# In which case can the term "singularity" and "isolated singularity" be used interchangeably?

I am right now going through complex analysis with the end of goal being able to use Cauchy's residue theorem. I understand that a point $$z$$ in the complex plane will be termed a "singularity" if a function $$f(z)$$ fails to be analytic at the point. But when will the same $$z$$ be termed "isolated singularity" ? I keep getting confused with this. For instance, consider this example from Section 74 of Churchill & Brown, 8th edition where the author says that the function $$f(z) = \frac{z+1}{z^2 +9}$$ has an isolated singularity at $$z=3i$$. My question is, why is $$z=3i$$ an isolated singularity? If I have to eventually use residue theorem on a certain function, my understanding says that I would follow these steps:

1. Identify if $$f(z)$$ has singularities.
2. Check if the singularities are isolated.
3. If the singularities are isolated, what category do they fall under?;removable, pole of order $$n$$, a simple pole or essential.

If I have to go by the definition of isolated singularity, then I would have to check it's region of convergence or specifically, find the deleted neighbourhood. Would this be a rigorous approach of doing things? feel like I am missing this part in my study.

I am referring to Zill & Shanahan and Churchill & Brown to study complex analysis because I am in engineering. Both the literature sources use these terms casually without justifying why a singularity would be termed isolated.

• It's an isolated singularity because a rational function is holomorphic except at the zeros of the denominator. You probably already have relevant results about, say, when the quotient of holomorphic functions is holomorphic. Commented Apr 9 at 12:41

## 1 Answer

A function may have multiple singularities, or even infinitely many singularities.

If you can find a small ball around a singularity which doesn't contain any other singularities, then that singularity is isolated. Said differently, it means there isn't any sequence of singularities which has that singularity as a limit.

In particular, if there are only finitely many singularities, then they are all isolated because the minimum distance between any two of them is a positive number (say $$\epsilon$$), so an open disk of radius $$\epsilon$$ centered at any of them doesn't contain any other of them.

This can fail if there are infinitely many singularities. For example, if a function has singularities at $$0,\frac12,\frac13,\frac14,\ldots$$, then the singularities $$\frac1n$$ are isolated (the disk of radius $$\frac1n-\frac1{n+1}$$ centered at $$\frac1n$$ contains no singularity other than $$\frac1n$$), but the singularity $$0$$ is not isolated since $$\lim\limits_{n\to\infty} \frac1n = 0$$ (every disk centered at $$0$$ contains $$\frac1n$$ for $$n$$ sufficiently large).