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

(Solution)
(Solution)
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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>
  
Let <math>z = a + bi</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 $z$ 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 $z=\cos 3^\circ + i\sing 3^\circ$ (Error compiling LaTeX. Unknown error_msg)

Using DeMoivre's theorem we can now find $z^{2000}+\frac 1{z^{2000}}$.

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