 Conjugation of a Möbius transformation with two distinct fixed points

For $f(z)=(az+b)/(cz+d)$ with distinct finite fixed points of a Möbius transformation $\alpha ,\beta$, the fixed-point equation gives $f(z)-\alpha =(a-c\alpha )(z-\alpha )/(cz+d)$. Dividing the two identities gives $k=(a-c\alpha )/(a-c\beta )$, which is nonzero because $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The identities extend over the Riemann sphere. This conjugates the transformation to a dilation.