# Understanding the Riemann Surface of the $\sqrt{z}$ [duplicate]

I have been trying all the day to understand the Riemann Surface of the $\sqrt z$. I Can not understand . Can anyone help me to understand how this picture is Riemann Surface of $\sqrt z$. The other questioner did not ask this question. He already could visualize my question's answer. So I request you not to try to block this question. This is the only source from where I get help. Please read that question.Intuitively understanding Riemann surfaces

Explanation in simple words will be highly appreciated.

## marked as duplicate by Chris Culter, Misha Lavrov, The Phenotype, Namaste, Ethan BolkerFeb 24 '18 at 1:06

• The two sheets interpenetrate. You do realize that the surface can't be embedded in $\mathbb R^3,$ right? – saulspatz Feb 23 '18 at 18:59
• The other questioner did not ask this question. He already could visualize my question's answer. So I request you not to try to block this question. This is the only source from where I get help. Please read that question.@ChrisCulter – cmi Feb 23 '18 at 19:04
• I could not understand anything about the Riemann surface of $\sqrt z$ @saulspatz – cmi Feb 23 '18 at 19:05
• The other questioner did not ask this question. He already could visualize my question's answer. So I request you not to try to block this question. This is the only source from where I get help. Please read that question. You may have read the first line of that question. U@ChrisCulter – cmi Feb 23 '18 at 19:16

I'll try to give you an intuitive idea; I don't know if I can. First of all, the problem is that if we define $\sqrt{Re^{i\theta}} = \sqrt{R}e^{i\theta/2},$ we have a problem as we wind around the origin. We don't get back to where we started. As we pass $\theta =\pi$ we switch from the "branch" where the square root has positive real part to the branch where it has negative real part.

The Riemann surface is a surface on which we can define the square root so that it is holomorphic. Both branches will be represented.

Now, you've see the picture where each branch is defined on a split plane, and the slits are glued together. That is what is shown in the picture. First, realize that it is not the graph of $f(z)=\sqrt{z}.$ All that is shown is the domain of the function. The surface itself cannot be embedded in three dimensional space, so you have to use some imagination.

Now, suppose we start out on the right, where you see the yellow. As we wind around the origin on the top sheet, going through the solid blue we get to a place where the sheet suddenly turns red. The line between the red and the blue represents the place where the sheets have been glued together. We now pass down to the lower sheet, also blue, and continue circling around the origin counterclockwise. Eventually, we get to the red part of the lower sheet. (Note that most of the circle in the lower sheet is hidden in the picture by the upper sheet.) Again we're at the place where the sheets are glued together, and we pass from the red on the lower sheet to the red on the upper sheet, and then to the yellow, where we started.

In more advanced courses, Riemann surfaces are defined rigorously, but starting out, they're just supposed to be an informal aid to understanding. If they don't help you, you can ignore them temporarily, and just go back to them from time to time until something clicks.

Ok so first of all the Riemann surface of $\sqrt{z}$ is $\mathbb{C}$.

(Although, more accurately, $\mathbb{P}^1$, as we look at compact surfaces.)

So you real problem to understand the covering map from $\mathbb{C}\to\mathbb{C}$. This map is given by $w\mapsto w^2$. So this maps the origin to the origin, and every other point has two preimages.So look at the unit circle and draw out for yourself how the $w^2$ map looks on the circle, basically looping around twice. The map on the Riemann surface is the same with rays drawn from the origin.