Solution: Using De Moivre's Theorem, $ z^n = \cos\left(n\theta\right) + i \sin\left(n\theta\right) $. Here, $ z = \cos\left(\frac\pi6\right) + i \sin\left(\frac\pi6\right) $, so $ z^6 = \cos\left(6 \cdot \frac\pi6\right) + i \sin\left(6 \cdot \frac\pi6\right) = \cos(\pi) + i \sin(\pi) $. Evaluating, $ \cos(\pi) = -1 $ and $ \sin(\pi) = 0 $, so $ z^6 = -1 + 0i $. - Nelissen Grade advocaten