If z + z^{-1} = 1,then z^{100} + z^{-100} is | vedclass

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(D) Given $z + z^{-1} = 1$,we have $z + \frac{1}{z} = 1$,which implies $z^2 - z + 1 = 0$.Using the quadratic formula,$z = \frac{1 \pm \sqrt{1 - 4}}{2}

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$-1$

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(D) Given $z + z^{-1} = 1$,we have $z + \frac{1}{z} = 1$,which implies $z^2 - z + 1 = 0$.Using the quadratic formula,$z = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ or $-\omega^2$,where $\omega$ is a complex cube root of unity.Case $1$: If $z = -\omega$,then $z^{100} + z^{-100} = (-\omega)^{100} + (-\omega)^{-100} = \omega^{100} + \frac{1}{\omega^{100}}$.Since $\omega^3 = 1$,$\omega^{100} = (\omega^3)^{33} \cdot \omega = \omega$.Thus,$z^{100} + z^{-100} = \omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.Case $2$: If $z = -\omega^2$,then $z^{100} + z^{-100} = (-\omega^2)^{100} + (-\omega^2)^{-100} = \omega^{200} + \frac{1}{\omega^{200}}$.Since $\omega^3 = 1$,$\omega^{200} = (\omega^3)^{66} \cdot \omega^2 = \omega^2$.Thus,$z^{100} + z^{-100} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$.In both cases,the result is $-1$.

If $z + z^{-1} = 1$,then $z^{100} + z^{-100}$ is equal to

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