Solution: Using De Moivre's Theorem, $ z^n = \cos\left(n\theta\right) + i \sin\left(n\theta\right) $. Here, $ z = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) $, so $ z^6 = \cos\left(6 \cdot \frac{\pi}{6}\right) + i \sin\left(6 \cdot \frac{\pi}{6}\right) = \cos(\pi) + i \sin(\pi) $. Evaluating, $ \cos(\pi) = -1 $ and $ \sin(\pi) = 0 $, so $ z^6 = -1 + 0i $.

Sat, 02 May 2026 12:11:17 +0000

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