Distinct fixed points of Möbius transformation I am attempting the following problem.

Let $f:\mathbb{C}\cup\{\infty\}\to \mathbb{C}\cup\{\infty\}$ be a Möbius transformation. Prove that if $f(\alpha)=\alpha,f(\beta)=\beta$ with $\alpha\neq\beta\in\mathbb{C}\cup\{\infty\}$, then $f'(\alpha)f'(\beta)=1$.

I know $f$ must have the form
$$
f(z)=\frac{az+b}{cz+d}
$$
and $f'$ has the form
$$
f'(z)=\frac{ad-bc}{(cz+d)^2},
$$
but I do not know how to proceed.
I wonder if I can use $f$ to define a composition $g$ of conformal maps from $\mathbb{D}\to\mathbb{D}$, such that $g$ has two distinct fixed points, one at $0$, then apply Schwarz's Lemma somehow. But I do not know how to do so. Thank you for any help!
 A: $$f'(\alpha)f'(\beta)=\frac{(ad-bc)^2}{(c\alpha+d)^2(c\beta+d)^2}$$
The denominator of this expands as
$$((c\alpha+d)(c\beta+d))^2=(c^2\alpha\beta+cd(\alpha+\beta)+d^2)^2$$
But by manipulating $f(z)=z$ we see that $\alpha,\beta$ are the solutions to $cz^2+(d-a)z-b=0$. Therefore, by Viète's relations,
$$\alpha+\beta=\frac{a-d}c\qquad\alpha\beta=-\frac bc$$
$$c^2\alpha\beta+cd(\alpha+\beta)+d^2=c^2\left(-\frac bc\right)+cd\left(\frac{a-d}c\right)+d^2=ad-bc$$
Finally
$$f'(\alpha)f'(\beta)=\frac{(ad-bc)^2}{(ad-bc)^2}=1$$
A: Fixed point of the Möbius transformation are given by $f(z)=z,$ and hence by the quadratic equation $$cz^2+(d-a)z−b= 0.$$ Assume $c\neq0,$ then $$\alpha+\beta=\dfrac{a-d}{c},\qquad\qquad \alpha\beta=-\dfrac{b}{c}.$$ Consider $$f'(\alpha)f'(\beta)=\dfrac{(ad-bc)^2}{(c\alpha+d)^2(c\beta+d)^2}=\dfrac{(ad-bc)^2}{(c^2\alpha\beta+cd(\alpha+\beta)+d^2)^2}.$$ From here I will leave the rest of the simplification for you.
Finally consider the case $c=0$ case separately, in which case $\infty$ is a fixed point.
However I believe that there must bee a simple geometrical reasoning for this nice formula.
