A Möbius map has four coefficients, but scaling all of them by the same number changes nothing, so it has three degrees of freedom. Choosing where three points go uses them all up: there is exactly one Möbius map for each choice.
Idea clave
Given distinct $z_1, z_2, z_3$, the map
$$S(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}$$
is the unique Möbius map with $S(z_1) = 0$, $S(z_2) = 1$, $S(z_3) = \infty$. (If one $z_j$ is $\infty$, drop the two factors containing it.) To send $z_1, z_2, z_3$ to $w_1, w_2, w_3$, solve
$$\frac{(w - w_1)(w_2 - w_3)}{(w - w_3)(w_2 - w_1)} = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}$$
for $w$. These quotients are cross-ratios, which Möbius maps preserve.
Por qué funciona
Check $S$ directly: the factor $z - z_1$ makes $S(z_1) = 0$, the factor $z - z_3$ in the bottom makes $S(z_3) = \infty$, and at $z = z_2$ everything cancels to $1$. Uniqueness: if $T$ also sends $z_1, z_2, z_3$ to $0, 1, \infty$, then $T \circ S^{-1}$ fixes $0$, $1$ and $\infty$; fixing $\infty$ and $0$ forces the form $kz$, and fixing $1$ forces $k = 1$. For general targets, if $S_z$ and $S_w$ send the triples to $0, 1, \infty$, the map is $S_w^{-1} \circ S_z$, which is what the cross-ratio equation says.
Worked examples
Ejemplo resuelto: To $0$, $1$, $infty$
Find the Möbius map with $T(1) = 0$, $T(i) = 1$, $T(-1) = \infty$, and compute $T(0)$.
- $T(z) = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)}$.
- $\frac{i + 1}{i - 1} = \frac{(1 + i)(-1 - i)}{2} = \frac{-2i}{2} = -i$, so $T(z) = -i\,\frac{z - 1}{z + 1}$.
- $T(0) = -i \cdot (-1) = i$.
Respuesta: $T(z) = -i\frac{z - 1}{z + 1}$, $T(0) = i$
Ejemplo resuelto: With $infty$ among the points
Find $T$ with $T(2) = 0$, $T(3) = 1$, $T(\infty) = \infty$.
- Fixing $\infty$ means no denominator: $T(z) = Az + B$.
- $T(2) = 0$ and $T(3) = 1$ give $T(z) = z - 2$.
Respuesta: $T(z) = z - 2$
Ejemplo resuelto: General targets
Find the Möbius map with $0 \mapsto 1$, $1 \mapsto i$, $\infty \mapsto -1$, and compute $T(-1)$.
- Cross-ratio with $z_3 = \infty$: the right side is $\frac{z - 0}{1 - 0} = z$. The left side is $\frac{(w - 1)(i + 1)}{(w + 1)(i - 1)} = -i\,\frac{w - 1}{w + 1}$.
- Solve $-i\frac{w - 1}{w + 1} = z$: $\frac{w - 1}{w + 1} = iz$, so $w = \frac{1 + iz}{1 - iz}$.
- $T(-1) = \frac{1 - i}{1 + i} = -i$. Check: $T(0) = 1$, $T(1) = \frac{1 + i}{1 - i} = i$, $T(\infty) = -1$.
Respuesta: $T(-1) = -i$
Three points determine a circle (or a line, if they are collinear). The map $S$ sends that circle onto the extended real line, since it sends the three points to $0$, $1$ and $\infty$, which lie on it. This is the quickest way to find which region a Möbius map sends onto a half-plane.
Error común
Pairing the points in the wrong order. The formula assigns roles: $z_1 \mapsto 0$, $z_2 \mapsto 1$, $z_3 \mapsto \infty$. Swapping two points gives a different map. In the first example, sending $1, -1, i$ to $0, 1, \infty$ instead gives $T(0) = \frac{-1 \cdot (-1 - i)}{-i \cdot (-2)} = \frac{1 + i}{2i}$, not $i$. Check your map on all three given points before using it.