Finding the Laurent expansion of $\frac{1}{\sin^3(z)}$ on $0<|z|<\pi$? How do you find the Laurent expansion of $\frac{1}{\sin^3(z)}$ on $0<|z|<\pi$? I would really appreciate someone carefully explaining this, as I'm very confused by this general concept! Thanks 
 A: There are a few ways to do this. One way that works here is to take the expansion for $1/sin(z)$ and cube it.
However, I think it's more instructive to look at how to invert power series in general. The first observation is that to invert a power series (around 0, for simplicity), it needs to have a nonzero constant term, or we can't invert it because it won't have an inverse around 0 (this is the inverse function theorem). The second observation is that we can invert any power series with a constant coefficient, by the following method (where we normalize the constant to 1, for simplicity):
Write $g(z)=1+a_1z+a_2z^2+...$ . We seek $1/g(z)$, or a function $f(z)=1+b_1z+b_2z^2+...$ such that $f(z)g(z)=1$. Writing out this last equation we have:
$$1=(1+a_1z+a_2z^2+...)(1+b_1z+b_2z^2+...)$$
From there is it an easy matter to multiply out and solve for the coefficients recursively - first write out the terms that give you the coefficient of $z$, solve for $b_1$, then write out the expression for $z^2$ in the product and use your value of $b_1$ to solve for $b_2$, and so on.
First, we must write out $sin^3 (z)$ as a power series. I leave it to you to show that this is $z^3 - (1/2)z^5+...$. Now we want
$$\frac{1}{z^3 - (1/2)z^5+...}=\frac{1}{z^3}\frac{1}{1-(1/2)z^2+...}$$
You should be able to invert the second term and multiply out to get your answer.
For further information (and more rigor), consult Serge Lang's book on complex analysis, where he gives a detailed treatment of power series. 
A: use this formula 
$$\sum _{k=1}^{\infty } (-1)^{3 k} \left(-\frac{x^3}{\pi ^3 k^3 (\pi  k-x)^3}-\frac{x^3}{\pi ^3 k^3 (\pi  k+x)^3}+\frac{3 x^2}{\pi ^2 k^2 (\pi  k-x)^3}-\frac{3 x^2}{\pi ^2 k^2 (\pi  k+x)^3}-\frac{x^3}{2 \pi  k (\pi  k-x)^3}-\frac{x^3}{2 \pi  k (\pi  k+x)^3}+\frac{x^2}{(\pi  k-x)^3}-\frac{x^2}{(\pi  k+x)^3}-\frac{\pi  k x}{2 (\pi  k-x)^3}-\frac{3 x}{\pi  k (\pi  k-x)^3}-\frac{\pi  k x}{2 (\pi  k+x)^3}-\frac{3 x}{\pi  k (\pi  k+x)^3}\right)+\frac{1}{x^3}+\frac{1}{2 x}=\csc ^3(x)$$
