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Suppose $z_1$, $z_2$ and $z_3$ are distinct points in $\hat{\mathbb {C}}$ . Show that $z$ is on the circle passing through $z_1$, $z_2$ and $z_3$ if and only if [$z$, $z_1$, $z_2$, $z_3$] is real or ∞.

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marked as duplicate by Martin R, Pragabhava, Chris Godsil, Harish Chandra Rajpoot, Daniel W. Farlow Dec 10 '15 at 0:55

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  • $\begingroup$ A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka ? $\endgroup$ – BCLC Jul 28 '18 at 12:52
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Sketch:

  1. Möbius transformations preserve the cross-ratio, $$ [z,z_1,z_2,z_3]=[T(z),T(z_1),T(z_2),T(z_3)]. $$
  2. There is a Möbius transformation sending $z_1 \mapsto 0$, $z_2 \mapsto 1$, $z_3 \mapsto \infty$.
  3. Since Möbius transformations are bijective and map circles to circles, $w$ is on the real line (${}\cup \{\infty\}$) if and only if $z$ is on the circle through $z_1,z_2,z_3$.
  4. $[w,0,1,\infty]=w$ (or $1/w$ or $1-w$ or some composition thereof, depending on the definition). Hence the cross-ratio is real iff $w$ is real or $\infty$, which proves what you want.
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  • $\begingroup$ Chappers, Why is w real if z is on the circle through z1,z2,z3?, $w=T(z)$? It seems you proved cross ratio is real iff z is on the circle. But what about real or ∞ iff z is on the circle ? $\endgroup$ – BCLC Jul 28 '18 at 13:36
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    $\begingroup$ @BCLC There's a reason it says "sketch". Möbius transformations are bijective on $\hat{\mathbb{C}}$, so $w$ could only be $\infty$ if it's one of the three points we already know are on the circle. $\endgroup$ – Chappers Jul 28 '18 at 18:56
  • $\begingroup$ thanks so much! ^-^ $\endgroup$ – BCLC Jul 29 '18 at 1:03
  • $\begingroup$ I think I got it. Is my answer wrong please? $\endgroup$ – BCLC Jul 29 '18 at 5:37

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