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Given distinct points $z_1, z_2, z_3, z_4 \in \mathbb{C}\cup\{\infty\}$, define the cross ratio $$(z_1, z_2, z_3, z_4) = \frac{(z_1 −z_3)(z_2 −z_4)}{(z_1 − z_4)(z_2 − z_3)};$$ if one of the points $z_k$ is $\infty$, cross out the two factors containing $z_k$.

  1. Show that the cross ratio is $f(z_1)$, where $f$ is the unique linear fractional transformation sending $z_2 \mapsto 1$, $z_3 \mapsto 0$, and $z_4 \mapsto \infty$.

  2. Given another four distinct points $w_1, w_2, w_3, w_4 \in \mathbb{C}\cup\{\infty\}$, show that there exists a linear fractional transformation sending $z_k \mapsto w_k$ for $k = 1, \dots, 4$ if and only if $(z_1, z_2, z_3, z_4) = (w_1, w_2, w_3, w_4)$.

  3. Show that $(z_1, z_2, z_3, z_4)$ is real if and only if the four points $z_1, z_2, z_3, z_4$ lie on a line or a circle.

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What are all those private-use unicode characters supposed to be? –  Chris Eagle Sep 20 '12 at 9:43
    
Also, what have you tried so far? –  Michael Albanese Sep 20 '12 at 11:47
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1 Answer

(a): The key is that any $f(z)=\frac{az+b}{cz+d}$ preserves the cross ratio. It can be proved directly, but follows in a more elegant manner from the following as well:

The best if we consider $\mathbb C^2$ instead of $\mathbb C$ (the complex projective plane) and identify each $z\in\mathbb C$ to any $(az,a)$ for $a\in\mathbb C\setminus\{0\}$. If $(a,b) = (\lambda\cdot a,\lambda\cdot b)$ then they represent the same element of $\mathbb C$.

It has more benefits, for example $\infty$ can be smoothly interpreted as $(1,0)$ (represented also by any $(a,0)$).

Lemma: Assume that ${\bf u},{\bf v}\in\mathbb C^2$ are given [representing $u,v\in\mathbb C$], and ${\bf w}={\bf u}+\alpha\cdot{\bf v}$, ${\bf z}={\bf u}+\beta\cdot{\bf v}$. Then $$(uvwz) = \displaystyle\frac\alpha\beta $$

Lemma: For $f$ as above, and if $z\in\mathbb C\cup\{\infty\}$ is represented by $\bf z$, $f(z)$ is represented by $$\left[\begin{array}{cc} a&b\\c&d \end{array}\right]\cdot {\bf z}$$

Ok, so, suppose we know $f$ preserves the cross ratio, and that $f(z_2)=1$, $f(z_3)=0$, $f(z_4)=\infty$. Then, $$(z_1 z_2 z_3 z_4) = (f(z_1),1,0,\infty) = \text{by def.}= \displaystyle\frac{f(z_1)-0}{1-0} = f(z_1)$$

One direction of (b) is exactly my first statement up there, and the other direction is the existence of such $f$: For given different $w_2,w_3,w_4$, it is relatively easy to construct an $f$ such that $f(w_2)=1,\ f(w_3)=0,\ f(w_4)=\infty$. The inverse of any $f$ as above can be given by inverting the matrix (note that we can discard the constant multiplier): $\left[\begin{array}{cc} d&-b\\-c&a \end{array}\right] $, i.e. $f^{-1}(z)=\displaystyle\frac{dz-b}{-cz+a}$.

For (c), you may need the equation of an arbitrary circle on the complex plain, say with centre $c\in\mathbb C$ and sugar $\varrho>0$. Then $z$ is on this circle iff $$|z-c|=\varrho \iff (z-c)(\bar z-\bar c) = \varrho^2 \iff \dots$$ and that being real for any $s\in\mathbb C$ means that $s=\bar s$.

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