AMC12 2019 B

Answer (D): For $0$, $z$, and $z^3$ to be vertices of an equilateral triangle, the distances that $z$ and $z^3$ are from $0$ must be equal, which implies that $|z|=1$. Let $z=\cos\theta+i\sin\theta$, where $0\le\theta<360^\circ$. Then by De Moivre’s Theorem, $z^3=\cos(3\theta)+i\sin(3\theta)$. For the three points to form an equilateral triangle, the angle from $z$ to $0$ to $z^3$ must be $60^\circ$, which implies that $2\theta=\pm60^\circ+k\cdot360^\circ$ for some integer $k$. It follows that $\theta$ must be one of $30^\circ$, $150^\circ$, $210^\circ$, or $330^\circ$. In each of these 4 cases, $0$, $z$, and $z^3$ are vertices of a triangle with two congruent sides and an included angle of $60^\circ$, so they are vertices of an equilateral triangle.