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!
 A: 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 $1/3$.
A: $$\begin{align}\frac{e^z}{z^3 \sin{z}} &= \frac1{z^4} \frac{\displaystyle 1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots}{\displaystyle 1-\frac{z^2}{6} + \frac{z^4}{120}+\cdots} \\ &= \frac1{z^4} \left (1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots \right )\left [1-\left (-\frac{z^2}{6} + \frac{z^4}{120}+\cdots\right )+ \left (-\frac{z^2}{6} + \frac{z^4}{120}+\cdots\right )^2+\cdots\right ]  \end{align}$$
Now, we need to pluck the coefficient of $1/z$ from the above.  To this effect, you should be able to see that we do not need the squared term in the brackets on the RHS, as its first term is $O(z^4)$, which cancels out the $1/z^4$ in front of everything.  
Thus, the residue is simply
$$\frac16 + \frac16 = \frac13 $$
