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Question Number 147405 by tabata last updated on 20/Jul/21
detirmine the residues f(z)=((cosz)/(z^2 (z−π)^3 ))
$${detirmine}\:{the}\:{residues}\:{f}\left({z}\right)=\frac{{cosz}}{{z}^{\mathrm{2}} \left({z}−\pi\right)^{\mathrm{3}} } \\ $$
Commented by Mrsof last updated on 20/Jul/21
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Answered by mathmax by abdo last updated on 20/Jul/21
f(z)=((cosz)/(z^2 (z−π)^3 )) the poles are o and π Res(f,0)=lim_(z→0) (1/((2−1)!)){z^2 f(z)}^((1)) =lim_(z→o) {((cosz)/((z−π)^3 ))}^((1)) =lim_(z→0) ((−sinz(z−π)^3 −3(z−π)^2 cosz)/((z−π)^6 )) =lim_(z→0) ((−sinz(z−π)−3cosz)/((z−π)^4 ))=((−3)/π^4 ) Res(f,π)=lim_(z→π) (1/((3−1)!)){(z−π)^3 f(z)}^((2)) =(1/2)lim_(z→π) {((cosz)/z^2 )}^((2)) =(1/2)lim_(z→π) {((−z^2 sinz−2zcosz)/z^4 )}^((1)) =−(1/2)lim_(z→π) {((zsinz+2cosz)/z^3 )}^((1)) =−(1/2)lim_(z→π) (((sinz+zcosz−2sinz)z^3 −3z^2 (zsinz+2cosz))/z^6 ) =−(1/2)lim_(z→π) (((zcosz−sinz)z−3(zsinz+2cosz))/z^4 ) =−(1/2)×(((−π)π−3(−2))/π^4 )=−(1/2)×((−π^2 +6)/π^4 ) =((π^2 −6)/(2π^4 ))
$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}−\pi\right)^{\mathrm{3}} }\:\:\mathrm{the}\:\mathrm{poles}\:\mathrm{are}\:\mathrm{o}\:\mathrm{and}\:\pi \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{0}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\mathrm{z}^{\mathrm{2}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{o}} \:\:\:\left\{\frac{\mathrm{cosz}}{\left(\mathrm{z}−\pi\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\frac{−\mathrm{sinz}\left(\mathrm{z}−\pi\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{z}−\pi\right)^{\mathrm{2}} \mathrm{cosz}}{\left(\mathrm{z}−\pi\right)^{\mathrm{6}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\:\:\frac{−\mathrm{sinz}\left(\mathrm{z}−\pi\right)−\mathrm{3cosz}}{\left(\mathrm{z}−\pi\right)^{\mathrm{4}} }=\frac{−\mathrm{3}}{\pi^{\mathrm{4}} } \\ $$$$\mathrm{Res}\left(\mathrm{f},\pi\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\pi\right)^{\mathrm{3}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \left\{\frac{\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} }\right\}^{\left(\mathrm{2}\right)} \:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\left\{\frac{−\mathrm{z}^{\mathrm{2}} \mathrm{sinz}−\mathrm{2zcosz}}{\mathrm{z}^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\left\{\frac{\mathrm{zsinz}+\mathrm{2cosz}}{\mathrm{z}^{\mathrm{3}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\:\frac{\left(\mathrm{sinz}+\mathrm{zcosz}−\mathrm{2sinz}\right)\mathrm{z}^{\mathrm{3}} −\mathrm{3z}^{\mathrm{2}} \left(\mathrm{zsinz}+\mathrm{2cosz}\right)}{\mathrm{z}^{\mathrm{6}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\:\frac{\left(\mathrm{zcosz}−\mathrm{sinz}\right)\mathrm{z}−\mathrm{3}\left(\mathrm{zsinz}+\mathrm{2cosz}\right)}{\mathrm{z}^{\mathrm{4}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(−\pi\right)\pi−\mathrm{3}\left(−\mathrm{2}\right)}{\pi^{\mathrm{4}} }=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{−\pi^{\mathrm{2}} +\mathrm{6}}{\pi^{\mathrm{4}} } \\ $$$$=\frac{\pi^{\mathrm{2}} −\mathrm{6}}{\mathrm{2}\pi^{\mathrm{4}} } \\ $$

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