Small questions regarding residue of $\frac{e^z}{\sin^2 z}$ at $z=k\pi$

Question

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?

$$\frac{e^{k\pi}}{2!} = a_0 -\frac{2}{3!}a_{-2}$$

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

$$ \color{#66f}{\large\,{\rm Res}_{\bracks{z\ =\ k\pi}}\bracks{% {\expo{z} \over \sin^{2}\pars{z}}}} = \color{#66f}{\large\expo{k\pi}} $$

Answer 2

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.

Answer 3

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.