$f$ 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.

$f$ has an essential singularity at $z_0$ if and only if neither of the following hold:
(1) $\displaystyle \lim_{z \rightarrow z_o} f(z) \in \mathbb{C}$ (removable singularity)
(2)  $\displaystyle \lim_{z \rightarrow z_o} |f(z)| = +\infty$ (pole)
Note that if $1/f$ satisfies (1) then $f$ satisfies (2) or (1) depending on whether the limit is 0 or not and if $1/f$ satisfies (2) then $f$ satisfies (1). So if $f$ has an essential singularity then so must $1/f$.
