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

Revision as of 19:34, 18 March 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 $\boxed{000}$ .

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

@@ Line 15: / Line 15: @@
 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}}

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

Revision as of 19:34, 18 March 2008

Problem

Solution