Looking for assistance with the following function..
i need to calculate the Residues of $$f(z)= \sin\left(\frac{z^2+1}{z}\right)$$
regrading my way:
$$\sin\left(\frac{z^2+1}{z}\right)=\sin(z)\cos(\frac{1}{z})+\cos(z)\sin(\frac{1}{z}) $$
$$= \left(z-\frac{z^3}{6}+...\right)\left(1-\frac{1}{2z^2}+...\right)+\left(1-\frac{z^2}{2}+...\right)\left(\frac{1}{z}-\frac{1}{6z^3}+...\right) $$
look on the coefficient of $$\frac{1}{z}$$
we get
$$ -\frac{7}{12}$$
it seems right?
Thanks!
I would do it this way. $$\begin{align} \sin\left(\frac{z^2+1}{z}\right)&=\sin\left(z+\frac1z\right)\\ &=\frac1{2i}\left(e^{i(z+1/z)}-e^{-i(z+1/z)}\right) \end{align}$$ Now, $$\begin{align} e^{i(z+1/z)}&=e^{iz}e^{1/z}\\ &=\sum_{n=0}^\infty\frac{(iz)^n}{n!}\sum_{m=0}^\infty\left(\frac iz\right)^m\frac1{m!}\\ \end{align}$$ so that $$\operatorname{Res}_{z=0}e^{iz}e^{i/z}=\sum_{n=0}^\infty\frac{i^n}{n!}\frac{i^{n+1}}{(n+1)!}=i\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}$$
A similar calculation gives $$\operatorname{Res}_{z=0}e^{-iz}e^{-i/z}=-i\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}$$ so that $$\operatorname{Res}_{z=0}\sin\left(\frac{z^2+1}{z}\right)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}=J_1(2)$$ where $J_1$ is the Bessel function of the first kind of order $1$.
Numerical computation gives approximately $0.5767248077568736$.
