Where is my mistake in classifying singularity and using power series and general singularity questions

Question

I have a few questions and they are also easy to ask in the context of the following example,

so I was asked to classify the singularity at $z=0$

of $$f(z)=z^{2}/(e^{z}-1)$$

My first thought was that the limit as z goes to $0$ is $0$ by L'Hospital rule.

So first question, does that automatically rule out that $z=0$ is a pole? My thought is yes.

So we could either have a removable , or a essential singularity.

Now here is the second part of my question

I know $$e^{z}=1+z+z^{2}/2!+z^{3}/3!+O(z^{4})$$

so

$$e^{z}-1=z+z^{2}/2!+z^{3}/3!+O(z^{4})$$

So $$f(z)=\frac{z^2}{z^{2}/2!+z^{3}/3!+(O(z^{4}))}$$

but that doesn't get me much because I know that dividing power series in general we want to avoid.

But now, looking at $f(z)$, is there something that an experienced eye can simply note which gives away which type of singularity we have? My thoughts are that it is removable, but I am not sure if that is true, and if so how to prove it.

So any advice?

Update: The book claims this is an essential singularity. Now im confused

Thanks

Answer 1

Don't forget that you can take $\;z\;$ very close to zero and since we're only interested to know how the Laurent (or power) series around that point behaves, we can "chop off" many summands and use, for example, the development for geometric series:

$$\frac{z^2}{e^z-1}=\frac{z^2}{\left(1+z+\frac{z^2}2+\ldots\right)-1}=\frac{z^2}{z\left(1+\frac z2+\ldots\right)}=$$

$$z\left(1-\frac z2+\frac{z^2}4-\ldots\right)=z-\frac{z^2}2+\ldots\implies$$

we have that $\;z=0\;$ is a removable singularity as we get above a power series (or, if you prefer, the function's principal part of its Laurent series about that singularity is zero).

I've found along the years this method very useful to evaluate residues, kinds of singularities, etc. for many functions.