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

Question

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.

Answer 1

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).