I am looking for universal covering spaces and I am now wondering if the Riemann surface for the square root $z^{1/2}$ (or even more general for $z^{1/n}$) is simply-connected and therefore a universal covering space for the punctured complex plane?
If so, is (the topology of) the Riemann surface for $log(z)$ homeomorphic to the Riemann surface for $z^{1/n}$? If both Riemann surfaces are simply-connected covering spaces for the punctured complex plane, then they clearly must be homeomorphic, but it doesn't seem obvious just from looking at them.
I am not familiar with complex analysis (I am specializing in geometry), but need an explanation for a result in geometry...so any help is greatly appreciated!