2019 AMC 12B Problems/Problem 17

Revision as of 15:54, 14 February 2019 by Dodgers66 (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 $\pi < \theta < 2\pi$ is identical, because $3\theta=\theta$ 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 , I did, -Dodgers66

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