I have a problem in understanding why a 1-1, onto analytic map which sends upper half plane to unit disk is of the form $f(z)=\frac {az+b}{cz+d}$ where $ad-bc \neq 0$. (I know the map can be refined further but only have problems in understanding this part.)

So please explain why we have to start our work with this form.

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    $\begingroup$ Have you studied cross ratio, orientation and symmetry principle? $\endgroup$ – Swapnil Tripathi Nov 8 '14 at 23:45
  • $\begingroup$ What does $f(x)=\dfrac1x$ do ? Now, imagine rotating the real axis around the origin. What does x become, and what does $f(x)=\dfrac1x$ become ? Now, how can we modify $f(x)=\dfrac1x$ so that instead of mapping $(1,\infty)$ to $(0,1)$, it maps $(0,\infty)$ to $(0,1)$ ? Now, as we again rotate the real axis around the origin, what does this new function become ? $\endgroup$ – Lucian Nov 9 '14 at 1:02

Claim 1. Every conformal map $f$ of $\mathbb D$ onto itself such that $f(0)=0$ is the rotation, $f(z)=e^{i\theta} z$.

Proof: Apply the Schwarz lemma to $f$ and to $f^{-1}$; conclude that equality holds, use the equality statement in the lemma.

Claim 2. For every $w\in \mathbb H$ (the upper halfplane) the map $\phi(z) = \dfrac{z-w}{z-\overline w}$ transforms $\mathbb H$ onto $\mathbb D$.

Proof. It's invertible, and the real axis goes to the unit circle.

Claim 3. Every conformal map $f$ of $\mathbb H$ onto $\mathbb D$ is a fractional linear transformation.

Proof: Apply Claim 1 to $$g(z) = \frac{f(z)-f(0)}{f(z)-\overline{f(0)}}$$ which maps $\mathbb D$ onto $\mathbb D$ by virtue of Claim 2.

Then use the fact that fractional linear transformations form a group (in particular, the composition of two such transformations is again of the same form).

Answer based on Which conformal maps UHP$\to$UHP extend continuously to the closure?


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