The cross ratio is a function where the Input are four complex numbers $z_1,z_2,z_3,z_4$ and the output is $\frac{z_3-z_1}{z_3-z_2}:\frac{z_4-z_1}{z_4-z_2}$. I have to Show that the cross ratio is invarient under a broken-linear Transformation which is of the form $T:\mathbb{C}\backslash \{-\frac{d}{c}\}\rightarrow\mathbb{C}, T(z)=\frac{az+b}{cz+d}$. I have already shown that I can split the broken-linear Transformation in two Linear Transformations and one Inversion such that $T=L_2\circ I\circ L_1$. I could Show that it the number reameains the same after a liner Transformation but I don't know how I can Show that the number also will be the same when I Invert $z_1,...,z_4$.

So I want to Show that

$\frac{z_3-z_1}{z_3-z_2}:\frac{z_4-z_1}{z_4-z_2}= \frac{z_3^{-1}-z_1^{-1}}{z_3^{-1}-z_2^{-1}}:\frac{z_4^{-1}-z_1^{-1}}{z_4^{-1}-z_2^{-1}}$

After some manipulations I got the Claim

Let $z_1=a,z_2=b,z_3=c,z_4=d$


Is there Maybe an easier way to Show the equality? Because this fraction Looks very complicated and I could have made an error.

Also what does the crossection mean ?

I have made a Picture with $4$ Points in order to understand the crosssection and the properties


$z_5$ is the crosssection and if the other Points go through a broken linear Transformation the Point remains the same. But what does $z_5$ geometrically mean for the Points $z_1,z_2,...,z_4$. The Formula says that under broken-linear Transformation $z_5$ remains the same. That means it is an equivalencerealtion. If I take a $z_5$ I will get 4 representatives for the respective equivalenceclass. Are there some Special equivalenceclasses?


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