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Difference between revisions of "2000 AIME II Problems/Problem 9"

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== Solution ==
 
== Solution ==
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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>.
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Using the quadratic equation on <math>z^2 - (2 \cos 3 )z + 1 = 0</math>, we have <math>z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>.
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There are other ways we can come to this conclusion. 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>. Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math>.
  
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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>.
 
  
 
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  

Revision as of 11:39, 6 September 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

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 $\boxed{000}$.

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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