# Complex cross-ratio and determinants

Suppose that $$z_1,z_2,z_3,$$ and $$w_1,w_2,w_3$$ are two triplets of complex numbers. Show that if $$w = T(z)$$ is the Möbius transformation such that $$T(z_j) = w_j$$ for $$j = 1,2,3$$ then it can be described by the equation:

$$\det\begin{bmatrix} 1&1&1&1\\ z&z_1&z_2&z_3\\ w&w_1&w_2&w_3\\ zw&z_1w_1&z_2w_2&z_3w_3\\ \end{bmatrix} = 0$$

I dont understand where the determinant comes from. I am assuming (since we are dealing with triplets of points) that we are supposed to use the cross-ratio, but I have never seen a determinant used together with cross-ratio in complex analysis. Any pointers on how to approach a problem like this? What theory am I missing?

Expand the determinant along the first column to see that the equation is of the form $$A + Bz + Cw + Dzw = 0 \iff w = \frac{-A-Bz}{C+Dz} =: S(z)$$ for some complex constants $$A, B, C, D$$. So the equation defines a Möbius transformation $$S$$. If $$(z, w) = (z_j, w_j)$$ for some $$j$$ then the determinant is zero because two columns of the determinant are equal. It follows that $$S(z_j) = w_j$$ for $$j=1,2,3$$, and since Möbius transformations are uniquely determined by their images at three distinct points, $$S = T$$.