1984 AIME Problems/Problem 8

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 1

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 $\boxed{\theta=160}$ .

Solution 2

Note that the substitution $y=z^3$ simplifies this to $y^2+y+1$ . Simply applying the quadratic formula gives roots $y_{1,2}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$ , which have angles of 120 and 240, respectively. This means $arg(z) = \frac{120,240}{3} + \frac{360n}{3}$ , and the only one between 90 and 180 is $\boxed{\theta=160}$ .

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1984 AIME Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See also