We know that a function $f : U\subset \mathbb{C}\rightarrow \mathbb{C}$ is said to have singularity at a point $z_0$ if $f(z)$ is not analytic at point $z_0$, although function may be analytic in some small neighborhood of $z_0$. Note that $U$ is any open subset of complex plane.

To check whether the function is analytic or not at the point, the point must be in the domain right? But while finding the singularity of function what we do is that we find those point at which function is not defined. Why do we do this ? Because by definition the points at which function is not defined those points are not domain. Hence why do we investigate for those points?

It might be that my question is very silly but I need to understand this.

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    $\begingroup$ The function's behavior near the bad point is of interest... $\endgroup$ Commented Jan 26, 2023 at 21:12
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    $\begingroup$ For example the radius of the Taylor series of the function at a point is the distance from that point to the closest singularity $\endgroup$
    – Conrad
    Commented Jan 26, 2023 at 22:23

1 Answer 1


Consider the function $f(z) = \frac{\sin(z)}{z}$.

This function is not defined at $z=0$, but it can be continuously extended to a new function $$g(z)= \begin{cases} f(z) \textrm{ if $z \neq 0$} \\ 1 \textrm{ if $z = 0$}\end{cases}$$

This new function $g$ is an entire function!

However, the function $f(z) = \frac{\cos(z)}{z}$ cannot be continuously extended to $z=0$. It is still interesting to investigate the nature of the singularity at $z=0$, such as by computing the Laurent expansion.

So, it is a bit of an abuse of language, but when we ask "is $z_0$ a singularity" for some point not in the domain of a function $f$, we are really asking about whether there is an analytic extension $g$ of $f$ with $z_0$ in the domain of $g$.

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    $\begingroup$ Or alternatively, you can say $f$ has a *removable singlarity"at $z_0$. $\endgroup$ Commented Jan 26, 2023 at 21:31
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    $\begingroup$ @RobertIsrael . +1. In the context of complex analysis, I think "removable singularity at $z_0$" usually means we can extend $f$ so that it is analytic and not merely continuous at $x_0$. E.g. let $\log$ be the principal branch of the complex logarithm. Let $f(z)=(1+z)\log (1+z)$ for $z\in dom(f)=\{z:Re(z)>-1\}.$ We can extend $f$ continuously to $z_0=-1$ with $f(z_0)=0$ but we cannot make $f$ analytic at $z_0.$ $\endgroup$ Commented Jan 26, 2023 at 22:23
  • $\begingroup$ @StevenGubkin Thank you for your answer. That's what I wanted to know. $\endgroup$ Commented Jan 27, 2023 at 8:17

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