TITLE: Can we define $z^{\frac{1}{2}}$ as a holomorphic function on $\mathbb{C}\left\backslash \left\{ 0\right\} \right.$?
QUESTION [2 upvotes]: Consider $$z^{\frac{1}{2}}:=e^{\frac{1}{2}(\log|z|+iarg(z))}.$$
We can see that, for example, $z^{\frac{1}{2}}$ can be defined as a holomorphic function near $z=\frac{1}{2}$, by chossing a very small neighbourhood of $z=\frac{1}{2}$, and define an appropriate $arg(z)$ to make it continuous there.
My question: Can $z^{\frac{1}{2}}$ be considered as a holomorphic function on $D\left\backslash \left\{ 0\right\} \right.$? Here $D$ is the unit disk in $\mathbb{C}$.
By holomorphic function I mean that a map $f:D\left\backslash \left\{ 0\right\} \right.\rightarrow \mathbb{C}$ satisfies the Cauchy-Riemann equation on $D\left\backslash \left\{ 0\right\} \right.$.
As answered below, we see that the answer for my question is negative. I would like to consider the following extra related question:
An extra question: similar question but this time we consider the domain $D\left\backslash B(0,\epsilon) \right.$, for a very small $\epsilon$.

REPLY [3 votes]: No, this is not possible. This function would be bounded in a punctured neighborhood of $0$, which would make $0$ a removable singularity of $z^{\frac{1}{2}}$. But then $0$ would also be a removable singularity of the derivative $\frac{1}{2z^{\frac{1}{2}}}$, which can't have a removable singularity at $0$ because it isn't bounded in a punctured neighborhood.
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