Here's the problem: suppose $f$ has an isolated singularity at $z_0$ and that $|f(z)|<C/|z-z_0|^{1/2}$ near $z_0$ for some constant $C$. Prove that $z_0$ is a removable singularity, and in fact $f$ is bounded near $z_0$.
So we have some $R$ such that $|f(z)|<C/|z-z_0|^{1/2}$ when $0<|z-z_0|<R$, and I know all I need to show is that $f$ is bounded in this disk because this will immediately imply $z_0$ is a removable singularity. But I'm totally stuck on how to even begin showing this is bounded.
It seems to me $|z-z_0|^{1/2}$ could become as small as we want as we approach $z_0$, so $C/|z-z_0|^{1/2}$ would then become arbitrarily large. But I guess my thought process must be wrong somewhere. Does anybody see what I'm missing?
Hint: Consider the function $(z-z_0)f(z).$