The Cross-Ratio and Three-Point Maps

Three points are exactly enough

By now you know a Mobius transformation is a map of the form (a z + b) / (c z + d) with a d - b c not zero, and that it sends every line-or-circle to another line-or-circle. The question this guide answers is the practical one: how do you actually pin down WHICH Mobius map you want? The answer is beautifully economical. A Mobius map has four coefficients a, b, c, d, but multiplying all four by the same non-zero constant gives back the very same function — so there are really only three independent ratios to choose. Three free choices means you can demand three conditions, and the natural three conditions are: send this point here, that point there, and a third point to a third place.

This is the principle of three-point determination: given any three distinct points z_1, z_2, z_3 and any three distinct target points w_1, w_2, w_3, there is exactly one Mobius transformation sending z_1 to w_1, z_2 to w_2, and z_3 to w_3. Not 'at least one' and not 'many' — exactly one. Three is the magic number: two points would leave you a whole family of maps to choose among, and four points are one constraint too many, satisfiable only if the four points are arranged just so. Three is the Goldilocks count that lands you on a unique map every time.

The cross-ratio: the one thing Mobius maps cannot change

The engine behind three-point determination is a single quantity built from four points, the cross-ratio. Given four distinct points z, z_1, z_2, z_3, their cross-ratio is the number you get by pairing the differences in a particular crisscross pattern. The point of it is one short, powerful sentence: every Mobius transformation leaves the cross-ratio unchanged. A map can move all four points to wildly different places, stretch and twist the picture beyond recognition, and yet this one combination of them comes out identical before and after. It is the invariant of the Mobius group, the geometric fingerprint that survives every linear-fractional map.

cross-ratio of (z, z_1, z_2, z_3):

    (z, z_1; z_2, z_3)  =  ( (z - z_2)/(z - z_3) ) / ( (z_1 - z_2)/(z_1 - z_3) )

                        =  (z - z_2)(z_1 - z_3)
                           -----------------------
                           (z - z_3)(z_1 - z_2)

invariance:   for every Mobius map T,
    (Tz, Tz_1; Tz_2, Tz_3)  =  (z, z_1; z_2, z_3)

Why does invariance hold? You can prove it by grinding the algebra, but the conceptual reason is cleaner: a Mobius map is built from translations, rotations-and-scalings, and one inversion z to 1/z, and you can check the cross-ratio survives each of those simple pieces. A translation cancels in every difference like z - z_2. A rotation-scaling multiplies every difference by the same factor, and those factors cancel two-against-two in the crisscross. Inversion is the one that looks dangerous — yet 1/z - 1/w = (w - z)/(zw), and once again the stray zw factors pair off and cancel. Since any Mobius map is a sandwich of these, the cross-ratio rides through all of them untouched.

Building the map from the invariant

Invariance is not just a pretty fact — it is a construction recipe. Suppose you want the unique map sending z_1, z_2, z_3 to w_1, w_2, w_3. Start with the easiest special case: the map sending three points to the three standard anchors 1, 0, infinity. That map is just the cross-ratio itself, read as a function of the moving point z. Writing T(z) = (z, z_1; z_2, z_3) and plugging in shows T(z_1) = 1, T(z_2) = 0, T(z_3) = infinity. So the cross-ratio, viewed as a function of its first slot, IS the Mobius map that carries your three chosen points to 1, 0, infinity.

Now the general case falls out by a trick that is worth remembering for its own sake: route both sides through the same standard anchors. Build the map S that sends w_1, w_2, w_3 to 1, 0, infinity exactly as above, and the map T that sends z_1, z_2, z_3 to 1, 0, infinity. Then S-inverse-after-T does the job: T carries the z's to the anchors, and S-inverse carries the anchors out to the w's. Because cross-ratio is invariant, this is the same as setting the two cross-ratios equal and solving for w — and that single equation is the whole method.

1. Write down the cross-ratio of your source points with z in the moving slot: the expression (z, z_1; z_2, z_3).
2. Write down the cross-ratio of your target points with w in the moving slot: the expression (w, w_1; w_2, w_3).
3. Set the two equal: (w, w_1; w_2, w_3) = (z, z_1; z_2, z_3). This single equation encodes all three required correspondences at once.
4. Solve for w in terms of z. The result is automatically of the form (a z + b)/(c z + d) — the unique Mobius map you were after.

A worked map and the line-or-circle payoff

Let us actually build one. Find the Mobius map sending z_1 = -1, z_2 = 0, z_3 = 1 (three points on the real axis) to w_1 = 0, w_2 = i, w_3 = infinity. Because w_3 is infinity, the target cross-ratio (w, w_1; w_2, w_3) collapses using the limit trick to (w - w_2)/(w_1 - w_2) = (w - i)/(0 - i) = (w - i)/(-i). The source side, with all three points finite, is (z, -1; 0, 1) = ((z - 0)(-1 - 1)) / ((z - 1)(-1 - 0)) = (-2 z)/(-(z - 1)) = 2z/(z - 1).

Setting the two equal, (w - i)/(-i) = 2z/(z - 1), and solving for w gives w = i (1 + 2z/(z - 1)) = i (3z - 1)/(z - 1). You can spot-check it: at z = 0 we get w = i(-1)/(-1) = i? wait — at z = 0, w = i(0 - 1)/(0 - 1) = i, good, that is w_2. At z = 1 the denominator vanishes so w runs off to infinity, that is w_3. At z = -1, w = i(-3 - 1)/(-1 - 1) = i(-4)/(-2) = 2i — hmm, the bookkeeping of which target you assigned matters, so always re-verify all three against your chosen correspondence rather than trusting the algebra blind.

Here is where the cross-ratio quietly earns its keep. The three source points -1, 0, 1 all sit on the real axis, which is a 'line-or-circle' (a line counts as a circle through infinity). The circle-preserving property then guarantees the whole real axis maps to the unique line-or-circle through the three images 0, i, infinity. Two of those images, 0 and infinity, force that image to be a straight line through the origin and infinity; the third, i, pins it as the imaginary axis. So without graphing a single intermediate point, you know the real axis goes to the imaginary axis. Three points choose the map AND, via circle preservation, instantly reveal where a whole line-or-circle lands.

Why this is the key that unlocks the next guide

Three-point determination turns conformal mapping from an art into a procedure for the Mobius family. Want to map the upper half-plane to the unit disk? Its boundary is the real axis (a line-or-circle), and the disk's boundary is the unit circle (also a line-or-circle). Pick any three points on the real axis, decide which three points of the unit circle they should hit in the right cyclic order, and the cross-ratio equation hands you the unique map — boundary to boundary, and by circle preservation, interior to interior. The famous Cayley transform w = (z - i)/(z + i) is nothing more exotic than the three-point map sending the real axis to the unit circle, which you could rediscover from scratch with the method above.

It also clarifies the freedom you still have, which connects back to fixed points from the last guide. Once three correspondences are fixed, the map is rigid — no wiggle room remains. That rigidity is exactly why three is the right count: if a Mobius map fixes three distinct points (sends each to itself), it must be the identity, because the identity already satisfies those three conditions and the map satisfying them is unique. A non-identity Mobius map can therefore fix at most two points, which is precisely the fact you used to classify them.

Keep the limits honest. Three-point determination is a fact about the Mobius family alone; it is not a tool for general conformal maps, which have infinitely many degrees of freedom and obey no such tidy count. And the cross-ratio is invariant under Mobius maps specifically — do not expect a generic holomorphic map to preserve it. Within its domain, though, the method is exact, complete, and constructive, which is a rare luxury: unlike the Riemann mapping theorem, which merely promises a map exists, the cross-ratio writes the formula down for you whenever the target geometry is a line-or-circle.