Difference between revisions of "2000 AIME II Problems/Problem 9"
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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 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> | ||
− | We have | + | 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\sing 3^\circ</math> | ||
− | + | Using DeMoivre's theorem we can now find <math>z^{2000}+\frac 1{z^{2000}}</math>. | |
== 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 17:48, 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 $z=\cos 3^\circ + i\sing 3^\circ$ (Error compiling LaTeX. Unknown error_msg)
Using DeMoivre's theorem we can now find .
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 |