Determine and classify all singular points Determine and find residues for all singular points $z\in \mathbb{C}$ for 
(i) $\frac{1}{z\sin(2z)}$ 
(ii) $\frac{1}{1-e^{-z}}$
Note: I have worked out (i), but (ii) seems still not easy.
 A: Related problems: (I), (II), (III). Just use the following test, a function $f(z)$ has a pole of order $m \in \mathbb{N} $ if
$$ \lim_{z\to z_0}(z-z_0)^m f(z) = c \neq 0, $$
where c is a finite number. For instance, in your case, $z=0$ is a pole of order $2$, since
$$ \lim_{z\to z_0}z^2 \frac{1}{z\sin(2z)} = \frac{1}{2}.$$ 
For finding residues see here.
A: Thank Mhenni Benghorbal for the hint! I have worked out (i) so far:
The singular points for (i) are $\frac{k\pi}{2}, k \in \mathbb{Z}$,
The case $k=0$ was justified by Mhenni,
For $k \neq 0$ the singular points are simple poles, since $$\lim_{z \to \frac{k\pi}{2}}\frac{z-\frac{k\pi}{2}}{z\sin(2z)}=\lim_{z \to \frac{k\pi}{2}}(-1)^k\frac{z-\frac{k\pi}{2}}{z\sin(2(z-\frac{k\pi}{2}))}=(-1)^k\frac{1}{k\pi}\neq0$$ which also gives the corresponding residues.
For (ii): CLearly the singularities are $2ki\pi, k\in\mathbb{Z}$, which are isolated poles (as their reciprocal corresponding to isolated zeros). 
But I have problem about computing the limit such as $$\lim_{z \to 0} \frac{z e^z}{e^z-1}$$.
Note that L-Hospitial's rule may not apply here, the method by series expansion is not applicable here either.?
