Prove $|z – z_0|^{-1 + \epsilon}$ is bounded

complex-analysis

This is the question from Stein and Shakarchi Complex Analysis.

Suppose that $f(z)$ is holomorphic in a punctured disc $D_r (z_0) – \{z_0 \}$. Suppose also that $|f(z)| \leq A |z – z_0|^{-1 + \epsilon}$ for some $\epsilon > 0$, and all $z$ near $z_0$. Show that the singularity of $f$ at $z_0$ is removable.

My instinct is to use Riemann's theorem on removable singularities and show that $f$ is bounded near $z_0$. My guess is that's why the inequality is given in the question, but $|z – z_0|^{-1 + \epsilon}$ doesn't seem to be necessarily bounded? As a real function at least it definitely isn't if $\epsilon = 0.5$.

Best Answer

Hint: Multiply both sides by $|z-z_0|$ and take the limit as $z$ tends to $z_0$.