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let-f-z-z-2-1-z-2-1-z-2-4-developp-f-at-integr-serie-

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Question Number 37354 by math khazana by abdo last updated on 12/Jun/18
let f(z) = ((z^2 +1)/((z^2 −1)(z^2 −4))) developp f at integr serie .
letf(z)=z2+1(z2−1)(z2−4)developpfatintegrserie.
Commented by math khazana by abdo last updated on 12/Jun/18
we have f(z)= ((z^2 −1 +2)/((z^2 −1)(z^2 −4))) = (1/(z^2 −4)) +(2/((z^2 −1)(z^2 −4))) =(1/(z^2 −4)) −(2/3){ (1/(z^2 −1)) −(1/(z^2 −4))} = (1/(z^2 −4)) +(2/(3(z^2 −4))) +(2/3) (1/(1−z^2 )) =(5/3) (1/(z^2 −4)) +(2/3) (1/(1−z^2 )) = ((−5)/(12)) (1/(1−((z/2))^2 )) +(2/3) (1/(1−z^2 )) so if ∣z∣<1 we get f(z)= −(5/(12)) Σ_(n=0) ^∞ ((z/2))^(2n) +(2/3)Σ_(n=0) ^∞ z^(2n) =−(5/(12)) Σ_(n=0) ^∞ (z^(2n) /4^n ) +(2/3)Σ_(n=0) ^∞ z^(2n) .
wehavef(z)=z2−1+2(z2−1)(z2−4)=1z2−4+2(z2−1)(z2−4)=1z2−4−23{1z2−1−1z2−4}=1z2−4+23(z2−4)+2311−z2=531z2−4+2311−z2=−51211−(z2)2+2311−z2soif∣z∣<1wegetf(z)=−512∑n=0∞(z2)2n+23∑n=0∞z2n=−512∑n=0∞z2n4n+23∑n=0∞z2n.

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