- #1
aruwin
- 208
- 0
Hello.
Can someone check if I got the answer right?
$f(z)=\frac{e^{-2z}}{(z+1)^2}$
My solution:
$f(z)=\frac{e^{-2z}}{(z+1)^2}$
$$Resf(z)_{|z=-1|}=\lim_{{z}\to{-1}}\frac{d}{dz}((z+1)^2\frac{e^{-2z}}{(z+1)^2})$$
$$\lim_{{z}\to{-1}}-2e^{-2z}=-2e^{2}$$
Can someone check if I got the answer right?
$f(z)=\frac{e^{-2z}}{(z+1)^2}$
My solution:
$f(z)=\frac{e^{-2z}}{(z+1)^2}$
$$Resf(z)_{|z=-1|}=\lim_{{z}\to{-1}}\frac{d}{dz}((z+1)^2\frac{e^{-2z}}{(z+1)^2})$$
$$\lim_{{z}\to{-1}}-2e^{-2z}=-2e^{2}$$