Transformation under rotation of Riemann sphere

Question

Suppose the Riemann sphere $S$ is rotated by the angle $\phi$ round the diameter whose end points have $a,-1/\bar{a} $ (which have antipodal preimages) as stereographic projections. Suppose moreover, $z$ and $\zeta$ are stereographic projections of points corresponding to each other under this rotation. What is the relationship between $z$ and $\zeta$ ?

Answer 1

The given rotation of $S$ appears as a Moebius transformation $T$ of the closed complex plane $\bar{\mathbb C}$. This $T$ will have the complex numbers $a$ and $-1/\bar a$ as fixed points. Therefore $z$ and $\zeta:=T(z)$ are related by a formula of the form $${\zeta -a\over \zeta+1/\bar a}=\lambda\ {z -a\over z+1/\bar a}\ ,\qquad(*)$$ where $\lambda$ is a certain complex constant. Since neither of the two fixed points is attracting one necessarily has $\lambda=e^{i\alpha}$.

In order to determine $\alpha$ we look at things in the immediate neighborhood of $a$. When $z\to a$ then the image point $\zeta=T(z)$ approaches $a$ as well, and we have $$T'(a)=\lim_{z\to a}{\zeta -a\over z-a}= e^{i\alpha}\lim_{z\to a}\ {\zeta+1/\bar a\over z+1/\bar a}=e^{i\alpha}\ .$$ As the stereographic projection is conformal we should have $\arg T'(a)=\phi$, and this leads to the conclusion that in fact $\alpha=\phi$.

Of course you now can solve $(*)$ for $\zeta$ in terms of $z$.