General Formula for an arbitrary Rotation of Sphere I am currently reading F. Klein's book "Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree" and in Part 1, Chapter 2, we desire to deduce a general formula for an arbitrary rotation of sphere. The idea is as follows: first we use stereographic projection to identify $S^2 = \{(\xi,\eta,\zeta)\in \mathbb{R}^3| \quad  \xi^2+\eta^2+\zeta^2=1\}$ with Riemann sphere $\mathbb{C} \cup \{\infty\}$ via the following formula: $z = x + i y = \frac{\xi+i\eta}{1-\zeta}$. If we further identify $\mathbb{C}\cup \{\infty\}$ with $\mathbb{CP}^1$, we can conclude every rotation will be represented by a fractional linear substitution.
Now, given a rotation which fixes antipodal points $(\xi,\eta,\zeta), (-\xi,-\eta,-\zeta)$ , which correspondences to $\frac{\xi+i\eta}{1-\zeta}, -\frac{\xi+i\eta}{1+\zeta}$ in $\mathbb{C}\cup \{\infty\}$  and rotate through an angle $\alpha$ counterclockwise. Above process can be decomposed into two steps: first we move $-\frac{\xi+i\eta}{1+\zeta}$ to $0$ and move $\frac{\xi+i\eta}{1-\zeta}$ to $\infty$, which corresponds to the following fractional linear substitution up to some constant:
$$
C\cdot \frac{z+\frac{\xi+i\eta}{1+\zeta}}{z -\frac{\xi+i\eta}{1-\zeta} }
$$
Then we get our new $0,\infty$ axis and then we rotate our $(\xi',\eta')$-plane (equatorial plane) by an angle $\alpha$ counterclockwise, which correspond to multipyling factor $e^{i\alpha}$.
Suppose $z'$ is the coordinate after the rotation, we must have:
$$
\frac{z'+\frac{\xi+i\eta}{1+\zeta}}{z' -\frac{\xi+i\eta}{1-\zeta} } = e^{i\alpha}\frac{z+\frac{\xi+i\eta}{1+\zeta}}{z -\frac{\xi+i\eta}{1-\zeta} } \quad(*)
$$
The author claimed that if we do the following change of notation:
$$
\xi \sin(\frac{\alpha}{2})=a, \quad \eta \sin(\frac{\alpha}{2})=b, \quad \zeta\sin(\frac{\alpha}{2})=c, \quad \cos(\frac{\alpha}{2})=d,
$$
Then $(*)$ can be rewritten as
$$
z' = \frac{(d+ic)z-(b-ia)}{(b+ia)z+(d-ic)} \quad (**)
$$
That is where I stuck. I think I understand the process of deduction of general rotation formula but I am not sure how to get the simple form $(**)$. Now the problem is purely elementary and I try to verify $(**)$ by brute force but it doesn't seem to be a correct way . I guess it will involve with some trigonometric formulas to simplify the computation. Could you please offer me some suggestions on how to start from $(*)$ to derive $(**)$? Thank you in advance.
 A: Letting $A:=\frac{\xi+i\eta}{1+\zeta}$ and $B:=\frac{\xi+i\eta}{1-\zeta}$ we obtain from $(*)$:
\begin{align*}
e^{-i\frac{\alpha}{2}}\ \frac{z^{\prime}+A}{z^{\prime}-B}&=e^{i\frac{\alpha}{2}}\ \frac{z+A}{z-B}\\
e^{-i\frac{\alpha}{2}}\left(z^{\prime}+A\right)\left(z-B\right)&=e^{i\frac{\alpha}{2}}(z+A)\left(z^{\prime}-B\right)\\
e^{-i\frac{\alpha}{2}}\left(zz^{\prime}+Az-Bz^{\prime}-AB\right)
&=e^{i\frac{\alpha}{2}}\left(zz^{\prime}-Bz+Az^{\prime}-AB\right)\tag{1}\\
\end{align*}
Extracting $z^{\prime}$ from (1) we obtain
\begin{align*}
\color{blue}{z^{\prime}}&=\frac{\left(Be^{i\frac{\alpha}{2}}+Ae^{-i\frac{\alpha}{2}}\right)z
+AB\left(e^{i\frac{\alpha}{2}}-e^{-i\frac{\alpha}{2}}\right)}
{\left(e^{i\frac{\alpha}{2}}-e^{-i\frac{\alpha}{2}}\right)z+Ae^{i\frac{\alpha}{2}}+Be^{-i\frac{\alpha}{2}}}\\
&\,\,\color{blue}{=\frac{\left(Be^{i\frac{\alpha}{2}}+Ae^{-i\frac{\alpha}{2}}\right)z
+2iAB\sin\left(\frac{\alpha}{2}\right)}
{2i\sin\left(\frac{\alpha}{2}\right)z+Ae^{i\frac{\alpha}{2}}+Be^{-i\frac{\alpha}{2}}}}\tag{2}
\end{align*}
Since we want to derive (**):
\begin{align*}
z' = \frac{(d+ic)z-(b-ia)}{(b+ia)z+(d-ic)}
\end{align*}
and the coefficient of $z$ of the denomintor in (2) is $2i\sin\left(\frac{\alpha}{2}\right)$ instead of $b+ia$, we consequently expand numerator and denominator of (2) with
\begin{align*}
\frac{b+ia}{2i\sin\left(\frac{\alpha}{2}\right)}
\end{align*}
We calculate the constant part of the denominator of (**) from (2) and obtain
\begin{align*}
&\color{blue}{\frac{b+ia}{2i\sin\left(\frac{\alpha}{2}\right)}}
\color{blue}{\left(Ae^{i\frac{\alpha}{2}}+Be^{-i\frac{\alpha}{2}}\right)}\\
&\quad=\frac{b+ia}{2i\sin\left(\frac{\alpha}{2}\right)}
\left(\frac{a+ib}{\sin\left(\frac{\alpha}{2}\right)+c}\,e^{i\frac{\alpha}{2}}
+\frac{a+ib}{\sin\left(\frac{\alpha}{2}\right)-c}\,e^{-i\frac{\alpha}{2}}\right)\tag{3}\\
&\quad=\frac{a^2+b^2}{2\sin\left(\frac{\alpha}{2}\right)}
\left(\frac{e^{i\frac{\alpha}{2}}}{\sin\left(\frac{\alpha}{2}\right)+c}
+\frac{e^{-i\frac{\alpha}{2}}}{\sin\left(\frac{\alpha}{2}\right)-c}\right)\\
&\quad=\frac{1}{2\sin\left(\frac{\alpha}{2}\right)}
\left(e^{i\frac{\alpha}{2}}\left(\sin\left(\frac{\alpha}{2}\right)-c\right)
+e^{-i\frac{\alpha}{2}}\left(\sin\left(\frac{\alpha}{2}\right)+c\right)\right)\tag{4}\\
&\quad=\frac{e^{i\frac{\alpha}{2}}+e^{-i\frac{\alpha}{2}}}{2}
-\frac{c}{\sin\left(\frac{\alpha}{2}\right)}\frac{e^{i\frac{\alpha}{2}}-e^{-i\frac{\alpha}{2}}}{2}\\
&\quad=\cos\left(\frac{\alpha}{2}\right)-ic\\
&\,\,\quad\color{blue}{=d-ic}
\end{align*}
according to the claim. The coefficients of the numerator in (**) can be calculated similarly.
Comment:

*

*In (3) we use
\begin{align*}\xi \sin\left(\frac{\alpha}{2}\right)=a,\ 
\eta \sin\left(\frac{\alpha}{2}\right)=b,\ 
\zeta \sin\left(\frac{\alpha}{2}\right)=c,\ 
\cos\left(\frac{\alpha}{2}\right)=d\text{.}
\end{align*}


*In (4) we use $a^2+b^2+c^2=\sin^2\left(\frac{\alpha}{2}\right)$ since $\xi^2+\eta^2+\zeta^2=1$. This implies \begin{align*}
a^2+b^2&=\sin\left(\frac{\alpha}{2}\right)^2-c^2\\
&=\left(\sin\left(\frac{\alpha}{2}\right)-c\right)\left(\sin\left(\frac{\alpha}{2}\right)+c\right)
\end{align*}
