Question

If $z_1+z_2+z_3=A,z_1+z_2\omega +z_3{\omega }^2=B$ and $z_1+z_2{\omega }^2+z_3\omega =C$, where $1,\omega ,{\omega }^2$ are the three cube root of unity, then ${\left|A\right|}^2+{\left|B\right|}^2+{\left|C\right|}^2=$

• A. $3({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2)$
• B. $2({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2)$
• C. $({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2)$
• D. none of these

Answer 1

The correct option is A $3({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2)$

We have, ${\left|A\right|}^2+{\left|B\right|}^2+{\left|C\right|}^2=A\overline{A}+B\overline{B}+C\overline{C}$ ....(1)

But $A\overline{A}=(z_1+z_2+z_3)(\overline{z_1}+\overline{z_2}+\overline{z_3})$

$=z_1\overline{z_1}+z_2\overline{z_2}+z_3\overline{z_3}+\overline{z_1}(z_2+z_3)+\overline{z_2}(z_3+z_1)+\overline{z_3}(z_1+z_2)$

$={\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2+\overline{z_1}(z_2+z_3)+\overline{z_2}(z_3+z_1)+\overline{z_3}(z_1+z_2)$

$B\overline{B}=(z_1+z_2\omega +z_3{\omega }^2)(\overline{z_1}+\overline{z_2\omega }+\overline{z_3{\omega }^2})$

$=(z_1+z_2\omega +z_3{\omega }^2)(\overline{z_1}+\overline{z_2}{\omega }^2+\overline{z_3{\omega }^2})$

$[\because \overline{\omega }={\omega }^2$ and $\overline{\left({\omega }^2\right)}=\omega ]$

$=z_1\overline{z_1}+z_2\overline{z_2}{\omega }^3+z_3\overline{z_3}{\omega }^3+\overline{z_1}(z_2\omega +z_3{\omega }^2)+\overline{z_2}(z_3{\omega }^4+z_1{\omega }^2)+\overline{z_3}(z_2\omega +z_2{\omega }^2)$

$={\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2+\overline{z_1}(z_2\omega +z_3{\omega }^2)+\overline{z_2}(z_1\omega +z_1{\omega }^2)+\overline{z_3}(z_1\omega +z_3{\omega }^2)$ ...(2)

similarly, $C\overline{C}={\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2+\overline{z_1}(z_2{\omega }^2+z_3\omega )$

$+\overline{z_2}(z_3\omega +z_1{\omega }^2)+\overline{z_3}(z_2{\omega }^2+z_2\omega )$ ...(3)

adding (1),(2) and (3), we get

$A\overline{A}+B\overline{B}+C\overline{C}=3[{\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2]$

$+\overline{z_1}[z_2(1+\omega +{\omega }^2)+z_3(1+{\omega }^2+\omega )]$

$+\overline{z_2}[z_3(1+\omega +{\omega }^2)+z_1(1+\omega +{\omega }^2)]$

$+\overline{z_3}[z_2(1+\omega +{\omega }^2)+z_2(1+{\omega }^2+\omega )]$

$=3[{\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2][\because 1+\omega +{\omega }^2=0]$

$\therefore$ From (1) and (2), we conclude

${\left|A\right|}^2+{\left|B\right|}^2+{\left|C\right|}^2=3[{\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2].$