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

$$findLourantseriesof$$ $$$$ $$f\left(z\right)=\frac{1}{1-z+z^2},0<\mid z-1\mid <1$$

Answered by Olaf_Thorendsen last updated on 25/Jun/21

$$\mathrm{Laurent}\mathrm{serie}o\mathrm{f}f\mathrm{in}a:$$ $$f\left(z\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{a}_n{(z-a)}^n$$ $$f\left(z\right)=\frac{1}{1-z+z^2}$$ $$f\left(z\right)=\frac{1}{\frac{3}{4}+{(z-\frac{1}{2})}^2}$$ $$f\left(z\right)=\frac{4}{3}.\frac{1}{1+{\left(\frac{2}{\sqrt{3}}(z-\frac{1}{2})\right)}^2}$$ $$f\left(z\right)=\frac{4}{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\left[\frac{2}{\sqrt{3}}(z-\frac{1}{2})\right]}^{2n}$$ $$\mathrm{The}\mathrm{Laurent}\mathrm{serie}o\mathrm{f}f\mathrm{in}\frac{1}{2}\mathrm{is}:$$ $$f\left(z\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{2}^{2n+2}}{3^{n+1}}{(z-\frac{1}{2})}^{2n}$$ $${\mathrm{a}}_n=\frac{{(-1)}^n{2}^{2n+2}}{3^{n+1}}$$

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