Residue $\frac{e^z}{z^3\sin(z)}$ I want to find the residue of $$\frac{e^z}{z^3\sin(z)}$$ and get $$ \frac 1 {3!} \lim_{z \to 0} \left( \frac{d^3}{dz^3} \left(\frac{ze^z}{\sin(z)} \right)\right) = \frac{1}{3}$$
Can anyone confirm this?  I tried using the Laurent Series, but I didn't know how to compute it.
Oh, I was able to compute the Laurent Series and confirm it was $\frac{1}{3}$
I don't know how to close this question though!

Near zero,
$$ e^z = 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + O(z^4). $$
Also,
$$\sin{z}=z-\frac{1}{6}z^3 + O(z^5),$$
so the denominator is
$$ z^{-4} \left(1-\frac{1}{6}z^2+ O(z^4)\right)^{-1} = \frac{1}{z^4}\left( 1-\frac{z^2}{6} +O(z^4) \right), $$
using the binomial theorem.
Hence the Laurent series has principal part
$$ \frac{e^z}{z^3\sin{z}} = \frac{1}{z^4}\left( 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + O(z^4) \right)\left( 1+\frac{z^2}{6} +O(z^4) \right) \\
  = \frac{1}{z^4} + \frac{1}{z^3} + \frac{2}{3z^2} + \frac{1}{3z} + O(1), $$
and the residue is $1/3$.
