Solution: Let $w = z^2$, so $w^2 + w + 1 = 0$. Solving gives $w = \frac-1 \pm \sqrt-32 = e^\pm 2\pi i/3$. 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 - Groen Casting
Mar 01, 2026
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