4

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?

math110
  • 93,304
  • 3
    After a membership of almost two years, you should know that people here usually ask: What have you tried? And why do we have to guess which numbers are real and which numbers are complex? – Han de Bruijn Jan 08 '17 at 18:12

2 Answers2

0

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.

kotomord
  • 1,814
0

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.

Andreas
  • 15,175