I am interesting in finding the four parameters of a Möbius transformation given three complex points $z_1$, $z_2$, and $z_3$, and their images $w_1$, $w_2$, and $w_3$. Wikipedia provides the following equations which seem to work:
$$w=\frac{az+b}{cz+d}$$
$$a=\det\begin{bmatrix}
z_1w_1 & w_1 & 1\\
z_2w_2 & w_2 & 1\\
z_3w_3 & w_3 & 1\end{bmatrix}$$
$$b=\det\begin{bmatrix}
z_1w_1 & z_1 & w_1\\
z_2w_2 & z_2 & w_2\\
z_3w_3 & z_3 & w_3\end{bmatrix}$$
$$c=\det\begin{bmatrix}
z_1 & w_1 & 1\\
z_2 & w_2 & 1\\
z_3 & w_3 & 1\end{bmatrix}$$
$$d=\det\begin{bmatrix}
z_1w_1 & z_1 & 1\\
z_2w_2 & z_2 & 1\\
z_3w_3 & z_3 & 1\end{bmatrix}$$
Unfortunately, Wikipedia does not provide a reference for this approach. Do you know of any book or publication which I could use as a source to understand this approach?
Best Answer
Sorry I don't know any references, but I can derive the result for you.
One method is to use invariance of the cross-ratio. You write $[w,w_1,w_2,w_3] = [z,z_1,z_2,z_3]$ and solve for $w$. Checking that this is equivalent to the formulas you wrote might be tedious, so I'll use a different method.
To get the formulas you wrote, first write down the linear equations $c, -a, d, -b$ must satisfy. We obtain a homogenous system with matrix $$ \begin{pmatrix} z_1 w_1 & z_1 & w_1 & 1 \\ z_2 w_2 & z_2 & w_2 & 1 \\ z_3 w_3 & z_3 & w_3 & 1 \\ \end{pmatrix}. $$ By the choice of the $z_i$'s and $w_i$'s (which I presume have been chosen to be triples of distinct numbers), the solutions $(c, -a, d, -b)$ are determined up to proportionality. Hence the set of solutions is a line through the origin, and the system has rank 3. The problem is to select a particular nonzero solution in a neat way.
In general, given a homogeneous system with matrix $$ \begin{pmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ \end{pmatrix}, $$ one obtains a solution by letting $x_j$, for $j = 1, 2, 3, 4$, be $(-1)^{j - 1}$ times the determinant obtained by deleting the $j$th column. (This is the solution for $(c, -a, d, -b)$ given on the Wikipedia page.)
To prove that this is indeed a solution, note that for any $i = 1, 2, 3$, the square matrix $$ \begin{pmatrix} a_{i1} & a_{i2} & a_{i3} & a_{i4} \\ a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ \end{pmatrix}, $$ is singular because it has a repeated row. Now the $i$th equation to be verified amounts to saying that the determinant of the above matrix, expanded along the first row, is zero.
It is not true in general that the solution $(x_1, x_2, x_3, x_4)$ obtained will be nonzero. But if the original $3 \times 4$ system has rank 3, then this will in fact be the case. For if the determinants defining the $x_j$'s were all zero, then every $3 \times 3$ submatrix of the original system would have zero determinant, showing that its rank is $< 3$.