Determine three complex numbers α , β ,γ such that α=β^( 2) but β ≠ α^( 2) β = γ^( 2) but γ ≠ β^( 2) γ = α^( 2 ) but α ≠ γ^( 2)


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Question Number 1540 by Rasheed Soomro last updated on 17/Aug/15

$Determine three complex numbers α, β, γ such that$ $α = β^{2} b u t β \neq α^{2}$ $β = γ^{2} b u t γ \neq β^{2}$ $γ = α^{2} b u t α \neq γ^{2}$
	Answered by 123456 last updated on 17/Aug/15
	${ ((α=β^2 ∧β≠α^2 )),((β=γ^2 ∧γ≠β^2 )),((γ=α^2 ∧α≠γ^2 )) :}≡ { ((α=β^2 ∧β≠γ)),((β=γ^2 ∧γ≠α)),((γ=α^2 ∧α≠β)) :} α=β^2 =(γ^2 )^2 =[(α^2 )^2 ]^2 =α^(2×2×2) =α^8 α^8 −α=0 α(α^7 −1)=0 α=0∨α^7 =1 α=0⇒γ=0⇒γ^2 =α α=e^(((2π)/7)kı) ,k∈Z_7 γ=α^2 =e^(((4π)/7)kı) β=γ^2 =e^(((8π)/7)kı) (α,β,γ)=(e^(((2π)/7)kı) ,e^(((8π)/7)kı) ,e^(((4π)/7)kι) ) β≠α^2 ⇒e^(((8π)/7)kı) ≠e^(((4π)/7)kı) (k≠0) γ≠β^2 ⇒e^(((4π)/7)kı) ≠e^(((16π)/7)kı) =e^(((2π)/7)kı) (k≠0) α≠γ^2 ⇒e^(((2π)/7)kı) ≠e^(((8π)/7)kı) (k≠0) (α,β,γ)=(e^(((2π)/7)kı) ,e^(((8π)/7)kı) ,e^(((4π)/7)kı) ),k∈Z_7 \{0}$
	${\begin{cases} α = β^{2} \land β \neq α^{2} \\ β = γ^{2} \land γ \neq β^{2} \\ γ = α^{2} \land α \neq γ^{2} \end{cases} \equiv {\begin{cases} α = β^{2} \land β \neq γ \\ β = γ^{2} \land γ \neq α \\ γ = α^{2} \land α \neq β \end{cases}$ $α = β^{2} = {(γ^{2})}^{2} = {[{(α^{2})}^{2}]}^{2} = α^{2 \times 2 \times 2} = α^{8}$ $α^{8} - α = 0$ $α (α^{7} - 1) = 0$ $α = 0 \lor α^{7} = 1$ $α = 0 \Rightarrow γ = 0 \Rightarrow γ^{2} = α$ $α = e^{\frac{2 π}{7} k ı}, k \in Z_{7}$ $γ = α^{2} = e^{\frac{4 π}{7} k ı}$ $β = γ^{2} = e^{\frac{8 π}{7} k ı}$ $(α, β, γ) = (e^{\frac{2 π}{7} k ı}, e^{\frac{8 π}{7} k ı}, e^{\frac{4 π}{7} k ι})$ $β \neq α^{2} \Rightarrow e^{\frac{8 π}{7} k ı} \neq e^{\frac{4 π}{7} k ı} (k \neq 0)$ $γ \neq β^{2} \Rightarrow e^{\frac{4 π}{7} k ı} \neq e^{\frac{16 π}{7} k ı} = e^{\frac{2 π}{7} k ı} (k \neq 0)$ $α \neq γ^{2} \Rightarrow e^{\frac{2 π}{7} k ı} \neq e^{\frac{8 π}{7} k ı} (k \neq 0)$ $(α, β, γ) = (e^{\frac{2 π}{7} k ı}, e^{\frac{8 π}{7} k ı}, e^{\frac{4 π}{7} k ı}), k \in Z_{7} ∖ {0}$
	Commented by 123456 last updated on 17/Aug/15

	$(ω, ω^{4}, ω^{2})$ $(ω^{2}, ω, ω^{4})$ $(ω^{3}, ω^{5}, ω^{6})$ $(ω^{4}, ω^{2}, ω)$ $(ω^{5}, ω^{6}, ω^{3})$ $(ω^{6}, ω^{3}, ω^{5})$
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