If z = frac{sqrt{3} + i}{2},then the value of | vedclass

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(A) Given $z = \frac{\sqrt{3}}{2} + i\frac{1}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)$.Using De Moivre's theorem,$z^{69}

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$-i$

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(A) Given $z = \frac{\sqrt{3}}{2} + i\frac{1}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)$.Using De Moivre's theorem,$z^{69} = \left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right)^{69}$.$z^{69} = \cos\left(\frac{69\pi}{6}\right) + i\sin\left(\frac{69\pi}{6}\right)$.$z^{69} = \cos\left(\frac{23\pi}{2}\right) + i\sin\left(\frac{23\pi}{2}\right)$.Since $\frac{23\pi}{2} = 11\pi + \frac{\pi}{2}$,we have $\cos\left(11\pi + \frac{\pi}{2}\right) = 0$ and $\sin\left(11\pi + \frac{\pi}{2}\right) = \sin\left(\pi + \frac{\pi}{2}\right) = -1$.Therefore,$z^{69} = 0 + i(-1) = -i$.

If $z = \frac{\sqrt{3} + i}{2}$,then the value of $z^{69}$ is

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