$f$ has an essential singularity in $z_0$. What about $1/f$? Let $\Omega$ be a non-empty, open subset of $\mathbb C$. Consider an holomorphic function $f:\Omega \setminus \{z_0\} \to \mathbb C$ and suppose we know $z_0$ is an essential singularity of $f$. 
I am wondering what we can say about the function $\tilde{f}:=\frac{1}{f}$ and its singularity in $z_0$. Do you know any theorem that answers to this question?
Actually, I can't prove anything, since I do not know the answer: I've studied some examples. For instance, if you take $f(z)=e^{\frac{1}{z}}$ then $\tilde{f}$ will still have an essential singularity, hasn't it? 
On the other side, if we take $f(z)=\sin(\frac{1}{z})$ then I think that $z_0=0$ becomes a limit point of poles for $\tilde{f}$ (so we can't classify it, because it isn't an isolated singularity). 
Wha do you think? Do you know any useful theorem concerning this? Thank you in advance.

So we are given essential singularity of $f$ in $z_0$.
If the function $g=1/f$ have a pole in $z_0$ of order $m$, then $f=1/g$ will have a removable singularity. In particular this will be zero of order $m$. Contradiction, hence $z_0$ is an essential singularity of $1/f$.
