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Question Number 63079 by mathmax by abdo last updated on 28/Jun/19
let f(z) =((sin(2z))/z^n ) with n integr natural calculate Res(f,0)
$${let}\:{f}\left({z}\right)\:=\frac{{sin}\left(\mathrm{2}{z}\right)}{{z}^{{n}} }\:\:\:\:{with}\:{n}\:{integr}\:{natural}\: \\ $$$${calculate}\:{Res}\left({f},\mathrm{0}\right) \\ $$
Commented by mathmax by abdo last updated on 03/Jul/19
0 is pole of f at ordre n ⇒Res(f,0) =lim_(z→0) (1/((n−1)!)){ z^n f(z)}^((n−1)) =lim_(z→0) (1/((n−1)!)) {sin(2z)}^((n−1)) we have sin^((1)) (2z) =2cos(2z)=2 sin(2z+(π/2)) let suppose that sin^((n)) (2z) =2^n sin(2z+((nπ)/2)) ⇒sin^((n+1)) (2z) =2^(n+1) cos(2z+((nπ)/2)) =2^(n+1) sin(2z +(n+1)(π/2)) so the relation is true at ordr n+1 ⇒ Res(f,0) =lim_(z→0) (1/((n−1)!)) ×2^(n−1) sin(2z+(((n−1)π)/2)) =(2^(n−1) /((n−1)!))sin((((n−1)π)/2)) with n≥1 .
$$\mathrm{0}\:{is}\:{pole}\:{of}\:{f}\:{at}\:{ordre}\:{n}\:\Rightarrow{Res}\left({f},\mathrm{0}\right)\:={lim}_{{z}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\:{z}^{{n}} {f}\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\left\{{sin}\left(\mathrm{2}{z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$${we}\:{have}\:{sin}^{\left(\mathrm{1}\right)} \left(\mathrm{2}{z}\right)\:=\mathrm{2}{cos}\left(\mathrm{2}{z}\right)=\mathrm{2}\:{sin}\left(\mathrm{2}{z}+\frac{\pi}{\mathrm{2}}\right)\:{let}\:{suppose}\:{that} \\ $$$${sin}^{\left({n}\right)} \left(\mathrm{2}{z}\right)\:=\mathrm{2}^{{n}} {sin}\left(\mathrm{2}{z}+\frac{{n}\pi}{\mathrm{2}}\right)\:\Rightarrow{sin}^{\left({n}+\mathrm{1}\right)} \left(\mathrm{2}{z}\right)\:=\mathrm{2}^{{n}+\mathrm{1}} {cos}\left(\mathrm{2}{z}+\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}^{{n}+\mathrm{1}} {sin}\left(\mathrm{2}{z}\:+\left({n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right)\:\:{so}\:{the}\:{relation}\:{is}\:{true}\:{at}\:{ordr}\:{n}+\mathrm{1}\:\Rightarrow \\ $$$${Res}\left({f},\mathrm{0}\right)\:={lim}_{{z}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:×\mathrm{2}^{{n}−\mathrm{1}} {sin}\left(\mathrm{2}{z}+\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{2}^{{n}−\mathrm{1}} }{\left({n}−\mathrm{1}\right)!}{sin}\left(\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:\:\:{with}\:{n}\geqslant\mathrm{1}\:. \\ $$

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