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I need to explicitly determine every isomorphism of the first quadrant of the complex plane.

Considering the set $\{ z\in\mathbb{C}:\Re(z)>0\text{ and }\Im(z)>0\}$ as the first quadrant, I know that I can map this (using Möbius Transformations) set into the unit disc (this is stated by the Riemann Conformal Mapping Theorem). Using the disc automorphisms I already know I can produce several automorphisms of the first quadrant.

The question is: how can I proof that every automorphism of the first quadrant into the unit disc is a Möbius transform (so I can explicitly determine all of them)?

Thanks in advance.

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  • $\begingroup$ Take a look at this: math.stackexchange.com/questions/52338/… $\endgroup$ – Betty Mock Nov 14 '13 at 1:44
  • $\begingroup$ This was exactly what I was looking for. I would even ask for a moderator to mark this as duplicate. Thanks! $\endgroup$ – Marra Nov 14 '13 at 2:58
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how can I proof that every automorphism of the first quadrant into the unit disc

Let's get the terminology right. The prefix auto means "self-". In mathematical context it usually indicates a bijection $X\to X$ with some additional properties (here, being holomorphic). Thus, you can have an "automorphism of $X$", but not an "automorphism of $X$ into $Y$".

A holomorphic bijection $f:X\to Y$ could in principle be called an isomorphism, but on this occasion the terminology of abstract algebra did not catch up in analysis. One speaks of biholomorphic maps, or conformal maps.

Using the disc automorphisms I already know I can produce several automorphisms of the first quadrant.

Not just several, all of them. This is a rather general fact: if you know the automorphisms of $Y$ (say, the set of all holomorphic bijections $\phi:Y\to Y $), and have just one isomorphism $f:X\to Y$, then you get all automorphisms of $X$ in the form $f^{-1}\circ \phi\circ f$, where $\phi$ runs through the automorphisms of $Y$. The reason is simple: if $\psi$ is any automorphism of $X$, then the map $\phi:= f\circ \psi\circ f^{-1}$ is an automorphism of $Y$, and you have $\psi = f^{-1}\circ \phi\circ f$.

Note that nothing in the preceding paragraph depends on what formula $f$ has, or what the word "holomorphic" even means; this is pure abstract nonsense.

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Let f(z) = z^2.

Let g(z) be the inverse of f(z) along the principle branch of the square root.

Any automorphism of the upper half plane has the form: h(z) = (az+b) / (cz+d), with a,b,c,d real and ad-bc > 0.

Then any automorphism of the first quadrent is given by a composition of these three functions: g(h(f(z))).

See the solution at: http://sertoz.bilkent.edu.tr/courses/math302/2011/hwk2-sol.pdf

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  • $\begingroup$ Please MathJax. It will make it much easier to parse what you have written (the fractions in the Moebius transformation, in particular). $\endgroup$ – Xander Henderson Dec 3 '17 at 22:39

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