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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 $\epsilon$.

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    $\begingroup$ No. I gave an explanation why in this answer: math.stackexchange.com/a/3783606/774222 $\endgroup$ – Rivers McForge Aug 15 '20 at 23:04
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    $\begingroup$ The function is not even continuous on all of $\mathbb C\setminus \{0\}$ $\endgroup$ – Matematleta Aug 15 '20 at 23:06
  • $\begingroup$ I 've editted my question with an extra part. I think it would be quite different now. $\endgroup$ – Hana Aug 15 '20 at 23:21
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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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  • $\begingroup$ Thks, would you have a look at my extra question. $\endgroup$ – Hana Aug 15 '20 at 23:18
  • $\begingroup$ @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. $\endgroup$ – Vercassivelaunos Aug 16 '20 at 4:17
  • $\begingroup$ Thks, I got it now. $\endgroup$ – Hana Aug 16 '20 at 12:05

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