# Riemann Surface of $w^{2}=\sqrt{1-z^{2}}$

I'm working in the problem of finding branch points and build the Riemann Surface of the following complex function:

$$w(z)=\sqrt{1-z^{2}} \,\, .$$

I'm reading lots of texts about how to do this, but I'm not able to do it. How to indentify the branch points and how to build the Riemann surfaces step-by-step of this function?

Greetings!

Since the square root has two possible values, your Riemann surface will have two sheets. Branch points of the square root are at $0$ and at $\infty$, so in this case the finite branch points are at the points where $1-z^2 = 0$, i.e., at $z=\pm 1$. At both of these points you have a simple zero, so analytic continuation around them will get you to the other sheet. Now you can make a branch cut from $-1$ to $+1$ and glue the two sheets together along this cut. Visualizing this on the Riemann sphere and opening up the cuts, you will get two spheres with disks removed, glued along their respective boundaries, which will again get you a sphere.
• Thank you for answering @LukasGeyer. But there still is a doubt: How could I make an analytic continuatinuation around them? To simple square root, I have to put it in the polar form and after set the $arg(z)+2\pi$, but I can't do this in this case. – Herr Schrödinger Nov 30 '15 at 13:27
• @WaynerKlën: If a function has a simple zero at $z_0$, it can be written as $(z-z_0)g(z)$ with $g(z_0) \ne 0$, and then locally $g(z)$ has an analytic square root, whereas $z-z_0$ has the standard two-sheeted Riemann surface of its square root around $z_0$. – Lukas Geyer Nov 30 '15 at 21:25
• How could you affirm that there's a branch point at infinity @LukasGeyer ? Doing the derivative analysis it doesn't look like a problem to me, unlike at $0$, where is a problematical point... – Herr Schrödinger Dec 1 '15 at 17:37
• @WaynerKlën: You could do that, too, it is essentially the same. There are two singularities $\pm 1$ on the Riemann sphere, and any cut connecting these two will work. Yours just goes through $\infty$. – Lukas Geyer Dec 5 '15 at 7:46