Contour integration in the complex plane gone wrong

Question

Considering a function of complex variable $z$: $$f(z)=\frac{e^z}{z}$$ and a contour integral: $$\oint_C dz f(z)$$ such that the contour $C$ encircles the origin counterclockwise, it is clear from the residue theorem that the result of the integration is: $$\oint_C dz f(z)=2\pi i ~\text{Res}_{z=0}[f(z)]=2\pi i$$ If we map the complex plane onto a punctured sphere, the contour $C$ can be continuously deformed to $C'$ such that it encircles the point at infinity clockwise. This suggests that there should be a residue at infinity which gives us the same result as the one at the origin. Performing the substitution: $$z=\frac{1}{w}~~~,~~~dz=-\frac{dw}{w^2}$$ we get $$-\oint_{C''}\frac{dw}{w^2}f\left(\frac{1}{w}\right)=-\oint_{C''}dw\frac{e^{1/w}}{w}$$ where $C''$ now is a small clockwise circle around the origin in $w$. However, instead of giving the same result as above, the residue seems to diverge in this case. Something even more strange happens if we consider: $$\oint_{C'''}dz \frac{e^z}{z(z-1)}=2\pi i(-1+e)$$ and then make the same kind of substitution here: $$-\oint_{C''''}dw \frac{e^{1/w}}{(1-w)}$$ where $C''''$ is now supposed to encircle the point $w=0$, but there isn't even a pole there! Evidently, something goes very wrong here. Please, point out my errors.

Answer 1

The Laurent series expansion of $e^{1/w}/w$ about $w=0$, using the Maclaurin series for $\exp$, is $\dfrac{1}{w} + \dfrac{1}{1! w^2} + \dfrac{1}{2! w^3} + \ldots$ and the residue, obtained as usual from the $w^{-1}$ term, is $1$.

Similarly for your other integral: $w = 0$ is not a pole but an essential singularity, and using the Taylor expansion of $\exp$ you can get the residue at $w=0$.