TITLE: Vertices of equilateral triangle inscribed in the unit circle
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: You are given three vectors $z_i\in{\mathbb C}$ summing to $0$; therefore they can be regarded as the "directed sides" of a triangle. Since all three of them have length $1$ this triangle is equilateral. It follows that $z_{i+1}$ is obtained from $z_i$ by a $120^\circ$ rotation, or in terms of complex numbers: $z_{i+1}=e^{2\pi i/3} z_i$.