Suppose \(f\) has an isolated singularity at \(a\in\mathbb{C}\) and \(\operatorname{Re} f\) is bounded above in a deleted neighborhood of \(a\). Prove that the singularity is removable.
Solution
Choose \(M\in\mathbb R\) and \(r>0\) such that \[ \operatorname{Re}f(z)\le M \qquad(0<|z-a|<r). \] Set \[ g(z)=e^{f(z)-M}. \] Then \(g\) is holomorphic on the punctured disk and \[ |g(z)|=e^{\operatorname{Re}f(z)-M}\le1. \] Hence the isolated singularity of \(g\) at \(a\) is removable. Let \(\widetilde g\) denote the holomorphic extension.
We claim \(\widetilde g(a)\ne0\). If instead \(\widetilde g(a)=0\), then for some \(m\ge1\), \[ \widetilde g(z)=(z-a)^m h(z),\qquad h(a)\ne0. \] On the punctured disk, \[ \frac{g'(z)}{g(z)}=f'(z), \] while the factorization gives \[ \frac{g'(z)}{g(z)}=\frac{m}{z-a}+\frac{h'(z)}{h(z)}. \] Thus \(f'\) has residue \(m\) at \(a\). But the derivative of a holomorphic function on a punctured disk has residue \(0\): in the Laurent expansion of \(f\), differentiating termwise produces no \((z-a)^{-1}\) term. This is a contradiction.
Therefore \(\widetilde g(a)\ne0\). Shrinking the disk if necessary, \(\widetilde g\) is nowhere zero there, so it has a holomorphic logarithm \(L\). On the punctured disk, \[ e^{L(z)}=\widetilde g(z)=e^{f(z)-M}. \] Hence \(f(z)-M-L(z)\) takes values in \(2\pi i\mathbb Z\) and is continuous, so it is constant on the punctured disk. Therefore \(f\) agrees there with a holomorphic function \(M+L+2\pi i k\), and the singularity at \(a\) is removable.