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$.

  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ Now I understand, thank you very much. $\endgroup$ – Natalio Sep 8 '18 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.