{left( frac{sqrt{3} + i}{2} right)^6} + {left | vedclass

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(A) Let $z_1 = \frac{\sqrt{3} + i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.Then $z_1^6 = (e^{i\pi/6})^6 =

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$-2$

Vedclass · Answer

(A) Let $z_1 = \frac{\sqrt{3} + i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.Then $z_1^6 = (e^{i\pi/6})^6 = e^{i\pi} = -1$.Let $z_2 = \frac{i - \sqrt{3}}{2} = \frac{-\sqrt{3} + i}{2} = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = e^{i5\pi/6}$.Then $z_2^6 = (e^{i5\pi/6})^6 = e^{i5\pi} = \cos(5\pi) + i\sin(5\pi) = -1$.Therefore,$z_1^6 + z_2^6 = -1 + (-1) = -2$.

${\left( \frac{\sqrt{3} + i}{2} \right)^6} + {\left( \frac{i - \sqrt{3}}{2} \right)^6}$ is equal to

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