alleryUsing De Moivre's theorem, evaluate $\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right)^6$.
A$1$
B$i$
C$-i$
D$-1$
Answer & Solution
Correct answer: D. $-1$
De Moivre's theorem: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$.
With $\theta = \dfrac{\pi}{6}$ and $n = 6$: $\cos\pi + i\sin\pi = -1 + 0i = -1$.
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