Find and classify all isolated singularities of $f(z)=\frac{3z^2-1}{z(z^2+1)}$ Given
$$f(z)=\dfrac{3z^2-1}{z(z^2+1)}$$
I also have to find, classify, and compute the residue at each singularity. So, the first thing I've done is expand this to get a better view of whats going on
$$f(z)=\frac{3(z+\dfrac{\sqrt{3}}{3})(z-\dfrac{\sqrt{3}}{3})}{z(z-i)(z+i)}$$
From this, we obviously have singularities, I'm still a bit confused on the classifications of them. I think this is only when we have undefined function values, not necessarily a $\frac{0}{0}$ situation. Even still, we have problems at $z=\pm i, 0$. We get $0$ when $z=\pm\frac{\sqrt{3}}{3}$, but I don't think those count as singularities. Looking at the $z=\pm i, 0$ solutions, we cannot factor and reduce any of them, so they're not removable. I don't know how to distinguish between poles and essential.
From there, calculating the residues of each one is also a bit confusing. I think that's done by computing this limit
$$\lim_{z\to z_0}f(z)(z-z_0)$$ for each $z_0$. In this case I believe would just be the three, $\pm i, 0$?
 A: Let $\displaystyle f(z)=\frac{3z^{2}-1}{z(z^{2}+1)}$,
Finding the singularities:

*

*$\displaystyle f(z)=\frac{3z^{2}-1}{z(z-i)(z+i)}$ so singularities at $z=0, z=i$ and $z=-i$.

Classifying each singularities:
Using the following fact:

$z_{0}$ is an isolated singularity and $\displaystyle \lim_{z\to 0}(z-z_{0})^{n}f(z)=A\not=0$ so $z_{0}$ is a pole of order $n$ and when $n=1$, $z_0$ is a simple pole.


*

*$\boxed{z_0=0}:$ $\displaystyle \lim_{z\to 0}(z-0)^{1}\left(\frac{3z^{2}-1}{z(z-i)(z+i)}\right)=-1\not=0$ so $z_{0}=0$ is a simple pole.


*$\boxed{z_1=i}:$ $\displaystyle \lim_{z\to i}(z-i)^{1}\left(\frac{3z^{2}-1}{z(z-i)(z+i)}\right)=2\not=0$ so $z_{1}=i$ is a simple pole.


*$\boxed{z_2=-i}:$ $\displaystyle \lim_{z\to -i}(z+i)^{1}\left(\frac{3z^{2}-1}{z(z-i)(z+i)}\right)=2\not=0$ so $z_{2}=-i$ is a simple pole.
Note that since $0,\pm i$ they are simple poles so cannot be essential singularities. Is there singularities at infinity?
Residue at each singularity:
Using the following fact:

If $z_{0}$ is a simple pole, then $\displaystyle {\rm Res}(f,z_{0})=\lim_{z\to z_{0}}(z-z_{0})f(z)$.


*

*$\boxed{z_{0}=0}:$ $\displaystyle {\rm Res}(f,0)=\lim_{z\to 0}(z-0)f(z)=-1$.


*$\boxed{z_{1}=i}:$ $\displaystyle {\rm Res}(f,i)=\lim_{z\to i}(z-i)f(z)=2$.


*$\boxed{z_{2}=-i}:$ $\displaystyle {\rm Res}(f,-i)=\lim_{z\to -i}(z+i)f(z)=2$.
