Is there a nice solution using complex numbers/roots of unity for the following question:
The set of points that satisfy both $x^3-3y^2x\ge 3x^2y-y^3$ and $x+y=-1$ is a line segment. Find the length of this line segment.
This question was posted here and some nice solutions were provided using conventional algebra.
The question here is whether there is a solution which uses the complex numbers/roots of unity approach, as this question was found in Zeitz's The Art and Craft of Problem Solving under the section on this topic.
From Gerry Myerson's comment, we note that the inequality is
$$ \Re(( x-\mathrm i y)^3) + \Im ((x-\mathrm i y)^3) > 0$$ or equivalently
$$ \Re ((x+\mathrm i y)^3 ) > \Im((x+\mathrm i y)^3).$$ writing $x+\mathrm i y = r e^{\mathrm i \theta}$, we need to know when
$$ r^3 cos(3\theta) > r^3 \sin(3\theta).$$ Let me skip the high school trigonometry work, this gives you 3 regions of angles
$$ \theta \in \left(\frac{-\pi}4, \frac{\pi}{12}\right) + \frac{2k\pi}3, \quad k = 0,1,2.$$
Draw a graph:
now its clear we just need to find these 2 intersection points. I'll leave that part to you.