Consider the function $$f(z)=\frac{z}{e^\frac{1}{z}-1}$$
I'm checking the points $z_0=\frac{1}{2n\pi i}$
Now, we can see that $\lim_{z \to \ z_0} f(z) =\infty$. Hence, I claim that these points are singularities of $f(z)$.
As we can see, for larger and larger values of $n$ the singularities tend to 'cluster' together, about $z_0=0$.
So, this is clearly a non-isolated singularity.
However, note that $\lim_{z \to \ z_0} (z-z_0)f(z)$ does exist. Hence, we have $z=z_0$ is a first order pole.
Does it mean that this is an example of a first order pole and a non-isolated singularity at the same time.
Or is it that, all the infinite first order poles cluster about $z=0$, and so, $0$ is a non-isolated singularity, but not a pole, while all the $z_0$ are first order poles.
Best Answer
Those poles are each isolated singularities: there is a small disc $D_n$ centered at $1/(2\pi ni)$ that contains no other point of that form, and $f(z)$ is analytic on $D_n$ except at its center, which is what an isolated singularity means.
Maybe you meant to ask if the origin is a non-isolated singularity. It is, but it is not a pole.
The function has different behavior as $z \to 0$ along the real axis (it tends to $0$ from both directions) or as $z \to 0$ along the imaginary axis (it does not tend to 0 since it gets very large around the points $1/(2\pi in)$. So as $z \to 0$, $|f(z)|$ does not have a finite or infinite limit.
If $g(z)$ is an analytic function on a punctured neighborhood of a complex number $a$, then the possible behavior of the function near $a$ has three mutually exclusive possibilities.
The value $|g(z)|$ is bounded on a punctured neighborhood of $a$, in which case the Riemann removable singularities theorem tells us $g$ has a limit, say $L$, as $z \to a$, and if we declare $g(a) := L$ then $g$ is analytic at $a$.
For some positive integer $m$, $|(z-a)^mg(z)|$ is bounded on a punctured neighborhood of $a$. If $m$ is the smallest such positive integer, then $(z-a)^mg(z)$ has a nonzero limit as $z \to a$ and $g$ has a pole of order $m$ at $a$.
For each positive integer $m$, $(z-a)^mg(z)$ is not bounded on any punctured neighborhood of $a$. In this case $a$ is called an essential singularity of $g(z)$.
The page here gives a viewpoint on these possibilities in terms of a Laurent expansion for $g(z)$ at $a$. Your function does not have an essential singularity at $z=0$ since the function is not analytic on a punctured neighborhood of $0$.