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Consider $f: \mathbb{C} \to \bar{\mathbb{C}}$, $z \mapsto \exp\left(\frac{1}{z}\right)$. Is it analytic in a neighborhood of $0$?

I feel like it should be (since $z \mapsto \frac{1}{z}$ is a chart, and $z \mapsto \exp(z)$ is analytic near $0$), but I lack confidence because I never seriously studied Riemann surfaces before.

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A meromorphic function only has poles? – copper.hat Nov 6 '12 at 2:02
@copper.hat If you say so :) – Alexei Averchenko Nov 6 '12 at 2:07
Analyticiy of $z\mapsto \exp(z)$ at $0$ only gives you that $f$ can be extended to $\overline C\setminus\{0\}$. – Jonas Meyer Nov 6 '12 at 2:07
Alexei, sorry, I was sloppy. What I meant was is that a meromorphic function is analytic except at some set (which has no limit point), and it only has poles at these points. The point being that it cannot have an essential singularity. – copper.hat Nov 6 '12 at 2:48
up vote 3 down vote accepted

Note that your function is not even defined at $z = 0$. You can try to define it at $z = 0$ to be, say, $\infty$, but this is an essential singularity, so you won't get a continuous extension. So it can't be holomorphic.

In fact, a holomorphic function $f : \mathbb{C} \rightarrow \bar{\mathbb{C}}$ (as a map between Riemann surfaces) is the same as a meromorphic function or the constant map $z \mapsto \infty$.

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