Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 12 / 4 / Solution

For the unrestricted zero sequence, introduce the Weierstrass elementary factorsFor $|u|\le 1/2$, cancellation of the first $p$ terms of the logarithm power series gives a holomorphic logarithm of the factor:Define the canonical productFix any compact disc $|z|\le R$. Since $|z_n|\to \infty$, eventually $|z/z_n|\le 1/2$ uniformly on this disc. The tail logarithms are then bounded by $2(1/2{)}^{n+1}$, a summable bound independent of $z$. Their sum is holomorphic by locally uniform convergence of holomorphic functions, and its exponential is nowhere zero. Multiplication by the finite initial product therefore gives an entire function. Each factor has a simple zero exactly at its specified $z_n$, the exponential factor has none, and a tail beginning beyond any given compact disc is nonvanishing there. Thus $f$ has precisely the prescribed zeros and no others, each simple, and $f(0)=1$. This proves the required existence without an assumption on ${\sum }_n|z_n{|}^{-1}$.

For the final, unheaded product request, suppose now that ${\sum }_n|z_n{|}^{-1}<\infty$. On $|z|\le R$, all sufficiently late factors have $|z/z_n|\le 1/2$, so part (a) bounds their principal complex logarithms by $2R/|z_n|$. The logarithms converge absolutely and uniformly; infinite product convergence from logarithmic tails proves thatis entire and again has exactly the prescribed simple zeros. For its finite partial products,Passing to the limit provesIn particular $P$ is an entire function of exponential type. This argument uses a valid modulus estimate and does not require the false logarithm assertion in part (c).