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I'm confused about the Riemann surface needed to properly describe the function $f(z)=\sqrt{z^2}$ on the complex plane.

In other words, I am not interested in the definition on the real axis, $\sqrt{x^2}=|x|$ .

It seems to me that this function is well defined on the whole complex plane alone (unlike, say, $\sqrt{z}$), because we always have

\begin{equation} \sqrt{z^2}=\exp(\frac12(\log(|z|^2e^{2i \arg(z)})))=|z|e^{i \arg(z)}. \end{equation}

Thus, when $z$ is such that $\arg(z)\mapsto \pi-\epsilon, \epsilon<<1$, we have $\sqrt{z^2}\mapsto |z|e^{i (\pi-\epsilon)}\mapsto -|z|$, and similarly, when z is such that $\arg(z) \mapsto -\pi + \epsilon$, we still have $|z|e^{i (-\pi+\epsilon)} \mapsto -|z|$.

Is what I'm saying correct, or am I screwing something up? Is it true that $\sqrt{z^2}$ has no branch cuts?

Cameron L. Williams
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sillyQsman
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  • I think the problem sneaks in when you say "the" function $f(z) =\sqrt{z^2}$. In your definition, you use one of the two branches of $\sqrt{}$ and realize that the composed function is identical to $f(z) = z$ everywhere except on (the preimage under $z \mapsto z^2$ of) a possible branch cut, but hence can be made analytic everywhere. If you had used the other branch of $\sqrt{}$, you would have gotten $f(z) = -z$. – Torsten Schoeneberg Nov 27 '24 at 17:40
  • $\sqrt z$ is well defined on the whole complex plane. It's not continuous on the whole plane, but it is well defined. – jjagmath Nov 27 '24 at 17:57
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    @jjagmath: And $\sqrt{-1}$ is $i$, or $-i$? Both can be well defined, on the entire complex plane. – Torsten Schoeneberg Nov 28 '24 at 00:00
  • There is a convention to use the symbol $\sqrt{\ }$ to refer to the principal branch of the multivalued inverse of the square function, that means for example that $\sqrt{-1} = i$. – jjagmath Nov 28 '24 at 01:39
  • In general, the definition is $\sqrt{z} = \sqrt{|z|} \left(\cos(\tfrac{\arg(z)}{2}) + i \sin(\tfrac{\arg(z)}{2})\right)$ where $\arg(z)$ is chosen to be in $(-\pi,\pi]$. – jjagmath Nov 28 '24 at 01:43

1 Answers1

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You are sort of right, there are no branch cuts needed... however, your description of the "Riemann surface" is not complete.

On a problem like this, it greatly helps to get rid of that pesky square root sign, convert the problem into a polynomial equation, and then examine the solution set of that equation. That solution set is, formally, the "Riemann surface" that you are looking for. But there are issues...... which I'll get to in a moment.

The way you do this symbolically is take the "equation" $w=\sqrt{z^2}$ and square both sides, to get the equation $$w^2 = z^2 $$ This has the highly desirable side effect of avoiding all of those annoying discussions about principal branches of the square root.

But it also has a formal advantage, namely a formal definition of the "Riemann surface", which is just the solution set of the equation $w^2=z^2$. That "Riemann surface" is the union of two $1$-dimensional subspaces of $\mathbb C^2$, namely the subspace $w=z$ and the subspace $w=-z$. The intersection of these two subspaces is the origin $(0,0) \in \mathbb C^2$.

There's a few issues to be discussed here. This solution set is not exactly a "Riemann surface" --- hence the square quotes that I have been using. The point is that every neighborhood of $(0,0)$ is disconnected, hence no neighborhood of $(0,0)$ is modelled on an open disc in $\mathbb C$ as is ordinarily required for a Riemann surface. The point $(0,0)$ is therefore a kind of "singularity" of the Riemann surface structure. This is a pretty simple kind of singularity and it has its own terminology: it is called a node, and the solution set itself is called a noded Riemann surface.

Also, no branch cuts are needed at all to understand this example. Part of the reason for this is that the complement of $(0,0)$ in $w=z$ and the complement of $(0,0)$ in $w=-z$ are disjoint from each other; they form the two connected components of the complement of $(0,0)$ in the whole solution set, each equivalent to the Riemann surface $\mathbb C - \{(0,0)\}$ as one can see by projecting to the $z$-plane.

Lee Mosher
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  • Thank you! That is a very different way of thinking about it for me. I viewed Riemann surfaces here as ways to 'make an otherwise multivalued function single valued'. ie when trying to understand a function with square roots etc, I would look for the Riemann surface underlying the equation to understand the domain of that function. But if I understood your response, the answer to those two questions might be distinct? Namely that if I want a natural, single valued domain where $f(z)=\sqrt{z^2}$, then the complex plane is enough, whereas the (noded) Riemann surface gives the full solution space – sillyQsman Nov 27 '24 at 18:12
  • Well, part of the issue with the expression $f(z)=\sqrt{z^2}$ is this: What's the difference between $f(1)$ and $f(-1)$? You can declare that when $z$ is negative then you'll use the negative square root and when $z$ is positive you'll use the positive square root. But then I'll ask you: What is the difference between $f(i)$ and $f(-i)$? And you may have to squirm around again before hacking some answer together... That's why I am proposing that the correct formal statement of the problem does not even start until you get rid of that square root sign. – Lee Mosher Nov 27 '24 at 18:38
  • Another reason for thinking about Riemann surfaces as solution sets of (well-defined) equations of the form $F(z,w)=0$ --- often polynomial equations in the variables $z,w$ --- is that it unveils a spectacular new theory. – Lee Mosher Nov 27 '24 at 18:41