Show that if a Mobius transformation has 3 fixed points then it is the identity map. I have that any non trivial Mobius transformation has at most 2 fixed points since f(z)-z=0 has at most 2 roots. But I cannot deduce why it must then be the identity.

Suppose $Tz=(az+b)/(cz+d)$.  The fixed points of $T$ must satisfy the equation:
$$\begin{align}
\frac{az+b}{cz+d}=z & \iff az+b=cz^2+dz\\
& \iff cz^2+(d-a)z+b=0
\end{align}$$
If this quadratic equation has more than 3 solutions, the polynomial must be identically zero. So, $c=0,a=d,b=0$.
Therefore $Tz=az/a=z$.
This argument works when either $a\ne0$ or $d\ne0$.  If either one is zero, then $a=b=c=d=0$ which is impossible.
