# Is $\sqrt{z}$ a meromorphic function?

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). –  Qiaochu Yuan Dec 16 '12 at 23:58
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 $\alpha\in\mathbb C-\{0\}$. –  Mario Carneiro Dec 17 '12 at 0:00
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! :) –  paul garrett Dec 17 '12 at 1:52

It is worth emphasizing that the description of a function includes its domain. Changing the domain from entire plane to slit plane or to Riemann surface entails changing the function. There is nothing strange in the fact that some of the resulting functions are holomorphic while others are not.

In particular, on the appropriate Riemann surface $\Sigma$ the function $\sqrt{z}$ is holomorphic: indeed, it is a biholomorphism between $\Sigma$ and $\mathbb C$ which gives $\Sigma$ its complex manifold structure. This function has a zero of order $1$ at the point over $z=0$. Accordingly, $1/\sqrt{z}$ is meromorphic on $\Sigma$, with pole of order $1$ (not $1/2$) at the origin.

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Why are the root and pole of order 1 though? While $\lim_{x \to 0} z \left(1/\sqrt{z}\right)$ exists it is equal to zero. Do we say that the function $1/\sqrt{z}$ has a first order pole, but a residue of $0$? This is the part that really confuses me as the literature normall defines poles to have integer order. –  ObsessiveMathsFreak Dec 17 '12 at 4:32
@ObsessiveMathsFreak Sorry, I added to the confusion by using the same letter for the coordinate on $\Sigma$ as on $\mathbb C$. They are different things, which is why Matt E used $w$ for the coordinate on $\Sigma$ in his answer. As he said, on the surface $1/\sqrt{z}$ becomes $1/w$: a single pole at $0$. –  user53153 Dec 17 '12 at 4:56
Ah, I see now. For $f(z)=z^n$, the argument principle gives $2\pi i N=\oint f'(z)/f(z) dz = \oint n/z dz$, but for $n=1/2$ in $\sqrt{z}$ you have to loop around twice to make the contour closed, so the order really is $1$. I'm still not sure about the $\omega$ argument though. –  ObsessiveMathsFreak Dec 17 '12 at 5:02

Since the question and another answer mention the "Riemann surface" on which $\sqrt{z}$ becomes meromorphic, it might be worth making this more explicit.

If we let $\Sigma$ denote the Riemann surface over which $\sqrt{z}$ becomes single-valued, then $\Sigma$ is just a copy of the Riemann sphere. If we let $w$ denote the coordinate on $\Sigma$, then the map from $\Sigma$ to the usual Riemann sphere (the one with coordinate $z$) is given by $z = w^2$. So on $\Sigma$ the function $\sqrt{z}$ just becomes the coordinate function $w$ (and so $1/\sqrt{z}$ becomes $1/w$).

So there is nothing very mysterious happening here. Without invoking the somewhat mystical-sounding language of Riemann surfaces (not that this language isn't valuable, it's just that sometimes it can be more obfuscating than clarifying), one can describe the situation as follows:

The function $\sqrt{z}$ is not a meromorphic function of $z$: it is branched at $0$, and also at $\infty$. But if we make the substituation $z = w^2$, the resulting function $w ( = \sqrt{w^2})$ is meromorphic as a function of $w$. That's all.

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I don't quite follow. I understand that the function is not meromorphic on $\Sigma$, but are you saying that because $z=\omega^2$ and $\omega$ is globally meromorphic, then that must mean that $z$ is globally meromorphic on some surface? –  ObsessiveMathsFreak Dec 17 '12 at 4:53
One can extend the sqrt function from the double covering of $\mathbb C^2$ (which, as you observe, is "just" $\mathbb C\setminus0$) to the complete sphere, giving an holomorphic function, no? (In other words, the singularities of your meromorphic function are avoidable) –  Mariano Suárez-Alvarez Dec 17 '12 at 6:55
@MarianoSuárez-Alvarez: Dear Mariano, Yes, $w$ has no poles as a function of $w$ (other than at $\infty$). I was not really trying to distinguish between holomorphic and meromorphic in this answer, but rather between holomorphic/meromorphic and "non-trivially branched". Cheers, –  Matt E Dec 17 '12 at 12:54
(Upvoting and) seconding @Matt E's answer, in parsing texts and other literature, local holomorphy and global holomorphy are often insufficiently distinguished, thus understandably leading to confusions.(Years ago, it took me a while to understand (a) that my confusion was reasonable (b) how to resolve it.) Locally, except at $0$, there are two holomorphic square roots. Globally, which itself asks "on what open set? ... perhaps on what Riemann surface (complex manifold!?!)?"... the first (and archetypical) point is that, indeed, there is no square root of "z" on the complex plane/line. Ok. But, again archetypically/cliched-ly, on any simply-connected open subset of the complex plane/line not containing $0$, there is a (global) square root.
When one launches oneself into "Riemann surfaces", there is already some cognitive dissonance, reasonably-enough. The first point is that a given algebraic relation/function "$f(z,w)=0$" defines a finite-degree covering of "the Riemann sphere". The critical point for the question is that this can achieve the effect that a "function" only locally definable/holomorphic on $\mathbb C$ can become globally definable. Indeed, the cognitive troubles are amplified by the idea/fact that a given not-globally-definable function "defines" a Riemann surface... (This cracks me up... or not, given the many hours I labored to parse this cryptic mythology. :)
You are correct. I should have clarified that I am asking whether the function is globally meromorphic(Though in fact, I am actually interested myself in whether it is locally meromorphic at $z=0$). Regarding Riemann functions, I am working under the assumption that for any algebraic function, there will be a corresponding Riemann surface which will make the function single valued and (hopefully) meromorphic. –  ObsessiveMathsFreak Dec 17 '12 at 4:49