“…If f is defined by [10] f (z) f (z) = a + e −2az , f (z) f (z) = a 2 + e −4az , then f and f − a 2 f have no zeros at all in the plane. For an example of finite order, define a zero-free function f ∈ U 2 by setting [7] f (z) f (z) = −16z 2 + 8z + 2, f (z) − 12f (z) f (z) = 256z 3 (z − 1), so that f − 12f has only real zeros. The proof of Theorem 1.4 uses machinery developed in [25,28] for the Wiman conjecture, and refinements from [5,6,22], but departs from the earlier methods in several significant steps.…”