Assume that \(f : \mathbb{C} \to \mathbb{C}\) is an entire function, not identically equal to 0 and let \(\mathcal{Z} = \{z \in \mathbb{C} : f(z) = 0\}\). Prove that if \(\mathcal{Z}\) is unbounded, then \(f\) has an essential singularity at \(\infty\).
Solution
Consider \(g(w)=f(1/w)\) for \(w\ne0\). It is holomorphic on the punctured plane. The type of the isolated singularity of \(f\) at infinity is, by definition, the type of \(g\) at zero. Since the zero set is unbounded, choose distinct zeros \(z_n\) with \(|z_n|\to\infty\) and \(z_n\ne0\). Then \(w_n=1/z_n\) are distinct, tend to zero, and satisfy \(g(w_n)=0\).
1. The singularity of \(g\) at zero is not removable.
If it were removable, \(g\) would have a holomorphic extension \(G\) to a disk about zero. The zeros \(w_n\) would accumulate at the interior point zero. The identity theorem would give \(G=0\) on that disk [@SS03]. Thus \(f(z)=g(1/z)=0\) for all sufficiently large \(|z|\). Applying the identity theorem to the entire function \(f\) would give \(f=0\) on \(\mathbb C\), contrary to the hypothesis.
2. The singularity is not a pole and hence is essential.
If zero were a pole of order \(m\geq1\), one could write \(g(w)=w^{-m}h(w)\) near zero, with \(h\) holomorphic and \(h(0)\ne0\) [@SS03]. Continuity would make \(h\), and hence \(g\), nonzero in a sufficiently small punctured disk. This contradicts the sequence of zeros \(w_n\to0\).
An isolated singularity is removable, a pole, or essential [@SS03]. The first two cases have been excluded, so \(g\) has an essential singularity at zero. Equivalently, \(f\) has an essential singularity at infinity.