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

Question

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 $2 \mathbb{C}$.

Answer 1

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.

Answer 2

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.

Answer 3

(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. :)

Eventually, one may discover that a "global" definition of a "function", e.g. defined by ODEs or by algebraic equations, that has "problems" about being pieced together globally, as in "covering space theory", "admit" a covering of the usual complex plane on which the "multi-valued-ness" pseudo-problems go away...

In summary: the traditional descriptions of the situations are pretty wacky, in my opinion!!!!! But, in fact, especially from our current viewpoint, it's not so crazy.