Question Number 144414 by mohammad17 last updated on 25/Jun/21
$${find}\:{Lourant}\:{series}\:{of}\: \\ $$ $$ \\ $$ $${f}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{1}−{z}+{z}^{\mathrm{2}} }\:\:\:\:,\mathrm{0}<\mid{z}−\mathrm{1}\mid<\mathrm{1} \\ $$
Answered by Olaf_Thorendsen last updated on 25/Jun/21
![Laurent serie of f in a : f(z) = Σ_(n=0) ^∞ a_n (z−a)^n f(z) = (1/(1−z+z^2 )) f(z) = (1/((3/4)+(z−(1/2))^2 )) f(z) = (4/3).(1/(1+((2/( (√3)))(z−(1/2)))^2 )) f(z) = (4/3)Σ_(n=0) ^∞ (−1)^n [(2/( (√3)))(z−(1/2))]^(2n) The Laurent serie of f in (1/2) is : f(z) = Σ_(n=0) ^∞ (((−1)^n 2^(2n+2) )/3^(n+1) )(z−(1/2))^(2n) a_n = (((−1)^n 2^(2n+2) )/3^(n+1) )](Q144416.png)
$$\mathrm{Laurent}\:\mathrm{serie}\:{o}\mathrm{f}\:{f}\:\mathrm{in}\:{a}\:: \\ $$ $${f}\left({z}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} \left({z}−{a}\right)^{{n}} \\ $$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{z}+{z}^{\mathrm{2}} } \\ $$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{4}}+\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$ $${f}\left({z}\right)\:=\:\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)^{\mathrm{2}} } \\ $$ $${f}\left({z}\right)\:=\:\frac{\mathrm{4}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right]^{\mathrm{2}{n}} \\ $$ $$\mathrm{The}\:\mathrm{Laurent}\:\mathrm{serie}\:{o}\mathrm{f}\:{f}\:\mathrm{in}\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{is}\:: \\ $$ $${f}\left({z}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{3}^{{n}+\mathrm{1}} }\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{n}} \\ $$ $$\mathrm{a}_{{n}} \:=\:\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{3}^{{n}+\mathrm{1}} } \\ $$