Suppose \(f\) is meromorphic. Show that if \(z_0\) is a removable singularity of \(f\), then it is also a removable singularity of \(f'\). Conversely, if \(z_0\) is removable for \(f'\), then it is also removable for \(f\).
Solution
Let \(f\) be holomorphic on a punctured disk \(\DD_\eps^*(z_0)\).
If \(z_0\) is removable for \(f\), let \(F\) be the holomorphic extension of \(f\) to \(\DD_\eps(z_0)\). Then \(F'\) is holomorphic on \(\DD_\eps(z_0)\) and equals \(f'\) on \(\DD_\eps^*(z_0)\), so \(z_0\) is removable for \(f'\).
For the converse, if \(z_0\) is removable for \(f'\), let \(g\) be the holomorphic extension of \(f'\) to \(\DD_\eps(z_0)\), which exists by Riemann’s removable singularity theorem. Since \(g\) is holomorphic on the disk, it has a primitive \(F(z) \da \int_{z_0}^z g(\xi) \dxi\) there. Now \(G\da F - f\) satisfies \(G' = g - f' \equiv 0\) on the connected set \(\DD_\eps^*(z_0)\), so \(G\equiv c\) is constant and \(f(z) = F(z) - c\). In particular, \[ \lim_{z\to z_0} f(z) = F(z_0) - c \] exists, so \(z_0\) is removable for \(f\).