# 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 modulus-argument (polar) form, giving $z=r\text{cis}(\theta)$ for some $r$ and $\theta$. Thus, by De Moivre's Theorem, $z^3=r^3\text{cis}(3\theta)$. Since the distance from $0$ to $z$ is $r$, and the triangle is equilateral, the distance from $0$ to $z^3$ must also be $r$, so $r^3=r$, giving $r=1$. (We know $r \neq 0$ since the problem statement specifies that $z$ must be nonzero.)

Now, to get from $z$ to $z^3$, which should be a rotation of $120^{\circ}$ if the triangle is equilateral, we multiply by $z^2 = r^2\text{cis}(2\theta)$, again using De Moivre's Theorem. Thus we require $2\theta=\pm\frac{\pi}{3} + 2\pi k$ (where $k$ can be any integer). If $0 < \theta < \frac{\pi}{2}$, we must have $\theta=\frac{\pi}{6}$, while if $\frac{\pi}{2} \leq \theta < \pi$, we must have $\theta = \frac{5\pi}{6}$. Hence there are $2$ values that work for $0 < \theta < \pi$. By symmetry, the interval $\pi \leq \theta < 2\pi$ will also give $2$ solutions. The answer is thus $2 + 2 = \boxed{\textbf{(D) }4}$.

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

## Solution 2

For the triangle to be equilateral, the vector from $z$ to $z^3$, i.e $z^3 - z$, must be a $60^{\circ}$ rotation of the vector from $0$ to $z$, i.e. just $z$. Thus we must have $$\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)$$

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

Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of $z$ is $\boxed{\textbf{(D) }4}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 