Prove that a certain Möbius transformation looks like $f(z)=(az+b)/(\bar{a}z+\bar{b})$.

Question

I have some trouble proving this:

A Möbius transformation $f$ is such that $f(\mathbb{R}\cup\{\infty\})=S^{1}$ iff it looks like $$ f(z)=\frac{az+b}{\bar{a}z+\bar{b}} $$ with $a$ and $b$ in $\mathbb{C}$ and not zero.

When I want to prove that $f$ has to look like above, I start with $f(z)=(az+b)/(cz+d)$, and try to obtain that $c=\bar{a}$ and $d=\bar{b}$, but the further I've reached is that $|a|=|c|$ and $|b|=|d|$, by evaluating $f$ at $0$ and $\infty$, taking the norm and making it equal to 1. Does anyone know what else I can do?
Thanks c:

Answer 1

Here is another way to prove it: Since for $x$ real we have $f(x)\in S^1$, then it follows $1/f(x) = \overline{f(x)}$, and we get that the identity $$ \frac{ \overline{a}z + \overline{b}}{\overline{c}z+\overline{d}} = \frac{cz+d}{ az + b},$$ holds throughout the real line, then by analytical identity it must also hold on the intersection of the domains, but two Möbius transformations only are equal if there exists a factor $\lambda$ such that the respective coefficients of the transformations differ by such a factor. Then $c=\lambda \overline{a}$ and $d=\lambda \overline{b}$ (also, it follows that $|\lambda|=1$ by the calculations you made), substitute the coefficients and now just multiply the fraction by another number $e^{i\theta}$ such that the new coefficients fulfill the requeriment of the problem (you will need to solve a simple linear equation for $\theta$).

Also, the statement as you wrote it is slightly wrong, it should say "... iff it can be put in the form..."

Answer 2

So you found $|a|=|c|$ and $|b|=|d|$. Note that $f$ remains the same function if we replace $a,b,c,d$ with $\lambda a, \lambda b,\lambda c,\lambda d$ for some $\lambda\ne 0$. We can choose $\lambda$ such that $\lambda b=\overline{\lambda d}$, namely $\lambda=\sqrt{\frac {\overline d}b}$ (whereby incidentally $\lambda\in S^1$). In other words, we may assume wlog that $d=\overline b$. Now consider $\frac {az+b}{cz+\overline b}$ for small real $z$.