let z from C and f(z)= ((2z)/((z−1)(2z +1))) developp f at integr serie.


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Question Number 35220 by abdo.msup.com last updated on 16/May/18

$l e t z f r o m C a n d f (z) = \frac{2 z}{(z - 1) (2 z + 1)}$ $d e v e l o p p f a t i n t e g r s e r i e .$
Commented by abdo mathsup 649 cc last updated on 18/May/18
$let decompose f(z)= f(z)= (a/(z−1)) + (b/(2z+1)) a =lim_(z→1) (z−1)f(z) = (2/3) b =lim_(z→−(1/2)) (2z+1)f(z)= ((−1)/(−(3/2))) =(2/3) f(z)= (2/3){ (1/(z−1)) +(1/(2z+1))} =−(2/3) (1/(1−z)) + (2/3) (1/(1+2z)) so for ∣z∣<(1/2) f(z) =−(2/3) Σ_(n=0) ^∞ z^n +(2/3) Σ_(n=0) ^∞ (−1)^n 2^n z^n =(2/3) Σ_(n=0) ^∞ {(−1)^n 2^n −1}z^n .the radius of convergence is R=(1/2) .$
$l e t d e c o m p o s e f (z) =$ $f (z) = \frac{a}{z - 1} + \frac{b}{2 z + 1}$ $a = {l i m}_{z \to 1} (z - 1) f (z) = \frac{2}{3}$ $b = {l i m}_{z \to - \frac{1}{2}} (2 z + 1) f (z) = \frac{- 1}{- \frac{3}{2}} = \frac{2}{3}$ $f (z) = \frac{2}{3} {\frac{1}{z - 1} + \frac{1}{2 z + 1}}$ $= - \frac{2}{3} \frac{1}{1 - z} + \frac{2}{3} \frac{1}{1 + 2 z} s o f o r ∣ z ∣< \frac{1}{2}$ $f (z) = - \frac{2}{3} \sum_{n = 0}^{\infty} z^{n} + \frac{2}{3} \sum_{n = 0}^{\infty} {(- 1)}^{n} 2^{n} z^{n}$ $= \frac{2}{3} \sum_{n = 0}^{\infty} {{(- 1)}^{n} 2^{n} - 1} z^{n} . t h e r a d i u s o f$ $c o n v e r g e n c e i s R = \frac{1}{2} .$
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