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Question Number 21307 by Tinkutara last updated on 20/Sep/17
Let z_1 , z_2 , z_3 be complex numbers such that (i) ∣z_1 ∣ = ∣z_2 ∣ = ∣z_3 ∣ = 1 (ii) z_1 + z_2 + z_3 ≠ 0 (iii) z_1 ^2 + z_2 ^2 + z_3 ^2 = 0 Prove that for all n ≥ 2, ∣z_1 ^n + z_2 ^n + z_3 ^n ∣ ∈ {0, 1, 2, 3}.
$$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{ii}\right)\:{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{iii}\right)\:{z}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{z}_{\mathrm{2}} ^{\mathrm{2}} \:+\:{z}_{\mathrm{3}} ^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:{n}\:\geqslant\:\mathrm{2}, \\ $$$$\mid{z}_{\mathrm{1}} ^{{n}} \:+\:{z}_{\mathrm{2}} ^{{n}} \:+\:{z}_{\mathrm{3}} ^{{n}} \mid\:\in\:\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right\}. \\ $$
Commented by Tinkutara last updated on 25/Sep/17
help pls
$$\mathrm{help}\:\mathrm{pls} \\ $$
Commented by Tinkutara last updated on 24/Sep/17
help pls
$$\mathrm{help}\:\mathrm{pls} \\ $$
Commented by Tinkutara last updated on 26/Sep/17
help pls
$$\mathrm{help}\:\mathrm{pls} \\ $$
Commented by Tinkutara last updated on 27/Sep/17
Let z_k =cos θ_k +isin θ_k ; k∈{1,2,3} Given that z_1 ^2 +z_2 ^2 +z_3 ^2 =0 cos 2θ_1 +cos 2θ_2 =−cos 2θ_3 sin 2θ_1 +sin 2θ_2 =−sin 2θ_3 Squaring and adding, 2+2cos 2(θ_1 −θ_2 )=1 θ_1 −θ_2 =kπ±(π/3) So let z_1 ^n =cos nθ+isin nθ z_2 ^n =cos n((π/3)+θ)+isin n((π/3)+θ) z_3 ^n =cos n(((2π)/3)+θ)+isin n(((2π)/3)+θ) ∴∣z_1 ^n +z_2 ^n +z_3 ^n ∣=(√(3+2Σ_(cyc) cos n(θ_1 −θ_2 ))) Now I only got 0 in all cases. Can someone continue after this?
$${Let}\:{z}_{{k}} =\mathrm{cos}\:\theta_{{k}} +{i}\mathrm{sin}\:\theta_{{k}} ;\:{k}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$${Given}\:{that}\:{z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta_{\mathrm{1}} +\mathrm{cos}\:\mathrm{2}\theta_{\mathrm{2}} =−\mathrm{cos}\:\mathrm{2}\theta_{\mathrm{3}} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta_{\mathrm{1}} +\mathrm{sin}\:\mathrm{2}\theta_{\mathrm{2}} =−\mathrm{sin}\:\mathrm{2}\theta_{\mathrm{3}} \\ $$$${Squaring}\:{and}\:{adding}, \\ $$$$\mathrm{2}+\mathrm{2cos}\:\mathrm{2}\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right)=\mathrm{1} \\ $$$$\theta_{\mathrm{1}} −\theta_{\mathrm{2}} ={k}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$$${So}\:{let}\:{z}_{\mathrm{1}} ^{{n}} =\mathrm{cos}\:{n}\theta+{i}\mathrm{sin}\:{n}\theta \\ $$$${z}_{\mathrm{2}} ^{{n}} =\mathrm{cos}\:{n}\left(\frac{\pi}{\mathrm{3}}+\theta\right)+{i}\mathrm{sin}\:{n}\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$$${z}_{\mathrm{3}} ^{{n}} =\mathrm{cos}\:{n}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+\theta\right)+{i}\mathrm{sin}\:{n}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+\theta\right) \\ $$$$\therefore\mid{z}_{\mathrm{1}} ^{{n}} +{z}_{\mathrm{2}} ^{{n}} +{z}_{\mathrm{3}} ^{{n}} \mid=\sqrt{\mathrm{3}+\mathrm{2}\underset{{cyc}} {\sum}\mathrm{cos}\:{n}\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right)} \\ $$$${Now}\:{I}\:{only}\:{got}\:\mathrm{0}\:{in}\:{all}\:{cases}.\:{Can} \\ $$$${someone}\:{continue}\:{after}\:{this}? \\ $$
Commented by Tinkutara last updated on 22/Mar/18
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