Question Number 148724 by Sozan last updated on 30/Jul/21

find laurent series f(z)=(1/(z^2 −z+1)) ,0<∣z−1∣<1

Answered by mathmax by abdo last updated on 30/Jul/21

the way of this kind is to use changement z−1=y ⇒ f(z)=ϕ(y)=(1/((y+1)^2 −(y+1)+1))=(1/(y^2 +2y+1−y)) =(1/(y^2 +y+1)) roots of y^2 +y+1=0 Δ=1−4=−3 ⇒z_1 =((−1+i(√3))/2)=e^((2iπ)/3) and z_2 =((−1−i(√3))/2)=e^(−((2iπ)/3)) ⇒ϕ(y)=(1/((y−z_1 )(y−z_2 )))=(1/(z_1 −z_2 ))((1/(y−z_1 ))−(1/(y−z_2 ))) =(1/(2i.((√3)/2)))((1/(y−z_1 ))−(1/(y−z_2 ))) we have ∣(y/z_1 )∣=∣y∣<1 ∣(y/z_2 )∣=∣y∣<1 ⇒ϕ(y)=(1/(i(√3))){ (1/(z_1 ((y/z_1 )−1)))−(1/(z_2 ((y/z_2 )−1)))} =(1/(i(√3)))((1/z_2 )×(1/(1−(y/z_2 )))−(1/z_1 )×(1/(1−(y/z_1 )))) =(e^((2iπ)/3) /(i(√3)))Σ_(n=0) ^∞ (y^n /z_2 ^n )−(e^(−((2iπ)/3)) /(i(√3)))Σ_(n=0) ^∞ (y^n /z_1 ^n ) =(1/(i(√3)))e^((2iπ)/3) Σ_(n=0) ^∞ e^((i2nπ)/3) y^n −(1/(i(√3)))e^((−2iπ)/3) Σ_(n=0) ^∞ e^(−((i2nπ)/3)) y^n =(1/(i(√3)))Σ_(n=0) ^∞ e^((i(2n+1)π)/3) (z−1)^n −(1/(i(√3)))Σ_(n=0) ^∞ e^(−((i(2n+1)π)/3)) (z−1)^n =(1/(i(√3)))Σ_(n=0) ^∞ 2isin((((2n+1)π)/3))(z−1)^n ⇒ f(z)=(2/( (√3)))Σ_(n=0) ^∞ sin((((2n+1)π)/3))(z−1)^n