Find the residues of $\frac{e^z-1}{\sin^2(z)}$ for $z=k \pi$, $k \in \mathbb{Z}$

I'm trying to find the residue of $\frac{e^z-1}{\sin^2(z)}$ for $z=k \pi$ around $z=k \pi$, $k \in \mathbb{Z}$. I know that the order of the pole is $2$, and I've therefore tried using the formula:

$$ \frac{1}{(2-1)!} \left((z-k \pi)^2 \frac{e^z-1}{\sin^2(z)}\right)^{(2-1)} = \left((z-k \pi)^2 \frac{e^z-1}{\sin^2(z)}\right)'.$$

But when I try this, all I get is:

$$\underset{z \rightarrow k \pi}{\text{lim}}\left((z-k \pi)^2 \frac{e^z-1}{\sin^2(z)}\right)' \\ = \underset{z \rightarrow k \pi}{\text{lim}}\left(\frac{2(z-k \pi)(e^z-1)+(z-k \pi)^2e^z}{\sin^2z} - \frac{(z-k \pi)^2(e^z-1)2 \cos z}{\sin^3z}\right).$$

And I don't really know what to do from here. Can anyone help me out?

Solution 1:

Near $k\pi$, you have\begin{align}e^z-1&=e^{k\pi}e^{z-k\pi}-1\\&=e^{k\pi}-1+e^{k\pi}(z-k\pi)+\frac{e^{k\pi}}2(z-k\pi)^2+\frac{e^{k\pi}}{3!}(z-k\pi)^3+\cdots\end{align}and\begin{align}\frac{\sin^2(z)}{(z-k\pi)^2}&=\frac{\sin^2(z-k\pi)}{(z-k\pi)^2}\\&=1-\frac13(z-k\pi)^2+\cdots\end{align}So, if$$(z-k\pi)^2\frac{e^z-1}{\sin^2(z)}=\frac{e^z-1}{\frac{\sin^2(z)}{(z-k\pi)^2}}=a_0+a_1(z-k\pi)+a_2(z-k\pi)^2+\cdots,\tag1$$you have\begin{multline}e^{k\pi}-1+e^{k\pi}(z-k\pi)+\frac{e^{k\pi}}2(z-k\pi)^2+\cdots=\\=\left(1-\frac13(z-k\pi)^2+\cdots\right)\left(a_0+a_1(z-k\pi)+a_2(z-k\pi)^2+\cdots\right)=\\=a_0+a_1(z-k\pi)+\left(a_2-\frac{a_1}3\right)(z-k\pi)^2+\cdots\end{multline}and therefore $a_1=e^{k\pi}$. But it follows from $(1)$ that$$\frac{e^z-1}{\sin^2(z)}=\frac{a_0}{(z-k\pi)^2}+\frac{a_1}{z-k\pi}+a_2+\cdots$$and therefore$$\operatorname{res}_{z=k\pi}\left(\frac{e^z-1}{\sin^2(z)}\right)=a_1=e^{k\pi}.$$