Can we have a non-isolated pole? 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.
 A: 
Hence, we have $z = 1/(2\pi i n)$ 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?

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 $0$.
A: By definition, a pole must be an isolated singularity. (We define three types of isolated singularities: removable, pole, essential.) As you noted, $z = 0$ is not isolated, so we do not classify it as a pole.
The other singularities are first order poles, which can be seen by considering the Laurent series expansion about a given singularity.
A: Meromorphic functions are ratios of analytic functions.   Hence their singularities are isolated. For such functions,  and at such points,  there will be a Laurent series (sometimes more than one,  depending upon the radii).
Since, as you have correctly noted,  $0$ is not an isolated singularity of $\dfrac z{e^{\frac 1z}-1}$, the function is not meromorphic there.  Hence no Laurent series.
Each of the $\dfrac 1{2\pi ki}$ is an isolated singularity,  otoh.

There are two types of non-isolated singularities that can occur for complex functions of a single variable:

*

*cluster points

*natural boundaries

See the discussion over here.
