Question

If $\omega =e^{\frac{i2\pi }{3}}$ and $a,b,c,x,y,z$ are non-zero complex numbers such that $a+b+c=x,a+b\omega +c{\omega }^2=y,a+b{\omega }^2+c\omega =z$. Then, the value of $\frac{\left(\right|x{|}^2+|y{|}^2+|z{|}^2)}{\left(\right|a{|}^2+|b{|}^2+|c{|}^2)}$.

• A. $2$
• B. $3$
• C. $1$
• D. $4$

Answer 1

The correct option is B

$3$

Explanation for the correct option:
Step 1: Required Information

When $\omega =e^{\frac{i2\pi }{3}}$, this means that $\omega$ is a cube root of unity.

Also, when $\omega$ is a cube root of unity then we know that ${\omega }^3=1\text{and}1+\omega +{\omega }^2$.

Step 2: Simplification required for the answer

Here, we can rewrite the expression $\left(\right|x{|}^2+|y{|}^2+|z{|}^2)$ as $x\overline{x}+y\overline{y}+z\overline{z}$. Then from the given, we have

$\begin{array}{rcl}x\overline{x}+y\overline{y}+z\overline{z}& =& \left(a+b+c\right)\left(\overline{a+b+c}\right)+\left(a+b\omega +c{\omega }^2\right)\left(\overline{a+b\omega +c{\omega }^2}\right)+\left(a+b{\omega }^2+c\omega \right)\left(\overline{a+b{\omega }^2+c\omega }\right)\end{array}$

Now, one can write that,

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

Step 3:Simplification for the RHS of the above equation

$\begin{array}{rcl}\left(a+b+c\right)\left(\overline{a+b+c}\right)+\left(a+b\omega +c{\omega }^2\right)\left(\overline{a+b\omega +c{\omega }^2}\right)+\left(a+b{\omega }^2+c\omega \right)\left(\overline{a+b{\omega }^2+c\omega }\right)& =& \left(a+b+c\right)\left(\overline{a}+\overline{b}+\overline{c}\right)+\left(a+b\omega +c{\omega }^2\right)\left(\overline{a}+\overline{b}{\omega }^2+\overline{c}\omega \right)\\ & & +\left(a+b{\omega }^2+c\omega \right)\left(\overline{a}+\overline{b}\omega +\overline{c}{\omega }^2\right)\\ & =& a\overline{a}+a\overline{b}+a\overline{c}+b\overline{a}+b\overline{b}+b\overline{c}+c\overline{a}+c\overline{b}+c\overline{c}+a\overline{a}+\\ & & a\overline{b}{\omega }^2+a\overline{c}\omega +b\overline{a}\omega +b\overline{b}{\omega }^3+b\overline{c}{\omega }^2+c\overline{a}{\omega }^2+c\overline{b}{\omega }^4\\ & & +c\overline{c}{\omega }^3+a\overline{a}+a\overline{b}\omega +a\overline{c}{\omega }^2+b\overline{a}{\omega }^2+b\overline{b}{\omega }^3+b\overline{c}{\omega }^4\\ & & +c\overline{a}\omega +c\overline{b}{\omega }^2+c\overline{c}{\omega }^3\\ & =& 3{\left|a\right|}^2+a\overline{b}\left(1+\omega +{\omega }^2\right)+a\overline{c}\left(1+\omega +{\omega }^2\right)+\\ & & b\overline{a}\left(1+\omega +{\omega }^2\right)+{\left|b\right|}^2+{\left|b\right|}^2{\omega }^3+{\left|b\right|}^2{\omega }^3+\\ & & b\overline{c}\left(1+\omega +{\omega }^2\right)+c\overline{a}\left(1+\omega +{\omega }^2\right)+c\overline{b}\left(1+\omega +{\omega }^2\right)+\\ & & {\left|c\right|}^2+{\left|c\right|}^2{\omega }^3+{\left|c\right|}^2{\omega }^3\end{array}$

Since, ${\omega }^3=1\text{and}1+\omega +{\omega }^2$

$\begin{array}{rcl}\left(a+b+c\right)\left(\overline{a+b+c}\right)+\left(a+b\omega +c{\omega }^2\right)\left(\overline{a+b\omega +c{\omega }^2}\right)+\left(a+b{\omega }^2+c\omega \right)\left(\overline{a+b{\omega }^2+c\omega }\right)& =& 3{\left|a\right|}^2+3{\left|b\right|}^2+3{\left|c\right|}^2\end{array}$

From equation $\left(1\right)$, we have

$\left({\left|x\right|}^2+{\left|y\right|}^2+{\left|z\right|}^2\right)=3\left({\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2\right)$

Divide both the sides by ${\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2$, we get

$\frac{\left({\left|x\right|}^2+{\left|y\right|}^2+{\left|z\right|}^2\right)}{{\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2}=3$

Hence, the correct option is (B).