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