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Let $z,z_2,z_3,z_4$ be four points on the extended plane. Their cross-ratio $(z,z_2,z_3,z_4)$ by definition is the image $Tz$ of $z$ under the Möbius transformation $T$ that sends $z_2,z_3,z_4$ to $0,1,\infty$ respectively.

According to L. Ahlfors on Complex Analysis, 3rd edition, page 79, to prove that the cross-ratio of four points is a real number exactly when the four points lie on a line or a circle, it suffices to show that "the image of the real axis under any [Möbius transformation] is either a circle or a straight line." He says this is obvious since $Tz=(z,z_2,z_3,z_4)$ is real exactly when it is on the image of the real line under the transformation $T^{-1}$ which is also a Möbius transformation.

He says in his book that "indeed, $Tz=(z,z_2,z_3,z_4)$ is real on the iamge of the real axis under the transformation $T^{-1} $ and nowhere else." How does the theorem follow?

brainjam
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superAnnoyingUser
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  • See also this question: http://math.stackexchange.com/questions/1257294/four-complex-numbers-z-1-z-2-z-3-z-4-lie-on-a-generalized-circle-if-and-only-i/1257375#1257375 – EthanAlvaree Apr 29 '15 at 11:17

1 Answers1

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Every Möbius transformation is an automorphism of the extended plane. So $Tz\in \mathbb{R}\cup\{\infty\}$ if and only if $z\in T^{-1}(\mathbb{R}\cup\{\infty\})$. Since Möbius transformations map circles (in the extended plane) to circles, that means $z$ lies on a circle passing through $z_1,z_2,z_3$. But through any three points in the extended plane, there passes only one circle, so

$$Tz\in\mathbb{R}\cup\{\infty\}\iff z \text{ lies on the unique circle passing through } z_1,z_2,z_3.$$

Daniel Fischer
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  • Indeed, if $T^{-1}z$ is real, it lies on the cirlcle passing through $z_1,z_2,z_3$, since $T^{-1}z_i$ are also real, thus they are on the circle (line) as $T^{-1}z$. – superAnnoyingUser Jul 05 '14 at 13:04