find-Lourant-series-of-f-z-1-1-z-z-2-0-lt-z-1-lt-1-

Question Number 144414 by mohammad17 last updated on 25/Jun/21

f i n d L o u r a n t s e r i e s o f

f (z) = \frac{1}{1 - z + z^{2}}, 0 <∣ z - 1 ∣< 1

Answered by Olaf_Thorendsen last updated on 25/Jun/21

Laurent serie o f f in a :

f (z) = \underset{n = 0}{\sum^{\infty}} a_{n} {(z - a)}^{n}

f (z) = \frac{1}{1 - z + z^{2}}

f (z) = \frac{1}{\frac{3}{4} + {(z - \frac{1}{2})}^{2}}

f (z) = \frac{4}{3} . \frac{1}{1 + {(\frac{2}{\sqrt{3}} (z - \frac{1}{2}))}^{2}}

f (z) = \frac{4}{3} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {[\frac{2}{\sqrt{3}} (z - \frac{1}{2})]}^{2 n}

The Laurent serie o f f in \frac{1}{2} is :

f (z) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} 2^{2 n + 2}}{3^{n + 1}} {(z - \frac{1}{2})}^{2 n}

a_{n} = \frac{{(- 1)}^{n} 2^{2 n + 2}}{3^{n + 1}}