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

Solution 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.

Solution 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.