Thanks, Gareth! Nice proof (below), makes complete sense. (I had got as far as: If z exp(-z) omits the value c, then z exp(-z) - c = exp(g(z)) for some entire function g(z), which looked quite impossible. But I didn't see how to finish the proof.) —Dan I wrote: ----- The "little theorem" of Picard states that an entire function is surjective or omits just one complex value, or else it is a constant. Clearly the function f(z) = z exp(-z) is not a constant. Does it omit a value? In more generality, what about entire functions of the form g(z) = P(z) exp(Q(z)) where P and Q are polynomials? ----- Gareth McCaughan replied (slightly edited for my readability): ----- An entire function that's never zero can be written as exp(F(z)) where F is entire. So an entire function that omits the value u can be written as u + exp F(z) where F(z) is entire. If z exp(-z) = u + exp(F(z)) then exp(F(z)) = z exp(-z) - u equals -u for _exactly one_ choice of z, namely z=0. Hence there is only one z for which F(z) is _any_ logarithm of -u. But there are lots of logarithms of -u, so F(z) omits almost all of them, which is impossible for entire F(z). (What if u=0? Impossible, because z exp z definitely doesn't omit the value 0.) If we have P exp Q (P,Q polynomials) instead, we can still do more or less the same. ... ... ... -----