Small questions regarding residue of $\frac{e^z}{\sin^2 z}$ at $z=k\pi$ Could someone check the correctness following and answer the small questions?

Calculate the residue of $$f(z) = \frac{e^z}{\sin^2 z}$$ at $z=k\pi \;(k \in \mathbb{Z})$

I classify $z=k\pi$ as a pole of order 2.
I find it difficult to use the limit calculation so I try to calculate the Laurent-expansion around $z=k\pi$.
Let $t=z-k\pi$ then $z=t+k\pi$ and $$f(t) = \frac{e^{t+k\pi}}{\sin^2 (t+k\pi)} = \frac{e^{k\pi}e^t}{\sin^2 t}$$
Question 1: can I write $f(t)$ or should I write $f(t+k\pi)$?
Laurent-expansion around $t=0$.
$$\begin{align}
\frac{e^{k\pi}e^t}{\sin^2 t} & = a_{-2}t^{-2}+a_{-1}t^{-1}+\ldots\\
e^{k\pi}\left(1+t+\frac{t^2}{2!}+\ldots\right) &= \left( t-\frac{t^3}{3!}+\ldots\right)^2 \cdot (a_{-2}t^{-2}+a_{-1}t^{-1}+\ldots)\\
e^{k\pi}\left(1+t+\frac{t^2}{2!}+\ldots\right) &= \left( t^2 -\frac{2t^4}{3!}+\ldots\right) \cdot  (a_{-2}t^{-2}+a_{-1}t^{-1}+\ldots)
\end{align}$$
Meaning $$e^{k\pi} = a_{-2}\qquad e^{k\pi}  = a_{-1}$$
Concludes $$e^{k\pi}  = \operatorname*{res}_{t=0} f(t) = \operatorname*{res}_{z=k\pi} f(z)$$
Question 2: if I would want to calculate $a_0$, is the following right?
$$\frac{e^{k\pi}}{2!} = a_0 -\frac{2}{3!}a_{-2}$$
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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There are, at least, two ways to find the residue:


*
*
\begin{align}
{\expo{z} \over \sin^{2}\pars{z}}&=
{\expo{k\pi} \over \sin^{2}\pars{z}}
+\dsc{\expo{k\pi}\pars{z - k\pi} \over \sin^{2}\pars{z}}
+{\expo{k\pi}\pars{z - k\pi}^{2} \over 2\sin^{2}\pars{z}} + \cdots
\\[5mm]&=
{\expo{k\pi} \over \sin^{2}\pars{z}}
+\dsc{\color{#00f}{\expo{k\pi}}
\bracks{{z - k\pi \over \sin\pars{z - k\pi}}}^{2}\,\color{magenta}{1 \over z - k\pi}}
+{\expo{k\pi}\pars{z - k\pi}^{2} \over 2\sin^{2}\pars{z}} + \cdots
\end{align}


*

*
\begin{align}
&\lim_{z\ \to\ k\pi}\totald{}{z}
\bracks{\pars{z - k\pi}^{2}\,{\expo{z} \over \sin^{2}\pars{z}}}
=\color{#00f}{\expo{k\pi}}\ \underbrace{\lim_{z\ \to\ 0}\totald{}{z}
\bracks{z^{2}\expo{z} \over \sin^{2}\pars{z}}}_{\dsc{1}}
\end{align}



$$
\color{#66f}{\large\,{\rm Res}_{\bracks{z\ =\ k\pi}}\bracks{%
{\expo{z} \over \sin^{2}\pars{z}}}} = \color{#66f}{\large\expo{k\pi}}
$$
A: Your approach is correct. Perhaps more direct is
$$
\begin{align}
\frac{e^{k\pi}(1+z+z^2/2+O(z^3))}{z^2(1-z^2/3+O(z^4))}
&=\frac{e^{k\pi}(1+z+5z^2/6+O(z^3))}{z^2}\\
&=\color{#C00000}{e^{k\pi}}\left(\frac1{z^2}+\color{#C00000}{\frac1z}+\frac56+O(z)\right)
\end{align}
$$
Thus, the residue at $z=k\in\mathbb{Z}$ is $e^{k\pi}$.
In the end, everything is about computing the coefficient of $z^{-1}$, which is what you've done.
A: You are making everything a bit too lenghty. Since $e^z$ is an entire function and $\sin^2(z)$ is a periodic function with period $\pi$, the residue is just $e^{k\pi}$ times the residue of $\frac{e^z}{\sin^2 z}$ at $z=0$, that is:
$$\frac{d}{dz}\left.\frac{z^2 e^z}{\sin^2 z}\right|_{z=0}=[z](1+z)=1$$
since both $\frac{z}{\sin z}$ and $\frac{z^2}{\sin^2 z}$ are $1+O(z^2)$ in a neighbourhood of the origin.
