Solution: By De Moivre's Theorem, $z^6 = \cos(6\theta) + i\sin(6\theta) = -1 + 0i$. This implies $6\theta = \pi + 2\pi k$ for integer $k$. Solving for $\theta$ gives $\theta = \frac{\pi}{6} + \frac{\pi k}{3}$. The principal solution in $[0, 2\pi)$ is $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$. The smallest positive $\theta$ is $\boxed{\dfrac{\pi}{6}}$.

Sun, 03 May 2026 12:30:22 +0000

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