Prove this three complex are $z_{1}=z_{2}=z_{3}$ When I deal with a geometric problem, get the following algebraic problems:
Assmue that 
$$H(p,q)=\dfrac{\omega p}{\omega-1+a(\omega p-q)},a>0$$
where $\omega^3=1,\omega\neq 1$. If
$$H(z_{1},z_{2}),H(z_{2},z_{3}),H(z_{3},z_{1})\in\mathbb{R},|z_{1}|=|z_{2}|=|z_{3}|=1$$
show that
$$z_{1}=z_{2}=z_{3}.$$
How to solve this problem?
 A: It is not a full answer.
It looks like false.
Set in $\mathbb{C}^3$ $(z_1, z_2, z_3) | |z_1|=|z_2|=|z_3|=1$ without $z_1=z_2=z_3$ and $w-1 + a(wz_i-z_j) \ne 0$ is linear connected.  (need to strict proof, maybe i'll edit it later)
$f(z_1, z_2,z_3) = im(H(z_1,z_2)H(z_2, z_3)H(z_3, z_1))$ is continious.
So, if we can find point $(z_1,z_2,z_3)$ with the negative $im(H(z_1,z_2)H(z_2, z_3)H(z_3, z_1))$ and point $(z_1,z_2,z_3)$ with the positive $im(H(z_1,z_2)H(z_2, z_3)H(z_3, z_1))$ ,
we can find point  $(z_1,z_2,z_3)$ with $H(z1,z2)H(z2,z3)H(z3,z1) \in \mathbb{R}$  on the way from first point to the second point
Let $z_2 =wz_1$, $z_3=wz_2$ 
$H(z1,z2)H(z2,z3)H(z3,z1) = Cz_1^3$, $C \ne 0$ - so we can create this 2 points.
A: Here is a counterexample which however is somewhat special.
If we select $w = \exp(i 2 \pi/3)$ and $a = \sqrt 3$ and $z_1 = z_3 =\exp(i5  \pi / 6)$ and $z_2 = \exp(i 7  \pi / 6)$ 
then we have 
$H(z_2,z_3) = H(z_3,z_1) = 1/\sqrt 3$ and 
$1/ H(z_1,z_2) = 0$.  
Hence in a situation where NOT all $z_i$ are equal we have that all $H(p,q)$ are real, however  $H(z1,z2)$ becomes infinite.
