I'm trying to understand the following observation in my textbook:
...in the sense that for any rotation $\phi$ of $S$, the conjugate map $$C_\infty\xrightarrow[]{\pi}S\xrightarrow[]{\phi}S\xrightarrow[]{\pi^{-1}}C_\infty$$ is a $\sigma$ isometry. In fact, the rotations of $S$ coincide (in this sense) with the class of Mobius transformations of the form $z \rightarrow \frac{az-\overline{c}}{cz+\overline{a}}$$\,\,\,\,\,\,\,$ $|a|^2+|c|^2=1.$
I'm trying to reason out why they are all of that form. Consider the following:
$$\frac{az+b}{cz+d}=\pi(\phi(\pi^{-1}(z)))$$
where $$\pi(z)=\frac{z-z_0}{z+\frac{1}{\overline{z_0}}}$$
This gives me the following
$$\frac{a\pi(z)+b}{c\pi(z)+d}=\frac{\phi(z)-z_0}{\phi(z)-\frac{1}{\overline{z_0}}} $$
Before I go through the torture of algebraic manipulation on the above, my goal is to show through a bunch of manipulation that $a=a,b=\overline{b}, c=c, d=\overline{a}$.
Let's put things straight: The Riemann sphere is an abstract idea implementing in a correct way the addition of a point $\infty$ to ${\mathbb C}$. There are several models of this Riemann sphere, one just being your ${\mathbb C}_\infty:={\mathbb C}\cup\{\infty\}$ (together with certain exception handling rules), and another one being $S^2$ inheriting the conformal structure from its embedding in ${\mathbb R}^3$ and being related to $\bar{\mathbb C}$ via the stereographic projection $$\sigma:\quad(x_1,x_2,x_3)\mapsto{x_1+ix_2\over 1-x_3}\ ,$$ resp. $\sigma^{-1}$.
We take it for granted that rotations of $S^2$, being conformal maps, are conjugated via $\sigma$ to certain special Moebius transformations $$T:\quad\bar{\mathbb C}\to\bar{\mathbb C}\ .$$ The question now is to characterize these transformations $T$.
A rotation of $S^2$ which is different from the identity has two antipodal fixed points which are neither attracting nor repelling. We therefore have to find the Moebius transformations $T$ with two fixed points that are carried by $\sigma^{-1}$ to antipodal points on $S^2$, and have a "scaling factor" $\lambda$ of absolute value $1$.
Two points $p$, $q\in{\mathbb C}$ are carried to antipodal points by $\sigma^{-1}$ iff $q=-{1\over \bar p}$. It follows that the Moebius transformations $T:\>z\mapsto z'$ in question can be written as $${z'-p\over z'+1/\bar p}=e^{i\phi}\>{z-p\over z+1/\bar p}\ .\tag{1}$$ Here $e^{i\phi}$ is the "scaling factor" mentioned above. Now solve $(1)$ for $z'$, and you will obtain $T$ in the form $$z'=T(z)={az+b\over cz+d}$$ with certain coefficients $a$, $b$, $c$, $d$. Looking at the resulting formula you will be able to recognize the forms given in your text, after multiplying all coefficients with ${e^{-i\phi/2}\over1+|p|^2}$.
Conversely you should show that a $T$ having coefficients as in your text has fixed points $p$ and $-1/\bar p$ and a "scaling factor" $e^{i\phi}$.