Given 3 distinct complex numbers z2,z3,z4, prove that the map $z \to [z,z_{2},z_{3},z_{4}]$ is a fractional linear transformation mapping $z_{2} \to {1},z_{3} \to 0, z_{4} \to \infty$.

I'm really not sure how to go about this exercise and think I need help with it. What do they mean by mapping $z$ to the cross ratio of $z,z_{2},z_{3},z_{4}$.

I'm only familiar with the $S(z) = \frac{(az+b)}{(cz+d)}$ form of a möbius transformation. I'm pretty sure I read that möbius transformation is preserve by cross ratio.

$(z_1,z_2,z_3,z_4) := \frac{(z_1 - z_3)(z_2 - z_4)}{(z_2 - z_3)(z_1 - z_4)}$


Shoving the cross ratio into the standard definition of a möbius transformation seems tedious.

So, I think of a function (which is of the form of a möbius transformation) that would map $z_{2} \to 1, \; z_{3} \to 0, \; z_{4} \to \infty$.

Initially, $f(z) = \frac{z-z_{3}}{z - z_{4}}$ seem like it would work, but nay since $z_{2}$ is not mapped properly.

Then, we add some generality to the function by introducing $a,b \in \mathbb{C}$.

$g(z) = \frac{a(z-z_{3})}{b(z - z_{4})}$. Then for some $a$ and $b$, $g(z_{1}) = 1$.

So we showed that a möbius transformation mapped to what we want. Not the particular mapping $z \to (z, z_2, z_3, z_4)$ though...

  • $\begingroup$ Can you also write down the definition of the cross ratio in the question? Thanks. $\endgroup$ – user99914 Oct 2 '15 at 7:30
  • $\begingroup$ @JohnMa I believe I have answered the question in the comments of the Answer by Groups $\endgroup$ – Yulo Oct 2 '15 at 8:00
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    $\begingroup$ $f(z)=[z,z_2,z_3,z_4]=\frac{(z-z_3)(z_2-z_4)}{(z-z_4)(z_2-z_3)}$, which is correct answer. Just the definition of $[z,z_2,z_3,z_4]$. $\endgroup$ – Groups Oct 2 '15 at 8:44
  • $\begingroup$ ....wow. I made this a lot more complicated than it was. Thank you for the insight you gave earlier. $\endgroup$ – Yulo Oct 2 '15 at 8:53
  • $\begingroup$ Your calculation are correct. However, what you got at the end (after you find out $a, b$ will be the same as the mapping $z\mapsto [z, z_2, z_3, z_4]$ (you should try and see that) $\endgroup$ – user99914 Oct 2 '15 at 9:26

Answer is in your question itself.

The map is $$z\mapsto [z,z_2,z_3,z_4]=\frac{...?...}{...?...}$$ Therefore, $z_1\rightarrow ...?...$ and so on.

  • $\begingroup$ Well, to keep with the mapping, let $f(z) = \frac{z - z_{3}}{z - z_{4}}$. This maps $z_3 \to 0$ and $z_4 \to 0$. This map isn't flexible enough though. Is this correct idea to go with? $\endgroup$ – Yulo Oct 2 '15 at 7:48
  • $\begingroup$ So, $g(z) = \frac{(az - az_3)}{ (bz - bz_4) }$. $a,b \in \mathbb{C}$. Then for some $a,b$, $g(z_2) = 1$ $\endgroup$ – Yulo Oct 2 '15 at 7:51
  • $\begingroup$ $g(z)$ is of the form of a fractional linear transformation which maps $z_2 \to 1$, $z_3 \to 0$, and $z_4 \to \infty$. We are done. Do you agree? $\endgroup$ – Yulo Oct 2 '15 at 7:55
  • $\begingroup$ @Groups it's added in the question $\endgroup$ – user99914 Oct 2 '15 at 8:15
  • $\begingroup$ @Groups I thought I completed the answer... see my edit $\endgroup$ – Yulo Oct 2 '15 at 8:39

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