Why is a Möbius transform uniquely determined based on known mappings of three points? Why can the Möbius transform $\displaystyle{\frac{az+b}{cz+d}}$ be uniquely determined if one is aware that $\{p_1,p_2,p_3\}$ maps to $\{q_1,q_2,q_3\}$ where $p_1,\dots,q_3\in\Bbb{C}$?
Seeing as there are four unknowns $a,b,c,d$, shouldn't the mappings be known for four points instead of three?
 A: There is a symmetry.  Double $a$,$b$,$c$ and $d$, and you get the same transform.  So there are only three relevant numbers: $a/b$,$a/c$ and $a/d$
A: Mobius transformations can be represented by 2 x 2 matrices, given by
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix} \mapsto \frac{az + b}{cz + d}
$$
Well, sort of. There is a map $M_{2\times 2}(\mathbb{R}) \to Aut(\mathbb{H})$ (where  $Aut(\mathbb{H})$ is the collection of fractional linear transformations), which looks as you say like we should have four real parameters. However, there is a kernel to this map! Specifically, diagonal matrices all map to the identity (which you should check), and so the group of fractional linear transformations is really
$$
M_{2\times 2}(\mathbb{R})/(\lambda \cdot Id) = PSL_2(\mathbb{R})
$$
which is three dimensional. Hence you only need three points to determine a Mobius tranformation.
A: You can rewrite as $\displaystyle\frac{a}{c}\cdot\frac{z+ba^{-1}}{z+dc^{-1}}=A\frac{z+B}{z+C}$
A: Let $S(z)$ and $R(z)$ be two Möbius transformations which satisfy
\begin{gather*}
S(q_j) = p_j\\
R(q_j) = p_j.
\end{gather*}
Möbius transformations are rational functions, the transformation
$$S(z) - R(z)$$
has order at most 2, then it can have at most two zeros, but you know it has 3, so $S(z)-R(z)=0$ for all complez $z$. 
And that's why 3 points are enough to characterize a Möbius transformations.
