let-z-from-C-and-f-z-2z-z-1-2z-1-developp-f-at-integr-serie-

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

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} .