Vertices of equilateral triangle inscribed in the unit circle 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. 
 A: $|z_1+z_2|^2=|z_3|^2\Rightarrow z_1\bar{z_2}+z_2\bar{z_1}=-1\Rightarrow|z_1|^2+|z_2|^2-z_1\bar{z_2}-z_2\bar{z_1}=|z_1-z_2|^2=3$
So $|z_1-z_2|=\sqrt{3}$ and similarly you have $|z_2-z_3|=|z_3-z_1|=\sqrt{3}.$
A: 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: $z_{i+1}=e^{2\pi i/3} z_i$. 
A: 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 $-1/2$. 
A: Consider the points $0$, $z_1$, $z_1+z_2$, and $z_1+z_2+z_3=0$ of the complex plane (the first and last are the same). By what is given, they form an equilateral triangle with side $1$, and one vertex at the origin. Then it is geometrically obvious that either $z_1+z_2=\exp\left(\frac{\pi\mathbf i}3\right)z_1$ or $z_1+z_2=\exp\left(\frac{5\pi\mathbf i}3\right)z_1$, and from this it follows that either
$$
 z_2=\Bigl(\exp\left(\frac{\pi\mathbf i}3\right)-1\Bigr)z_1=\exp\left(\frac{2\pi\mathbf i}3\right)z_1
 \quad\text{and} \quad z_3=\exp\left(\frac{4\pi\mathbf i}3\right)z_1,
$$
or
$$
  z_2=\exp\left(\frac{4\pi\mathbf i}3\right)z_1\quad\text{and} \quad z_3=\exp\left(\frac{2\pi\mathbf i}3\right)z_1.
$$
