Non-isolated singularity of $\frac{1}{z^2(1+e^{1/z})}$ I was doing a practice question and the solution provided indicated that the function $$\frac{1}{z^2(1+e^{1/z})}$$ Had an non-isolated singularity at the origin. If there was a singularity there I can see why it is non-isolated but I am not convinced there is even a singularity there. This is because the denominator seems to go to $\infty$ as $z\rightarrow 0$. Who is right me or the answer provided, please can you explain either way.
 A: Just for fun, here is a plot of $|f(z)|$ in the region $\Re(z), \Im(z) \in [-1/b, 1/b]$ for $b > 1$.  Note that the vertical axis is automatically rescaled in this animation, so as we "zoom" into a neighborhood of $0$, the height that is being plotted varies somewhat.  We can clearly see that the singularity at $z = 0$ is not isolated owing to a progression of singularities along the imaginary axis.

A: It's a non-isolated singularity. The $1+e^{1/z}$ term will be zero whenever $e^{1/z} = -1$.
Now the exponential function along the imaginary axis is $e^{j\phi} = \cos\phi + j\sin\phi$ and therefore $e^{j(2n+1)\pi} = -1$. So if $z_n=1/(j(2n+1)\pi)$ we have that $1+e^{1/z}=0$ and such can be found arbitrarily near the origin. That is we have singularities at $z_n$ and $\lim_{n\to\infty} z_n = 0$.
The concept of singularity is maybe not so simple to define as there are some various situations that should be covered - they are not only poles. The cases are (ordered from least severe to most severe case)


*

*Removable singularity: a case where you can define a value in the point that makes the singularity go away, for example $z/z$ at origin.

*Pole: a singularity that can be handled by multiplying with a polynomial, for example $1/z$ at the origin. This case means that the limit will be $\infty$.

*Essential singularity: an isolated singularity that is not able to be handled by multiplying with polynomial, $e^{1/z}$ is of this kind, they are characterized by the Laurent series having infinite number of negative terms.

*Non-isolated singularity: a situation when you have a sequence of singularities $z_n$ and a limit of this $z = \lim z_n$, your function is an example of this. Since it's non-isolated you can't have a laurent series around (just) that singularity.


The two last cases will have that the function will oscillate wildly near the singularity instead of going to infinity. 
