Let's sy I have something like $f(z)=\frac{z^6+1}{z^2+1}$. I know that the singularity is at $z=\pm i$ and that it ends up being a $\frac{0}{0}$ case, but is that an indication for what type of singularity it is? A similar example on a wiki page said that this type of case is a removable singularity and the case of $\frac{1}{0}$ like for $f(z)=\frac{1}{\sin(z)}$ is a simple pole.
Is that the only way by which one can tell which type of singularity it is?
I would appreciate any help.
Note that as long as $z \neq \pm i$, we have $$ f(z) = \frac{z^6+1}{z^2 + 1} = \frac{(z^2 + 1)(z^4 - z^2 + 1)}{z^2 + 1} = z^4 - z^2 + 1 $$ which means the singularities are very much removable.
You can't be too certain about the type of pole from a $\frac00$ expression, though. For instance, $\frac{z^3}z$ has a pole of degree $2$ at the origin.