Question

Let $\omega =e^{\frac{2\pi i}{3}}$ and $a,b,c,x,y,z$ be non-zero complex numbers such that
$a+b+c=x$, $a+b\omega +c{\omega }^2=y$ and $a+b{\omega }^2+c\omega =z$.
Then the value of $\frac{{\left|x\right|}^2+{\left|y\right|}^2+{\left|z\right|}^2}{{\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2}$ is:

• A. $4$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is D $3$

$\omega =e^{\frac{2i\pi }{3}}$

$\Rightarrow \omega$ is one of the cube roots of unity.

Other roots are ${\omega }^2$ and $1$

$\therefore 1+\omega +{\omega }^2=0$ and ${\omega }^3=1$

$|x{|}^2=x\overline{x}=(a+b+c)(\overline{a}+\overline{b}+\overline{c})=|a{|}^2+|b{|}^2+|c{|}^2+a\overline{b}+a\overline{c}+b\overline{a}+b\overline{c}+c\overline{a}+c\overline{b}$

$|y{|}^2=(a+b\omega +c{\omega }^2)(\overline{a}+\overline{b}{\omega }^2+\overline{c}\omega )=|a{|}^2+|b{|}^2+|c{|}^2+a\overline{b}{\omega }^2+a\overline{c}\omega +\overline{a}b\omega +b\overline{c}{\omega }^2+\overline{a}c{\omega }^2+\overline{b}c\omega$

$|z{|}^2=(a+b{\omega }^2+c\omega )(\overline{a}+\overline{b}\omega +\overline{c}{\omega }^2)=|a{|}^2+|b{|}^2+|c{|}^2+a\overline{b}\omega +a\overline{c}{\omega }^2+\overline{a}b{\omega }^2+b\overline{c}\omega +\overline{a}c\omega +\overline{b}c{\omega }^2$

$\therefore |x{|}^2+|y{|}^2+|z{|}^2=3(|a{|}^2+|b{|}^2+|c{|}^2)+a\overline{b}(1+{\omega }^2+\omega )+a\overline{c}(1+\omega +{\omega }^2)+\overline{a}b(1+\omega +{\omega }^2)+b\overline{c}(1+{\omega }^2+\omega )$

$+c\overline{a}(1+\omega +{\omega }^2)+\overline{b}c(1+\omega +{\omega }^2)$

$=3(|a{|}^2+|b{|}^2+|c{|}^2)$

$\therefore \frac{|x{|}^2+|y{|}^2+|z{|}^2}{|a{|}^2+|b{|}^2+|c{|}^2}=3$