# Can we define $z^{\frac{1}{2}}$ as a holomorphic function on $\mathbb{C}\left\backslash \left\{ 0\right\} \right.$?

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 $$\epsilon$$.

• No. I gave an explanation why in this answer: math.stackexchange.com/a/3783606/774222 – Rivers McForge Aug 15 '20 at 23:04
• The function is not even continuous on all of $\mathbb C\setminus \{0\}$ – Matematleta Aug 15 '20 at 23:06
• I 've editted my question with an extra part. I think it would be quite different now. – Hana Aug 15 '20 at 23:21

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.
• @Hana: then it would be defined on an annulus centered at zero, so it would have a Laurent series expansion around zero. The square of this series would be just $z$, a power series, so the original Laurent series should also be a power series (can probably prove it using Cauchy product formula). But then it could be analytically extended to the whole disc, which should probably lead to a similar contradiction as above. – Vercassivelaunos Aug 16 '20 at 4:17