= Solution
Write the <Möbius transformation> as
$$
f(z)=\frac{az+b}{cz+d},
\qquad ad-bc\ne0.
$$
A finite <fixed point of a Möbius transformation> satisfies
$$
cz^2+(d-a)z-b=0.
$$
If three distinct points of the Riemann sphere are fixed, this quadratic is identically zero, with the usual interpretation when one point is infinity. Hence $c=b=0$ and $d=a$, so $f(z)=z$. A nonidentity transformation's fixed-point equation is a nonzero polynomial of degree at most two on the sphere, and therefore it has one repeated fixed point or two distinct fixed points.
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