Singularity and residue theorem

[MissP.25_5]

Hello.
Can someone check if I got the answer right?

Find the singularity and the residue.

$f(z)=\frac{e^{-2z}}{(z+1)^2}$

My solution:

$f(z)=\frac{e^{-2z}}{(z+1)^2}$
$$Resf(z)_{|z=-1|}=\lim_{{z}\to{-1}}\frac{d}{dz}((z+1)^2\frac{e^{-2z}}{(z+1)^2})$$

$$\lim_{{z}\to{-1}}-2e^{-2z}=-2e^{2}$$

 

[vanhees71]

A check with Mathematica tells me that your solution is correct :-).