What are the necessary and sufficient conditions for the roots of the equation $z^{2}+az+b$, and $0$, to be the vertices of an equilateral triangle?

Question

What are the necessary and sufficient conditions for the roots of the equation $z^{2}+az+b$, and $0$, to be the vertices of an equilateral triangle?

Answer 1

Let $z_1$ and $z_2$ be the two solutions. Then we want $b=z_1z_2\ne0$ and $${z_1\over z_2}\in\{e^{i\pi/3},e^{-i\pi/3}\}=\left\{{1\over2}+i{\sqrt{3}\over2},{1\over2}-i{\sqrt{3}\over2}\right\}\ .$$ This means that ${z_1\over z_2}$ should be a solution of the equation $t^2-t+1=0$. This implies $$z_1^2-z_1z_2+z_2^2=0\ ,$$ and therefore $$(z_1+z_2)^2=3z_1z_2\ .$$ It follows that $$a^2=3b\ne0$$ is a necessary condition for $\triangle(0,z_1,z_2)$ to be equilateral, and as the argument can be reversed, this condition is also sufficient.