# conformal map disc with two removed points

I need to find all the bijective conformal maps from $D = \{ |z| < 1, z\neq \pm 1/2 \}$ onto itself.

Since this set is not simply connected, I think that the $180°$ degree rotation is the only non-trivial conformal map, but I am not completely sure.

• What is your argument that $z\mapsto -z$ and the identity are the only automorphisms of $D$? – Daniel Fischer May 13 '15 at 17:10
• One path to a solution is to argue that any such map can be extended continuously to a map from the open unit disk to itself (i.e., that it has "removable singularities" at $\pm\frac12$). Then you can use what you know about the automorphisms of the unit disk. – Greg Martin May 13 '15 at 17:43
• @Greg Martin Very interesting is there a theorem with a name corresponding to this fact? – tired May 14 '15 at 13:35
• I doubt that there is. – Greg Martin May 14 '15 at 16:33
• @GregMartin Wikipedia and Wolfram MathWorld call it Riemann's theorem on removable singularities – user147263 May 17 '15 at 3:28

By Riemann's theorem on removable singularities any such map extends to a holomorphic map of unit disk $\Delta$ into $\Delta$. The extended map is still injective: indeed, if a holomorphic map attains some value $w_0$ at $k\ge 1$ points, it attains all neighboring values at least $k$ times. (This follows from Rouché theorem, or from the argument principle.)
This, the extended map is an automorphism of $\Delta$, i.e., a function of the form $$f(z) = \gamma\frac{z-a}{1-\bar az},\qquad |a|<1,\ |\gamma|=1$$ Of these, only the identity map and $f(z)=-z$ fix the set $\{\pm 1/2\}$. (A geometric way to see this is to use the fact that $f$ is an isometry in the hyperbolic metric; in particular, it must send the geodesic $[-1/2,1/2]$ onto itself isometrically. Hence $f(0)=0$, etc.)