# Understanding a mistake regarding removable and essential singularity.

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 $z_0$.

• 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. – Natalio Sep 8 '18 at 21:51
• Ok but which step in my "proof" is wrong. – Natalio Sep 8 '18 at 21:55
• Have a look at the answer. – 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.
• But I picked $z\in B_\varepsilon(0)$ then show that $f$ is bounded in the open set of $B_\varepsilon(0)$. – Natalio Sep 8 '18 at 22:07
• $|z|<\epsilon$ doesn't imply $|f(z)|<|f(\epsilon)|.$ – mfl Sep 8 '18 at 22:08