Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 9 4 Solution Created 2026-10-03 Updated 2026-10-14
First construct a holomorphic function without requiring boundedness. Separate the finitely many zeros at zero into , and enumerate the remaining prescribed zeros as . Since , there is no interior accumulation point and every prescribed point has finite multiplicity. Using the Weierstrass elementary factors, setFor a compact disc , choose with . Eventually . The power series identitygives an absolutely summable uniform bound. The infinite product convergence from logarithmic tails therefore makes the tail holomorphic and nonvanishing on this compact set. The finitely many initial factors have exactly their prescribed zeros, with the prescribed multiplicities. Hence has exactly the required zero sequence. This is a direct canonical product construction for zeros escaping to the disk boundary.
For a nonzero bounded holomorphic function, Jensen's formula bounds after separating the zero at zero. Thus the Blaschke condition is necessary; the Blaschke product construction makes it sufficient. Use the curvature-minus-one convention for Hyperbolic distance in the Poincare disc:The latter lies between and . Consequently the precise bounded-function criterion isThis counts multiplicities; the finitely many occurrences at zero cause no problem.
For the requested hyperbolic zero estimate for a Blaschke product, the pseudohyperbolic distance identity isIf is not a zero, . The real function has and , so . Thus . Apply this to each factor, including factors at zero, and then pass from finite products to the Blaschke product by locally uniform convergence:At a zero of the left side is zero, so the inequality is immediate; the logarithmic calculation at is unnecessary. The unimodular constant contributes nothing. Away from the zeros the product is nonzero, so the same inequality also bounds the sum and shows it is finite there.