MHB Analytic is UHP to unit disc

[Dustinsfl]

Prove that the most general analytic isomorphism of the open upper half plane, $\mathcal{H}$, onto the open unit disc is of the form
$$
T(z) = e^{i\varphi}\frac{z - a}{z - \bar{z}}
$$
for some $\varphi\in\mathbb{R}$ and some $a\in\mathbb{C}$ with $\text{Im}(a) > 0$

I need some guidance here. Opalg keeps suggestion to multiple by the conjugate so as a stab in the dark should I multiple by the conjugate here as well?

 

[Tinyboss]

Prove that the most general analytic isomorphism of the open upper half plane, $\mathcal{H}$, onto the open unit disc is of the form
$$
T(z) = e^{i\varphi}\frac{z - a}{z - \bar{z}}
$$
for some $\varphi\in\mathbb{R}$ and some $a\in\mathbb{C}$ with $\text{Im}(a) > 0$

I need some guidance here. Opalg keeps suggestion to multiple by the conjugate so as a stab in the dark should I multiple by the conjugate here as well?

Are you sure that shouldn't be $\overline{z}-a$ in the denominator?

 

[Opalg]

Are you sure that shouldn't be $\overline{z}-a$ in the denominator?

It should be $z-\bar{a}$. (If $T(z)$ has any $\bar{z}$s then it won't be analytic.)

 

[Dustinsfl]

Yes it was supposed to be \bar{z}. How should I proceed then?

 

[Dustinsfl]

Yes it was supposed to be \bar{z}. How should I proceed then?

I meant \bar{a} sorry.

 

[Opalg]

Given an analytic isomorphism $T(z)$ of $\mathcal{H}$ onto the open unit disc, let $a = T^{-1}(0)$, and define $f(z) = \dfrac{z - a}{z - \bar{a}}$. Notice that any point $z\in \mathcal{H}$ is closer to $a$ than it is to $\bar{a}$, and therefore $f$ maps $\mathcal{H}$ into the open unit disc (and in fact you should show that $f$ maps $\mathcal{H}$ onto the open unit disc). Thus $f(z)$ is an analytic isomorphism from $\mathcal{H}$ onto the open unit disc. Therefore the map $x\mapsto f(H^{-1}(z))$ is an analytic isomorphism from the open unit disc onto itself that fixes the point 0. Now use Schwarz's lemma.

 

[Dustinsfl]

Why have we disregarded $e^{i\varphi}$?

 

[Opalg]

Why have we disregarded $e^{i\varphi}$?

That comes in at the end of the proof, when you apply Schwarz's lemma.

 

[Dustinsfl]

I know that $\psi:\mathcal{H}\to\mathbb{D}$ by $\psi(z):\dfrac{z-i}{z+i}$ maps the upper half plane into the unit disc.

The analytic isomorphism of $\varphi:\mathbb{D}\to\mathbb{D}$ are $\varphi(z)=e^{i\theta}\dfrac{a-z}{1-\bar{a}z}$.

By function composition $\psi\circ\varphi$, I obatin
$$
e^{i\theta}\frac{a(z+i)-(z-i)}{z+i-\bar{a}(z-i)}.
$$

How does this simplify down to
$$
e^{i\theta}\frac{z-a}{z-\bar{a}}
$$
though?

I know that the composition is supposed to yield the correct result but no matter the order, I can't get it.

 

[Dustinsfl]

Let $f:\mathcal{H}\to\mathbb{D}$ by $f(z) = \dfrac{z - i}{z + i}$.
By Lang (p. 215), we see that $f$ maps the upper half plane into the unit disc.
Let $g:\mathbb{D}\to\mathbb{D}$ by $g(z) = e^{i\theta}\dfrac{z - a}{\bar{a}z - 1}$.
Then
$$
(g\circ f)(z) = e^{i\theta}\frac{z(1 - a) - i(1 + a)}{z(\bar{a} - 1) - i(\bar{a} + 1)} = e^{i\theta}\left(\frac{1 - a}{\bar{a} - 1}\right)\frac{z - i \left(\frac{1 + a}{1 - a}\right)}{z - i \left(\frac{\bar{a} + 1}{\bar{a} - 1}\right)}.
$$
So let $\alpha = i\dfrac{1 + a}{1 - a}$.
Then $\overline{\alpha} = i\dfrac{1 + \bar{a}}{\bar{a} - 1}$.
Thus
$$
(g\circ f)(z) = e^{i\theta}\left(\frac{1 - a}{\bar{a} - 1}\right)\frac{z - \alpha}{z - \overline{\alpha}}.
$$
Since $\left|e^{i\theta}\left(\dfrac{1 - a}{\bar{a} - 1}\right)\right| = 1$, this expression lies on the unit circle and can be written as $e^{i\varphi}$.
So
$$
(g\circ f)(z) = e^{i\varphi}\frac{z - \alpha}{z - \overline{\alpha}} = T(z).
$$