Intuition/references for why Möbius transformations are uniquely defined by three points? Assume that I have a Möbius transformation
$$M(z)=\frac{az+b}{cz+d}$$
with real coefficients $a$, $b$, $c$, and $d$, and complex argument $z$. I have been interested in gaining intuition on why this transformation is uniquely defined by three points $(z_1,z_2,z_3)$ and their images $(w_1,w_2,w_3)$. I combed the StackExchange and found this answer by user Lehs, which suggests the reason is that the equation above can be rewritten as:
$$M(z)=\frac{a}{c}\frac{z+\frac{b}{a}}{z+\frac{d}{c}}=A\frac{z+B}{z+C}$$
where
$$A=\frac{a}{c}$$
$$B=\frac{b}{a}$$
$$C=\frac{d}{c}$$
This sort of explains how the four coefficients $a$, $b$, $c$, and $d$ can be reduced to three coefficients $A$, $B$, and $C$, but leaves me with two new questions:

*

*Why is Möbius transformation usually expressed in terms of $a$, $b$, $c$, and $d$ if there is a seemingly more succinct way of representing this transformation?


*An earlier comment in the same question also mentions 'symmetry', and suggests that while the transformation is uniquely determined, the coefficients $a$, $b$, $c$, and $d$ are not. I can see that multiplying the four coefficients with the same non-zero factor would yield the same $A$, $B$, and $C$. Does this mean that $A$, $B$, and $C$ uniquely define a transformation?
If you have any other comments which may aid my intuition, I would greatly appreciate it. Also, if you have any references you could suggest with regards to intuition about the Möbius transformation, or which detail any of the results I collected here, that would be amazing.
 A: If $T, S$ have same images at 3 unique points, then $T\circ S^{-1}$ is identity map because Mobius transform has at most 2 fixed points (quadratic equation from $\frac{az+b}{cz+d}= z$, has at most 2 solutions). Then its uniquely defined.
A: 
I have been interested in gaining intuition on why this transformation is uniquely defined by three points and their images.

A little bit of intuition comes from counting dimensions. While there are four variables written for a transformation, there are really only three degrees of freedom since they can all be multiplied by a scalar without changing the transformation. Similarly, there are three degrees of freedom in specifying three points on the complex projective line. (I will make the executive decision to assume we are talking about complex coefficients, with real coefficients as a possible special case.) This intuition is useful, however not enough to turn into a proof - in principle, it only guarantees there are "discretely" many orbits (acting on $3$-tuples of distinct points) and discrete stabilizers, since discrete sets are $0$-dimensional.
One way to show $\mathrm{GL}_2\mathbb{C}$ (or, WLOG, $\mathrm{PSL}_2\mathbb{C}$) acts sharply $3$-transitively is that only the identity fixes all three points $0,\infty,1$ and we can arrange for any three points $a,b,c$ to become $0,\infty,1$ via
$$ M(z)=\frac{z-a}{z-b}\cdot\frac{c-b}{c-a}, $$
called a cross-ratio. The factor $z-a$ ensures $a\mapsto 0$, the factor $z-b$ ensures $b\mapsto\infty$, and the second fraction is a normalizaton factor that ensures $c\mapsto1$. The ratios must be suitably interpreted using the rules $\infty/\infty=1$ and anything$/0=\infty$, which means $a,b,c$ (distinct) can also be e.g. $\infty$.

Why is Möbius transformation usually expressed in terms of a, b, c, and d if there is a seemingly more succinct way of representing this transformation?

Not every transformation can be written that way, actually. But even ignoring that, the form with $a,b,c,d$ highlights the fact that Mobius transformations originate from matrix transformations. It's no coincidence that composition of Mobius transformations is represented by matrix multiplication! One starts with $2\times2$ matrices acting on $\mathbb{C}^2$ (or $\mathbb{R}^2$, if you wish), and then forms the projective line $\mathbb{CP}^1$ (or Riemann sphere $\mathbb{C}\cup\{\infty\}$) by simply turning a pair of coordinates into a fraction, $(w,z)\mapsto w/z$.

Does this mean that A, B, and C uniquely define a transformation?

Technically, no. If $c=0$ then $A=C=\infty$ but then there is no way to figure out what coefficient should be out in front of $(z+B)$. But if $c\ne0$ then $A,B,C$ do uniquely determine $M$.
