# Difference between revisions of "2019 AMC 12B Problems/Problem 17"

## Problem

How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle? $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$

## Solution

Convert $z$ and $z^3$ into $$r\text{cis}\theta$$ form, giving $$z=r\text{cis}\theta$$ and $$z^3=r^3\text{cis}(3\theta)$$. Since the distance from $0$ to $z$ is $r$, the distance from $0$ to $z^3$ must also be $r$, so $r=1$. Now we must find $$\text{cis}(2\theta)=60$$. From $0 < \theta < \pi/2$, we have $$\theta=\frac{\pi}{2}$$ and from $\pi/2 < \theta < \pi$, we see a monotonic decrease of $$\text{cis}(2\theta)$$, from $180^{\circ}$ to $0^{\circ}$. Hence, there are 2 values that work for $0 < \theta < \pi$. But since the interval $\pi < \theta < 2\pi$ is identical, because $3\theta=\theta$ at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. $\boxed{D}$

-FlatSquare

Someone pls help with LaTeX formatting, thanks , I did, -Dodgers66