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

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}) - 3 z_{1} z_{2} z_{3} \in R .

Prove that \max (\arg z_{1}, \arg z_{2}, \arg z_{3}) ⩾

\frac{π}{6} . Where 0 < \arg (z_{1}), \arg (z_{2}), \arg (z_{3})

< 2 π .

Answered by revenge last updated on 24/Sep/17

L e t z_{k} = \cos θ_{k} + i \sin θ_{k}; k \in {1, 2, 3}

N o w 2 (\sin θ_{1} + \sin θ_{2} + \sin θ_{3}) = 3 \sin (θ_{1} + θ_{2} + θ_{3})

L e t i t b e 2 Σ \sin θ = 3 \sin Σ θ

\Rightarrow \frac{\sin Σ θ}{Σ \sin θ} = \frac{2}{3} \dots (1)

S i n c e b y {J e n s e n}^{'} s i n e q u a l i t y, \sin (\frac{Σ θ}{3}) ⩾ \frac{Σ \sin θ}{3}

T a k i n g Σ θ = \frac{π}{2} s o t h a t L H S = \frac{1}{2}, w e g e t Σ \sin θ ⩽ \frac{3}{2}

A n d f r o m (1) \Rightarrow Σ \sin θ = \frac{3}{2}

T h i s i s o n l y p o s s i b l e w h e n θ_{1} = θ_{2} = θ_{3}; a l s o Σ θ = \frac{π}{2}

S o m a x i m u m v a l u e o f θ i s \frac{1}{3} (\frac{π}{2}) = \frac{π}{6} .

Commented by Tinkutara last updated on 24/Sep/17

Thank you very much Sir!