Supoose that
$f(z)=\frac{e^\frac{1}{z}}{1-2z}$.
Show that
$\oint_{\lvert z \rvert =1} f(z)dz = -\pi i$.
Now, I used the Laurent series of $e^\frac{1}{z}$ and the Taylor series of $\frac{1}{1-2z}$ to compute the $a_{-1}$ and then by the multiplication of the two series we have that the $a_{-1}$ is a series which is basically $\frac{e^2-1}{2}$, which is the residue at zero. Also, the residue at half is just $e^2$. But if we add them and multiply by $2\pi i$ then we will not get the desired result. I cannot figure out what is my mistake. Any help would be greatly appreciated.