The wikipedia entry Mobius transformation states the following.
The subgroup of all Möbius transformations that map the open disk $D:= \{z :|z| < 1\}$ to itself consists of all transformations of the form $$f(z) = e^{i\phi} \frac{z + b}{\bar bz + 1}$$ with $\phi\in R$, and $|b| < 1$, $\bar b$ being the complex conjugate of $b$. This is equal to the group of all biholomorphic (or equivalently: bijective, angle-preserving and orientation-preserving) maps $D\to D$.
How does one prove that?