Find a biholomorphic map from the region $\Omega=\{z=x+iy:x^2+y^2<4 \text{ and } x+y<2\}\setminus\{0\}$ onto the unit punctured disk $D^*=D(0,1)\setminus\{0\}$.
My idea is to map the line x+y=2 to a circle thru $\infty$ on the Riemann sphere. Since circular arcs are mapped onto lines or circles, the only possibility is the arc mapped onto the great circle as described above.
However after this i am at a loss to what i should do. The issue seems to be the deleted point. Any hints?
On
Let $\Omega'$ be the domain $\Omega$ with the point $0$ thrown in again. $\Omega'$ is bounded by two "circles" making intersecting at $45^\circ$ at $z=2$ and $z=2i$. The Moebius transformation $$z\mapsto w:={z-2\over z-2i}$$ maps these two "circles" onto two "circles" intersecting at $0$ and $\infty$, hence onto two lines. It follows that $\Omega'$ is mapped onto a wedge with opening angle ${3\pi\over4}$. Open up this wedge with $w\mapsto w^{4/3}$ (suitably adjusted) and you arrive at a half plane $H$. Under these two maps the point $z=0$ has gone to some point $c\in H$. Now choose another Moebius transformation that maps $H$ onto $D$ whereby $c$ is mapped to $0$.
This is impossible. Any biholomorphic function is at least a homeomorphism, and homeomorphisms induce isomorphisms of the fundamental group. But the fundamental groups of your domains are very different: $D(0,1)\setminus\{0\}$ has fundamental group $\mathbb{Z}$, being homotopic to $S^1$ (just retract along straight lines through the origin); and $\Omega$ is contractible.