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.
Key idea
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.
Why it works
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
Worked example: 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 $T(z) = -i\,\frac{z - 1}{z + 1}$.
- $T(0) = -i \cdot (-1) = i$.
Answer: $T(z) = -i\frac{z - 1}{z + 1}$, $T(0) = i$
Worked example: 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 $T(z) = z - 2$.
Answer: $T(z) = z - 2$
Worked example: 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: $T(0) = 1$, $T(1) = \frac{1 + i}{1 - i} = i$, $T(\infty) = -1$.
Answer: $T(-1) = -i$
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.
Common mistake
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.