TITLE: Linear fractional transformation fixing origin and preserving all distances
QUESTION [10 upvotes]: Prove that the most general transformation which leaves the origin fixed and preserves all distances is either a rotation or a rotation followed by reflection in the real axis.

We represent a (linear fractional) transformation by $f(z)=\dfrac{az+b}{cz+d}$ for some complex constants $a,b,c,d$ with $ad-bc\neq 0$.
Since the origin is fixed, we have $0=f(0)=\dfrac{b}{d}$, so that $b=0$, and $f(z)=\dfrac{az}{cz+d}$.
Now, the transformation preserves all distances. So $$|z_1-z_2|=|f(z_1)-f(z_2)|=\left|\dfrac{az_1}{cz_1+d}-\dfrac{az_2}{cz_2+d}\right| = \frac{|a||d||z_1-z_2|}{|cz_1+d||cz_2+d|}$$ for all complex values $z_1,z_2$, so that $|cz_1+d||cz_2+d|=|ad|$. Since this holds for all $z_1,z_2$, we have that $|cz+d|$ must be constant, and that constant is equal to $\sqrt{|ad|}$. Also, when $z_0$, $|cz+d|=|d|$. So $\sqrt{|ad|}=|d|$ implies $|a|=|d|$. 
What can we do next? I don't see how to get to the rotation/reflection part.

REPLY [4 votes]: If you are considering only linear fractional transformations, then $0 \mapsto 0$ and $\infty \mapsto \infty$ (preserves all distances) imply $f(z)=az$. Since the transformation preserves all distances we have $a=e^{i\alpha}$, i.e. $f$ is a rotation.
But if you don't require $f$ to be a linear fractional transformation then let $e^{i\alpha}:=f(1)$. Since $f$ preserves all distances there are only two possibilities for each $re^{i\varphi}$:
$$re^{i\varphi} \mapsto re^{i(\alpha+\varphi)} \quad \text{ or } \quad re^{i\varphi} \mapsto re^{i(\alpha-\varphi)}.$$
The first mapping is the rotation by $\alpha$, the second one is the rotation by $-\alpha$ followed by reflection in the real axis.