On the zeros of a meromorphic function and its derivatives
Let \(f\) be a meromorphic function of finite order with only finitely many simple zeros. Suppose that \(f''\) has only finitely many zeros. It is shown that then \(f\) has the form \(f(x)=R(z)\exp P(z)\) where \(P\) is is a polynomial and \(R\) is a rational function. Another result says that a meromorphic function \(f\) of finite order has this form if there exists a polynomial \(h\) with \(h\not\equiv 1\) and \(h\not\equiv (n+1)/ n\) for all positive integers \(n\) such that \(ff''-h(f')^2\) has only finitely many zeros.
scientific article; zbMATH DE number 5077522