# Proposition 3.10 from Conway's book

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!

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.

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 $$S(z_1) = x_* \in \mathbb{R}$$.

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.$$\qquad\text{Q.E.D.}$$