How do I calculate the residue of $\sin(z+1/z)$?

Question

How do I do this about $\displaystyle z=0$ ?.

I tried creating a Laurent expansion and extracting it from there but I wasn't sure how to isolate the $\displaystyle 1/z$ expression. $$ \mbox{I got}\quad \left(\, z + {1 \over z}\,\right) -{1 \over 6}\left(\, z + {1 \over z}\,\right)^{3} + \cdots $$

Any help is appreciated !.

Answer 1

The coefficient of $1/z$ in $z+\frac1{z}$ is $1$.

The coefficient of of $1/z$ in $\left ( z+\frac1{z}\right )^3 = 3$

In general, the coefficient of $1/z$ in $\left ( z+\frac1{z}\right )^{2 k+1}$ may be found from the expansion

$$\left ( z+\frac1{z}\right )^{2 k+1} = z^{2 k+1} + \binom{2 k+1}{1} z^{2 k-1} + \cdots +\binom{2 k+1}{k+1} \frac1{z} + \cdots \frac1{z^{2 k+1}} $$

so really, we want the sum

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)!} \binom{2 k+1}{k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (k+1)!} = J_1(2)$$

where $J_1$ is the Bessel function of the first kind of first order.

Answer 2

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The residue of $\ds{\sin\pars{z + {1 \over z}}}$ is given by:

\begin{align}&\color{#66f}{\large% \,{\rm Res}_{z\ =\ 0}\pars{\sin\pars{z + {1 \over z}}}}\ =\ \overbrace{\oint_{\verts{z}\ =\ 1}\sin\pars{z + {1 \over z}} \,{\dd z \over 2\pi\ic}}^{\dsc{z}\ \ds{\equiv}\ \dsc{\expo{\ic\theta}}} \\[5mm]&={1 \over 2\pi} \int_{-\pi}^{\pi}\sin\pars{2\cos\pars{\theta}}\expo{\ic\theta}\,\dd\theta ={1 \over \pi}\int_{0}^{\pi}\sin\pars{2\cos\pars{\theta}}\cos{\theta}\,\dd\theta =\color{#66f}{\large\,{\rm J}_{1}\pars{2}}\approx{\tt 0.5767} \end{align}

$\ds{\,{\rm J}_{\nu}}$ is a Bessel Function of the First Kind.