Difference between revisions of "2000 AIME II Problems/Problem 9"
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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\ | + | Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math> |
Using DeMoivre's theorem we have <math>z^{2000} = \cos 6000^\circ + i\isin 6000^\circ</math> | Using DeMoivre's theorem we have <math>z^{2000} = \cos 6000^\circ + i\isin 6000^\circ</math> | ||
| + | 6000 = 16(360) + 240, so <math>z^{2000} = \cos 240^\circ + i\isin 240^\circ</math> | ||
| − | + | We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math> | |
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== See also == | == 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}} | ||
Revision as of 17:58, 3 January 2008
Problem
Given that 
is a complex number such that 
, find the least integer that is greater than 
.
Solution
Note that if z is on the unit circle in the complex plane, then 
and 
We have 
and 
Alternatively, we could let 
and solve to get 
Using DeMoivre's theorem we have $z^{2000} = \cos 6000^\circ + i\isin 6000^\circ$ (Error compiling LaTeX. Unknown error_msg) 6000 = 16(360) + 240, so $z^{2000} = \cos 240^\circ + i\isin 240^\circ$ (Error compiling LaTeX. Unknown error_msg)
We want 
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 | ||
