Answer to Question #308381 in Complex Analysis for Nivedita

Question

find the residue of f(z)=z^2+2z/(z+1)^2(z+4)at its poles

Answer 1

The poles of $f(z)=\dfrac {z^2+2z}{(z+1)^2(z+4)}$ are obtained from $(z+1)^2(z+4) = 0$.

We get, $z = -1$ is a pole of order 2 and $z = -4$ is a simple pole.

$\begin{aligned} Res\{f(z),-1\} &= \lim_{z=-1}\dfrac{d}{dz}\left\{(z+1)^2~f(z)\right\}\\ &=\lim_{z=-1}\dfrac{d}{dz}\left\{(z+1)^2 \dfrac{(z^2+2z)}{(z+1)^2(z+4)}\right\}\\ &=\lim_{z=-1}\dfrac{d}{dz}\left\{ \dfrac{z^2+2z}{z+4}\right\}\\ &=\lim_{z=-1} \left\{\dfrac{z^2+8z+8}{(z+4)^2}\right\}\\ &= \frac{1}{9}\\ Res\{f(z),-4\} &= \lim_{z=-4}\left\{(z+4)~f(z)\right\}\\ &=\lim_{z=-4}\left\{(z+4) \dfrac{(z^2+2z)}{(z+1)^2(z+4)}\right\}\\ &=\lim_{z=-4}\left\{ \dfrac{z^2+2z}{(z+1)^2}\right\}\\ &= 0 \end{aligned}$