Solution: Let $w = z^2$, so $w^2 + w + 1 = 0$. Solving gives $w = \frac{-1 \pm \sqrt{-3}}{2} = e^{\pm 2\pi i/3}$. Thus, $z^2 = e^{\pm 2\pi i/3}$, so $z = e^{\pm \pi i/3 + \pi i k/2}$ for $k = 0, 1$. The roots have angles $\pi/3, 5\pi/3, 7\pi/3, 11\pi/3$, but

Tue, 05 May 2026 22:46:11 +0000

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