AIME 1984 · 第 8 题

Solution 1

We shall introduce another factor to make the equation easier to solve. 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 $m$ (the ninth degree roots of unity). Now we simply need to find the root within the desired range that satisfies our original equation $x^6 + x^3 + 1 = 0$.

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 $r^6+r^3+1=3$. (When we multiplied by $r^3 - 1$ at the beginning, we introduced some extraneous solutions, and the solution with $120^\circ$ was one of them.) This leaves $\boxed{\theta=160}$.

Solution 2

The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$. Applying the quadratic formula gives roots $y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$, which have arguments of $120$ and $240,$ respectively. We can write them as $z^3 = \cos 240^\circ + i\sin 240^\circ$ and $z^3 = \cos 120^\circ + i\sin 120^\circ$. So we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above! For $\cos 240^\circ + i\sin 240$ we have $(\cos 240^\circ + i\sin 240^\circ)^{1/3}$ $\Rightarrow$ $\cos 80^\circ + i\sin 80^\circ, \cos 200^\circ + i\sin200^\circ,$ and $\cos 320^\circ + i\sin320^\circ.$ Similarly for $(\cos 120^\circ + i\sin 120^\circ)^{1/3}$, we have $\cos 40^\circ + i\sin 40^\circ, \cos 160^\circ + i\sin 160^\circ,$ and $\cos 280^\circ + i\sin 280^\circ.$ The only argument out of all these roots that fits the description is $\theta = \boxed{160}$

Note: We can add $120$ to the angles of the previous solutions to get new solutions because De Moivre's formula says that $(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta$ and $\frac{360}{3} = 120$. ~programmeruser

~ blueballoon

Solution 3

As in Solution 2, make the substitution $u=z^3$. Then we are left with $u^2+u+1=0$, and we would have the solutions $e^{\frac{2\pi}{3}i},e^{\frac{4\pi}{3}i},e^{\frac{8\pi}{3}i},e^{\frac{10}{3}\pi}$. The latter two solutions are obtained by adding one extra revolution around the unit circle. (Notice how we omitted $e^{\frac{6\pi}{3}i}$, since this would yield $e^{2i\pi}=1$ and does not satisfy the equation.) Now, we substitute back, which gives us $z^3=e^{\frac{2\pi}{3}i},e^{\frac{4\pi}{3}i},e^{\frac{10}{3}\pi}\implies z=e^{\frac{2\pi}{9}i},e^{\frac{4\pi}{9}i},e^{\frac{8\pi}{9}i},e^{\frac{10\pi}{9}i}$. The only root in the range $\frac{\pi}{2}<\theta<\pi$ is achieved when $\theta=\frac{8\pi}{9}=\boxed{160^\circ}$. Note that there are 6 solutions to $z$ in this equation, and they are obtained by simply adding more revolutions around the unit circle and dividing by 3.