Let $T_{1} = \frac{az+b}{cz+d}$, $T_{2}=\frac{a'z+b'}{c'z+d'}$ be two Möbius transformation. Prove that $T_{1}(z)=T_{2}(z)$, $\forall z\in\mathbb{C}_{\infty}$ if and only if exsists $\lambda\neq 0$ such that $a=\lambda a'$, $b=\lambda b'$, $c=\lambda c'$, $d=\lambda d'$.
$(\Leftarrow)$ It's clear that $z_{0} = -\frac{b}{a}$ it's such that $T_{0}(z_{0})=0$, then $T_{1}(z_{0})=0$ and the same argument works to see that if $z_{1,2}$ are such that if $T_{1}(z_{1})=\infty$, then $T_{2}(z_{1})=\infty$ and, if $T_{1}(z_{2})=1$, then $T_{2}(z_{2}) =1$. A Möbius transformation has at most two fixed point so $T_{1}=T_{2}$.
I have some problems to prove the other implication. Using the same argument, I have that $-\frac{b}{a} = -\frac{b'}{a'}$ but I don't know how to prove the existence of $\lambda$. Can someone help me? Thanks before!
Idea: if $T_1 = T_2$:
$$ac'z^2 + (ad' + bc')z + bd' = (az + b)(c'z + d') = (a'z + b')(cz + d) = a'cz^2 + (a'd + b'c)z + b'd.$$ Equal polynomials have the same coefficients. Start form $bd' = b'd$ and define $\lambda = b/b' = d/d'$ (what happens if $b' = 0$ or $d' = 0$?).