Find residue of $f(z) = \frac{\sin z}{(z^2+1)^2}$ at $z = \infty$

Question

Find residue of $f(z) = \dfrac{\sin z}{(z^2+1)^2}$ at $z = \infty$. Then this is the same as finding the residue at $z=0$ for $\dfrac{-1}{z^2}f(1/z)= \dfrac{-z^2 \sin 1/z}{(z^2+1)^2}$

$z = 0$ is a pole. But $z^n \sin 1/z \to \infty$ when $z \to 0, n = 1,2,3...$ Also the denominator is not $0$ when $z=0$. So I can't use famous formulas. Now, $-z^2\sin 1/z = \sum_{n = 0}^\infty {\dfrac{(-1)^{n-1}}{(2n+1)!z^{2n-1}}}$. But what I should do with $(z^2+1)^2$ I don't know. Also, the answer is $1/2e$ which suggests me that there is something tricky. Should I evaluate an integral over some circle? Also, I know that the residue at $z=\infty$ of $g(z) = \dfrac{\cos z}{(z^2+1)^2}$ is $0$, so $-\dfrac{1}{2\pi i}\int_{|z|=R}{g(z)dz}=0$, can this fact be of help? Any hints will be appreciated, thanks in advance.

Answer 1

The function $\frac{-z^2}{(z^2+1)^2}$ is holomorphic near $z=0$. So you can expand it as a power series at $z=0$: $$ \frac{-z^2}{(z^2+1)^2}=a_0+a_1z+a_2z^2+\cdots\tag{1} $$ where you can work out the coefficients by looking at the Taylor expansion of $\frac{z}{(z^2+1)^2}$ at $z=0$ and then multiply by $-z$.

Using the expansion of $\sin(z)$, you can write the Laurent series for $\sin(\frac1z)$ as $$ \frac1z-\frac{1}{3!z^3}+\frac{1}{5!z^5}-\cdots\tag{2} $$ Now multiply the two series (1) and (2) to find the coefficient of $\frac1z$: $$ (a_0-\frac{a_2}{3!}+\frac{a_4}{5!}-\frac{a_6}{7!}+\cdots)\frac1z $$

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To find out the exact coefficients, observe that: \begin{align} \frac{-z^2}{(1+z^2)^2} &= \frac{z}2(\frac{1}{1+z^2})' =\frac{z}2(1-z^2+z^4+\cdots)'\\ &=\frac{z}2(-2z+4z^3-6z^5+\cdots)\\ &=\frac12(-2z^2+4z^4-6z^6+\cdots) \end{align} So $$ \begin{align} (a_0-\frac{a_2}{3!}+\frac{a_4}{5!}-\frac{a_6}{7!}+\cdots) &=\frac12\left(\frac{2}{3!}+\frac{4}{5!}+\frac{5}{6!}+\cdots\right)\\ &=\frac12\left(\frac{3-1}{3!}+\frac{5-1}{5!}+\frac{6-1}{6!}+\cdots\right)\\ &=\frac12\left(\frac{1}{2!}-\frac{1}{3!} +\frac{1}{4!}-\frac{1}{5!} +\frac{1}{6!}-\frac{1}{7!} +\cdots\right)\\ &=\frac12\sum_{n=0}^\infty\frac{(-1)^n}{n!}=\frac{1}{2e}\;. \end{align} $$

Answer 2

Using $$ \frac{1}{(1+z^2)^2}=\sum_{n=1}^\infty(-1)^nnz^{2n-2},\sin \left(\frac1z\right)=\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{(2n-1)!z^{2n-1}} $$ one has \begin{eqnarray} \dfrac{-1}{z^2}f\left(\frac1z\right)&=&\dfrac{-z^2 \sin \left(\frac1z\right)}{(z^2+1)^2}=-\sum_{n=1}^\infty(-1)^nnz^{2n}\cdot \sum_{n=1}^\infty(-1)^{n-1}\frac{1}{(2n-1)!z^{2n-1}}. \end{eqnarray} Now you can obtain the coefficient of $\frac1z$ which is easy.