# Reference request: parameters of Möbius transformation by three points and images?

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?

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,

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

and

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

$$\cdots$$

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...)