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Question on answer to this (*), a duplicate of this and related to this

(*) Exer 3.24 in A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

What I understand:

  1. Based on Exer 2.14 or Exer 3.10, $T(z)$ refers to an arbitrary Möbius of the form $$\frac{az+b}{cz+d}, ad-bc \ne 0.$$

  2. $\forall T$, Möbius, $$[z,z_1,z_2,z_3] = [T(z),T(z_1),T(z_2),T(z_3)]$$

  3. Recall that the cross ratio itself is Möbius by Prop 3.12. Choose $T(z)$ to be $[z,z_1,z_2,z_3]$. Observe that $$T(z_1)=0, T(z_2)=1, T(z_3)=\infty,$$ which is pointed out in Prop 3.12.

  4. Denote $w:=T(z)$. Observe that $$w = T(z) = [z,z_1,z_2,z_3] = [T(z),T(z_1),T(z_2),T(z_3)] = [w,0,1,\infty].$$

  5. By (4), $$w \in \mathbb R \iff [z,z_1,z_2,z_3] \in \mathbb R$$

  6. Recall that $\forall T$, Möbius, $T$ is bijective and maps clines to clines by Prop 3.1 and Thm 3.4.

  7. By (6), $$w \in \mathbb R \iff z \in C(z_1,z_2,z_3)$$

  8. By (5) and (7), $$\therefore, [z,z_1,z_2,z_3] \in \mathbb R \iff z \in C(z_1,z_2,z_3)$$ QED


Questions:

  1. Which part of the paraphrasing is wrong, and why?

  2. Why is (7) true please?

I believe this is the crux of my misunderstanding. I don't believe I understand the explanations in the answers in the linked questions.

  1. Actually, it seems like we needed to have proved that $$[z,z_1,z_2,z_3] \in \mathbb R \cup \{\infty\} \iff z \in C(z_1,z_2,z_3).$$ Am I wrong? Was that done? If $[z,z_1,z_2,z_3] = \infty$, then $z=z_3 \in C(z_1,z_2,z_3)$. Not sure about converse. Oh actually, I think what may have been meant by 'real line' is $\mathbb R \cup \{\infty\}$? In which case, I think #4 in the answer should have been 'iff w is on the real line' rather than 'w is real' ?

  2. Btw, can we get somewhere with using Cauchy-Riemann? I'm thinking if $f$ is real-valued then it is constant wherever it is holomorphic by Exer 2.19. Then the cross ratio $[z,z_1,z_2,z_3]$ equals some constant real number $c$. If we would solve for $z$ in terms of $c,z_1,z_2,z_3$, then could we do something with it? I mean, other than manually computing the equation of the circle and then plugging $z$ into it.

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    $\begingroup$ Functions of the form that $T$ has send the set of circles and lines to the set of circles and lines. Since $z$ is in the circle passing through $z_1,z_2,z_3$, and these three points are sent to the real line, to the points $0,1,\infty$, then the point $z$ would also be sent to that line. $\endgroup$
    – user578878
    Jul 28, 2018 at 13:49
  • $\begingroup$ @nextpuzzle thanks....soooo if the 3 points' image is in R then the circle between them has an image in R and so if z is on the circle, then the image of z is real valued? Why/why not? And what about the converse? If w is in R, then why is the preimage in the circle passing through the preimages of 0,1,infty? $\endgroup$
    – BCLC
    Jul 28, 2018 at 17:14
  • $\begingroup$ @nextpuzzle I think I got it. Is my answer wrong please? $\endgroup$
    – BCLC
    Jul 29, 2018 at 5:37

1 Answer 1

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I think I got it: w is the image of z under T. Thus...

$\leftarrow$

If z is on the circle, then the image under T is on the real line. Why? T maps 3 distinct points to 2 finite points and $\infty$. Thus, T maps the points to a subset of the real line. I conclude this is the whole real line either because $\infty$ is there or because an empty or proper subset would not be a line. Now because T maps clines to clines bijectively, it maps the entire circle passing through the 3 points to entire real line. $\therefore$, if z is in the circle, its image is on the real line. But the image under T is the cross ratio.

$\rightarrow$

Now suppose the cross ratio, which is the image of z under T, is on the real line. We know the whole circle is mapped to the whole real line under T. Since T is bijective and maps clines to clines, the preimage of T(z), w/c is z, is on the circle.

QED

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