If omega is a complex cube root of unity,then | vedclass

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(C) Given that $\omega$ is a complex cube root of unity,we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.We simplify the expression: $(3 + 5\ome

Vedclass · Accepted Answer

$-4$

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(C) Given that $\omega$ is a complex cube root of unity,we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.We simplify the expression: $(3 + 5\omega + 3\omega^2)^2 + (3 + 3\omega + 5\omega^2)^2$.Using $3 + 3\omega + 3\omega^2 = 3(1 + \omega + \omega^2) = 3(0) = 0$,we can rewrite the terms:First term: $(3 + 3\omega + 3\omega^2 + 2\omega)^2 = (0 + 2\omega)^2 = 4\omega^2$.Second term: $(3 + 3\omega + 3\omega^2 + 2\omega^2)^2 = (0 + 2\omega^2)^2 = 4\omega^4$.Since $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$,the sum is $4\omega^2 + 4\omega$.Using $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.Therefore,$4\omega^2 + 4\omega = 4(\omega^2 + \omega) = 4(-1) = -4$.

If $\omega$ is a complex cube root of unity,then $(3 + 5\omega + 3\omega^2)^2 + (3 + 3\omega + 5\omega^2)^2 = $

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