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

See also

2000 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AIME Problems and Solutions

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