Question about the correspondence between unitary Möbius transformations and quaternions One of the main theorems about the classification of Möbius transformations states that pure rotations of the Riemann sphere (without translation and dilatation) correspond to unitary Möbius transformation; that is, Möbius transformations of the form:
$$f(z) = \frac{\alpha z +\beta}{-\bar{\beta}z+\bar{\alpha}} = \frac{(a+bi) z +(c+di)}{-(c-di)z+(a-bi)}$$
where $\lvert\alpha\rvert^2+\lvert\beta\rvert^2 = a^2+b^2+c^2+d^2 = 1$.
I was wondering how to correspond a unit quaternion to a given unitary Mobius transformation, as both can be identified with rotation of the Riemann sphere. To calculate the corresponding quaternion, I outlined the following procedure:

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*First, find the fixed points $z_1$, $z_2$ of the given unitary Mobius transformation by solving the quadratic equation: $z=f(z)\implies -\bar{\beta}z^2+(\bar{\alpha}-\alpha)z-\beta=0$. This step actually enables to find the axis of rotation of the Riemann sphere, which correspond to the vector part of the desired quaternion $q$. The reason for this is that the fixed points of $f(z)$ are exactly the images of the intersection of the axis with the Riemann sphere (the poles of the rotation), under the stereographic projection.

*Secondly, as explained in the first step, one needs to find the pre-images of $z_1$, $z_2$ under the stereograpic projection. This can be done by the equation: $z_1 = \cot(\frac{1}{2}\theta)e^{i\phi}\implies \lvert z_1\rvert=\cot(\frac{1}{2}\theta), arg(z_2) = \phi$, where $\theta$, $\phi$ are the zenith angle and the azimuth of the pre-image.  This completes the calculation of the axis direction (which we denote by $\eta_1$), and the vector part of $q$.

*Now the problem is to find the angle of rotation $\gamma$, which correspond to the real part of the $q$. To do this, apply $f(z)$ to the simplest complex number $z = 1$ to get $f(1) = \frac{\alpha  +\beta}{-\bar{\beta}+\bar{\alpha}}$. Now apply the inverse stereographic projection (which we denote $S^{-1}$) to find the point on the sphere that is projected to $f(1)$. We now have three known vectors: $\eta_1$, $\eta_2 = (1,0,0)$, $\eta_3 = S^{-1}(f(1))$.

*As a last step, calculate the length of difference vector: $l = \lvert\eta_3-\eta_2\rvert$. Since $l$ is a chord in a circle whose radius is $r = \sqrt{1-(\eta_1 \cdot \eta_2)^2}$ and central angle $\gamma$, one can solve for $\gamma$ by the equation $l = 2r\sin(\gamma/2)$. The desired quaternion is: $q = \cos(\gamma/2)+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)\sin(\gamma/2)$.

Although this procedure might appear practical, the algebra I got is so complicated that I didn't succeed in finding a closed form expression for $q$ in terms of $a$, $b$, $c$, and $d$. Therefore, my questions are:

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*Are there any known results on the correspondence between unitary Möbius transformation and quaternions?

*Is there a conceptually more transparent, and simpler algebraic, way of deriving $q$? If the answer is yes, what is the resulting formula?

Remark:
I've already asked this question on Math StackExchange and didn't get any responses except a comment that  the angle of rotation is easily computed by the derivative of the Mobius transformation at one of the fixed points $f'(z_1)$.
 A: After getting some very useful comments by user10354138 from math stackexchange, i succeeded in finding a solution, which is remarkably simple. According to his comments, the angle of rotation $\gamma$ can be computed by taking the derivative of Mobius transformation $f(z)$ at one of the fixed points of the transformation; that is: $e^{i\gamma} = f'(z_1)$. First, we need to find the fixed point:

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*the resulting quadratic equation is: $-(c-di)z^2-2biz-(c+di)=0\implies z_{1,2} = \frac{-d + ci}{c^2+d^2} (-b\pm \sqrt{1-a^2})$.

The calculation of the derivative yields:
$ e^{i\gamma}=f'(z_1)= \frac{2a+\sqrt{(2bi)^2-4(c^2+d^2)}}{2a-\sqrt{(2bi)^2-4(c^2+d^2)}} = \frac{2a+2i\sqrt{1-a^2}}{2a-2i\sqrt{1-a^2}} = \frac{a+i\sqrt{1-a^2}}{a-i\sqrt{1-a^2}}$.
According to several trigonometric identities, this means that $cos(\gamma/2) = a$. So the real part of the desired quaternion $q$ is $a$. Now, the norm of $z_1$ is:
$cot(\theta/2)=Norm(z_1) = \frac{-b+\sqrt{1-a^2}}{\sqrt{c^2+d^2}}\implies cos\theta = \frac{cot^2(\theta/2)-1}{cot^2(\theta/2)+1} = -\frac{b}{\sqrt{1-a^2}}$. Since the representation of $q$ is:
$$q = cos(\gamma/2)+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)sin(\gamma/2) = a+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)\sqrt{1-a^2}$$, one gets that the $k$-component of $q$ is $-b$. In addition, since $\frac{\eta_{1y}}{\eta_{1x}} = -\frac{d}{c}$, and $\eta_{1x}^2+\eta_{1y}^2+\eta_{1z}^2 = 1$, one gets the final result that the desired quaternion is:
$$q = a + ci - dj -bk$$.
