solve inside C (x−(1/x))^3 +(x−(1/x))^2 +(x−(1/x))+1 =0


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Question Number 96211 by mathmax by abdo last updated on 30/May/20

$solve inside C {(x - \frac{1}{x})}^{3} + {(x - \frac{1}{x})}^{2} + (x - \frac{1}{x}) + 1 = 0$
	Answered by mr W last updated on 30/May/20

	$l e t t = x - \frac{1}{x}$ $t^{3} + t^{2} + t + 1 = 0$ $(t - 1) (t^{3} + t^{2} + t + 1) = 0$ $\Rightarrow t^{4} - 1 = 0$ $\Rightarrow t^{4} = 1$ $\Rightarrow t_{k} = \cos \frac{2 k π}{4} + i \sin \frac{2 k π}{4} w i t h k = 1, 2, 3$ $(k = 0 i s n o t r o o t o f o r i g i n a l e q n .)$ $x - \frac{1}{x} = t$ $x^{2} - t x - 1 = 0$ $\Rightarrow x = \frac{t}{2} \pm \sqrt{{(\frac{t}{2})}^{2} + 1}$ $. . . . . .$
	Answered by mathmax by abdo last updated on 30/May/20
	$let x−(1/x) =z (e)⇒z^3 +z^2 +z +1 =0 ⇒((1−z^4 )/(1−z)) =0 (z≠1) ⇒ z^4 =1 and z ≠1 let z =r e^(iθ) ⇒r^4 =1 and 4θ =2kπ ⇒r=1 and θ =((kπ)/2) so the roots are z_k =e^(i((kπ)/2)) and k ∈{1,2,3} z_1 =e^((iπ)/2) =i ,z_2 =−1 ,z_3 =e^(i((3π)/2)) =e^(i(2π−(π/2))) =−i x−(1/x) =i ⇒x^2 −1 =ix ⇒x^2 −ix−1=0 Δ =−1+4 =3 ⇒x_1 =((i+(√3))/2) x_2 =((i−(√3))/2) x−(1/x) =−1 ⇒x^2 −1 =−x ⇒x^2 +x−1 =0 Δ =5 ⇒x_3 =((−1+(√5))/2) and x_4 =((−1−(√5))/2) x−(1/x) =−i ⇒x^2 −1 =−ix ⇒x^2 +ix−1 =0 Δ =−1+4 =3 ⇒x_5 =((−i+(√3))/2) and x_6 =((−i−(√3))/2) the roots of this equation are the complex (x_i )_(1≤i≤6)$
	$let x - \frac{1}{x} = z (e) \Rightarrow z^{3} + z^{2} + z + 1 = 0 \Rightarrow \frac{1 - z^{4}}{1 - z} = 0 (z \neq 1) \Rightarrow$ $z^{4} = 1 and z \neq 1 let z = r e^{i θ} \Rightarrow r^{4} = 1 and 4 θ = 2 k π \Rightarrow r = 1 and θ = \frac{k π}{2}$ $so the roots are z_{k} = e^{i \frac{k π}{2}} and k \in {1, 2, 3}$ $z_{1} = e^{\frac{i π}{2}} = i, z_{2} = - 1, z_{3} = e^{i \frac{3 π}{2}} = e^{i (2 π - \frac{π}{2})} = - i$ $x - \frac{1}{x} = i \Rightarrow x^{2} - 1 = ix \Rightarrow x^{2} - ix - 1 = 0$ $Δ = - 1 + 4 = 3 \Rightarrow x_{1} = \frac{i + \sqrt{3}}{2} x_{2} = \frac{i - \sqrt{3}}{2}$ $x - \frac{1}{x} = - 1 \Rightarrow x^{2} - 1 = - x \Rightarrow x^{2} + x - 1 = 0$ $Δ = 5 \Rightarrow x_{3} = \frac{- 1 + \sqrt{5}}{2} and x_{4} = \frac{- 1 - \sqrt{5}}{2}$ $x - \frac{1}{x} = - i \Rightarrow x^{2} - 1 = - ix \Rightarrow x^{2} + ix - 1 = 0$ $Δ = - 1 + 4 = 3 \Rightarrow x_{5} = \frac{- i + \sqrt{3}}{2} and x_{6} = \frac{- i - \sqrt{3}}{2}$ $the roots of this equation are the complex {(x_{i})}_{1 ⩽ i ⩽ 6}$
	Answered by MJS last updated on 30/May/20

	${(x - \frac{1}{x})}^{3} + {(x - \frac{1}{x})}^{2} + (x - \frac{1}{x}) + 1 = 0$ $\Rightarrow$ $x^{6} + x^{5} - 2 x^{4} - x^{3} + 2 x^{2} + x - 1 = 0$ $(x^{4} - x^{2} + 1) (x^{2} + x - 1) = 0$ $and this is very easy to solve$
	Commented by mr W last updated on 30/May/20

	$g o o d i d e a!$
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