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Let-z-1-z-2-z-3-be-complex-numbers-not-all-real-such-that-z-1-z-2-z-3-1-and-2-z-1-z-2-z-3-3z-1-z-2-z-3-R-Prove-that-max-arg-z-1-arg-z-2-arg-z-3-pi-6-Wher

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Question Number 21294 by Tinkutara last updated on 19/Sep/17
Let z_1 , z_2 , z_3 be complex numbers, not all real, such that ∣z_1 ∣ = ∣z_2 ∣ = ∣z_3 ∣ = 1 and 2(z_1 + z_2 + z_3 ) − 3z_1 z_2 z_3 ∈ R. Prove that max(arg z_1 , arg z_2 , arg z_3 ) ≥ (π/6) . Where 0 < arg(z_1 ), arg(z_2 ), arg(z_3 ) < 2π.
$$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{not} \\ $$$$\mathrm{all}\:\mathrm{real},\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{2}\left({z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \right)\:−\:\mathrm{3}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \:\in\:{R}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{max}\left(\mathrm{arg}\:{z}_{\mathrm{1}} ,\:\mathrm{arg}\:{z}_{\mathrm{2}} ,\:\mathrm{arg}\:{z}_{\mathrm{3}} \right)\:\geqslant \\ $$$$\frac{\pi}{\mathrm{6}}\:.\:\mathrm{Where}\:\mathrm{0}\:<\:\mathrm{arg}\left({z}_{\mathrm{1}} \right),\:\mathrm{arg}\left({z}_{\mathrm{2}} \right),\:\mathrm{arg}\left({z}_{\mathrm{3}} \right) \\ $$$$<\:\mathrm{2}\pi. \\ $$
Answered by revenge last updated on 24/Sep/17
Let z_k =cos θ_k +isin θ_k ;k∈{1,2,3} Now 2(sin θ_1 +sin θ_2 +sin θ_3 )=3sin (θ_1 +θ_2 +θ_3 ) Let it be 2Σsin θ=3sin Σθ ⇒((sin Σθ)/(Σsin θ))=(2/3) ...(1) Since by Jensen′s inequality, sin (((Σθ)/3))≥((Σsin θ)/3) Taking Σθ=(π/2) so that LHS=(1/2), we get Σsin θ≤(3/2) And from (1)⇒Σsin θ=(3/2) This is only possible when θ_1 =θ_2 =θ_3 ; also Σθ=(π/2) So maximum value of θ is (1/3)((π/2))=(π/6).
$${Let}\:{z}_{{k}} =\mathrm{cos}\:\theta_{{k}} +{i}\mathrm{sin}\:\theta_{{k}} ;{k}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$${Now}\:\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{1}} +\mathrm{sin}\:\theta_{\mathrm{2}} +\mathrm{sin}\:\theta_{\mathrm{3}} \right)=\mathrm{3sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} +\theta_{\mathrm{3}} \right) \\ $$$${Let}\:{it}\:{be}\:\mathrm{2}\Sigma\mathrm{sin}\:\theta=\mathrm{3sin}\:\Sigma\theta \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\Sigma\theta}{\Sigma\mathrm{sin}\:\theta}=\frac{\mathrm{2}}{\mathrm{3}}\:…\left(\mathrm{1}\right) \\ $$$${Since}\:{by}\:{Jensen}'{s}\:{inequality},\:\mathrm{sin}\:\left(\frac{\Sigma\theta}{\mathrm{3}}\right)\geqslant\frac{\Sigma\mathrm{sin}\:\theta}{\mathrm{3}} \\ $$$${Taking}\:\Sigma\theta=\frac{\pi}{\mathrm{2}}\:{so}\:{that}\:{LHS}=\frac{\mathrm{1}}{\mathrm{2}},\:{we}\:{get}\:\Sigma\mathrm{sin}\:\theta\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${And}\:{from}\:\left(\mathrm{1}\right)\Rightarrow\Sigma\mathrm{sin}\:\theta=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${This}\:{is}\:{only}\:{possible}\:{when}\:\theta_{\mathrm{1}} =\theta_{\mathrm{2}} =\theta_{\mathrm{3}} ;\:{also}\:\Sigma\theta=\frac{\pi}{\mathrm{2}} \\ $$$${So}\:{maximum}\:{value}\:{of}\:\theta\:{is}\:\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{6}}. \\ $$
Commented by Tinkutara last updated on 24/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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