I'm trying to show that three points in the complex plane $A,B,C$ with affixes denoted by $a,b,c$ form a direct equilateral triangle iff $a+b\omega+c\omega^2=0$, where $\omega=\exp(\frac{2\pi}{3}i)$, that is $\omega$ is a cube root of unity ($\omega^3=1$). Now, I know this question has been asked before, but I'm still asking it because I think there are some things I'm misunderstanding, and I'm not entirely sure my argument is quite correct.
One thing I was able to see, is that multiplication by some complex $z$, which amounts to scaling and rotating the triangle doesnt change the equality which makes sense, rotating and scaling an equilateral triangle still yields an equilateral triangle. Similarly, since $1+\omega+\omega^2=0$, it's easy enough to see that if we translate $a,b,c$ by some $z$ we still have $(a-z)+(b-z)\omega+(c-z)\omega^2=0\quad (1)$. Again, geometrically, this makes sense since moving the triangle around doesn't change the fact that it's equilateral. So proving the equivalence for a specific equilateral triangle should be enough to prove if for all equilateral triangle (right?).
Specifically, taking $z=a$ in $(1)$ we get $(b-a)\omega+(c-a)\omega^2=0$, and also diving by $b-a$ we end up with $\omega+\frac{c-a}{b-a}\omega^2=0$. For this to be true, we must have $\frac{c-a}{b-a}=-1/\omega=-\omega^2$. That is, $\left|c-a\right|=\left|b-a\right|=1$ and $\arg\left(\frac{b-a}{c-a}\right)=\frac{\pi}{3}$ thus $\widehat{BAC}=\pi/3$. So, I think this allows me to conclude that if $a+b\omega+c\omega^2=0$ then $\Delta ABC$ is indeed equilateral.
To prove the other direction, let's assume $\Delta ABC$ is indeed equilateral. Then $\arg\left(\frac{c-a}{b-a}\right)=\frac{\pi}{3}$ and $|b-a|=|c-a|$. Again, since every equilateral triangle can be obtained with some rotation and scaling of another one, we can assume WLOG $|b-a|=|c-a|=1$, thus $\frac{c-a}{b-a}$ lies on the unit circle, so that $\frac{c-a}{b-a}=\exp\left(i\frac{\pi}{3}\right)$. Multiplying both sides by $\omega^2$ then we have $$\frac{c-a}{b-a}\omega^2 = \exp\left(i\frac{\pi}{3}\right)\cdot\exp\left(i\frac{4\pi}{3}\right)=\exp\left(i\frac{5\pi}{3}\right)=\exp\left(-\frac{\pi}{3}i\right)=-\exp\left(-\frac{\pi}{3}i+\pi i\right)=-\omega.$$
Thus, $\omega+\frac{c-a}{b-a}\omega^2=0$ which in turn implies $(b-a)\omega+(c-a)\omega^2=0$. Finally, translating this by $a$ (that's still okay right?) yields $a+b\omega+c\omega^2=0$.
Best Answer
We know that $1+w+w^2=0$, if $a,b,c$ are complex numbers representing the vertices $A,B.C$ of a triangle ABC, then $$a+bw+cw^2=0 \implies a+bw+c(-1-w)=0 $$ $$\implies a-c+(b-c)w=0 \implies \frac{b-c}{a-c}=w=e^{2i\pi/3}=e^{2i\pi/3} e^{-2i\pi}$$ $$\frac{b-c}{a-c}=1.e^{-i\pi/3}$$ AS per Coni's law $$\frac{BC}{AC}=\frac{|BC|}{|AC|}e^{i\theta}.$$ The angle between BC and AC is $\pi/3$ and $|BC|=AC|,$ consequently the triangle ABC is equilaterral.