Let the Möbius transform associated to the matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ be defined as $\mu_A:\mathbb C\to\mathbb C:z\mapsto\frac{az+b}{cz+d}$ provided $\det A\neq 0$.
It is straightforward to verify that $\mu_A\circ\mu_B=\mu_{AB}$. I was wondering if there is a more intuitive (and preferably elementary) way to see why we have this identity without having to do the calculation.
I was thinking of viewing $\frac{az+b}{cz+d}$ as a 'formal' fraction; that is, just another notation for $\begin{pmatrix}az+b\\cz+d\end{pmatrix}$ and then trying to find out if $AB\begin{pmatrix}z\\1\end{pmatrix}$ corresponds to the usual composition $\mu_A\circ\mu_B$. I can't see it. There should be a deeper reason for this.