- Show that if f is an entire function of finite order that omits two values, then f is constant. This result remains true for any entire function and is known as Picard’s little theorem.
[Hint: If f misses a, then \(f ( z ) - a\) is of the form \(e ^ { p ( z ) }\) where p is a polynomial.]
Solution
Suppose the entire function \(f\) has finite order and omits two distinct values \(a,b\in\mathbb C\).
Since \(f-a\) is an entire finite-order function with no zeros, Hadamard’s factorization theorem implies \[ f(z)-a=e^{P(z)} \] for some polynomial \(P\).
If \(P\) were nonconstant, choose any logarithm \(\lambda\) of the nonzero number \(b-a\), so \[ e^\lambda=b-a. \] A nonconstant complex polynomial assumes every complex value: by the fundamental theorem of algebra, \(P(z)-\lambda\) has a root. Thus there would be \(z_0\) with \(P(z_0)=\lambda\), and then \[ f(z_0)-a=e^{P(z_0)}=b-a, \] so \(f(z_0)=b\), contradicting that \(f\) omits \(b\).
Therefore \(P\) is constant, and consequently \(f\) is constant.