Laurent Series of $f(z) = \sin {\frac{1}{z}}$ at $z=0$

I am attempting to find the Laurent expansion at $z=0$ for $$f(z) = \sin {\frac{1}{z}}$$ where $z$ is a complex number. I rewrote the function as $$f(z)=\frac{ e^{iz^{-1}} - e^{-iz^{-1}}}{2i}$$ $$=\frac{1}{2i} \sum_{n=0}^{\infty}[i^n-(-i)^n]\frac{z^{-n}}{n!} $$ What do I do next?

Solution 1:

Function $\sin z$ is holomorphic on $\mathbb{C}$, so its holomorphic in every punctured neighbourhood of $\infty$. Laurent series of $\sin z$ in $\mathbb{C}$ is $$ \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \ldots $$ and in particular it's the Laurent series of $\sin z$ for $z=\infty$. Thus the Laurent series of $\sin z^{-1}$ for $z=0$ is $$ \sin z^{-1} = z^{-1} - \frac{z^{-3}}{3!} + \frac{z^{-5}}{5!} - \frac{z^{-7}}{7!} + \ldots $$