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

(Solution)
(Solution)
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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%27s_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>
  

Revision as of 19:11, 3 January 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 $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$

Of course, we cannot have -1 as an answer on the AIME, but they asked for the smallest integer greater than this value, which is 0.

See also

2000 AIME II (ProblemsAnswer KeyResources)
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