the Bieberbach bound on the second coefficient
/ BEE-ber-bahkh /
Once you have normalized your univalent maps into the class S, the very first question is: how big can the next coefficient be? The series starts f(z) = z + a_2 z^2 + ... , and a_2 measures the first deviation from the plain identity map. Bieberbach answered this in 1916 with a single, clean number: the second coefficient can never exceed 2 in size. It is the foundation stone on which almost every later estimate in the theory rests.
Precisely: for every f in the class S, |a_2| <= 2, and the bound is sharp — equality holds exactly for rotations of the Koebe function k(z) = z + 2 z^2 + 3 z^3 + ... , whose a_2 is exactly 2. The proof is short and beautiful: given f in S, form the odd square-root transform h(z) = sqrt(f(z^2)) = z + (a_2/2) z^3 + ... , check it is again univalent, pass to its reciprocal-type expansion to land in the class Sigma, and apply the area theorem. The single inequality sum n|b_n|^2 <= 1 from the area theorem reads off as |a_2| <= 2.
The reason to care is leverage. From |a_2| <= 2 alone one derives the Koebe one-quarter theorem (every S-image covers a disk of radius 1/4) and, with a little more work, the growth and distortion theorems. Bieberbach also conjectured in the same breath that the pattern continues — |a_n| <= n for all n — which became the Bieberbach conjecture and stood open for 69 years. So the n = 2 case is a theorem from 1916; the general statement waited until de Branges in 1985. Keep that distinction clear: |a_2| <= 2 is elementary, |a_n| <= n is deep.
Test the Koebe function k(z) = z/(1 - z)^2. Expanding, k(z) = z + 2 z^2 + 3 z^3 + ... , so a_2 = 2 exactly. This is the extremal case: no function in S can have |a_2| larger than this, and any that achieves 2 is a rotation e^(-i theta) k(e^(i theta) z) of Koebe.
The Koebe function saturates the bound |a_2| <= 2 with a_2 = 2.
The bound |a_2| <= 2 (1916) is a theorem; the general |a_n| <= n was only a conjecture until de Branges proved it in 1985. Do not cite the n = 2 case as if it settled the whole conjecture.