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I'm trying to find all conformal automorphisms of the upper half plane $\{\Im[z] \gt 0\}$, known to be $f(z) = \frac{az + b}{cz + d}$ where $a, b, c, d$ are real and $ad - bc \gt 0$.

The main work is to show that automorphisms are rational functions with real coefficients. The thing I'm having trouble with is steps (1.) and (2.) below.

  1. Show that taking the limit as $z$ approaches the real line, we get a automorphism of $\{\Im[z] \ge 0\} \cup \{\infty\}$

  2. Then the the real line + $\infty$ maps to the real line + $\infty$. Extend to an automorphism of the entire complex plane + infinity by Schwartz Reflection.

  3. $\infty$ maps to a finite real number or else we have an entire function and the only automorphisms of the complex plane are $f(z) = az + b$ and we are done because $f(0)$ must be real and $f(1)$ must be real by (2.)

  4. So for one real number $r$, $f(r) = \infty$. $r$ cannot be an essential singularity or else we violate injectivity. So $f$ is a pole of order 1, because any higher order and we violate injectivity again.

  5. In the end we have a meromorphic function in the entire plane with one pole on the real line bounded at infinity. By Mittag Leffler we know that $f$ is a rational function. By injectivity, we know its degree is 1.

  6. Since it's real on the real line, we know its coefficients can be written real. Here we can appeal to the power series expansion of $f(z)(z-r)$ around $r$ which must have real coefficients since they correspond to derivatives of $f(z)(z-r)$. But such derivatives must be real since we can view the function as a real function of a real variable.

I know there is another way to do this with mappings of the unit disk, but I want to see if this way can work out.

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