How do I specify the singularities of $f(z):=\frac{z}{(z^2+2)^2(z^2-3z+2)}$ in $\Bbb{C}$?

Question

I have the following problem:

I need to specify all the singularities of the function $$f(z):=\frac{z}{(z^2+2)^2(z^2-3z+2)}$$ in $\Bbb{C}$.

We had the following definition:

We say that $f$ has an isolated singularity at $b$ if $f$ is analytic in $0<|z-b|<\epsilon$ for some $\epsilon>0$ and $f(b)$ is not defined. Write $$f(z)=\sum_{n\in \Bbb{Z}} a_n(z-b)^n$$Note the following:

a. if $a_n=0$ for $n<0$ then $f$ extends to be analytic at $b$ with $f(b)=a_0$. We say $b$ is a removable singularity

b. If $a_n=0$ for $n<n_0$ with $n_0>0$ and $a_{n_0}\neq 0$ we can write $$f(z)=(z-b)^{n_0} \sum_{n=0}^\infty a_{n_0+n}(z-b)^n$$and say $b$ is a zero of order $n$.

c. If $a_n=0$ for $n<-n_0$ with $n_0>0$ and $a_{-n_0}\neq 0$ then we can write $$f(z)=(z-b)^{-n_0} \sum_{n=0}^\infty a_{-n_0+n}(z-b)^n$$ and call $b$ a pole of order $n_0$

d. if $a_n\neq 0$ for infinitely many negative $n$ then $b is called an essential singularity.

What I don't understand is do I always first need to say if it is an isolated singularity or not and if yes then I can go through the points $a,...,d$? Or does isolated has nothing to do with the points $a,...,d$?

Because I my case I know that $$f(z)=\frac{z}{(z-2)(z-1)(z+\sqrt{2}i)^2(z-\sqrt{2}i)^2}$$ so I have some critical points in $B=\{0,1,2,\sqrt{2}i,-\sqrt{2}i\}$. But $f$ is not analytic in $0<|z-b|<\epsilon$ for some $\epsilon>0$ if $b\in B$. So I would say that non of this points are isolated singularities right? Now I intuitively would say that $2,1$ is a pole of order $1$ and $\sqrt{2}i,-\sqrt{2}i$ are poles of order $2$ in addition $0$ is a zero of order $1$. But I don't see how to work with the definition so I mean should I always write the Laurent series and then prove the definition?

Thanks for your help.

Answer 1

So I would say that non of this points are isolated singularities right?

No, all elements of $B$ are isolated points. More generally, any non-empty, finite subset of $\Bbb C$ consists of isolated points because you can draw discs around each point that do not contain any of the other points.

An example of a set that does contain a point that's not isolated is $\{1/n:n\in\Bbb N\}\cup\{0\}$. You cannot isolate 0 by any open neighborhood, because any open neighborhood will contain some (infinitely many, actually) numbers of the form $1/n$.

But $f$ is not analytic in $0<|z−b|<ϵ$ for some $ϵ>0$ if $b∈B$.

Not quite:

• $x=0$ is not a singularity.
• $f$ has poles at $1$, $2$, $\pm\sqrt{-2}$ and thus is not analytic there, but what you are saying that the singularities extend over a complete neighborhood, which is not the case.