Let \(f(z)\) be entire and assume values of \(f(z)\) lie outside a bounded open set \(\Omega\). Show without using Picard’s theorems that \(f(z)\) is a constant.
Solution
Choose \(w_0 \in \Omega\) and \(\varepsilon > 0\) with \(\overline{D_\varepsilon(w_0)} \subseteq \Omega\). Then \(\abs{f(z) - w_0} \geq \varepsilon\) for every \(z\), so \(g(z) \da 1/(f(z)-w_0)\) is entire and bounded by \(1/\varepsilon\). By Liouville, \(g\) is constant, hence \(f\) is constant.