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 

use this formula 
$$\sum _{k=1}^{\infty } (-1)^{3 k} \left(-\frac{x^3}{\pi ^3 k^3 (\pi  k-x)^3}-\frac{x^3}{\pi ^3 k^3 (\pi  k+x)^3}+\frac{3 x^2}{\pi ^2 k^2 (\pi  k-x)^3}-\frac{3 x^2}{\pi ^2 k^2 (\pi  k+x)^3}-\frac{x^3}{2 \pi  k (\pi  k-x)^3}-\frac{x^3}{2 \pi  k (\pi  k+x)^3}+\frac{x^2}{(\pi  k-x)^3}-\frac{x^2}{(\pi  k+x)^3}-\frac{\pi  k x}{2 (\pi  k-x)^3}-\frac{3 x}{\pi  k (\pi  k-x)^3}-\frac{\pi  k x}{2 (\pi  k+x)^3}-\frac{3 x}{\pi  k (\pi  k+x)^3}\right)+\frac{1}{x^3}+\frac{1}{2 x}=\csc ^3(x)$$
