CVCI2

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\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.

ze^{{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\,

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