Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 12B Problems/Problem 17"

(→‎Solution 4)
m (→‎Solution 4 (Quick and Easy))
Line 44: Line 44:
 
Therefore, we note that either <math>z^3 = z cis60^\circ{}</math> or <math>z cis -60^\circ{} = z^3</math>
 
Therefore, we note that either <math>z^3 = z cis60^\circ{}</math> or <math>z cis -60^\circ{} = z^3</math>
  
−
The first equation in <math>z</math> (meaning <math>z^3 = z cis60^\circ{}) gives us: </math>z^2 = cis 60^\circ{}<math>, so applying Demoivres Theorem gives us </math>z = \pm cis30^\circ{}<math>
+
The first equation in <math>z</math> (meaning <math>z^3 = z cis60^\circ{}</math>) gives us: <math>z^2 = cis 60^\circ{}</math>, so applying Demoivres Theorem gives us <math>z = \pm cis30^\circ{}</math>
  
−
The second equation in </math>z<math> (</math>z cis -60^\circ{} = z^3<math>) gives us </math>z^2 = cis -60^\circ{} = cis 300^\circ{}<math>, so again by applying DeMoivres Theorem we get </math>z = \pm cis 150^\circ{}<math>.
+
The second equation in <math>z</math> (<math>z cis -60^\circ{} = z^3</math>) gives us <math>z^2 = cis -60^\circ{} = cis 300^\circ{}</math>, so again by applying DeMoivres Theorem we get <math>z = \pm cis 150^\circ{}</math>.
  
−
Therefore, there are </math>\boxed{E) 4}<math> solutions for </math>z$.
+
Therefore, there are <math>\boxed{E) 4}</math> solutions for <math>z</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 11:52, 18 October 2019

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 $4$ solutions. $\text{cis}(2\theta)$ cycles twice, so $2\theta=\pm\frac{\pi}{3}+2n\pi$, where $n=0,1$. Then, $\theta=\frac{\pi}{6}$, $\frac{7\pi}{6}$, $\frac{5\pi}{6}$, $\frac{11\pi}{6}$. Substitute those values into $z$ and check that they are valid. $\implies\boxed{\textbf{(D) }4}$

(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 a 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 $z^3 = z cis60^\circ{}$ or $z cis -60^\circ{} = z^3$

The first equation in $z$ (meaning $z^3 = z cis60^\circ{}$) gives us: $z^2 = cis 60^\circ{}$, so applying Demoivres Theorem gives us $z = \pm cis30^\circ{}$

The second equation in $z$ ($z cis -60^\circ{} = z^3$) gives us $z^2 = cis -60^\circ{} = cis 300^\circ{}$, so again by applying DeMoivres Theorem we get $z = \pm cis 150^\circ{}$.

Therefore, there are $\boxed{E) 4}$ solutions for $z$.

Video Solution

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

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_17&oldid=110411"