2000 AIME II Problems/Problem 9

Problem

Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ .

Solution

Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$ , we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$ .

There are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\theta} = \cos \theta + i\sin \theta$ and $\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta$ . We have $z+\frac 1z = 2\cos \theta = 2\cos 3^\circ$ and $\theta = 3^\circ$ . Alternatively, we could let $z = a + bi$ and solve to get $z=\cos 3^\circ + i\sin 3^\circ$ .

Using De Moivre's Theorem we have $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ$ , $6000 = 16(360) + 240$ , so $z^{2000} = \cos 240^\circ + i\sin 240^\circ$ .

We want $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$ .

Finally, the least integer greater than $-1$ is $\boxed{000}$ .

Solution 2

Let $z=re^{i\theta}$ . Notice that we have $2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.$

$r$ must be $1$ (or else if you take the magnitude would not be the same). Therefore, $z=e^{i\frac{\pi}{\theta}}$ and plugging into the desired expression, we get $e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1$ . Therefore, the least integer greater is $\boxed{000}.$

~solution by williamgolly

Solution 3 Intuitive

For this solution, we assume that $z^{2000} + 1/z^{2000}$ and $z^{2048} + 1/z^{2048}$ have the same least integer greater than their solution. we have $z + 1/z = 2\cos 3$ . Since $\cos 3 < 1$ , $2\cos 3 < 2$ . If we square the equation $z + 1/z = 2\cos 3$ , we get $z^2 + 2 + 1/(z^2) = 4\cos^2 3$ , or $z^2 + 1/(z^2) = 4\cos^2 3 - 2$ . $4\cos^2 3 - 2$ is is less than $2$ , since $4\cos^2 3$ is less than $4$ . if we square the equation again, we get $z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2$ . since $4\cos^2 3 - 2$ is less than 2, $(4\cos^2 3 - 2)^2$ is less than 4, and $(4\cos^2 3 - 2)^2 -2$ is less than 2. However $(4\cos^2 3 - 2)^2 -2$ is also less than $4\cos^2 3 - 2$ . we can see that every time we square the equation, the right hand side gets smaller, and into the negatives. Since the smallest integer that is allowed as an answer is 0, thus smallest integer greater is $\boxed{000}.$

~ PaperMath

Solution 4

First, let $z = a+bi$ where $a$ and $b$ are real numbers. We now have that $a+bi + \frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}$ given the coniditons of the problem. Equating imaginary coefficients, we have that $b \left( 1 - \frac{1}{a^2+b^2}\right) = 0$ giving us that either $b=0$ or $|z| = 1$ . Let's consider the latter case for now.

We now know that $a^2+b^2=1$ , so when we equate real coefficients we have that $2a = 2 \cos{3^{\circ}}$ , therefore $a = \cos{3^{\circ}}$ . So, $b = \cos{3^{\circ}}$ and then we can write $z = \text{cis}(3)^{\circ}$

By De Moivre's Theorem, $z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}$ . The imaginary parts cancel, leaving us with $2 \cos{6000^{\circ}}$ which is $240 \pmod{360}$ . Therefore, we have it being $-1$ and our answer is $\boxed{000}$ .

Now, if $b=0$ then we have that $a+\frac{1}{a} = 2 \cos{3^{\circ}}$ . Therefore, $a$ is not real violating our conditions set above.

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search