If z^{2} + z + 1 = 0,z in mathbb{C},then left | vedclass

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

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$2$

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(C) Given $z^{2} + z + 1 = 0$,the roots are $z = \omega$ or $z = \omega^{2}$,where $\omega$ is a complex cube root of unity.Since $\omega^{3} = 1$,we have $\frac{1}{\omega} = \omega^{2}$ and $\frac{1}{\omega^{2}} = \omega$.Let $S = \sum_{n=1}^{15} \left( z^{n} + (-1)^{n} \frac{1}{z^{n}} \right)^{2} = \sum_{n=1}^{15} \left( z^{2n} + \frac{1}{z^{2n}} + 2(-1)^{n} \right)$.For $z = \omega$,$z^{2n} + \frac{1}{z^{2n}} = \omega^{2n} + \omega^{-2n} = \omega^{2n} + \omega^{n} = 2 \cos\left(\frac{2n\pi}{3}\right)$.Summing the terms:$\sum_{n=1}^{15} z^{2n} = \sum_{n=1}^{15} \omega^{2n} = 0$ (since sum of $3$ consecutive powers of $\omega$ is $0$ and $15$ is a multiple of $3$).$\sum_{n=1}^{15} \frac{1}{z^{2n}} = \sum_{n=1}^{15} \omega^{-2n} = 0$.$\sum_{n=1}^{15} 2(-1)^{n} = 2(-1 + 1 - 1 + 1 ... - 1) = 2(-1) = -2$.Thus,$|S| = |0 + 0 - 2| = |-2| = 2$.

If $z^{2} + z + 1 = 0$,$z \in \mathbb{C}$,then $\left| \sum_{n=1}^{15} \left( z^{n} + (-1)^{n} \frac{1}{z^{n}} \right)^{2} \right|$ is equal to

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