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Question Number 21294 by Tinkutara last updated on 19/Sep/17

$$\mathrm{Let}{z}_1,z_2,z_3\mathrm{be}\mathrm{complex}\mathrm{numbers},\mathrm{not}$$$$\mathrm{all}\mathrm{real},\mathrm{such}\mathrm{that}\mid z_1\mid =\mid z_2\mid =\mid z_3\mid =1$$$$\mathrm{and}2(z_1+z_2+z_3)-3z_1{z}_2{z}_3\in R.$$$$\mathrm{Prove}\mathrm{that}\max (\arg z_1,\arg z_2,\arg z_3)⩾$$$$\frac{\pi }{6}.\mathrm{Where}0<\arg \left(z_1\right),\arg \left(z_2\right),\arg \left(z_3\right)$$$$<2\pi .$$

Answered by revenge last updated on 24/Sep/17

$$Let{z}_k=\cos {\theta }_k+i\sin {\theta }_k;k\in \{1,2,3\}$$$$Now2(\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3)=3\sin ({\theta }_1+{\theta }_2+{\theta }_3)$$$$Letitbe2\mathrm{\Sigma }\sin \theta =3\sin \mathrm{\Sigma }\theta$$$$\Rightarrow \frac{\sin \mathrm{\Sigma }\theta }{\mathrm{\Sigma }\sin \theta }=\frac{2}{3}...\left(1\right)$$$$Sinceby{Jensen}^{\prime }sinequality,\sin \left(\frac{\mathrm{\Sigma }\theta }{3}\right)⩾\frac{\mathrm{\Sigma }\sin \theta }{3}$$$$Taking\mathrm{\Sigma }\theta =\frac{\pi }{2}sothatLHS=\frac{1}{2},weget\mathrm{\Sigma }\sin \theta ⩽\frac{3}{2}$$$$Andfrom\left(1\right)\Rightarrow \mathrm{\Sigma }\sin \theta =\frac{3}{2}$$$$Thisisonlypossiblewhen{\theta }_1={\theta }_2={\theta }_3;also\mathrm{\Sigma }\theta =\frac{\pi }{2}$$$$Somaximumvalueof\theta is\frac{1}{3}\left(\frac{\pi }{2}\right)=\frac{\pi }{6}.$$

Commented by Tinkutara last updated on 24/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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