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.
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*}