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The literature seems rather coy on this point.

While $\sqrt{z}$ is not meromorphic on the complex plane $\mathbb{C}$, can it be regarded as globally meromorphic on the appropriate Riemann surface (two branched copies of $\mathbb{C}$), or (equivalently?) locally meromorphic at $z=0$? Moreover, can the root of the function at $z=0$ be regarded as a zero of order $1/2$?

And moreover, is $1/\sqrt{z}$ also meromorphic on the surface, and can it be regarded as having a pole of order $1/2$?

EDIT: Clarified(?) that I was asking whether the function globally meromorphic on $2 \mathbb{C}$.

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    (Some branch of) $\sqrt{z}$ is meromorphic on $\mathbb{C}$ minus a ray through the origin (a choice of branch cut).2012-12-16
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    The definition I'm aware of says that these functions are not meromorphic, or at best are meromorphic on $\mathbb C-\{z:\alpha z\in[0,\infty)\}$ for your choice of $\alpha\in\mathbb C-\{0\}$.2012-12-17
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    This is a great question... Elephant in the room, and all that! :) Oddly, the fact that you'd have the sense/nerve to ask it cheered me up greatly! :)2012-12-17

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