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.

The following trick works for any region bounded by two circular arcs (or a circular arc and a line). 
Find the points of intersection of the arc and the line. (Here, they're $i$ and $-i$.) Now pick a Mobius transformation that takes one of those points to $0$ and the other to $\infty$; here $z \mapsto \frac{z-i}{z+i}$ works. Then the arc and the line go to two rays (because a Mobius transformation sends circles in $S^2$ to circles in $S^2$, and the only circles in $S^2$ that goes through both $0$ and $\infty$ is a line in $\Bbb C$), both starting at $0$ and going off the $\infty$. Your domain maps to the region bounded by these two rays.
Let's compute the rays. It suffices to find where a single point on each arc maps; if $z_0$ is on the arc, the ray will be $\{f(z_0)t : 0 \leq t < \infty\}$. I say we pick $0$ to be our point of choice on $Re(z) = 0$ and $1$ to be the point of choice for the circular arc. These are mapped to $-1$ and $-I$ respectively; so our two arcs are the negative real axis and the negative imaginary axis. I'd like the "lower" arc to be the positive real axis, so let's multiply by $-1$ to do this.
So we have a conformal map from your half-disc to the upper-right quadrant given by $z \mapsto -\frac{z-i}{z+i}$. The upper half-plane is nicer, so let's map to that by squaring; now we have a map to the upper half plane given by $z \mapsto \frac{(z-i)^2}{(z+i)^2}$. (For other regions bounded by rays that make different angles, you get to the upper half plane by a $z \mapsto z^\beta$ for the appropriate $\beta$.)
Now there's a standard map from the upper half plane to the unit disc given by $z \mapsto \frac{z-i}{z+i}$. Composing this with our last map gives us a map from the semi-disc to the unit disc, given by $$z \mapsto -i\frac{z^2+2z-1}{z^2-2z-1}.$$
