2000 AIME II Problems/Problem 9: Difference between revisions

Revision as of 17:36, 3 January 2008

Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ .

Note that if z is on the unit circle in the complex plane, then Let $z = a + bi$ and we have $z = e^{i\theta}$ and $\frac 1z= e^{-i\theta}$

@@ Line 3: / Line 3: @@
 == Solution ==
-{{solution}}
 Note that if z is on the unit circle in the complex plane, then
-Let <math>z = a + bi</math> and we have <math>z = e^i\theta</math>  and <math>\frac 1z= e^{-i\theta}</math>
+Let <math>z = a + bi</math> and we have <math>z = e^{i\theta}</math>  and <math>\frac 1z= e^{-i\theta}</math>
 == See also ==
 {{AIME box|year=2000|n=II|num-b=8|num-a=10}}

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