- Suppose f is entire and never vanishes, and that none of the higher derivatives of \(f\) ever vanish. Prove that if \(f\) is also of finite order, then \(f ( z ) = e ^ { a z + b }\) for some constants a and b.
Solution
Because \(f\) is entire, of finite order, and never vanishes, Hadamard’s factorization theorem gives \[ f(z)=e^{P(z)} \] for some polynomial \(P\).
Differentiating, \[ f'(z)=P'(z)e^{P(z)}. \] By hypothesis \(f'\) never vanishes, and \(e^{P(z)}\) never vanishes, so \(P'\) has no zeros. If \(\deg P\ge2\), then \(P'\) is a nonconstant polynomial and therefore has a complex zero by the fundamental theorem of algebra, a contradiction. Hence \(\deg P\le1\).
Thus \[ P(z)=az+b \] for constants \(a,b\in\mathbb C\), and \[ \boxed{f(z)=e^{az+b}}. \] Under the stated hypothesis that no positive-order derivative ever vanishes, necessarily \(a\ne0\); indeed, if \(a=0\), then \(f'\equiv0\).