Let \(f\) be entire. Show that \(f\) has a removable singularity at \(z_0 = \infty\) iff \(f\) is constant.
Solution
Suppose \(f\) is not constant. If \(z=\infty\) is removable, \(f\) is bounded in a neighborhood of \(\infty\), say by \(M_1\) on \(\abs{z} > R\). Now \(\abs{z} \leq R\) is a closed and bounded set, thus compact, and since \(f\) is continuous here it is bounded by the extreme value theorem, say by \(M_2\). Then \(\abs{f(z)} \leq \max(M_1, M_2)\) on \(\CC\) is entire and bounded, thus constant by Liouville, a contradiction. \(\contradiction\)
Conversely, if \(f\) is constant, \(f\) is bounded in every neighborhood of \(\infty\), making it a removable singularity.