Let \(f\) be an entire function such that \(f\) is nowhere zero and the set \[ f^{-1}(1) \] is finite. Prove that \(f\) is constant.
Solution
Suppose for contradiction that \(f\) is nonconstant.
1. The function \(f\) cannot be a polynomial.
A nonconstant complex polynomial has a zero by the fundamental theorem of algebra. This contradicts the hypothesis that \[ f(z)\ne0 \qquad(z\in\mathbb C). \] Hence a nonconstant \(f\) would have to be transcendental entire.
2. A transcendental entire \(f\) has an essential singularity at \(\infty\).
Consider \[ F(w)=f(1/w) \] near \(w=0\). If \(w=0\) were removable, then \(f\) would be bounded outside a disk and hence, by the maximum principle applied together with compactness of the disk, bounded on all of \(\mathbb C\); Liouville would make \(f\) constant. If \(w=0\) were a pole, then \(f\) would have polynomial growth at infinity and hence would be a polynomial. Both alternatives contradict 1 and the assumed nonconstancy. Thus \(w=0\), equivalently \(\infty\) for \(f\), is an essential singularity.
3. Great Picard gives the contradiction.
By Great Picard, in every neighborhood of the essential singularity at \(\infty\), the function \(f\) assumes every complex value with at most one exception.
The value \(0\) is omitted everywhere by hypothesis, so \(0\) must be the exceptional value. Therefore \(f\) takes the value \(1\) in every sufficiently small punctured neighborhood of \(\infty\); equivalently, for every \(R>0\) there exists \[ z_R, \qquad |z_R|>R, \qquad f(z_R)=1. \] Choosing \(R\to\infty\) produces infinitely many distinct points of \(f^{-1}(1)\), contradicting the assumption that this fiber is finite.
Hence \(f\) cannot be nonconstant, so \(f\) is constant.