Essential singularities of $z(\exp(1/z)-1)^{-1}$ I'm studying for a qualifying exam, so I'm trying to find efficient ways to answer difficult questions.  Here's an example: 
Find and classify the singularities of 
$$
f(z)=\frac{z}{e^{1/z}-1}
$$ Intuitively, $f$ should have an essential singularity at $z=0$ and $z=\infty$.  Is there an easy way to justify this? Computing the Laurent series seems difficult at first glance - or am I just being a wuss?    
 A: To me it seems it is certainly continuous at each point except $z=0$ and $\{e^{1/z}=1\}$. $\{e^{1/z}=1\}$ will give you poles, not an essential singularities. It remains to show that $z=0$ gives an essential singularity. Certainly it is an isolated point of discontinuity, so one need only show it is neither a pole, nor a removable discontinuity. To do so, find points $z_n$ arbitrarily close to $0$ with $|f(z_n)|$ bounded, and separately, show $f$ cannot have a limit at zero.
Edit : an alert user has pointed out that $z=0$ is not even an isolated singularity! Therefore, a priori it cannot be an essential singularity. So there cannot be any essential singularities of this function.
A: I believe that $z=0$ could be what is called a cluster point, a type of nonisolated singularity. So yes, you might not really need the Laurent series expansion, given that it does not exist. Here's the following link (jump to the section on nonisolated singularities): http://en.wikipedia.org/wiki/Isolated_singularity#Nonisolated_singularities 
I hope this helps.
