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2000 AIME II Problems/Problem 9: Difference between revisions

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==Solution 2==
==Solution 2==
Let <math>z=re^{i\theta}</math>. Notice that we have <math>2\cos(3\circ)=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{i\theta}.</math> Without loss of generality, suppose <math>r=1</math>. Therefore, <math>z=e^{i\frac{\pi}{\theta}}</math> and plugging into the desired expression, we get <math>e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}=0</math>. Therefore, the least integer greater is <math>\boxed{000}.</math>
Let <math>z=re^{i\theta}</math>. Notice that we have <math>2\cos(3\circ)=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{i\theta}.</math>  
 
Without loss of generality, suppose <math>r=1</math>. Therefore, <math>z=e^{i\frac{\pi}{\theta}}</math> and plugging into the desired expression, we get <math>e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=0</math>. Therefore, the least integer greater is <math>\boxed{000}.</math>
 
~solution by williamgolly


== See also ==
== See also ==

Revision as of 12:08, 14 July 2020

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

Without loss of generality, suppose $r=1$. Therefore, $z=e^{i\frac{\pi}{\theta}}$ and plugging into the desired expression, we get $e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=0$. Therefore, the least integer greater is $\boxed{000}.$

~solution by williamgolly

See also

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

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