Complex Analysis and Mobius Transformation.

[BrainHurts]

Homework Statement

If \phi \in \mathcal{M} (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 \mathcal{M} is a group.

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

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

My question would be this, if I define a map \phi = \phi_{2}^{-1} \circ \phi_{1} and show that that \phi_1 = \phi_2 is that enough to show that \phi is the identity?

 

[BrainHurts]

I think it is because the identity element and the inverse element is unique.