View source for 2019 AMC 12B Problems/Problem 17

==Problem==

How many nonzero complex numbers <math>z</math> have the property that <math>0, z,</math> and <math>z^3,</math> when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?

<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}</math>

==Solution 1==

Convert <math>z</math> and <math>z^3</math> into modulus-argument (polar) form, giving <math>z=r\text{cis}(\theta)</math> for some <math>r</math> and <math>\theta</math>. Thus, by De Moivre's Theorem, <math>z^3=r^3\text{cis}(3\theta)</math>. Since the distance from <math>0</math> to <math>z</math> is <math>r</math>, and the triangle is equilateral, the distance from <math>0</math> to <math>z^3</math> must also be <math>r</math>, so <math>r^3=r</math>, giving <math>r=1</math>. (We know <math>r \neq 0</math> since the problem statement specifies that <math>z</math> must be nonzero.)

Now, to get from <math>z</math> to <math>z^3</math>, which should be a rotation of <math>60^{\circ}</math> if the triangle is equilateral, we multiply by <math>z^2 = \text{cis}(2\theta)</math>, again using De Moivre's Theorem and <math>r=1</math>. Thus we require <math>2\theta=\pm\frac{\pi}{3} + 2\pi k</math> (where <math>k</math> can be any integer). If <math>0 < \theta < \frac{\pi}{2}</math>, we must have <math>\theta=\frac{\pi}{6}</math>, while if <math>\frac{\pi}{2} \leq \theta < \pi</math>, we must have <math>\theta = \frac{5\pi}{6}</math>. Hence there are <math>2</math> values that work for <math>0 < \theta < \pi</math>. By symmetry, the interval <math>\pi \leq \theta < 2\pi</math> will also give <math>2</math> solutions. The answer is thus <math>2 + 2 = \boxed{\textbf{(D) }4}</math>.

''Note'': Here's a graph showing how <math>z</math> and <math>z^3</math> move as <math>\theta</math> increases: https://www.desmos.com/calculator/xtnpzoqkgs.

==Solution 2 (Quick Look)==
As before, <math>r=1</math>. Represent <math>z</math> in polar form. By De Moivre's Theorem, <math>z^3=\text{cis}(3\theta)</math>. To form an equilateral triangle, their difference in angle must be <math>\frac{\pi}{3}</math>, so 
<cmath>\frac{\text{cis}(3\theta)}{\text{cis}(\theta)}=\text{cis}(2\theta)=\text{cis}(\pm\frac{\pi}{3})</cmath> 
From the polar form of <math>z</math>, we know that <math>0\geq\theta\leq2\pi</math>, so <math>\text{cis}(2\theta)</math> cycles in a circle twice. By contrast, <math>\pm\frac{\pi}{3}</math> represent <math>2</math> fixed, distinct points. Thus, <math>\text{cis}(2\theta)</math> intersects these points twice each<math>\implies\boxed{\textbf{(D) }4}</math>

''Visual: https://www.desmos.com/calculator/rnpxzns0jn''


 To be more rigorous, you can find the <math>4</math> solutions. <math>\text{cis}(2\theta)</math> cycles twice, so <math>2\theta=\pm\frac{\pi}{3}+2n\pi</math>, where <math>n=0,1</math>. Then, <math>\theta=\frac{\pi}{6}</math>, <math>\frac{7\pi}{6}</math>, <math>\frac{5\pi}{6}</math>, <math>\frac{11\pi}{6}</math>. Substitute those values into <math>z</math> and check that they are valid. <math>\implies\boxed{\textbf{(D) }4}</math>

(Solution by BJHHar)

==Solution 3==

For the triangle to be equilateral, the vector from <math>z</math> to <math>z^3</math>, i.e <math>z^3 - z</math>, must be a <math>60^{\circ}</math> rotation of the vector from <math>0</math> to <math>z</math>, i.e. just <math>z</math>. Thus we must have

<cmath>\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)</cmath>

Simplifying gives 
<cmath>z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)</cmath>
so
<cmath>z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)</cmath>

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 <math>z</math> is <math>\boxed{\textbf{(D) }4}</math>.

==Solution 4 (Quick and Easy) ==

Since the complex numbers <math>0,z,</math> and <math>z^3</math> form an equilateral triangle in the complex plane, we note that either <math>z^3</math> is a 60 degrees counterclockwise rotation about the origin from <math>z</math> or <math>z</math> is a 60 degrees counterclockwise rotation about the origin from <math>z^3</math>.

Therefore, we note that either <math>z^3 = z \text{cis} 60^\circ{}</math> or <math>z \text{cis}(-60^\circ{}) = z^3</math>

The first equation in <math>z</math> (meaning <math>z^3 = z \text{cis} 60^\circ{}</math>) gives us: <math>z^2 = cis 60^\circ{}</math>, which gives 2 solutions in <math>z</math>.

The second equation in <math>z</math> (which is <math>z \text{cis} (-60^\circ{}) = z^3</math>) gives us <math>z^2 = \text{cis} (-60^\circ{})</math>, which must give another 2 solutions in <math>z</math>.

Therefore, there are <math>\boxed{(D) 4}</math> solutions for <math>z</math>. (Professor-Mom)

Note: The motivation for this method came from an older AIME problem, namely https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8.

==Solution 5==

Let the <math>z = re^{i\theta}</math>, so <math>z^3 = r^3e^{3i\theta}</math>. To have an equilateral triangle, we must have <math>|z^3| = |z|</math>, so <math>r^3 = r</math>, so <math>r= 1</math>. 

Note that the angle between <math>z^3</math> and <math>z</math> is <math>3\theta - \theta = 2\theta</math>. Then, by the Law of Cosines,

<cmath> |z^3 - z|^2 = |z|^2 + |z^3|^2 - 2|z||z^3|\cos 2\theta.</cmath>

Since we have an equilateral triangle, it must be that <math>|z^3 - z| =|z| = |z^3| = r = 1</math>. Hence,
<cmath> 1 = 2 - 2\cos 2\theta = 2 - 2(1-2\sin^2\theta) = 4\sin^2\theta,</cmath>
<cmath> \sin^2\theta = \frac{1}{4}</cmath>
<cmath> \sin\theta = \pm\frac{1}{2}</cmath>
<cmath> \theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}.</cmath>

These <math>4</math> values of <math>\theta</math> correspond to <math>\boxed{(D)4}</math> distinct values of <math>z</math>.
~ brainfertilzer

==Solution 6 (Quick, rectangular form)==
Let <math>z = a + bi</math>, where <math>a</math> and <math>b</math> are real numbers. Then, since 0, <math>z</math>, and <math>z^3</math> are vertices of an equilateral triangle in the complex plane, expanding <math>z^3</math> gives the following : 
<cmath>(a + bi)^3 = (a^3 - 3ab^2) + i(3a^2b - b^3)</cmath>
Now, we let <math>z = z^3</math> and equate real and complex parts. It then follows that : 
<cmath>a^3 - 3ab^2 = a, 3a^2b - b^3 = b</cmath>
from which solving gives 
<cmath>a^2 - 3b^2 = 1 (1), 3a^2 - b^2 = 1 (2)</cmath>
We find (1) = (2)
<cmath>a^2 - 3b^2 = 3a^2 - b^2</cmath>
and
<cmath>a^2 = -b^2</cmath>
Focusing on (1) then gives 
<cmath>4a^2 = 1</cmath>
<cmath>a^2 = \frac{1}{4}</cmath>
Now, finding the square roots gives
<cmath>a = \pm\frac{1}{2}, b = \pm\frac{i}{2}</cmath>
And so there are <math>\boxed{(D)4}</math> possible solutions. 
~elpianista227

~edited by DRA777

==Video Solution==

For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc
==See Also==
{{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}}
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