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


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Question Number 21307 by Tinkutara last updated on 20/Sep/17
$Let z_1 , z_2 , z_3 be complex numbers such that (i) ∣z_1 ∣ = ∣z_2 ∣ = ∣z_3 ∣ = 1 (ii) z_1 + z_2 + z_3 ≠ 0 (iii) z_1 ^2 + z_2 ^2 + z_3 ^2 = 0 Prove that for all n ≥ 2, ∣z_1 ^n + z_2 ^n + z_3 ^n ∣ ∈ {0, 1, 2, 3}.$
$Let z_{1}, z_{2}, z_{3} be complex numbers such$ $that$ $(i) ∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = 1$ $(ii) z_{1} + z_{2} + z_{3} \neq 0$ $(iii) z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 0$ $Prove that for all n ⩾ 2,$ $∣ z_{1}^{n} + z_{2}^{n} + z_{3}^{n} ∣ \in {0, 1, 2, 3} .$
Commented by Tinkutara last updated on 25/Sep/17

$help pls$
Commented by Tinkutara last updated on 24/Sep/17

$help pls$
Commented by Tinkutara last updated on 26/Sep/17

$help pls$
Commented by Tinkutara last updated on 27/Sep/17
$Let z_k =cos θ_k +isin θ_k ; k∈{1,2,3} Given that z_1 ^2 +z_2 ^2 +z_3 ^2 =0 cos 2θ_1 +cos 2θ_2 =−cos 2θ_3 sin 2θ_1 +sin 2θ_2 =−sin 2θ_3 Squaring and adding, 2+2cos 2(θ_1 −θ_2 )=1 θ_1 −θ_2 =kπ±(π/3) So let z_1 ^n =cos nθ+isin nθ z_2 ^n =cos n((π/3)+θ)+isin n((π/3)+θ) z_3 ^n =cos n(((2π)/3)+θ)+isin n(((2π)/3)+θ) ∴∣z_1 ^n +z_2 ^n +z_3 ^n ∣=(√(3+2Σ_(cyc) cos n(θ_1 −θ_2 ))) Now I only got 0 in all cases. Can someone continue after this?$
$L e t z_{k} = \cos θ_{k} + i \sin θ_{k}; k \in {1, 2, 3}$ $G i v e n t h a t z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 0$ $\cos 2 θ_{1} + \cos 2 θ_{2} = - \cos 2 θ_{3}$ $\sin 2 θ_{1} + \sin 2 θ_{2} = - \sin 2 θ_{3}$ $S q u a r i n g a n d a d d i n g,$ $2 + 2 \cos 2 (θ_{1} - θ_{2}) = 1$ $θ_{1} - θ_{2} = k π \pm \frac{π}{3}$ $S o l e t z_{1}^{n} = \cos n θ + i \sin n θ$ $z_{2}^{n} = \cos n (\frac{π}{3} + θ) + i \sin n (\frac{π}{3} + θ)$ $z_{3}^{n} = \cos n (\frac{2 π}{3} + θ) + i \sin n (\frac{2 π}{3} + θ)$ $∴∣ z_{1}^{n} + z_{2}^{n} + z_{3}^{n} ∣= \sqrt{3 + 2 \sum_{c y c} \cos n (θ_{1} - θ_{2})}$ $N o w I o n l y g o t 0 i n a l l c a s e s . C a n$ $s o m e o n e c o n t i n u e a f t e r t h i s ?$
Commented by Tinkutara last updated on 22/Mar/18
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