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Given any meromorphic function $f: U \to \mathbb{C}, U \subseteq \mathbb{C}$ open, I want to prove or disprove the statement: for any isolated singularity $a \in U$, the function $g(z) = e^{f(z)}$ has the same kind of singularity (removable, pole or essential) at $a$.

I think all removable singularities simply get carried over, because, well, if $f$ can be holomorphically enhanced to include $a$, so can $g$.

For poles though, I'm not that sure anymore, although my guess would be that they also get carried over. If $a$ is a pole, then there exists a meromorphic function $h: U \to \mathbb{C}$ (that's defined at $a$) with $f(z) = (z - a) h(z)$, with $m$ the order of $a$. Therefore, $e^f = e^{(z-a)} e^{h}$. I don't know how to continue from there (I was aiming to reach something like $e^f = (z-a) h_2$ for a holomorphic function $h_2$ to show that $a$ is a pole for $e^f$ too), but I rather believe that poles get carried over too.

I don't know how to approach essential singularities though; if there's a function where things "can go wrong", then I believe it's with them. I haven't been able to come up with a counter-example though, so far. If I wanted to prove that essential singularities get carried over aswell, I would for example need to show that the image regarding $e^f$ each open set around $a$ is dense in $\mathbb{C}$. I don't know how to approach that, though. If the original statement is false, of course, then a simple counter-example would suffice.

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Hint: What kind of singularity does $e^{1/z}$ have at $0?$

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  • $\begingroup$ An essential one, because $e^{\frac{1}{z}} = 1 + 1/z + \frac{1}{z^2 2} + ...$ has infinitely many "negative" coefficients in the Laurent expansion? But $\frac{1}{z}$ only has a pole at $0$. Can it be disproven that easily? $\endgroup$ – moran Jan 13 '16 at 19:58
  • $\begingroup$ @moran Yes.$\,\,\,$ $\endgroup$ – zhw. Sep 9 '17 at 20:04

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