If z = frac{sqrt{3} + i}{-2},then z^{69} is e | vedclass

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(C) Given $z = \frac{\sqrt{3} + i}{-2} = -\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)$.We know that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.M

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$i$

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(C) Given $z = \frac{\sqrt{3} + i}{-2} = -\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)$.We know that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.Multiplying $z$ by $i$,we get $iz = -\left(i\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = -\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\omega$.Thus,$z = \frac{-\omega}{i} = i\omega$.Now,$z^{69} = (i\omega)^{69} = i^{69} \cdot \omega^{69}$.Since $i^{69} = i^{68} \cdot i = (i^4)^{17} \cdot i = 1^{17} \cdot i = i$ and $\omega^{69} = (\omega^3)^{23} = 1^{23} = 1$.Therefore,$z^{69} = i \cdot 1 = i$.

If $z = \frac{\sqrt{3} + i}{-2}$,then $z^{69}$ is equal to

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