What is the residue of $f(z)=e^{\frac{z-2}{z(z-1)}}$ at $z=0$?
I was trying to compute it using tricks shown here: Residue of $e^{1/z(z-1)}$ at $z=0$
I wrote $$\frac{z-2}{z(z-1)}=\frac1z-\frac{1}{z(z-1)}=\frac1z+2-\frac{z-1}{z}-\frac{z}{z-1}.$$ Thus we can write $$f(z)=e^{1/z+2}\sum_{n=-\infty}^{\infty} I_n(-2)(1-\tfrac{1}{z})^n,$$ where $I_n(x)$ are the modified Bessel functions of the first kind. However, writing $e^{1/z}$ as a series, we see that $I_n(-2)$ for negative $n$ can contribute towards the residue. So not sure if this helps.
Any help/suggestions are welcome.