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To obtain Riemann surface for $w = f(z) = \sqrt{z}$ we get two copies of $z$-planes with cuts. After they are joined $f(z)$ gives us a one-to-one correspondence between this Riemann surface and $w$-complex plane.

Now, Riemann surface for $w = f(z) = \sqrt{z^2}$ is obtained by joining two copies of $z$-planes at point $z = 0$. What I don't understand though is how a one-to-one correspondence between this Riemann surface and $w$-plane is built. I have an intuition that in variable $w$ one, too, has to take two copies of $w$-plane and join them at point $w = 0$. After that one gets a one-to-one correspondence between Riemann surface in $z$ and Riemann surface in $w$.

Is my intuition right? Thanks.

user75619
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  • If you "join two planes at a point", you don't get a surface. The Riemann surface of $\sqrt{z^2}$ is either $\mathbb{C}$ (if your surfaces are connected) or a disjoint union of two copies of $\mathbb{C}$ (if you insist on having a point corresponding to each possible value above the base). – Daniel Fischer Aug 07 '15 at 15:49
  • ok, let's say $z = 0$ is cut from the plane, no need to join the $z$-planes then. I'm more interested in what's going on in variable $w$. For a one-to-one correspondence, will I need to have a disjoint union of two copies of $w$-plane? I'm asking because textbook examples always seem to deal with Riemann surface-to-plane mapping, but in case of $\sqrt{z^2}$ I don't see how one could obtain one-to-one mapping this way. – user75619 Aug 07 '15 at 15:56

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