Equilateral Triangle Property

If $\omega=e^{2\pi i/3}$ , then $\omega(a+b+c)=(b+c+a)$ so $(\omega-1)(a+b+c)=0$ and $a+b+c=\boxed{0}$

Let be $a$ , $b$ and $c$ the complex numbers and their affixes are vertices of an equilateral triangle. They verify that

$b= 1 _{120^\circ} \cdot a$ and $c= 1 _{240^\circ} \cdot a$ . The affixes of $b$ and $c$ are the result of rotating the affix of $a$ the angles $120^\circ$ and $240^\circ$ respectively.

And we have

$a \cdot a^{-1} = 1$ , $b \cdot a^{-1} = 1 _{120^\circ}$ , $c \cdot a^{-1} = 1 _{240^\circ}$ .

If $\omega$ denote $\omega = b \cdot a^{-1} \Rightarrow w^{2} = c \cdot a^{-1}$ and $w^{3} = 1 = a \cdot a^{-1}$ .

Then $\{ 1, \omega, \omega^{2} \}$ are roots of the polynomial $z^{3} - 1$ .

Using the Cardano-Vieta formulas we have that the sum of the three roots is equal to the opposite of the grade 2 monomial coefficient of the polynomial $z^{3} - 1$ .

So $1 + \omega + \omega^{2} = 0$ due to 0 is the coefficient of $z^{2}$ monomial.

Then

$a \cdot a^{-1} + b \cdot a^{-1} + c \cdot a^{-1} = 0 \Rightarrow$

$(a + b + c) \cdot a^{-1} = 0$ and $a \neq 0 \Rightarrow$