Solution
= Solution
Take the circle with the usual positive, counterclockwise orientation. The only singularity is at zero, and the <Laurent series> gives
$$
ze^{1/z}=\sum_{n=0}^\infty\frac{z^{1-n}}{n!}.
$$
The coefficient of $z^{-1}$ is $1/2$, so the <residue theorem> gives
$$
\boxed{\oint_Cze^{1/z}\,dz=2\pi i\cdot\frac12=\pi i.}
$$
The <essential singularity> at zero still has a well-defined <residue>; the theorem does not require it to be a <pole>.