How to tell the type of a singularity without knowing the Laurent series?

180 Views Asked by At

I'm trying to classify singularities. So I need to figure out whether they're an isolated or non-isolated singularity. If they're isolated singularities is it then an essential or removable singularity, or is it a pole of some order? I find it very hard to approach this problem when I'm working with transcendental functions, where I don't know or can't seem to find the Laurent series, or can't easily see the order of the pole, like with rational functions.

For example consider the two functions:

a) $$f(z)=\cot(1/z)-1/z$$ b) $$g(z)=\cos\left(\frac{1}{\sin(1/z)}\right) $$


My "solution"/approach

a) f(z) clearly has a singularity at $z=0$ and since $\cot(1/z)=\frac{\cos(1/z)}{\sin(1/z)} $, it also have singularities when $\sin(1/z)=0 \implies z_n=1/n\pi,\; n\in\mathbb{Z}\setminus\{0\}$.
I can then show that $z=0$ is not an isolated singularity since $z_n\rightarrow0$ when $n\rightarrow\infty$.

The singularities $z_n$ are isolated singularities. But how do I actually figure out, whether they're an essential singularity or a pole? I know it's a pole first order because I tried calculating the limit: $$\lim _{z \rightarrow \frac{1}{m \pi}}\left(z-\frac{1}{m \pi}\right) f(z)=-\frac{1}{m^2\pi^2}$$ But that was just a trial and error solution. It would be a very bad approach if it was an essential singularity or a pole of a higher order. Is there some other way I could tell that this was a simple pole? I couldn't find the Laurent/taylor series for cot(z) in my book, so in that case I would have to calculate the laurent series myself, which I believe is quite hard for this function.

b) $g$ has a non-isolated singularity at $z=0$ and isolated singularities at $z_n=1/n\pi$ using the same approach. But how do I figure out what type of singularity $z_n$ is? I have already checked that the limit $$\lim _{z \rightarrow \frac{1}{n \pi}}f(z)$$ does not exist, which suggests it cannot be a removable singularity.
and that the limit $$\lim _{z \rightarrow \frac{1}{n \pi}}\left(z-\frac{1}{n \pi}\right)^{m} f(z)$$ gives zero for $1\leq m\leq 3$.

This suggests that it is either a pole of higher order or an essential singularity. But even if the laurent series is hard to find for $f$, I believe it's more or less impossible to find for $g$. So how do I show it's an essential singularity, if that's even the case?