pole, order and residue I was practising for an exam and I had some trouble with the following excersice: 
$$f(z)= \frac{1}{z \sin z}$$ 
a. Find the pole and its order. 
$$\frac{1}{z(z-z^3/3!+ z^5/5! + \cdots)}= \frac{1}{z^2(1-z^2/3!+ z^4/5! + \cdots)}$$
So the pole $z=0$ has order 2, but what about the other poles? $n\pi$?
b. Find the residue in this pole of $f$.
$$\lim_{z \to 0} \frac{\mathrm d}{\mathrm dz} \frac{z}{\sin z}= \lim_{z \to 0} \frac{\sin z-z \cos z}{\sin^2z}$$ and now I don't know how to continue.
Thanks in advance! 

DonAntonio answered all but your second question for part (a), so I'll discuss that.
To find the order of the poles at $z = n \pi$, compute
$$
\lim_{z \to n \pi} \frac{\sin(z)}{z - n \pi} \, .
$$
If the limit is nonzero, that means it's a simple pole.  If the limit is zero, compute 
$$
\lim_{z \to n \pi} \frac{\sin(z)}{(z - n \pi)^k}
$$
for larger values of $k$ until you get a nonzero limit.
