uniformization theorem - squares and circles I am trying to understand the uniformization theorem and get some intuition about it.
The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to one of the three domains: the open unit disk, the complex plane, or the Riemann sphere. 
Specifically, the Riemann mapping theorem states that every simply connected open subset of the complex plane that is different from the complex plane itself admits a conformal and bijective map to the open unit disk.
So, an open square is conformally equivalent to the open circle. But how do I find the bijection? 
What about a closed square - is it conformally equivalent to the open circle, and how?
What about an L-shape, made of two orthogonal rectangles? 
 A: Compilation of comments, expanded.
(1) In practical terms, it is slightly easier to work with upper half-plane instead of the open unit disk. The composition with $(z-i)/(z+i)$ then gives a map onto the disk. The Schwarz–Christoffel method gives a practical way to find a conformal map of upper half-plane to a polygon. 
(2) Freely downloadable program zipper by Donald Marshall computes and plots conformal maps using a sophisticated numerical algorithm. It can handle an L-shape, or far more complicated shapes: 

Zipper-generated images are very nice, though not as flashy as this one, linked to by brainjam. 
(3) Closed square is not allowed in the uniformization theorem. Conformal (or general holomorphic) maps are normally defined on an open set. While one may talk about boundary correspondence under conformal maps, it's understood in the sense of limits at the boundary. A conformal map of open square onto a disk has a continuous extension to the closed square, by Carathéodory's theorem, but I would not call the extended map conformal: the angles at the corners are not preserved.
