# Show that the cross-ratio is invarient under inversion

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

$$\frac{dab-dac-dbb+dbc-cab+cbb-cbc}{abcdc-abcdd}=\frac{dab-dac-dbb+dbc-aab+aac+abb-abc}{abcda-abcdd}$$

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?