Triple transitivity

A $2\times2$ complex matrix has four complex parameters. If we require the determinant $ad-bc$ to be 1, this allows three complex degrees of freedom. Thus, it is not surprising that prescribing the images of a Möbius transformation at three points determines the transformation completely. This is stated precisely as follows.

Triple Transitivity: Let $z_1$, $z_2$, $z_3$ be any three distinct points in the Riemann sphere, and let $w_1$, $w_2$, $w_3$ be another triple of distinct points in the Riemann sphere (distinct from each other but not necessarily from the $z_j$'s). Then there is precisely one conformal Möbius transformation $m(z)=\frac{az+b}{cz+d}$ such that $m(z_j)=w_j$ for each $j=1,2,3$.

Proof: This statement has both an existence and uniqueness part. First, let's prove there is a transformation that maps $z_1$, $z_2$, $z_3$ to 0, 1, $\infty$, respectively. This we can do by an explicit formula: $$T_1(z)=\left(\frac{z-z_1}{z-z_3}\right) \left(\frac{z_2-z_3}{z_2-z_1}\right).$$ Our claim may be verified by simply checking this formula for each of $z_1$, $z_2$, $z_3$.

Let $T_2(z)$ be the analogous transformation mapping $w_1$, $w_2$, $w_3$ onto 0, 1, $\infty$, respectively. Then $T_2^{-1}\circ T_1$ carries $z_1$, $z_2$, $z_3$ to $w_1$, $w_2$, $w_3$, respectively. This proves the existence of the map $m(z)$ claimed by the statement.

To prove uniqueness, assume there are two Möbius maps $m_1(z)$, $m_2(z)$ carrying $z_j$ to $w_j$ for $j=1,2,3$. Then $m_2^{-1}\circ m_1$ is a conformal map carrying $z_j$ to $z_j$ for $j=1,2,3$. Thus, $m_2^{-1}\circ m_1$ is a conformal Möbius transformation with at least three distinct fixed points. The only such transformation is the identity transformation. (Note this wouldn't be true if we allowed anti-conformal transformations such as reflections.) thus, $m_2^{-1}\circ m_1(z)=z$, from which by applying $m_2$ we deduce $m_1(z)=m_2(z)$. QED

This is an extremely practical tool for constructing Möbius maps that accomplish prescribed goals. For instance, suppose we wish to have a map that transforms the unit circle to the real line. Then we cook up a map sending $1$, $i$, $-1$ to 0, 1, $\infty$, respectively. This would be $$m(z)=\left(\frac{z-1}{z+1}\right)\left(\frac{i+1}{i-1}\right) = -i\,\frac{z-1}{z+1}.$$ Since three points determine a circle, and Möbius transformations carry circles to circles, this map does indeed map the unit circle to the real axis.