Finding an entire function such that $|f(z)| > e^{Im f(z)}$ and $f(0)=2$ & an analytic function that has an essential singularity at $z=0$

Question

Here are my problems:

1. Find an entire function $f(z)$ such that $|f(z)| > e^{\operatorname{Im} f(z)}$ for all $z \in \mathbb{C}$ and $f(0)=2$.

I am trying to guess $2\cos(z), 2e^z$, etc, but they are all failed in the end.

2. Let $f(z)$ be analytic in the punctured disk $0<|z|<1$, and $f(\frac 1n)=\sqrt{n}$ for every integer $n>1$. Prove that $f(z)$ has an essential singularity at $z=0$.

I am trying to use the definition of essential singularity to prove. But I don't know how to express the $f(z).$

Answer 1

1. You ask a lot: $2=|f(0)|< e^{Im f(0)}=e^0=1$. (For the new version: Have you tried $e^z+e^{-z}$?)

2. It would be a better idea to show that the singularity is neither removable nor a pole and then use the classification of singularities.

Answer 2

2) if there were a pole at zero $$ f(z)=\sum_{k=-N}^{\infty}a_kz^k, N>0 $$ then $f(1/n)\sim a_{-N}n^N$ but the assumption is $f(1/n)\sim n^{1/2}$