Singularities of $\frac{z^2-\pi^2}{\sin(z)}$ and $\frac{z}{e^z-1}$ 
Determine all singularities and their types of this functions:
  $$f(z)=\frac{z^2-\pi^2}{\sin(z)} \qquad\text{and}\qquad g(z)=\frac{z}{e^z-1}$$


For the first one I thought this:
Obviously there is a singularity at $z_0=0$. To determine the typ I did this:
$$\begin{align*}
f(z)=\frac{z^2-\pi^2}{\sin(z)}&=(z^2-\pi^2)\left(-\frac{1}{z}+\frac{3!}{z^3}±\dots\right)\\
&= \left(z-\frac{\pi^2}{z}\right)+\left(\frac{3!}{z}-\frac{3!\pi^2}{z^3}\right)±\dots\end{align*}$$
Unfortunately I don't how to get further. I suppose that $z_0$ is a pole. 

For the second one I have no idea to evaluate the typ. Sure I know that the singularity is at $z_1=0$ and I put in the series expansion of $e^z$, but this didn't help me. 

Any hints? Thank you a lot!
 A: Singularities of 
$\frac{z^2-\pi^2}{\sin(z)}$  are given by $z=k\pi $ with $ k\in \mathbb Z$ which are poles except $z=\pm \pi. $ where the singularity is apparent(fake singularity or removable singularity). Indeed $\sin z =0 
$ iff $z=\pi k$. But 
$$\lim_{z\to \pm \pi} \frac{z^2-\pi^2}{\sin z } =-\lim_{z\to \pm \pi} \frac{(z-\pi)(z+\pi)}{\sin( z\mp\pi )} = \mp 2\pi.$$
and the same for
$\frac{z}{e^z-1}$ are given by $z=2i\pi k $ with $ k\in \mathbb Z$ which are poles except $z=0. $ where the singularity is apparent(fake singularity or removable singularity). $e^z=1
$ iff $z=2i\pi k$. but 
$$\lim_{z\to 0} \frac{z}{e^z -1 } =1$$
A: For $\frac{z}{e^z - 1}$, remember that 
\begin{align}
\frac{1}{az + bz^2 + \ldots} 
&= \frac{1}{a} \cdot \frac{1}{z + (b/a)z^2 + \ldots} \\
&= \frac{1}{az} \cdot \frac{1}{1 + (b/a)z + \ldots} \\
&= \frac{1}{az} \left( 1 - (b/a)z + \ldots \right)\\
\end{align}
at least when the series in the denominator (after the initial "1") converges to something whose modulus is less than $1$. 
A: Hints: 1.  We have $\lim_{z \to 0} z f(z)=- \pi^2.$
Can you now deduce that $0$ is a simple pole of $f$ ?


*We have $\lim_{z \to 0}g(z)=1$.


Can you now deduce that $0$ is a removable singularity of $g$ ?
