If z = frac{-1-i sqrt{3}}{2},then sum_{k=1}^{ | vedclass

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(C) Given $z = \frac{-1-i \sqrt{3}}{2} = \omega$,where $\omega$ is the complex cube root of unity. We know that $\omega^3 = 1$ and $1 + \omega + \omeg

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$4044$

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(C) Given $z = \frac{-1-i \sqrt{3}}{2} = \omega$,where $\omega$ is the complex cube root of unity. We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Also,$\frac{1}{\omega} = \omega^2$. The expression is $\sum_{k=1}^{2022} (\omega^k + \omega^{2k})^2 = \sum_{k=1}^{2022} (\omega^{2k} + \omega^{4k} + 2\omega^{3k})$. Since $\omega^{3k} = 1$,this becomes $\sum_{k=1}^{2022} (\omega^{2k} + \omega^k + 2)$. We can split the sum: $\sum_{k=1}^{2022} \omega^{2k} + \sum_{k=1}^{2022} \omega^k + \sum_{k=1}^{2022} 2$. Since $2022$ is a multiple of $3$,the sum of powers of $\omega$ over $3$ terms is $0$. Thus,$\sum_{k=1}^{2022} \omega^k = 0$ and $\sum_{k=1}^{2022} \omega^{2k} = 0$. The total sum is $0 + 0 + 2 	imes 2022 = 4044$.

If $z = \frac{-1-i \sqrt{3}}{2}$,then $\sum_{k=1}^{2022} \left(z^k + \frac{1}{z^k}\right)^2 = $

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