The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the [complex plane](https://artofproblemsolving.com/wiki/index.php/Complex_plane). Determine the degree measure of $\theta$.
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](https://artofproblemsolving.com/wiki/index.php/Absolute_value) $1$ and argument of the form $40m^\circ$ for integer $m$ (the ninth degree [roots of unity](https://artofproblemsolving.com/wiki/index.php/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 $\boxed{\theta=160}$.