2019 AMC 12B Problems/Problem 17

Revision as of 15:45, 14 February 2019 by Mathinator1 (talk | contribs) (Solution)

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

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 \[\text{cis}(2\theta)=60\]. From 0 < theta < pi/2, we have \[\theta=\frac{\pi}{2}\] and from pi/2 < theta < pi, we see a monotonic decrease of \[\text{cis}(2\theta)\], from 180 to 0. Hence, there are 2 values that work for 0 < theta < pi. But since the interval pi < theta < 2pi is identical, because 3theta=theta at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. $\boxed{D}$

-FlatSquare

Someone pls help with LaTeX formatting, thanks

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 12 Problems and Solutions