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2000 AIME II Problems/Problem 9: Difference between revisions

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== Solution ==
== Solution ==
Note that if z is on the unit circle in the complex plane, then <math>z = e^{i\theta} = \cos \theta + i\sin \theta</math>  and <math>\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta</math>
Note that if <math>z</math> is on the [[unit circle]] in the complex plane, then <math>z = e^{i\theta} = \cos \theta + i\sin \theta</math>  and <math>\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta</math>.


We have <math>z+\frac 1z = 2\cos \theta = 2\cos 3^\circ</math> and <math>\theta = 3^\circ</math>
We have <math>z+\frac 1z = 2\cos \theta = 2\cos 3^\circ</math> and <math>\theta = 3^\circ</math>. Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math>.
Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math>


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


We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math>
We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math>.


Finally, the least integer greater than <math>-1</math> is <math>\boxed{000}</math>.
Finally, the least integer greater than <math>-1</math> is <math>\boxed{000}</math>.


== See also ==
{{AIME box|year=2000|n=II|num-b=8|num-a=10}}
{{AIME box|year=2000|n=II|num-b=8|num-a=10}}
[[Category:Intermediate Algebra Problems]]

Revision as of 09:28, 30 August 2008

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

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 $\boxed{000}$.

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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