AMC12 2020 B

Answer (B): If $|z_1|=|z_2|=1$ and $z_1+z_2=0$, then $0$ is the midpoint between $z_1$ and $z_2$, which is to say that they are opposite each other on the unit circle. Thus $n=2$ has the property in question. If $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$, then let $w_k=\overline{z_1}z_k$. The angle between $w_j$ and $w_k$ is the argument of $\frac{w_j}{w_k}$. Because $\frac{w_j}{w_k}=\frac{z_j}{z_k}$, the angles between the $w$'s are the same as those between the $z$'s. Also, $w_1+w_2+w_3=\overline{z_1}(z_1+z_2+z_3)=0$. (The advantage of looking at the $w$'s rather than the $z$'s is that $w_1=1$.) Let $u_k$ and $v_k$ be the real and imaginary parts of $w_k$. Then $1+u_2+u_3=0$ and $v_2+v_3=0$. Now $u_3\ge -1$, which is to say that $-u_3\le 1$, so $u_2=-1-u_3\le 0$. Similarly, $u_3\le 0$. From $v_3=-v_2$ it follows that $v_2^2=v_3^2$, so $u_2^2=1-v_2^2=1-v_3^2=u_3^2$ and hence $u_2=u_3$, because they are both in the interval $[-1,0]$. But $1+u_2+u_3=0$, so $u_2=u_3=-\frac12$. Hence $w_1,w_2,$ and $w_3$ are the cube roots of unity, with equal angular spacing. Thus $n=3$ has the property in question. It remains to show that no $n>3$ has the property. Suppose that $n=2m\ge 4$. Take $m$ pairs of numbers on the unit circle of the form $z,-z$. Such a set of $n$ numbers sum to $0$, but in general will not be equally spaced on the unit circle. Suppose that $n=2m+3\ge 5$. Take the three cube roots of unity and $m$ pairs of numbers of the form $z,-z$. They will sum to $0$, but in general will not be equally spaced. There are just $2$ such values of $n$.