Note: I am asking about both the singularities separately. I am not confused about their relation to each other etc. So my post essentially has two questions because both are asking about singularities occurring in complex analysis making two separate questions in this situation seemed weird.

An essential singularity is an isolated singularity such that the Laurent expansion about it has the maximum negative power of $(z-z_0)$ as $-\infty$. Does this basically mean that this singularity cannot be removed in any limit and that's why it is essential unlike poles or removable singularities which can be made finite with limits?

A non-isolated singularity is defined as a singularity around which one cannot draw a contour circle of arbitrarily small radius such that there are no other singularities. Basically saying that the neighborhood of a non-isolated singularity is dirty (having other singularities). I do not understand how can one use this definition to conclude that $$sin(1/z)$$ has a non-isolated singularity at $z=0$.

Isn't it undefined only exactly at $z=0$ and any arbitrary small variation $\delta \epsilon$ in $z$ will yield a finite value of $sin(1/z)$?

I am asking about non-isolated singularities in relation to a particular example only because I think it will help me get a deeper understanding of this concept in general.

Please feel free to answer in a more general setting that perhaps brings forth the intuition, definiton etc more clearly for both the concepts

  • 3
    $\begingroup$ who told you that $\sin (1/z)$ has a non-isolated singularity at zero? $\endgroup$
    – Conrad
    Commented Apr 28, 2022 at 18:29

1 Answer 1


An analytic function $f$ with domain $D$ has an isolated singularity at $z_0$ if:

  1. $z_0\notin D$;
  2. there is some $r>0$ such that, for every $z\in\Bbb C$, $0<|z-z_0|<r\implies z\in D$.

So, the definition of isolated singularity has nothing to do with Laurent series. And poles are isolated singularities. Besides,$$\begin{array}{ccc}\Bbb C\setminus\{0\}&\longrightarrow&\Bbb C\\z&\mapsto&\sin\left(\frac1z\right)\end{array}$$has an isolated singularity at $0$.

On the other hand$$\begin{array}{ccc}\Bbb C\setminus\left(\{0\}\cup\left\{\pm\frac1{n\pi}\,\middle|\,n\in\Bbb N\right\}\right)&\longrightarrow&\Bbb C\\z&\mapsto&\frac1{\sin\left(\frac1z\right)}\end{array}$$has a non-isolated singularity at $0$.


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