Show that if \(f\) is entire and there exists a disc \(\DD_r(a)\) not intersecting \(f(\CC)\), then \(f\) must be constant.
Solution
Let \(g(z)\da f(z)-a\). Then \(\DD_r(0)\) does not intersect \(g(\CC)\), so \(\abs{g(z)} \geq r\) for all \(z\in\CC\). In particular \(g\) has no zeros, so \(G(z) \da {1\over g(z)}\) is entire, and \(\abs{G(z)} \leq \inverseof{r}\) for all \(z\). By Liouville, \(G\) is constant, so \(g\) and hence \(f\) are constant.