# Compute using residuals the integral of the following function over the positively oriented circle |z|=3, f(z)=(e^-z)/(z^2)

Compute using residuals the integral of the following function over the positively oriented circle $|z|=3$

My solution: The only singular point of $f$ in $|z|\le 3$ is $z=0$ (double pole) and its remainder is therefore
${\mathrm{Res}}_{z=0}f\left(z\right)=\underset{z\to 0}{lim}\frac{1}{\left(2-1\right)!}{\left(\frac{{e}^{-z}{z}^{2}}{{z}^{2}}\right)}^{\mathrm{\prime }}=\underset{z\to 0}{lim}-{e}^{-z}=-1$
Consequently, ${\int }_{|z=3|}f\left(z\right)=2\pi i{\mathrm{Res}}_{z=0}f\left(z\right)=-2\pi i.$
this right?
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RamPatWeese2w
That is correct, although it is simpler, in order to compute the residue, to note that
$\frac{{e}^{-z}}{{z}^{2}}=\frac{1-z+\frac{1}{2!}{z}^{2}-\frac{1}{3!}{z}^{3}+\cdots }{{z}^{2}}={z}^{-2}-{z}^{-1}+\frac{1}{2!}-\frac{1}{3!}z+\cdots$
and that therefore, by definition,
${\mathrm{res}}_{z=0}\left(\frac{{e}^{-z}}{{z}^{2}}\right)=-1.$