# Riemann Surface of $z=\sqrt{w}$

Today I started learning about Riemann surfaces. In Gamelin's Complex Analysis, Gamelin states that the Riemann surface of $$z=\sqrt{w}$$ is "essentially a sphere with two punctures corresponding to $$0$$ and $$\infty$$." How is this true? The surface does not look like a sphere to me.

• "essentially" means that $R$ is biholomorphic with the punctured plane which is obvious since by definition $\sqrt z$ is a global holomorphic function on $R$ with values in $\mathbb C^*$ and it is conformal and bijective; the fact that it doesn't look like that means nothing as for example a slit disc doesn't look like a disc but it is "essentially" a disc by RMT (and we can take infinitely many slits to make the picture look quite ugly, not to speak of weird Jordan curves of planar positive measure like Osgood curves whose insides are still "essentially" the unit disc in the parlance above) Apr 26, 2020 at 3:43
• See Alekseev, Abel's Theorem in Problems and Solutions. Nov 24, 2020 at 19:19