Determine all possible values of $\int_\limits{\gamma} \sin(\frac{1-z}{z})dz$ for a closed path $\gamma$ in $\mathbb{C}\setminus\{0\}$.
My thoughts on this exercise were that since $0$ is an essential singularity I can't calculate the residue through differentiation. But I know that the residue of a function is the $(-1)$-coefficient of the laurent series.
Therefore the value of the integral is $0$ for all paths which don't enclose the origin and $\nu_\gamma(0) c_{-1}$ for all other closed paths in $\mathbb{C}\setminus \{0\}$. Where $\nu_\gamma(0)$ is the winding number of $\gamma$ around $0$ and $c_{-1}$ is the $(-1)$-laurent coefficient.
I would appreciate an opinion on my thoughts or some help if I'm incorrect.
HINT:
$$\begin{align} \sin\left(\frac{1-z}{z}\right)&=\cos\left(1\right)\sin\left(\frac1z\right)-\sin\left(1\right)\cos\left(\frac1z\right)\\\\ &=\cos(1)\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!z^{2n+1}}-\sin(1)\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!z^{2n}} \end{align}$$