2019 AMC 12B Problem 17

Attempt Problem 17 of the 2019 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

How many nonzero complex numbers zz have the property that 0,0, z,z, and z3,z^3, when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
00
11
22
44
infinitely many
Answer: D
Concepts:complex numberequilateral triangle
Small hint:
Equilateral means ∣z∣=∣z3∣=∣z3−z∣|z|=|z^3|=|z^3-z|
Big hint:
∣z∣=∣z3∣|z|=|z^3| forces ∣z∣=1;|z|=1; then require ∣z2−1∣=1|z^2-1|=1
Solution:
The three points form an equilateral triangle iff ∣z∣=∣z3∣=∣z3−z∣.|z|=|z^3|=|z^3-z|. From ∣z∣=∣z3∣=∣z∣3|z|=|z^3|=|z|^3 we get ∣z∣=1.|z|=1.
Then ∣z3−z∣=∣z∣ ∣z2−1∣=∣z2−1∣,|z^3-z|=|z|\,|z^2-1|=|z^2-1|, so we need ∣z2−1∣=1.|z^2-1|=1. Writing z=eiθ,z=e^{i\theta}, ∣z2−1∣=2∣sin⁡θ∣=1,|z^2-1|=2|\sin\theta|=1, so ∣sin⁡θ∣=12.|\sin\theta|=\dfrac12.
This gives θ=30∘,150∘,210∘,330∘,\theta=30^\circ,150^\circ,210^\circ,330^\circ, four values of z,z, all yielding distinct vertices.
Thus, D is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.