Three distinct points $x, y, z$ lie on a unit circle of the complex plane and satisfy $x+y+z=0$. Then $x, y, z$ form the vertices of.
On
The points $0, x, y, x+y$ form a parallelogram but $x+y=-z$ which implies $|x+y|=|z|=|x|=|y|$. This means that both $0, x, x+y$ and $0, y, x+y$ form congruent equilateral triangles. Similarly, $x, -z, y, -x, z, -y$ form the vertices of a regular hexagon and hence $x, y, z$ form an equilateral triangle.
Another solution: Suppose $a, b, c$ are three points. Assume that thetriangle ($\triangle abc$) has centroid $d$ which is also the center of the circumcircle ($\bigcirc abc$). By using a complex linear transformation ($\, z \mapsto (z-d)/(c-d)\,$) we can assume (without loss of generality) that $$d=0,\quad c=1,\quad a+b=-1, |a|=|b|=1.$$ Now also $\,\bar{a}+\bar{b}=-1\,$ and this implies the diagonals of quadrilateral $\,a\bar{b}b\bar{a}\,$ bisect each other and hence is a parallelogram. This now implies that $\,a=\bar{b}\,$ and they are primitive curbe roots of unity. Hence $\triangle abc$ is equilateral.
Let's prove the answer is 2).
Write $x=e^{ia}$ etc. so $e^{ia}+e^{ib}+e^{ic}=0$ and $e^{i(b-a)}+e^{i(c-a)}=-1$. Equating imaginary parts, $\sin(b-a)=\sin(a-c)$; equating real ones, $\cos(b-a)+\cos(a-c)=-1$. The cosines are equal, rather than equal and opposite, and so each is $-\frac12$, whence $\sin(b-a)=\pm\frac{\sqrt{3}}{2}=\pm\sin\frac{2\pi}{3}$. Note in particular that $\cos(b-a)=-\frac12\implies b-a\ne\pm\frac{\pi}{3}$.
The angles between $x,\,y,\,z$ are $\frac{2\pi}{3}$ each, while the circumradii at the vertices split $\triangle xyz$ into isosceles triangles of base angles $\frac12\left(\pi-\frac{2\pi}{3}\right)=\frac{\pi}{6}$. Thus each internal angle of $\triangle xyz$ is $\frac{\pi}{3}$, as expected.