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

$l e t f (z) = \frac{z^{2} + 1}{(z^{2} - 1) (z^{2} - 4)}$ $d e v e l o p p f a t i n t e g r s e r i e .$
Commented by math khazana by abdo last updated on 12/Jun/18
$we have f(z)= ((z^2 −1 +2)/((z^2 −1)(z^2 −4))) = (1/(z^2 −4)) +(2/((z^2 −1)(z^2 −4))) =(1/(z^2 −4)) −(2/3){ (1/(z^2 −1)) −(1/(z^2 −4))} = (1/(z^2 −4)) +(2/(3(z^2 −4))) +(2/3) (1/(1−z^2 )) =(5/3) (1/(z^2 −4)) +(2/3) (1/(1−z^2 )) = ((−5)/(12)) (1/(1−((z/2))^2 )) +(2/3) (1/(1−z^2 )) so if ∣z∣<1 we get f(z)= −(5/(12)) Σ_(n=0) ^∞ ((z/2))^(2n) +(2/3)Σ_(n=0) ^∞ z^(2n) =−(5/(12)) Σ_(n=0) ^∞ (z^(2n) /4^n ) +(2/3)Σ_(n=0) ^∞ z^(2n) .$
$w e h a v e f (z) = \frac{z^{2} - 1 + 2}{(z^{2} - 1) (z^{2} - 4)}$ $= \frac{1}{z^{2} - 4} + \frac{2}{(z^{2} - 1) (z^{2} - 4)}$ $= \frac{1}{z^{2} - 4} - \frac{2}{3} {\frac{1}{z^{2} - 1} - \frac{1}{z^{2} - 4}}$ $= \frac{1}{z^{2} - 4} + \frac{2}{3 (z^{2} - 4)} + \frac{2}{3} \frac{1}{1 - z^{2}}$ $= \frac{5}{3} \frac{1}{z^{2} - 4} + \frac{2}{3} \frac{1}{1 - z^{2}}$ $= \frac{- 5}{12} \frac{1}{1 - {(\frac{z}{2})}^{2}} + \frac{2}{3} \frac{1}{1 - z^{2}} s o i f ∣ z ∣< 1 w e g e t$ $f (z) = - \frac{5}{12} \sum_{n = 0}^{\infty} {(\frac{z}{2})}^{2 n} + \frac{2}{3} \sum_{n = 0}^{\infty} z^{2 n}$ $= - \frac{5}{12} \sum_{n = 0}^{\infty} \frac{z^{2 n}}{4^{n}} + \frac{2}{3} \sum_{n = 0}^{\infty} z^{2 n} .$
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