Let ` omega= e^((ipi)/3) and a, b, c, x, y, z` be non-zero complex numbers such that `a+b+c = x, a + bomega + comega^2 = y, a + bomega^2 + comega = z`.Then, the value of `(|x|^2+|y|^2|+|y|^2)/(|a|^2+|b|^2+|c|^2)`

Options

A. 3
B. 6
C. 9
D. 1

Answer

A

Explanation

We have,
`x=a+b+c`
`x=a+b+c`
`rArr |x|^(2)-xbarx=(a+b+c)(bara+barb+barc)`
`rArr |x|^(2)=abara+bbar+cbarc+(abarb+barab)+(barbc+barbc)+(cbara+barca)` and, `y=a+bomega+comega^(2)`
`rArr |y^(2)|=ybary=(a+bomega+comega^(2))(bar(a+bomega+comega^(2)))=(a+bomega+comega^(2))(bara+barbomega^(2)+barcomega)`
`rArr |y|^(2)=abara+barbb+cbarc+(abarbomega^(2)+barabomega)+(bbarcomega^(2)+barbcomega)+(barcaomega+cbaraomega^(2))`
Similarly, `rArr |z|^(2)=|a|^(2)+|b|^(2)+|c|^(2)+(abarbomega+barabomega^(2))+(bbarcomega+barbcomega^(2))+(barcaomega^(2)+cbaraomega)`
`therefore |x|^(2)+|y|^(2)+|z|^(2)=3{|a|^(2)+|b|^(2)+|c|^(2)}`
`rArr (|x|^(2)+|y|^(2)+|z|^(2))/(|a|^(2)+|b|^(2)+|c|^(2))=3`