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

$f\text{}(z)={\displaystyle \frac{{e}^{-z}}{{z}^{2}}}$

My solution: The only singular point of $f$ in $\left|z\right|\le 3$ is $z=0$ (double pole) and its remainder is therefore

${\mathrm{Res}}_{z=0}f(z)={\displaystyle \underset{z\to 0}{lim}{\displaystyle \frac{1}{(2-1)!}{\left({\displaystyle \frac{{e}^{-z}{z}^{2}}{{z}^{2}}}\right)}^{\mathrm{\prime}}={\displaystyle \underset{z\to 0}{lim}-{e}^{-z}=-1}}}$

Consequently, ${\int}_{|z=3|}f(z)=2\pi i{\mathrm{Res}}_{z=0}f(z)=-2\pi i.$

this right?

$f\text{}(z)={\displaystyle \frac{{e}^{-z}}{{z}^{2}}}$

My solution: The only singular point of $f$ in $\left|z\right|\le 3$ is $z=0$ (double pole) and its remainder is therefore

${\mathrm{Res}}_{z=0}f(z)={\displaystyle \underset{z\to 0}{lim}{\displaystyle \frac{1}{(2-1)!}{\left({\displaystyle \frac{{e}^{-z}{z}^{2}}{{z}^{2}}}\right)}^{\mathrm{\prime}}={\displaystyle \underset{z\to 0}{lim}-{e}^{-z}=-1}}}$

Consequently, ${\int}_{|z=3|}f(z)=2\pi i{\mathrm{Res}}_{z=0}f(z)=-2\pi i.$

this right?