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! 

You had
$$\frac1{z^2\left(1-\frac{z^3}6+\mathcal O(z^4)\right)}=\frac1{z^2}\left(1+\frac{z^3}6+\frac{z^6}{36}+\ldots\right)=\frac1{z^2}+\frac z6+\ldots$$
so the residue is zero.
Also, using what you did and applying l'Hospital's rule:
$$\lim_{z\to 0}\frac{\sin z-z\cos z}{z^2}\stackrel{\text{l'H}}=\lim_{z\to 0}\frac{z\sin z}{2z}=0$$
