Difference between revisions of "1984 AIME Problems/Problem 8"

(Solution)
(Solution)
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Also,
 
Also,
  
From above, you notice that <math>z^6+z^3+1 = \frac {r^9-1}{r^3-1</math>}<math>. Therefore, the solutions are all of the ninth roots of unity that are not the third roots of unity. After checking, the only angle is </math>\boxed{\theta=160}$.
+
From above, you notice that <math>z^6+z^3+1 = \frac {r^9-1}{r^3-1}</math>. Therefore, the solutions are all of the ninth roots of unity that are not the third roots of unity. After checking, the only angle is <math>\boxed{\theta=160}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 10:10, 14 March 2010

Problem

The equation $z^6+z^3+1$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$.

Solution

If $r$ is a root of $z^6+z^3+1$, then $0=(r^3-1)(r^6+r^3+1)=r^9-1$. The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\circ$ for integer $m$.

This reduces $\theta$ to either $120^{\circ}$ or $160^{\circ}$. But $\theta$ can't be $120^{\circ}$ because if $r=\cos 120^\circ +i\sin 120^\circ$, then $r^3=1$ and $r^6+r^3+1=3$, a contradiction. This leaves $\boxed{\theta=160}$.

Also,

From above, you notice that $z^6+z^3+1 = \frac {r^9-1}{r^3-1}$. Therefore, the solutions are all of the ninth roots of unity that are not the third roots of unity. After checking, the only angle is $\boxed{\theta=160}$.

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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