Question

Given: $z_1+z_2+z_3=A;z_1+z_{2}w+z_3{w}^2=B;z_1+z_2{w}^2+z_3{w}^{}=C$ where $w$ is cube rott of unity
Prove: ${\left|A\right|}^2+{\left|B\right|}^2+{\left|C\right|}^2=3({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2)$

Answer 1

We
${\left|A\right|}^2+{\left|B\right|}^2+{\left|C\right|}^2=A\overline{A}+B\overline{B}+C\overline{C}....\left(1\right)$
$A\overline{A}=(z_1+z_2+z_3)(\overline{z_1}+\overline{z_2}+\overline{z_3})=z_1{z}_1+z_2{z}_2+z_3{z}_3+\overline{z_1}(z_2+z_3)+\overline{z_2}(z_3+z_1)+\overline{z_3}(z_1+z_2)....$
$B\overline{B}=(z_1+z_2\omega +z_3{\omega }^2)(\overline{z_1}+\overline{z_2\omega }+\overline{z_3{\omega }^2})=(z_1+z_2\omega +z_3{\omega }^2)(\overline{z_1}+\overline{z_2}{\omega }^2+\overline{z_3}\omega )[\because \overline{\omega }={\omega }^2;\left(\overline{{\omega }^2}\right)=\omega ]$
$=z_1\overline{z_1}+z_2\overline{z_2}{\omega }^3+z_3\overline{z_3}{\omega }^3+\overline{z_1}(z_2\omega +z_3{\omega }^2)+\overline{z_2}(z_3{\omega }^4+z_1{\omega }^2)+\overline{z_3}(z_1\omega +z_2{\omega }^2)={\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2+\overline{z_1}(z_2\omega +z_3{\omega }^2)+\overline{z_2}(z_3{\omega }^4+z_1{\omega }^2)+\overline{z_3}(z_1\omega +z_2{\omega }^2)....\left(2\right)$
Similarly
$C\overline{C}={\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2+\overline{z_1}\left({z_2{\omega }^2+z}_3\omega \right)+\overline{z_2}(z_3{\omega }^2+z_1{\omega }^{})+\overline{z_3}(z_1{\omega }^2+z_2{\omega }^{})....\left(3\right)$
Adding (1), (2) and (3), we get
$A\overline{A}+B\overline{B}+C\overline{C}=3[{\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2]+\overline{z_1}[z_2(1+\omega +{\omega }^2)+z_3(1+{\omega }^2+\omega )]+\overline{z_2}[z_2(1+\omega +{\omega }^2)+z_1(1+{\omega }^2+\omega )]+\overline{z_3}[z_2(1+\omega +{\omega }^2)+z_2(1+{\omega }^2+\omega )]$
From (1) and (2) we conclude
${\left|A\right|}^2+{\left|B\right|}^2+{\left|C\right|}^2=3[{\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2]$