2019 AMC 12B Problem 17

Below is the professionally curated solution for Problem 17 of the 2019 AMC 12B, 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.

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Concepts:complex number equilateral triangle

Difficulty rating: 1910

17.

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?

$0$

$1$

$2$

$4$

infinitely many

Solution:

The three points form an equilateral triangle iff $|z|=|z^3|=|z^3-z|.$ From $|z|=|z^3|=|z|^3$ we get $|z|=1.$

Then $|z^3-z|=|z|\,|z^2-1|=|z^2-1|,$ so we need $|z^2-1|=1.$ Writing $z=e^{i\theta},$ $|z^2-1|=2|\sin\theta|=1,$ so $|\sin\theta|=\dfrac12.$

This gives $\theta=30^\circ,150^\circ,210^\circ,330^\circ,$ four values of $z,$ all yielding distinct vertices.

Thus, D is the correct answer.

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