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

Revision as of 21:11, 30 November 2007

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 $\boxed{\theta=160}$ .

@@ Line 10: / Line 10: @@
 {{AIME box|year=1984|num-b=7|num-a=9}}
-[[Category:Intermediate Complex Numbers Problems]]
 [[Category:Intermediate Trigonometry Problems]]

1984 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1984 AIME Problems/Problem 8"

Revision as of 21:11, 30 November 2007

Problem

Solution

See also