# Let $0$ be an isolated singularity of $f$. Prove that if $|f(z)|\leq |z|^{-\alpha}$, this singularity is removable.

I'm doing this exercise and I must be doing something wrong. Here it goes:

Let $0$ be an isolated singularity of f. Prove that if $|f(z)|\leq |z|^{-\alpha}$, $\alpha\in(0,1)$, this singularity is removable.

The definition on my notes of removable singularity states that a singularity is removable if exists $M\geq0$ such that $|f(z)|\leq M$ for every $z\in\Omega$, being $\Omega$ the domain of definition where $f$ is holomorphic. Obviously, $|z|^{-\alpha}=M\geq 0$, so we can say that it is removable by definition, I see nothing to prove here.

Am I right or am I missing something?

• If $f$ is assumed to be holomorphic, then you should say so.
You argument is not correct because $\lim_{z \to 0} |z|^{-\alpha} = \infty$ for $\alpha > 0$, i.e. $|z|^{-\alpha}$ is not bounded by a constant $M > 0$ which is independent of $z$.
However, $$|z f(z)| \le |z|^{1-\alpha} \to 0$$ for $z \to 0$, and then Riemann's theorem implies that $z=0$ is a removable singularity.