TITLE: Singularities of $ \frac{z-1}{z^2 \sin z} $
QUESTION [5 upvotes]: Find all singularities of $$
\
\frac{z-1}{z^2 \sin z}
\
$$
Determine if they are isolated or nonisolated. 
This is not hard, it is $z = 0$ and $z = k\pi$.
But how do I:
For isolated singularities, determine if they are removable or nonremovable
and, if nonremovable, determine their order.
Do I need to expand this to a power series? If so, I have no idea where to attack...
Please help! Thanks!

REPLY [3 votes]: Since for any $\,k\in\Bbb Z-\{0\}\,$ we have that
$$\lim_{z\to k\pi}(z-k\pi)\frac{z-1}{z^2\sin z}\stackrel{\text{L'Hospital}}=\lim_{z\to k\pi}\frac{z-1}{2z\sin z+z^2\cos z}=$$
$$=\frac{k\pi -1}{2k\pi \sin k\pi+k^2\pi^2\cos k\pi}=\frac{(-1)^k(k\pi -1)}{k^2\pi^2}$$
all these singularities are isolated (in fact, simple poles).
About the case $\,k=0\,$:
$$\lim_{z\to 0}z^3\frac{z-1}{z^2\sin z}=\lim_{z\to 0}\frac{z}{\sin z}(z-1)=-1$$
Thus we have here a pole of order $\,3\,$