Solving for coefficients on a Laurent series I am having an issue with the following complex analysis problem. I am suppose to 
find the coefficients of $z^{-1}$, $z^{-2}$ and $z^{-3}$ in the  Laurent series for 
$\displaystyle \frac{1}{\sin z}$ around $z_0 = 0$ which is valid for $2\pi < |z| < 3\pi$.
One way I thought of doing it was to say 
$$ \frac{1}{\sin z} = \frac{1}{(z - 2\pi)(z - 3\pi)}\frac{(z - 2\pi)(z - 3\pi)}{\sin z} $$
Let $H(z) = \displaystyle \frac{(z - 2\pi)(z - 3\pi)}{\sin z} $.
Then we have 
$$ \frac{1}{\sin z} = H(z)\left[ \frac{A}{z - 2\pi} - \frac{B}{z - 3\pi} \right]
 = H(z)\left[ \sum_{k = 0}^\infty \frac{A(2\pi)^k}{z^{k + 1}} 
  + \sum_{k = 0}^\infty \frac{Bz^k}{(3\pi)^{k + 1}} \right]. $$
This seems to get me part of the way of where i need to go, but leaves me having
to find a series that works for $H(z)$. I was going to continue in this way, but I though
that perhaps it was giving me an entire series, and the problem seems to be suggesting
to only solve for 3 of the coefficients.
Another way to solve for the coefficients is 
$\displaystyle a_k = \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{(w - z_0)^{k + 1}} dw$, where $\gamma$ is a
circle, of say radius $R$, that is in the annulus. 
I tried to do this just for $a_{-1}$ and I got the following
$$ \int_0^{2\pi} \frac{Rie^{it}}{\sin(Re^{it})} dt = \int_{u(0)}^{u(2\pi)} \frac{1}{\sin(u)} du $$
where $u = Re^{it}$, so $u(0) = R$ and $u(2\pi) = R$. Thus I'm integrating from 
$R$ to $R$, so $a_{-1} = 0$.
However the $u$ substitution seems like something went wrong. I am not even sure if I can
actually do a $u$ substitution like that, but I can't see how to solve the integral any other way.
I am not sure what direction to go with this problem, and it seems like I am making it a lot
harder than it is suppose to be. Can anyone give me some direction on it?
Should I be attempting to solve the integrals, or do something like I was at the beginning?
 A: Using integrals is the way to go here. Let $\gamma$ be a path in the specified annulus with winding number 1. 
We have 
$$2\pi i a_{-1} = \int_\gamma \frac{1}{\sin z}\ dz.$$
We can evaluate this with the residue theorem. We have poles at 0, $\pm \pi$, and $\pm 2\pi$. Computing the residues, we see that the coefficient is 1. Finding the rest of the coefficients is very similar. 
A: Try not to memorize mathematics. Here is how you proceed. Check the order of the pole of your function $\frac{1}{\sin(z)}$ at the point where you want to find the expansion. In your case, the function has a pole of order one. That means, Laurent series has the form,
$$ \frac{1}{\sin(z)}=\frac{ a_{-1} }{z} + a_{0} + a_{1}z + a_{2} z^3+\dots $$
Look at the series, to get $a_{-1}$, you need to multiply both sides by $z$ and then take the limit as $z\to 0$. This gives $ a_{-1} = 1 $. To find $a_{0}$, do the following:
1) - multiply both sides by $z$, you get
$$ \frac{z}{\sin(z)} = a_{-1} + a_{0}z + a_{1} z^2 + a_{2} z^3 + \dots $$
2) - differentiate both sides w.r.t $z$ to kill the constant term $a_{-1}$ which gives
$$ \frac{d}{dz} \, \frac{z}{\sin(z)} = a_{0} + 2 a_{1}z + 3\,a_{2} z^2 + \dots\quad (1) $$
3) - take the limit as $ z\to 0 $ to both sides of the above equation, you get $a_{0}=0$.
To find $a_{1}$, just differentiate $(1)$ and take the limit as $z\to 0$, that is 
$a_{1}=\frac{1}{6}$. I think you can now find other coefficients.
