# $f$ 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.

## 3 Answers

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 $z_0$."

$f$ 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 $1/f$.

• Logan M, may you help here please? – BCLC Aug 6 '18 at 16:06
• Wow! Such a compact, clear, and concise answer. – StammeringMathematician Jan 22 at 16:36

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 $1/f$.