# 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 1

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 $$2\theta=\pm\frac{\pi}{3}$$the requirements for being an equilateral triangle. From $0 < \theta < \pi/2$, we have $$\theta=\frac{\pi}{6}$$ and from $\pi/2 < \theta < \pi$, we see a monotonic increase of $2\theta$, from $\pi$ to $2\pi$, or equivalently, from $-\pi$ to $0$. Hence, there are 2 values that work for $0 < \theta < \pi$. But since the interval $\pi < \theta < 2\pi$ also consists of $2\theta$ going from $0$ to $2\pi$, it also gives us 2 solutions. Our answer is $\boxed{\textbf{(D) 4}}$

Here's a graph of how $z$ and $z^3$ move as $\theta$ increases- https://www.desmos.com/calculator/xtnpzoqkgs

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

## Solution 2

To be equilateral triangle, we should have

$$\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(2\pi/3)$$

Simplify left side:

$$z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)$$

That is,

$$z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)$$

We have two roots for both equations, therefore the total number of solution for $z$ is $\boxed{\textbf{(D) }4}$

(By Zhen Qin)