TITLE: $f$ has an essential singularity in $z_0$. What about $1/f$?
QUESTION [14 upvotes]: 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.

REPLY [13 votes]: Your two examples essentially span all the possibilities. By the Big Picard Theorem, we know that $f$ assumes all but one value in $\Bbb{C}$ infinitely often, in any neighborhood of $z_0$. 
If the missed value is $0$, then $\frac{1}{f}$ has an isolated singularity at $z_0$, and it must clearly be essential (since otherwise $f$ itself would have a pole or removable singularity).
If the missed value is not $0$, or if there is no missed value, then $\frac{1}{f}$ will have a sequence of poles that converges to $z_0$, and hence the singularity at $z_0$ will not be isolated. So, thinking of $\frac{1}{f}$ as a holomorphic function on its largest possible domain (which will omit some sequence of points converging to $z_0$), the singularity at $z_0$ is technically unclassifiable.
On the other hand, even in this case you can regard $\frac{1}{f}$ as a meromorphic function from $\Omega \backslash \{z_0\}$ to the extended complex plane $\hat{\Bbb{C}}$, and when considered as such it will have an essential singularity at $z_0$. This is a natural enough thing to do that I suspect most people would be happy just saying "$\frac{1}{f}$ has an essential singularity at $z_0$."