# Classify Isolated Singularity

Problem Statement: Suppose that $f$ is analytic in the region $0 < |z| < 1$ and suppose there exists a constant $K$ such that $$|f(z)| \leq K \ |z|^{-\frac{1}{2}}$$ on $(0,1)$. What kind of isolated singularity does $f$ have at zero?

Attempted Solution: What I want to do is multiply each side by $|z|$, giving us

$$|z \ f(z)| \leq K \ |z|^{\frac{1}{2}}$$

which gives that $|g(z)| = |z \ f(z)| \leq K \ |z|^{\frac{1}{2}}$ is bounded in an open neighborhood around zero by $K$. By the Riemann Removability Condition, I should be able to state that since $g$ is bounded around $z = 0$, then it has a removable singularity. Then, by the definition of a non-essential singularity, since there exists $$g(z) = (z - 0) f(z)$$ such that $g$ has a removable singularity at $0$, then $f$ has a non-essential singularity at $0$.

Is this a valid statement? And if so, is there anymore that can be said about the singularity of $f$ at $0$? Can I say anything about this singularity being either removable or a pole?

Another thing I had considered was attempting to use Picard's Big Theorem to show that $f$ wouldn't have an essential singularity, as it would be dense in that $\mathbb{C}$, and I would be able to find values to contradict the given inequality. However, with the $|z|^{-\frac{1}{2}}$, I'm not sure if this has a singularity that is essential or not (I've seen things about it being a branch cut, etc., but that doesn't seem relevant here, as I'm not computing integrals). Anyways, any methods for solving this problem using a similar method would also be of interest to me.

You're on the right track: note that not only does $g(z)$ have a removable singularity at $z=0$, but in fact the extension of $g$ to a holomorphic function on $\{|z|<1\}$ must satisfy $g(0)=0$.
This means that we can write $g(z)=zh(z)$ with $h$ holomorphic on $\{|z|<1\}$, but since also $g(z)=zf(z)$ it follows that $f$ has a removable singularity at $z=0$.
• When you say that the extension of $g$ to a holomorphic function on $\{|z| < 1 \}$ must satisfy g(0) = 0, does this come from taking $\lim_{z \rightarrow 0} g(z) \leq \lim_{z \rightarrow 0} K |z|^{\frac{1}{2}}$? Aug 9 '16 at 23:12