Find a conformal map from semi-disc onto unit disc This comes straight from Conway's Complex Analysis, VII.4, exercise 4.

Find an analytic function $f$ which maps $G:=$ {${z: |z| < 1, Re(z) > 0}$} onto $B(0; 1)$ in a one-one fashion.

$B(0;1)$ is the open unit disc.
My first intuition was to use $z^2$, which does the job splendidly, except for the segment $(-1,0] \subset B(0;1)$.  Under $z^2$, the pre-image for this segment is the segment $[-i,i]$, which is not in $G$.
My next thought is to modify $z^2$, something like $a(z-h)^2+k$.  I've yet to work out the details, but my gut tells me this isn't the right idea.
I've been teaching myself conformal maps in preparation for a qualifying exam.  So, if there's a shockingly basic, obvious solution... please patronize me.

As a supplement to the fine answers provided, here are the pictures of the conformal maps themselves:

We start with the right half of the unit disk, map it to the upper right quadrant, square this to the upper half-plane, and finally rotate onto the unit disk.

Per requests for code - the basic pattern to generate the image of a polar region $a\leq r \leq b$ and $\alpha \leq \theta \leq \beta$ under a function $f$ in Mathematica is:
ParametricPlot[{Re[f[r*Exp[I*theta]]], Im[f[r*Exp[I*theta]]]},
  {r, a, b}, {theta, alpha, beta}, Mesh -> True]

The pictures above may be generated with the following code block:
phi1[z_] = -(z - I)/(z + I);
f[z_] = z^2;
phi2[z_] = (z - I)/(z + I);
nestedFunctions = 
  Composition @@@ Table[Take[{phi2, f, phi1}, -k], {k, 0, 3}];
pics = Table[
  ParametricPlot[{Re[g[r*Exp[I*theta]]], Im[g[r*Exp[I*theta]]]},
      {r, 0, 1}, {theta, -Pi/2, Pi/2}, 
      Mesh -> True, PlotRange -> 2, ImageSize -> 360],
    {g, nestedFunctions}];
Grid[Partition[pics, 2]]

You can find a lot more information on visualizing complex function with Mathematica in this notebook.
