Prove that if $z_{1}+z_{2}+z_{3}=0$ and $|z_{1}|=|z_{2}|=|z_{3}|=1$ then the points $z_{1},z_{2},z_{3}$ are the vertices of an equilateral triangle inscribed in the unit circle $|z|=1$.
My idea was the following: since the sum of the numbers has to be $0$ and they have equal modulus, the interior angles between them should be equal to $120º$.
Later I could use the Inscribed Angles Theorem to prove that
$$\arg \left(\frac{z_{j}-z_{i}}{z_{j}-z_{k}}\right)=\frac{1}{2}\arg\left(\frac{z_{i}}{z_{k}}\right)$$
and since all $\arg(z_{i}/z_{k})=120º$, all angles would be equal. So the triangle formed would be equilateral.
I am wondering if this is correct and if would there be a more organized way to prove this.
Best Answer
Multiplying through by $z_1^{-1}$ will rotate the numbers without changing any angles or distances, so you may assume $z_1=1$. Then $z_2$ and $z_3$ must have cancelling imaginary parts but can't be additive inverses; since they are on the unit circle, this forces them to be conjugate. So, they have the same real part, which must be $-1/2$ to cancel $z_1$. But you know where the numbers are on the unit circle that have real part $-1/2$.