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(D) Given $z^2 + z + 1 = 0$,the roots are $z = \omega$ or $z = \omega^2$,where $\omega$ is a cube root of unity.Since $\omega^3 = 1$ and $1 + \omega +

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$12$

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(D) Given $z^2 + z + 1 = 0$,the roots are $z = \omega$ or $z = \omega^2$,where $\omega$ is a cube root of unity.Since $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,we have $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.Calculating each term:$1$. $\left( z + \frac{1}{z} \right)^2 = (-1)^2 = 1$$2$. $\left( z^2 + \frac{1}{z^2} \right)^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$$3$. $\left( z^3 + \frac{1}{z^3} \right)^2 = (1 + 1)^2 = 2^2 = 4$$4$. $\left( z^4 + \frac{1}{z^4} \right)^2 = (\omega + \omega^2)^2 = (-1)^2 = 1$$5$. $\left( z^5 + \frac{1}{z^5} \right)^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$$6$. $\left( z^6 + \frac{1}{z^6} \right)^2 = (1 + 1)^2 = 2^2 = 4$Sum $= 1 + 1 + 4 + 1 + 1 + 4 = 12$.

If $z^2 + z + 1 = 0$,where $z$ is a complex number,then the value of $\left( z + \frac{1}{z} \right)^2 + \left( z^2 + \frac{1}{z^2} \right)^2 + \left( z^3 + \frac{1}{z^3} \right)^2 + \dots + \left( z^6 + \frac{1}{z^6} \right)^2$ is

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