1
$\begingroup$

Assume that $z_1, z_2, z_3, z_4$ lie on a line or circle C.

By what I just mention above, there exist a mobius tran $T$ that takes the real numbers $x_2, x_3, x_4$ to $z_2, z_3,z_4$ respectively. Since there is only one such line or circle and since we know that Mobius take $\mathbb{R}_{\infty}$ to a line or circle, we see that there exist a real number $x_1$ such that $T(x_1) = z_1$.

Since the cross ratio is preserved under a Mobius trans, then $ (x_1,x_2,x_3,x_4)=(T(x_1),T(x_2),T(x_3),T(x_4))=(z_1,z_2,z_3,z_4)$

But since $x_i$ is all real numbers, its cross ratio is also real, which show that the cross ratio of $z_i$ is also real.

$\endgroup$
6
  • $\begingroup$ Do you already know that the cross-ratio is preserved under Möbius transformations? $\endgroup$ Commented Jan 6, 2016 at 6:18
  • $\begingroup$ @David yes, i also learn that there exist a unique mobius trans betweeen three distinct points on the plane. but i didn't learn that mobius take circles to circles yet. $\endgroup$ Commented Jan 6, 2016 at 6:40
  • $\begingroup$ Khoa ta, you should leave the original question, and add the new information after the word Edit. What's left doesn't make sense on its own. $\endgroup$ Commented Jan 6, 2016 at 18:15
  • $\begingroup$ Yes, this is fine. I was suggesting $1, 0, \infty$ and you've taken $x_2, x_3, x_4$. Note that $x_1$ could also be $\infty$. $\endgroup$ Commented Jan 6, 2016 at 18:18
  • $\begingroup$ @David thank you for reply, I'm new to this site, if $x_1 $ is $\infty$, then does it still hold true that the cross ratio of $x_i$ is real?? $\endgroup$ Commented Jan 6, 2016 at 18:22

1 Answer 1

Reset to default
1
$\begingroup$

Assume that $z_1$, $z_2$, $z_3$ and $z_4$ lie on a line or circle $C$. Let $\phi$ be the Möbius transformation taking $1, 0, \infty$ to $z_2$, $z_3$, $z_4$, respectively. The image of $\mathbb{R} \cup \{\infty\}$ under $\phi$ is a line or circle passing through $z_2$, $z_3$ and $z_4$. Since there is only one such line or circle, it must be $C$. Therefore $\phi^{-1}(z_1) \in \mathbb{R} \cup \{\infty\}$. But $(z_1, z_2, z_3, z_4) = \phi^{-1}(z_1)$.

$\endgroup$
4
  • $\begingroup$ thank you very much, i got it now, i'm reading the text by myself so the only help I can get is thru this site, thank u again $\endgroup$ Commented Jan 6, 2016 at 7:18
  • $\begingroup$ sorry to bother you again, but regard what I ask you, I have another way of showing that, can you check to see if my reasoning is correct??, $\endgroup$ Commented Jan 6, 2016 at 17:52
  • $\begingroup$ Write your argument as an edit to the question. $\endgroup$ Commented Jan 6, 2016 at 17:57
  • $\begingroup$ sorry for taking so long, I accidentally edit your comment instead of mine, thank you again, $\endgroup$ Commented Jan 6, 2016 at 18:13

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.