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Complex Analysis and Mobius Transformation.

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data

    If [itex]\phi \in \mathcal{M} [/itex] (group of all linear fractional transformations or Mobius Transformations has three fixed points, then it must be the identity. (The proof should exploit the fact that [itex]\mathcal{M} [/itex] is a group.





    3. The attempt at a solution

    Hi all,

    So the book I'm reading is "Complex Made Simple" by David C. Ullrich. and this is Exercise 8.5 if anyone has done this.

    Anyway this is what I was thinking.

    Let [itex]\phi_j \in \mathcal{M} [/itex] where [itex] j = 1,2[/itex] and let [itex]\phi_j(z_i) = z_i [/itex] for [itex] i = 1,2,3[/itex]

    So we see that [itex] z_i = \phi_{j}^{-1}(z_i)[/itex], we know that [itex]\phi_{j}^{-1}[/itex] since their elements of the group [itex] \mathcal{M} [/itex]

    My question would be this, if I define a map [itex]\phi = \phi_{2}^{-1} \circ \phi_{1}[/itex] and show that that [itex]\phi_1 = \phi_2[/itex] is that enough to show that [itex]\phi[/itex] is the identity?
     
  2. jcsd
  3. Oct 27, 2013 #2
    I think it is because the identity element and the inverse element is unique.
     
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