determining a Mobius map from three points

Suppose you want a Mobius transformation that sends one trio of points to another chosen trio — say it should carry these three corners of a region to these three target spots. How much freedom is there, and how do you actually build the map? The clean answer is that the trio nails it down completely: there is exactly one Mobius transformation sending any three distinct points to any three other distinct points. No more, no fewer — three is the magic number.

Why three? A Mobius map has the form (a z + b)/(c z + d), and scaling the four coefficients together changes nothing, so there are really three independent complex degrees of freedom. Each 'send this point to that point' condition uses up one degree, so three conditions fix the map uniquely. Uniqueness also follows from fixed points: if two maps both did the job, then one composed with the inverse of the other would fix three points, and a non-identity Mobius map can fix at most two — so the two maps must be identical.

To build it explicitly, use the cross-ratio. To send z_1, z_2, z_3 to w_1, w_2, w_3, write down the equation (w, w_1; w_2, w_3) = (z, z_1; z_2, z_3) — equate the cross-ratios — and solve for w as a function of z. Because the cross-ratio is Mobius-invariant, the resulting w(z) automatically maps the three source points to the three targets. This is the standard machine for producing every named conformal map between model regions: you choose three convenient boundary points (often including infinity) on each region and let the formula do the rest. A practical note: the three points must be DISTINCT, and you must respect orientation — sending the boundary's three points in the right cyclic order is what guarantees the interior maps to the interior, not to the exterior.