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The questions:

(i) Given the functions

$$ \phi_{a} (z) = \frac{z - a}{1 - \overline{a} z} \quad \text{and} \quad \rho_{\alpha} (z) = e^{i \alpha} \cdot z$$

where $|a| <1$ and $\alpha \in \mathbb{R}$, compute $$( \phi_{a} \circ \rho_{\alpha} - \rho_{\alpha} \circ \phi_{a}) (z)$$

(ii) Suppose $$ f(z) = \phi_{a} \circ \rho_{\alpha} (z) \quad \text{and} \quad g(z) = \phi_{b} \circ \rho_{\beta} (z) $$

belong to $Aut(D)$, the group of holomorphic automorphisms of the unit disc. Find $c$ and $\gamma$, where $|c| < 1 $ and $ \gamma \in \mathbb{R}$, such that $$ f \circ g = \phi_{c} \circ \rho_{\gamma}$$

(iii) If $\mathbb{H} = \{ z \in \mathbb{C} \mid Im(z) > 0 \}$ is the upper half plane, use your knowledge of the elements of $Aut(D)$ to describe the elements of $Aut(\mathbb{H})$ as fractional linear transformations.

Where I am having trouble:

(i) The solution I have is very messy, and I'm unsure of what the point of computing it is. What are we trying to show? $$( \phi_{a} \circ \rho_{\alpha} - \rho_{\alpha} \circ \phi_{a}) (z) = \frac{(e^{i\alpha}a - a) + (a\overline{a} - e^{2i\alpha}a\overline{a})z + (e^{2i\alpha}\overline{a} - e^{i\alpha}\overline{a})z^{2}}{(1 - e^{i \alpha}\overline{a}z)(1 - \overline{a}z)}$$

(ii) I tried to compute this however it was a mess of variables and there wasn't any obvious way to separate them into the form: $$ \frac{e^{i\gamma}z - c}{1 - e^{i\gamma}\overline{c}z}$$

(iii) I imagine this part would be quite straightforward but I need a hint on how to make the jump from $Aut(D)$ to $Aut(H)$.

  • did you try starting from $\phi_a(e^{ic} \phi_b(z))$ ? $f \circ g(e^{-\beta} z)$ reduces to that, and after expanding as the quotient of two polynomials, it should be related to (i) – reuns Mar 23 '16 at 07:02
  • For vanderlylic and @user1952009 The question is very very similar to http://math.stackexchange.com/q/141735 See its answer. – Jean Marie Mar 23 '16 at 14:43
  • @JeanMarie I have read the question/answer and it wasn't really helpful at all. Multiplying the matrix representations still doesn't help with the mess of variables and constants that follow in part (ii), and it doesn't provide any further information on part (i). – vanderlylic Mar 23 '16 at 22:42

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