep 19 '18 at 16:31
• @Clip yes, that's also correct. I usually prefer not to calculate Laurent series if possible, but it's also a way to check the type of singularity. – Mark Sep 19 '18 at 16:34

Hint: If you had one such $f,$ then $f(z)\cos (2\pi i/z)$ would also exhibit this behavior.

• I don't know how this helps my case. – Friedrich Sep 19 '18 at 16:13
• @Clip Does $f(z)\cos (2\pi i/z)$ look like it has a pole at $0?$ – zhw. Sep 19 '18 at 16:16
• Now I understand that $cos(2πi/z)$ has an essential singularity as $z$ approaches zero. – Friedrich Sep 19 '18 at 16:20