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


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

$f i n d l a u r e n t s e r i e s f (z) = \frac{1}{z^{2} - z + 1}, 0 <∣ z - 1 ∣< 1$
	Answered by mathmax by abdo last updated on 30/Jul/21
	$the way of this kind is to use changement z−1=y ⇒ f(z)=ϕ(y)=(1/((y+1)^2 −(y+1)+1))=(1/(y^2 +2y+1−y)) =(1/(y^2 +y+1)) roots of y^2 +y+1=0 Δ=1−4=−3 ⇒z_1 =((−1+i(√3))/2)=e^((2iπ)/3) and z_2 =((−1−i(√3))/2)=e^(−((2iπ)/3)) ⇒ϕ(y)=(1/((y−z_1 )(y−z_2 )))=(1/(z_1 −z_2 ))((1/(y−z_1 ))−(1/(y−z_2 ))) =(1/(2i.((√3)/2)))((1/(y−z_1 ))−(1/(y−z_2 ))) we have ∣(y/z_1 )∣=∣y∣<1 ∣(y/z_2 )∣=∣y∣<1 ⇒ϕ(y)=(1/(i(√3))){ (1/(z_1 ((y/z_1 )−1)))−(1/(z_2 ((y/z_2 )−1)))} =(1/(i(√3)))((1/z_2 )×(1/(1−(y/z_2 )))−(1/z_1 )×(1/(1−(y/z_1 )))) =(e^((2iπ)/3) /(i(√3)))Σ_(n=0) ^∞ (y^n /z_2 ^n )−(e^(−((2iπ)/3)) /(i(√3)))Σ_(n=0) ^∞ (y^n /z_1 ^n ) =(1/(i(√3)))e^((2iπ)/3) Σ_(n=0) ^∞ e^((i2nπ)/3) y^n −(1/(i(√3)))e^((−2iπ)/3) Σ_(n=0) ^∞ e^(−((i2nπ)/3)) y^n =(1/(i(√3)))Σ_(n=0) ^∞ e^((i(2n+1)π)/3) (z−1)^n −(1/(i(√3)))Σ_(n=0) ^∞ e^(−((i(2n+1)π)/3)) (z−1)^n =(1/(i(√3)))Σ_(n=0) ^∞ 2isin((((2n+1)π)/3))(z−1)^n ⇒ f(z)=(2/( (√3)))Σ_(n=0) ^∞ sin((((2n+1)π)/3))(z−1)^n$
	$the way of this kind is to use changement z - 1 = y \Rightarrow$ $f (z) = φ (y) = \frac{1}{{(y + 1)}^{2} - (y + 1) + 1} = \frac{1}{y^{2} + 2 y + 1 - y}$ $= \frac{1}{y^{2} + y + 1} roots of y^{2} + y + 1 = 0$ $Δ = 1 - 4 = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{2 i π}{3}} and z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{2 i π}{3}}$ $\Rightarrow φ (y) = \frac{1}{(y - z_{1}) (y - z_{2})} = \frac{1}{z_{1} - z_{2}} (\frac{1}{y - z_{1}} - \frac{1}{y - z_{2}})$ $= \frac{1}{2 i . \frac{\sqrt{3}}{2}} (\frac{1}{y - z_{1}} - \frac{1}{y - z_{2}}) we have ∣ \frac{y}{z_{1}} ∣=∣ y ∣< 1$ $∣ \frac{y}{z_{2}} ∣=∣ y ∣< 1 \Rightarrow φ (y) = \frac{1}{i \sqrt{3}} {\frac{1}{z_{1} (\frac{y}{z_{1}} - 1)} - \frac{1}{z_{2} (\frac{y}{z_{2}} - 1)}}$ $= \frac{1}{i \sqrt{3}} (\frac{1}{z_{2}} \times \frac{1}{1 - \frac{y}{z_{2}}} - \frac{1}{z_{1}} \times \frac{1}{1 - \frac{y}{z_{1}}})$ $= \frac{e^{\frac{2 i π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} \frac{y^{n}}{z_{2}^{n}} - \frac{e^{- \frac{2 i π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} \frac{y^{n}}{z_{1}^{n}}$ $= \frac{1}{i \sqrt{3}} e^{\frac{2 i π}{3}} \sum_{n = 0}^{\infty} e^{\frac{i 2 n π}{3}} y^{n} - \frac{1}{i \sqrt{3}} e^{\frac{- 2 i π}{3}} \sum_{n = 0}^{\infty} e^{- \frac{i 2 n π}{3}} y^{n}$ $= \frac{1}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{\frac{i (2 n + 1) π}{3}} {(z - 1)}^{n} - \frac{1}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{- \frac{i (2 n + 1) π}{3}} {(z - 1)}^{n}$ $= \frac{1}{i \sqrt{3}} \sum_{n = 0}^{\infty} 2 isin (\frac{(2 n + 1) π}{3}) {(z - 1)}^{n} \Rightarrow$ $f (z) = \frac{2}{\sqrt{3}} \sum_{n = 0}^{\infty} \sin (\frac{(2 n + 1) π}{3}) {(z - 1)}^{n}$
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