Embedding the Riemann surface for $\sqrt z$ in $\mathbb{R}^3$

Question

I've only ever seen the Riemann Surface for $\sqrt z$ described as below, with the $z$-axis typically being the $x$ or $y$ coordinate of $\sqrt z$.

This embedding into $\mathbb{R}^3$ has intersection points however. For example, if our $z$-axis represents the $x$-coordinate, then both $i$ and $-i$ have the same $x$-coordinate ($0$), so they will be projected onto the same point in $\mathbb{R}^3$.

It seems however, that if we use the $z$-axis to represent the angle, then we can't have intersections (both roots of any number have a different angle). Am I missing something? In my mind, it seems like this picture would look like a small portion of the Riemann Surface for $\log z$, growing from $0$ to $2\pi$. However, this creates a boundary... and I'm really confused at this point. I hope some of you could follow this train of thought.

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$It sounds as if you're asking, "Can the Riemann surface of the square root be embedded into $\Reals^{3}$?", but the picture you seek is constrained by criteria you aren't quite sure how to articulate?

There is no two-valued function $f:\Cpx \to \Reals$ whose values at each non-zero point have the form $\{u, -u\}$ and whose graph in $(\Cpx \setminus\{0\}) \times \Reals \subset \Reals^{3}$ is a connected, embedded surface.

The "helicoidal" surface you envision has a boundary (as you say), so it's not an embedding of the Riemann surface of the square root.

There do (trivially!) exist embeddings of the Riemann surface of the square root into $\Reals^{3}$, but they're unlikely to satisfy because they aren't graph-like in the sense above: Instead, they're essentially graphs of the squaring function, or embeddings of the abstract Riemann surface, namely $\Cpx$, as a surface (e.g., a plane) in $\Reals^{3}$.