How to actually plot Riemann Surfaces? I took a course in complex variables and I remember reading something about Riemann surfaces on Wikipedia. There are even some examples in this page:

I am curious about the following: How do we actually plot them? The definition in there doesn't look like anything I could translate into points $(x,y,z)$. I've been trying to guess for a while how to plot the Riemann surface for $f(z)=z^{\frac{1}{2}}$ but with no success. I think we need to compute the branches for $f(z)=z^{\frac{1}{2}}$ and we can do that with the following formula:
$$f_k(z)=|z|^{\frac{1}{n} } \left( \cos\left(\frac{\arg z + 2 k \pi}{n}\right) + i\sin\left(\frac{\arg z + 2 k \pi}{n}\right) \right) \quad k=0,1,2,..,n-1 $$
So in our case, we have $f_0$ and $f_1$ with $n=2$. Now I tried to plot the following sets of points:
$$(a,b,\arg f_0(a+bi))\qquad (a,b,\arg f_1(a+bi))$$
And I got this:
$\quad \quad \quad \quad \quad \quad \quad $
Which kinda looks like it can be cut and glued like the figure I gave previously. Is it correct?

EDIT:
I had some progress on it, I read on Agarwal's Introduction to Complex Analysis:




I tried to do as he describes and am plotting the function $z=w^2$, with the restrictions he gave in the text. By plotting the pairs $(\Re(z),\Im(z),\arg(z))$ I obtained this:

I don't know if this is the correct result but it still doesn't look like the one I found on Wikipedia:

 A: Modulo details about domains and images, if we view a function $f$ as a "maps-to" relation $y = f(x)$, the same set of points defines the inverse function $x = f^{-1}(y)$. More correctly, this works if and only if $f:X \to Y$ is a bijection. In that situation, each $x$ in $X$ corresponds to precisely one $y$ in $Y$, there exists an inverse mapping $f^{-1}:Y \to X$, and for all $x$ in $X$ and all $y$ in $Y$, $y = f(x)$ if and only if $x = f^{-1}(y)$.
This all holds for complex-valued functions. For example, the set of points $(z, w)$ in $\mathbf{C}^{2}$ satisfying $w = z^{2}$ represents the squaring function $f(z) = z^{2}$. Formally, the same set of points represents the complex square root. To make this correct we must restrict to an open set $Z$ in the $z$-plane where $f$ is injective, and take $W = f(Z)$ to be the image. But if we do not restrict, the "complex parabola" with equation $w = z^{2}$ may be viewed as the "Riemann surface of the square root" by interpreting $w$ as the independent variable.
If we write $z = re^{i\theta}$ in polar form, the graph of the squaring map may be viewed as the set of points
$$
(z, w) = (z, z^{2}) = (re^{i\theta}, r^{2}e^{2i\theta})
\leftrightarrow (r\cos\theta, r\sin\theta, r^{2}\cos(2\theta), r^{2}\sin(2\theta)).
$$
Up to permutation of coordinates in real four-space (swapping $z$ and $w$), this parametrizes the Riemann surface of the square root. (The calculations to do this in Cartesian, writing $z = x + iy$ and $w = u + iv$, are left as a pleasant exercise. The resulting plot, the Wikipedia diagram at the top of the question, is nice in its own way!)
This viewpoint sheds light on branch cuts and how sheets connect across branch cuts. A branch of square root or a sheet of the Riemann surface is a non-empty open subset $U$ of the graph $\{(z, w) : w = z^{2}\}$ that "passes the vertical line test in $w$" in the sense that the projection $(z, w) \to w$ is injective when restricted to $U$. A standard choice is to fix a ray from $0$ in the $w$-plane, remove the real parabola sitting over this ray when we project to the $w$-plane, and to take $U$ to be one of the two connected components.
We can play the same game with the complex logarithm: Here, the graph $w = \exp z$ may be parametrized by
$$
(z, w) = (x, y, e^{x}\cos y, e^{x}\sin y).
$$
The Riemann surface of $\log$ as pictured at the top of the post is obtained by discarding the real part of $z$. If we restrict $y$ to an interval $(\theta_{0}, \theta_{0} + 2\pi)$ (i.e., we restrict $z$ to a horizontal band of height $2\pi$) we obtain a branch of logarithm. Alternatively, we can visualize the complex logarithm by the parametrization
$$
(x, y) \mapsto (e^{x}\cos y, e^{x}\sin y, x + y).
$$
This depicts both the "parking garage" behavior of branches and the "log-ish" singularity near $0$.

