# constructing Riemann surface of $\sqrt{z-1/z}$

I am trying to construct the Riemann surface of the function $\sqrt{z-1/z}$. I rewrite the function as $\sqrt{\frac{z^2 - 1}{z}}$, from which I can see that the function has branch points at 0,1,-1 and $\infty$. But I can't figure out the number of Riemann sheets, how to find branch cuts and finally, how to form the Riemann surface by gluing the sheets. I would appreciate if someone could give me a somewhat detailed explanation. Thanks in advance.

• Since the square root has two values, there will be two sheets. There are several options for the branch cuts, you just have to make sure that there is a single-valued branch of the square root on the plane minus the cuts. In this case, any pair of cuts, each connecting two singularities, will work. E.g., cutting along $[-1,0]$ and $[1,\infty]$ is one possibility. – Lukas Geyer Jun 23 '15 at 20:28

Write the function as $P(z,w)=1-z^2+z w^2=0$. The branch table is:
$$\left( \begin{array}{cccc} 1 & 0 & \left( \begin{array}{cc} 1 & 2 \\ \end{array} \right) & \\ 2 & -1.0000000 & \left( \begin{array}{cc} 1 & 2 \\ \end{array} \right) & \\ 3 & 1.0000000 & \left( \begin{array}{cc} 1 & 2 \\ \end{array} \right) & \\ 4 & \infty & \left( \begin{array}{cc} 1 & 2 \\ \end{array} \right) & \\ \end{array} \right)$$
By Riemann-Hurwitz, we have $g=1-2+1/2(1+1+1+1)=1$, so that the Riemann surface is a torus. Since it's a double-cover with four branch points all fully-ramified, how about taking two Riemann-spheres and making on each sphere, one cut from 0 to 1 and another from -1 to infinity. Then open the cuts and glue the cuts together into a torus. Won't that work?