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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

{\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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