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