Prove that for a fractional linear transformation $f(z)$ such that $f(\Bbb{R})=\Bbb{R}$ we have $f(z)=\frac{az+b}{cz+d}$ with $a,b,c,d\in \Bbb{R}$

Question

Let $f(z)$ be a fractional linear transformation such that $f(\Bbb{R})=\Bbb{R}$. I want to show that $f(z)=\frac{az+b}{cz+d}$ $a,b,c,d\in \Bbb{R}$.

We have given a solution in class, where they distinguish into the following cases

1. $c,d\neq 0$
2. $c=0$
3. $d=0$

If I have the cases it is clear to me how they proceeded but I don't see how we get to those cases and why they are sufficient.

Thanks for your help.

Answer 1

I will replace the assumption by $$f(\mathbb{R}\cup \{\infty\})=\mathbb{R}\cup \{\infty\}$$ We have $ad-bc\neq 0,$ as otherwise the function $f$ is constant.

1. $a,c\neq 0.$

Thus $$f(x)={a\over c}\ {x+{b\over a}\over x+{d\over c}}$$

Substituting $x=\infty $ gives $\alpha :={a\over c}\in \mathbb{R}.$ Next the values $0$ and $\infty $ are attained at $\beta:=-{b\over a}$ and $\gamma:=-{d\over c},$ respectively. Hence $\beta,\,\gamma \in \mathbb{R}.$ Thus $$f(x)=\alpha {x-\beta\over x-\gamma}\qquad \alpha\,,\beta\,,\gamma \in\mathbb{R}$$

2. $c=0$

Then $a\neq 0$ and $$f(x)=f(0)\,(1-x)+f(1)\,x$$ Hence the coefficients are real numbers.

3. $a=0$

Then $c\neq 0.$ For $g(x)=(f(x))^{-1}$ we get $$g(x)=g(0)\,(1-x)+g(1)\,x$$ Thus the coefficients of $g(x),$ and consequently of $f(x),$ are real