a Blaschke factor
/ BLAHSH-kuh /
A Blaschke factor is the simplest non-trivial automorphism of the disk: a single, tailor-made map whose whole job is to send one chosen interior point of the disk to the center, while keeping the disk exactly in place. If you want to 'recenter' the disk so that some off-center point a becomes the new origin — without leaving the disk and without bending any angles — the Blaschke factor for a is the tool that does it.
For a point a with |a| < 1, the Blaschke factor is B_a(z) = (z - a) / (1 - a-bar z). Notice what it does at key points: B_a(a) = 0 (the chosen point goes to the center), and B_a(0) = -a. The magic is that it maps the disk onto the disk: on the boundary circle |z| = 1 one can verify |B_a(z)| = 1 (the numerator and denominator have equal modulus there, because 1 - a-bar z is, up to a unit factor, the conjugate-reflection of z - a), so the unit circle goes to the unit circle, and continuity plus the maximum principle drop the interior into the interior. Its zero is the single simple zero at z = a, and its only pole sits at z = 1/a-bar, safely outside the disk. Up to multiplying by a rotation e^(i theta), Blaschke factors are the building blocks of all disk automorphisms.
Blaschke factors are the atoms of disk geometry. Multiply several of them together — one for each of finitely many zeros — and you get a finite Blaschke product, a holomorphic self-map of the disk with exactly those zeros and modulus 1 on the boundary. Push to infinitely many zeros (subject to the Blaschke condition that the zeros do not crowd the boundary too fast) and you get the infinite Blaschke products central to bounded-function theory and Hardy spaces. A common misconception to head off: a Blaschke factor is NOT a translation of the disk — there is no holomorphic translation of the disk, since the only rigid motions available are these Mobius maps. It moves a to 0, but it warps every other point in a definite, non-uniform way.
With a = 1/2, B(z) = (z - 1/2)/(1 - z/2). Then B(1/2) = 0 as designed, and B(0) = -1/2. On the boundary, B(1) = (1/2)/(1/2) = 1, B(-1) = (-3/2)/(3/2) = -1, B(i) = (i - 1/2)/(1 - i/2): computing |i - 1/2| = sqrt(5)/2 and |1 - i/2| = sqrt(5)/2 gives |B(i)| = 1 — exactly on the unit circle, as it must be.
B_a sends a to 0 and keeps the boundary circle pinned to |z| = 1.
Watch the conjugate: it is a-bar, not a, in the denominator. Writing (z - a)/(1 - a z) by mistake gives a map that does NOT preserve the unit disk unless a is real. The conjugate is exactly what guarantees |B_a(z)| = 1 on the boundary.