Question #308381

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


1
Expert's answer
2022-03-10T03:21:04-0500

The poles of f(z)=z2+2z(z+1)2(z+4)f(z)=\dfrac {z^2+2z}{(z+1)^2(z+4)} are obtained from (z+1)2(z+4)=0(z+1)^2(z+4) = 0.


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


Res{f(z),1}=limz=1ddz{(z+1)2 f(z)}=limz=1ddz{(z+1)2(z2+2z)(z+1)2(z+4)}=limz=1ddz{z2+2zz+4}=limz=1{z2+8z+8(z+4)2}=19Res{f(z),4}=limz=4{(z+4) f(z)}=limz=4{(z+4)(z2+2z)(z+1)2(z+4)}=limz=4{z2+2z(z+1)2}=0\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}


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