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Question Number 118647 by cantor last updated on 18/Oct/20

Answered by mathmax by abdo last updated on 18/Oct/20

f(z)=((z^2 −2z)/((z+1)^2 (z^2 +4))) −1 is double poles for f so Res(f,−1) =lim_(z→−1) (1/((2−1)!)){(z+1)^2 f(z)}^((1)) =lim_(z→−1) {((z^2 −2z)/(z^2 +4))}^((1)) =lim_(z→−1) (((2z−2)(z^2 +4)−2z(z^2 −2z))/((z^2 +4)^2 )) =lim_(z→−1) (((−4)(5)+2(3))/5^2 ) =((6−20)/(25)) =−((14)/(25))

$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} −\mathrm{2z}}{\left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)}\:\:\:−\mathrm{1}\:\mathrm{is}\:\mathrm{double}\:\mathrm{poles}\:\mathrm{for}\:\mathrm{f}\:\mathrm{so} \\ $$$$\mathrm{Res}\left(\mathrm{f},−\mathrm{1}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{1}} \:\left\{\frac{\mathrm{z}^{\mathrm{2}} −\mathrm{2z}}{\mathrm{z}^{\mathrm{2}} +\mathrm{4}}\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{1}} \frac{\left(\mathrm{2z}−\mathrm{2}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)−\mathrm{2z}\left(\mathrm{z}^{\mathrm{2}} −\mathrm{2z}\right)}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{1}} \:\:\:\frac{\left(−\mathrm{4}\right)\left(\mathrm{5}\right)+\mathrm{2}\left(\mathrm{3}\right)}{\mathrm{5}^{\mathrm{2}} }\:=\frac{\mathrm{6}−\mathrm{20}}{\mathrm{25}}\:=−\frac{\mathrm{14}}{\mathrm{25}} \\ $$

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