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It is a well known fact that $0$ is an essential singularity of the function $e^{1/z}$ therefore it is not essential nor a pole.

On the other hand we know the Rieamann's continuation theorem which states that an holomorphic function $f:U\setminus \{z_0\}\to \mathbb{C}$ has a removable singularity at $ z_0$ if and only if $f(z) $ is bounded in some neighborhood of $ z_0$.

I'm confussed because of the following calculation: If we pick $\varepsilon>0 $ and $ z$ such that $|z|<\varepsilon $ then $$ |z e^{1/z} |< \varepsilon \sum^\infty_{k=0} \left|\frac{1}{k!z^k} \right|=\varepsilon \sum^\infty_{k=0} \frac{1}{k!\varepsilon^k} =\varepsilon e^{1/\varepsilon}$$ The previous means that $ 0$ is a removable singularity of $ ze^{1/z}$ which implies that $ 0$ is a pole of order 1 of $ e^{1/z}$ which cannot be since the well known fact of the begining.

Off course I'm wrong in something but I couldn't tell in what. Thank you all in advance

A (possibly) useful definition My definition of a pole is: $z_0 $ is a pole of order $ m$ of the holomorphic function $f:U\setminus \{z_0\}\to \mathbb{C}$ if $m $ is the minimum integer such that $(z-z_0)^mf(z) $ has a removable singularity at $ z_0$.

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    $\begingroup$ Sure, but how what I wrote is wrong. I mean, you say "it is wrong because it is not true: take the example $ (0,1)$" but I want to know where is the proof I posted wrong. $\endgroup$ – Natalio Sep 8 '18 at 21:51
  • $\begingroup$ Ok but which step in my "proof" is wrong. $\endgroup$ – Natalio Sep 8 '18 at 21:55
  • $\begingroup$ Have a look at the answer. $\endgroup$ – mfl Sep 8 '18 at 22:07
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The first equality in the displayed equation is wrong. You replaced $|z|$ by $\epsilon$, but $|z|$ can be arbitrarily small, and hence the summands can be arbitrarily large.

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  • $\begingroup$ But I picked $ z\in B_\varepsilon(0)$ then show that $ f$ is bounded in the open set of $ B_\varepsilon(0)$. $\endgroup$ – Natalio Sep 8 '18 at 22:07
  • $\begingroup$ $|z|<\epsilon$ doesn't imply $|f(z)|<|f(\epsilon)|.$ $\endgroup$ – mfl Sep 8 '18 at 22:08
  • $\begingroup$ @Natalio: I'm afraid I don't follow you. You asked for the specific step in your "proof" that's wrong. I pointed out the specific step. Please refer to that specific step if you disagree. $\endgroup$ – joriki Sep 8 '18 at 22:09
  • $\begingroup$ You are helping me a lot because you pointed that step but I don't understant why the step you pointed out is wrong. $\endgroup$ – Natalio Sep 8 '18 at 22:11
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    $\begingroup$ Now I understand, thank you very much. $\endgroup$ – Natalio Sep 8 '18 at 22:30

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