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Question Number 37300 by math khazana by abdo last updated on 11/Jun/18

let f(z)=(((1−z^2 )e^(2z) )/z^3 ) calculate Res(f, 0)

$${let}\:{f}\left({z}\right)=\frac{\left(\mathrm{1}−{z}^{\mathrm{2}} \right){e}^{\mathrm{2}{z}} }{{z}^{\mathrm{3}} } \\ $$$${calculate}\:{Res}\left({f},\:\mathrm{0}\right) \\ $$

Commented by prof Abdo imad last updated on 15/Jun/18

0 is a triple pole of f so Res(f,0) =lim_(z→0) (1/((3−1)!)){z^3 f(z)}^((2)) =lim_(z→0) (1/2){ (1−z^2 )e^(2z) }^((2) ) but {(1−z^2 )^ e^(2z) }^((1)) = −2z e^(2z) +2(1−z^2 )e^(2z) =(−2z +2−2z^2 )e^(2z) ⇒ {(1−z^2 )e^(2z) }^((2)) =(−4z −2)e^(2z) +2(−2z+2−2z^2 )e^(2z) =(−4z −2 −4z +4 −4z^2 )e^(2z) =(−4z^2 −8z +2)e^(2z) ⇒ Res(ϕ,0) = (1/2) 2 =1 .

$$\mathrm{0}\:{is}\:{a}\:{triple}\:{pole}\:{of}\:{f}\:\:{so} \\ $$$${Res}\left({f},\mathrm{0}\right)\:={lim}_{{z}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{{z}^{\mathrm{3}} \:{f}\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left(\mathrm{1}−{z}^{\mathrm{2}} \right){e}^{\mathrm{2}{z}} \right\}^{\left(\mathrm{2}\right)\:} {but} \\ $$$$\left\{\left(\mathrm{1}−{z}^{\mathrm{2}} \right)^{} \:{e}^{\mathrm{2}{z}} \right\}^{\left(\mathrm{1}\right)} =\:−\mathrm{2}{z}\:{e}^{\mathrm{2}{z}} \:\:+\mathrm{2}\left(\mathrm{1}−{z}^{\mathrm{2}} \right){e}^{\mathrm{2}{z}} \\ $$$$=\left(−\mathrm{2}{z}\:+\mathrm{2}−\mathrm{2}{z}^{\mathrm{2}} \right){e}^{\mathrm{2}{z}} \:\:\Rightarrow \\ $$$$\left\{\left(\mathrm{1}−{z}^{\mathrm{2}} \right){e}^{\mathrm{2}{z}} \right\}^{\left(\mathrm{2}\right)} =\left(−\mathrm{4}{z}\:−\mathrm{2}\right){e}^{\mathrm{2}{z}} \:\:+\mathrm{2}\left(−\mathrm{2}{z}+\mathrm{2}−\mathrm{2}{z}^{\mathrm{2}} \right){e}^{\mathrm{2}{z}} \\ $$$$=\left(−\mathrm{4}{z}\:−\mathrm{2}\:−\mathrm{4}{z}\:+\mathrm{4}\:−\mathrm{4}{z}^{\mathrm{2}} \right){e}^{\mathrm{2}{z}} \\ $$$$=\left(−\mathrm{4}{z}^{\mathrm{2}} \:−\mathrm{8}{z}\:+\mathrm{2}\right){e}^{\mathrm{2}{z}} \:\Rightarrow \\ $$$${Res}\left(\varphi,\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{2}\:=\mathrm{1}\:. \\ $$$$ \\ $$

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