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Theorem: The cross ratio $(z_1,z_2,z_3,z_4)$ is real if and only if the four points lie on a circle or on a straight line.

We need only show that the image of the real axis under any linear transformation is either a circle or a straight line. Indeed, $Tz=(z,z_2,z_3,z_4)$ is real on the image of the real axis under the transformation $T^{-1}$ and nowhere else.

I don't understand this paragraph. Why is it that $Tz$ is real on the image of the real axis under $T^{-1}$, and why is it real nowhere else? And why is it enough to show only this?

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The cross ratio, as a function of the first argument, is a Möbius transformation,

$$T \colon z \mapsto (z,z_2,z_3,z_4).$$

We have $T(z_2) = 0$, $T(z_3) = 1$, $T(z_4) = \infty$ (or some permutation thereof, depends on the used definition of the cross ratio). So if we know that Möbius transforms map circles or straight lines to circles or straight lines, we know that $T$ maps the circle or straight line passing through the three points $z_2,\,z_3,\,z_4$ to the real line (plus infinity). Since Möbius transformations are bijective, $T(z)$ is real (or $\infty$) if and only if $z = T^{-1}(Tz)$ lies on the circle or straight line passing through $z_2,\,z_3,\,z_4$, which is the image of the real line (plus $\infty$) under $T^{-1}$.

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