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Question Number 56146 by gunawan last updated on 11/Mar/19

Given complex number

z_{1}, z_{2}, and z_{3} satiesfied z_{1} + z_{2} + z_{3} = 0

and ∣ z_{1} ∣=∣ z_{2} ∣=∣ z_{3} ∣= 1 . Prove that

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 0

Commented by MJS last updated on 11/Mar/19

z_{1}, z_{2}, z_{3} are the solutions of

z^{3} = \cos 3 α + i \sin 3 α

z^{3} = e^{3 i α} \Rightarrow z_{1} = e^{i α}; z_{2} = e^{i (α - \frac{2 π}{3})}; z_{3} = e^{i (α + \frac{2 π}{3})}

Answered by MJS last updated on 11/Mar/19

we can choose z_{3} = 1 + 0 i

z_{1} = a + b i

z_{2} = c + d i

(1) a + c + 1 = 0

(2) b + d = 0

(3) a^{2} + b^{2} = 1

(4) c^{2} + d^{2} = 1

(1) \Rightarrow c = - (a + 1)

(2) \Rightarrow d = - b

(3) a^{2} + b^{2} = 1

(4) a^{2} + 2 a + b^{2} + 1 = 1

(4) - (3) 2 a + 1 = 0 \Rightarrow a = - \frac{1}{2} \Rightarrow c = - \frac{1}{2}

\Rightarrow b^{2} = \frac{3}{4} \Rightarrow b = \pm \frac{\sqrt{3}}{2} \Rightarrow d = \mp \frac{\sqrt{3}}{2}

\Rightarrow z_{1} = - \frac{1}{2} \pm \frac{\sqrt{3}}{2}; z_{2} = - \frac{1}{2} \mp \frac{\sqrt{3}}{2}

\Rightarrow z_{1}^{2} = z_{2}; z_{2}^{2} = z_{1} \Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 0

we have got the roots of unity \Rightarrow for any

given z_{3} = \cos α + i \sin α we get

z_{1, 2} = \cos (α \pm \frac{2 π}{3}) + i \sin (α \pm \frac{2 π}{3})

because {it}^{'} s just a rotation by α

the sum of the squares is zero which can be

shown by adding them and using trigonometric

formulas for additions etc .

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