1984 AIME Problems/Problem 8

Revision as of 21:27, 26 March 2007 by Scorpius119 (talk | contribs) (solution added)


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


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 or 160. But $\theta$ can't be 120 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 $\theta=160$.

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