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:


$$a=\det\begin{bmatrix} z_1w_1 & w_1 & 1\\ z_2w_2 & w_2 & 1\\ z_3w_3 & w_3 & 1\end{bmatrix}$$

$$b=\det\begin{bmatrix} z_1w_1 & z_1 & w_1\\ z_2w_2 & z_2 & w_2\\ z_3w_3 & z_3 & w_3\end{bmatrix}$$

$$c=\det\begin{bmatrix} z_1 & w_1 & 1\\ z_2 & w_2 & 1\\ z_3 & w_3 & 1\end{bmatrix}$$

$$d=\det\begin{bmatrix} z_1w_1 & z_1 & 1\\ z_2w_2 & z_2 & 1\\ z_3w_3 & z_3 & 1\end{bmatrix}$$

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?


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 $< 3$.


The equations can be written as a linear system of three equations in three unknowns

$$z_k\frac ad+\frac bd-z_kw_k\frac cd=w_k$$ for $k=1,2,3$.

Solve the system by Cramer's rule. Every unknown will be the ratio of two determinants, and you can set $d$ equal to the common denominator. Then $a,b,c$ are the determinants at the numerators.

In particular,






For understanding how to do it (for generic point sets), as opposed to obtaining a formula, I think it is clearest to do it in stages. The basic idea is to reduce a set $z_1,z_2,z_3$ of $3$ points to the canonical $0,1,\infty$ (on $\mathbb P^1$, the Riemann sphere), by linear fractional transformations, as follows. First, apply $z\to 1/(z-z_3)$ to send $z_3$ to $\infty$. Then (relabelling) send (the new) $z_1\to 0$ by $z\to z-z_1$. This stabilizes the previous effort sending $z_3\to \infty$. Then, preserving $0$ and $\infty$, apply $z\to z/z_2$ to send (the new) $z_2$ to $1$. (At each step, any obstruction is an indication of insufficient generic-ness...)


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