Residue of $f(z)=\frac{e^{1/z}\cosh(\frac{\pi z}2)}{z^2+1} $ at $z=0$

Question

The exercise asks me to calculate $$\int_{\vert z-1\vert=2} \frac{e^{1/z} \cosh(\frac{\pi z}2)}{z^2+1} dz $$ I tried to use the residue theorem and I looked for the singularity of the function and I found that $z=\pm i$ are removable singularities and $z=0$ is an essential singularity; Because of that the integral is equal to $2\pi i\operatorname{Res}_f(0)$ but I can’t find a way to calculate this residue. I tried to use the Laurent series and find that $$f(z)=\sum_{n=0}^\infty \frac 1{n!} z^{-n} \sum_{k=0}^\infty \frac{z^{2k}}{(2k)!}\left(\frac{\pi}2\right)^{2k} \sum_{l=0}^\infty (-1)^l z^{2l} = \sum_{j=-\infty}^{+\infty} C_j z^j$$ where $$C_j=\sum_{2k+2l-n=j} \frac 1{n!} \frac 1 {(2k)!} \left(\frac {\pi}2\right)^{2k}(-1)^l$$ here i tried to calculate $C_{-1}$ without success. Please help me