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 (Quick Look)

As before, $r=1$ . Represent $z$ in polar form. By De Moivre's Theorem, $z^3=\text{cis}(3\theta)$ . To form an equilateral triangle, their difference in angle must be $\frac{\pi}{3}$ , so $\frac{\text{cis}(3\theta)}{\text{cis}(\theta)}=\text{cis}(2\theta)=\text{cis}(\pm\frac{\pi}{3})$ From the polar form of $z$ , we know that $0\geq\theta\leq2\pi$ , so $\text{cis}(2\theta)$ cycles in a circle twice. By contrast, $\pm\frac{\pi}{3}$ represent $2$ fixed, distinct points. Thus, $\text{cis}(2\theta)$ intersects these points twice each $\implies\boxed{\textbf{(D) }4}$

Visual: https://www.desmos.com/calculator/rnpxzns0jn

To be more rigorous, you can find the  solutions.  cycles twice, so , where . Then, , , , . Substitute those values into  and check that they are valid.

(Solution by BJHHar)

Solution 3

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 $\boxed{\textbf{(D) }4}$ .

Solution 4 (Quick and Easy)

Since the complex numbers $0,z,$ and $z^3$ form an equilateral triangle in the complex plane, we note that either $z^3$ is a 60 degrees counterclockwise rotation about the origin from $z$ or $z$ is a 60 degrees counterclockwise rotation about the origin from $z^3$ .

Therefore, we note that either $z^3 = z \text{cis} 60^\circ{}$ or $z \text{cis}(-60^\circ{}) = z^3$

The first equation in $z$ (meaning $z^3 = z \text{cis} 60^\circ{}$ ) gives us: $z^2 = cis 60^\circ{}$ , which gives 2 solutions in $z$ .

The second equation in $z$ (which is $z \text{cis} (-60^\circ{}) = z^3$ ) gives us $z^2 = (\text{cis} -60^\circ{})$ , which must give another 2 solutions in $z$ .

Therefore, there are $\boxed{(D) 4}$ solutions for $z$ . (Professor-Mom) (Minor edit: mathbw225

Note: The motivation for this method came from an older AIME problem, namely https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8.

Video Solution

For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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