Suppose \(f\) is holomorphic with \(z_0 = 0\) an isolated singularity, and suppose there is some neighborhood of \(0\) on which \[ \abs{f(z)} \leq \abs{z}^{-{ 1\over 2}} .\] Show that \(z_0\) is removable.
Warning: Riemann’s removable singularity theorem won’t apply to \(z^{1\over 2}f(z)\) since \(z^{1\over 2}\) is highly singular at \(z=0\).
Solution
Using the inequality, \[ \abs{(z-0)f(z)} \leq \abs{z}^{1\over 2}\convergesto{\abs{z}\to 0}0 ,\] so \(z=0\) is removable by Riemann’s removable singularity theorem.