CVCI2

$\oint_{|z|=5} ze^{3/z} dz \,$

There is a singularity at 0, so $\oint_{|z|=5} ze^{3/z} dz =2\pi i \mathrm{Res}_{z=0}\,$ .

In general, $\mathrm{Res}_{z=a}\,$ is the coefficient in front of the $\frac{1}{z-a}\,$ term in the Laurent expansion of the integrand.

$z e^{3/z} = z \sum_{k=0}^\infty \frac{(3/z)^k}{k!}\,$

When $k=2\,$ the term from the sum is $\frac{9}{2}z^{-1}\,$ so $\mathrm{Res}_{z=0}=\frac{9}{2}\,$ .

Finally,

$\oint_{|z|=5} ze^{3/z} dz = 2\pi i \frac{9}{2} = 9\pi i\,$