Looking at the function $f(z) = \frac{z^2+1}{z^2(z+1)}$, I have found the singularities to be at $z=0$ and $z=-1$.
My question is if they are removable. I expanded this into the Laurent series $\frac{z^2+1}{z^2}-\frac{z^2+1}{z}+(z^2+1)-z(z^2+1)+...$, but from here I am not completely clear on how to tell if the singularity is removable. I believe that we are supposed to find the common coefficient, and if that coefficient is $0$ for all terms then a singularity is removable. If so, what is the common coefficient and which singularity does that make removable? $z=-1$?
The singularities at $z=0$ and $z=-1$ are not removeable. They are pole singularities.
If a function $f$ has a removeable singularity at $z=z_0$, then there is an analytic function at $z_0$ that is equal to $f$ for all $z\ne z_0$.
A point $z_0$ is a pole of $f$ if there exists a function $g$ that is analytic at $z_0$, with $g(z_0)\ne0$, such that $f(z) = g(z) / (z − z_0)^n$ for some positive integer $n$.
The smallest $n$ for which $f(z)=g(z)/(z-z_0)^n$ is called the order of the pole. Here, the pole at $z=0$ is a second order pole while the pole at $z=-1$ is a first order pole. That is, we have
$$\lim_{z\to 0}z^2 \frac{z^2+1}{z^2(z+1)}=1$$
and
$$\lim_{z\to -1}(z+1)\frac{z^2+1}{z^2(z+1)}=2$$