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Definition: If $z_1\in \mathbb{C}_{\infty}$ then $(z_1,z_2,z_3,z_4)$ (the cross ratio of $z_1,z_2,z_3$ and $z_4$) is the image of $z_1$ under the unique Mobius transformation which takes $z_2$ to $1$, $z_3$ to $0$ and $z_4$ to $\infty$.

Proposition: Let $z_1,z_2,z_3,z_4$ be four distinct points on $\mathbb{C}_{\infty}$. Then $(z_1,z_2,z_3,z_4)$ is a real number iff all four points lie on a circle.

Proof: Let $S:\mathbb{C}_{\infty}\to \mathbb{C}_{\infty}$ be defined by $Sz=(z,z_2,z_3,z_4) $; then $S^{-1}(\mathbb{R})=$the set of $z$ such that $(z,z_2,z_3,z_4)$ is real. So it is enough to show that the image of $\mathbb{R}_{\infty}$ under a Mobius transformation is a circle.

This is the excerpt from Conway's book on Complex Analysis of one variable.

Can anyone please explain why it is enough to show that $S(\mathbb{R}_{\infty})=\Gamma$ - circle? I have spent about one hour trying to understand it and write down something but I failed to understand it.

Would be very grateful for detailed help, please!

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I think to catch every degenerate case you have to view $\infty$ as a real number.

By the first remark of the proof, one has to show that $S^{-1}(\mathbb{R_\infty})$ is exactly the set of points $z$ such that $z, z_2, z_3, z_4$ lie on a circle. We know that $z_2, z_3, z_4 \in S^{-1}(\mathbb{R_\infty})$ since their images are $1, 0, \infty$ respectively and we also know that $S^{-1}$ is a Möbius transform, too. So if $S^{-1}(\mathbb{R_\infty})$ is a circle it is the unique circle through $z_2, z_3$ and $z_4$, completing the proof.

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By "circle" I mean a circle or straight line.

Suppose any Möbius transformation transforms $\mathbb{R}_\infty$ into a "circle".

Define $S(z) = (z,z_2,z_3,z_4)$. Note that $S^{-1}$ is also a Möbius transformation.

Suppose $S(z_1) \in \mathbb{R}$. Then $S(z_1), 1,0,\infty \in \mathbb{R}_\infty$, and $S^{-1}$ transforms $\mathbb{R}_\infty$ into a "circle", and $S^{-1}(S(z_1))=z_1, S^{-1}(1)=z_2, S^{-1}(0)=z_3, S^{-1}(\infty)=z_4$, hence the $z_k$ lie on a "circle".

Now suppose the $z_k$ lie on a "circle". Again, $S^{-1}$ is a Möbius transformation and hence $S^{-1} ( \mathbb{R}_\infty)$ is a "circle". Since the circle is uniquely defined by $z_2,z_3,z_4$ and $z_1$ lies on the "circle" we see that $z_1 = S^{-1}(x_*)$ for some $x_* \in \mathbb{R}_\infty$.

Now note that $S(z_3) = \infty$ and $z_1 \neq z_3$ we see that $x_* \in \mathbb{R}$ (that is, not the extended line) and so $S(z_1) = x_* \in \mathbb{R}$.

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Proposition: The image of $\mathbb{R}_{\infty}$ under a Mobius transformation is a circle.

Proof of proposition is given in Conway's textbook . Let's prove the following using the proposition.


$z_1,z_2,z_3,z_4$ be four distinct points on $\mathbb{C}_{\infty}$. Then $(z_1,z_2,z_3,z_4)$ is a real number iff all four points lie on a circle.

Proof:$(\implies\text{direction})$ Suppose $(z_1,z_2,z_3,z_4)$ is a real number. Define $Sz=(z,z_2,z_3,z_4)\;$(same as above). Then $S^{-1} $ is again a Mobius transformation. Note that $S^{-1}(\Bbb R_{\infty})=\{z\in\Bbb C_{\infty}\mid (z,z_2,z_3,z_4)\in\Bbb R_{\infty}\}.$ Then by above proposition, $S^{-1}(\Bbb R_{\infty})$ is a circle. In particular, $z_1,z_2,z_3,z_4$ lies on a circle.

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$(\impliedby\text{direction})$Suppose all the four points lie on a circle. We need to show that $(z_1,z_2,z_3,z_4)$ is a real number. Then again using the above proposition, we can find a Mobius transformation $T $ such that $T^{-1}(z_j)\in\Bbb R_{\infty},\;j\in\{1,2,3,4\}$. Now using the invariant property of cross ratio under Mobius transformation, we can write $$(z_1,z_2,z_3,z_4)=(T^{-1}z_1,T^{-1}z_2,T^{-1}z_3,T^{-1}z_4).$$ Note that the RHS contains only real numbers. Hence $(z_1,z_2,z_3,z_4)$ is a real number.$\qquad\text{Q.E.D.}$

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