Equilateral triangle condition

Condition. With $j=e^{2i\pi/3}$ (rotation by $2\pi/3$ about the origin), $ABC$ is equilateral with direct orientation iff $c-a=j^2(b-a)$ or $c-a=j(b-a)$ (depending on orientation). Let's use $c-a=j(b-a)$: $c-a=jb-ja$, so $c=a+jb-ja=a(1-j)+jb$. Since $1-j=j^2$ (because $1+j+j^2=0$), we get $c=aj^2+bj=a j^2+bj$. Hmm.

Let's write the condition $c-a=j(b-a)$, i.e. $c-a-jb+ja=0$, i.e. $a(j-1)-jb+c=0$. Multiply by $j^2$: $a(j^2-j^3)-j^3 b+j^2 c=a(j^2-1)-b+j^2 c=0$. Since $j^2-1=j^2-1=-(1-j^2)=-j(1+j)$... this is getting messy. Let me use the standard: $ABC$ equilateral iff $a+b\omega+c\omega^2=0$ where $\omega=j^2$ (or $j$), depending on orientation.

Conclusion. The condition is $a+bj+cj^2=0$ (or $a+bj^2+cj=0$ for the other orientation).