# Prove that $z_0$ is a removable singularity

Let $z_0$ be an isolated singularity of a holomorphic function $f$. Suppose that there are $A, \epsilon > 0$ such that for all $z$ sufficiently close to $z_0$ we have $$|f(z)| \leq \frac{A}{|z - z_0|^{1 - \epsilon}}\ .$$ Prove that $z_0$ is a removable singularity.

Can someone please show me how to do this problem? Thank you.

• A good start is to state the exact definition of a removable singularity. The answer will follow. – Alex R. Aug 6 '15 at 1:05
• Should be $A,\epsilon>0$, yes? – David Aug 6 '15 at 1:06
• I imagine that you mean $A, \epsilon> 0$? – mathcounterexamples.net Aug 6 '15 at 1:06
• yeah, you are right. $\epsilon$ should be greater than 0 – Alexander Aug 6 '15 at 1:13

If not, define $$h(z) = \begin{cases} (z-z_0)^2f(z) &\mbox{if } z \neq z_0 \\ 0 & \mbox{if } z=z_0 \end{cases}$$ and prove that $h$ is holomorphic in a neighborhood of $z=z_0$. Then consider the Taylor series of $h$ about $z_0$. Notice that $h(z_0)=h^\prime(z_0)=0$.
• The way I used to do this problem is compute the residue of $f(z)$ – Alexander Aug 6 '15 at 4:32