clarification on Riemann surfaces and branch cuts

Question

This semester I am going to take Complex Analysis. I started watching the lectures, and the professor said that branch cuts and Riemann surfaces solve the problem of multivaluedness of complex functions. he didn't say why, nor explain what is a branch and why each sheet in the Riemann surface corresponds to a branch in the function. I tried thinking about why the branch cut and Riemann surface solve the problem of multivaluedness, but I feel uncomfortable about my explanations. I will add them to this question, and ask you not to tag this question as a duplicate, since I didn't understand most of the answers given to related questions and another clarification would definitely help.

My explanations: In both of the next two points, I will talk about a concrete function, $w(z)=z^{1/2}$.

1. Branch cut: If we look at the function $w(z)=z^2$, it maps the complex plane two times, and so when we look at the inverse function,$w(z)=z^{1/2}$, we will have a problem of multivaluedness. in the inverse function, we get two branches (possible "solutions"/"representations", from what I can guess that branches mean) - $w1=r^{\frac{1}{2}}e^{\frac{i\theta}{2}}$ and $w2=r^{\frac{1}{2}}e^{\frac{i\theta+2\pi}{2}}$ ($0<\theta<2\pi$). But now it is a multivalued function - there are two possible outputs for each point in the complex plane. so we might say that we choose the first branch as the value. but there is now another problem - if we, for example. complete a full revolution around the origin, the angle $\theta$ moves from $0$ to $2\pi$, and $\theta/2$ moves from $0$ to $\pi$, so we get a discontinuity in the function w1 - even though we returned to the same point, at $\theta=0$, $w1(0)=r^{\frac{1}{2}}$, and at $\theta=2\pi$, $w1(2\pi)=r^{\frac{2\pi}{2}}=-r^{\frac{1}{2}}$ - a "jump discontinuity". So we define a branch cut (again, the professor didn't define it, so what I understood might be wrong) as a positive x-axis, and its definition is a line at which a discontinuity occurs in the function. If we go around that line, the function will be continuous. but how does the branch cut solve the problem of multivaluedness? In my understanding, it is the choice of one of the branches that solved this problem, not the branch cut.

2. I understood how the construction of the Riemann surface is made, but as in the first point, I don't get how the Riemann surface solves the problem of multivaluedness. To me it allows the function to vary continuously and return to its starting value when we complete a revolution around the origin starting at (1,0), but I can't see how it helps with the multivaludness problem. Also, why each sheet in the Riemann surface corresponds to a branch in the function? is it because each copy of the complex plane is in a different angle interval (the first between $0$ to $2\pi$, and the second is between $2\pi$ to $4\pi$), which then if look at $w(0)=r^{\frac{1}{2}}$, each brunch is found on a different sheet because on the different angle interval that matches $w1$ and $w2$ mentioned above?

Thank you for reading and for any insights!

Answer 1

$\DeclareMathOperator{\im}{Im}$In the hope a picture is worth 750 words (with ~250 actual words below):

At top is the $w$-plane. The open half-planes shown, $0 < \im w$ (green) and $\im w < 0$ (blue), each map under squaring onto the slit plane with the non-positive real axis removed.

At bottom is the surface parametrized by $z = w^{2}$ in the first two coordinates, and by $\operatorname{Re} w$ in the third coordinate. Projecting vertically (not shown) represents the squaring map. The colors connote sheets of the Riemann surface of the square root assuming we make our branch cut on the non-positive real axis. The entire surface is the "total square root," a "two-valued" rather than single-valued function of $z$.

Loosely, $z = w^{2}$ and $w = \pm\sqrt{z}$ are equivalent. More literally, removing the non-positive reals separates the surface into two "cross-glued" sheets: The graphs of (the real parts of) $w_{1}$ (top sheet) and $w_{2}$ (bottom sheet).

The jump discontinuity at the cut is not visible in the bottom picture because the imaginary part is projected away (to get the surface into real three-space) and the jump is pure imaginary. In the top diagram, however, the jump is manifest: When $z$ circles the origin, $w = \sqrt{z}$ makes only a half-circle in one half-plane, and jumps by a sign at the cut.

(Bonus: When $z$ circles the origin, a choice of "lift" to the $w$-plane illustrates analytic continuation.)

In either picture, we see that when $z$ circles the origin twice, the square root "returns to its initial value." Either picture may be viewed as the Riemann surface of square root, a domain on which "the square root is single-valued." In other words, the $w$-plane itself is the Riemann surface of $\sqrt{z} = \sqrt{w^{2}}$.

Answer 2

Forgetting the complications of the plane for a moment if we consider the map $f:x \rightarrow x^2$ on the real numbers we have the image being the non-negative reals. We can think of this geometrically as gluing the line by folding it in half. We can find the points that are glued together by considering the preimage of each singleton. Note that in general the preimages of each point in the codomain form a partition of domain here having exactly two points in each partition.

Now in the plane the same thing must happen to the real line on the map $f:z \rightarrow z^2$. Here we see the imaginary axis get mapped to the negative real axis where it also meets. If we consider the open half plane with real part greater than zero its image will be at some point on the plane that isn't the negative real axis. Similarly when the real part is less than zero it will be some point that isn't on the negative real axis (why?). So we have these two sheets separated by the imaginary axis. Something similar will happen with $f:z \rightarrow z^3$ however now we'll have three sheets and we'll have three cuts along three rays in the domain. See if you can figure out where those sheets are and what their branch cuts must be.