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.