This is not hard to prove with the group of conformal automorphisms of the open unit disk int(D^2) about the origin. (Which can be seen to be isomorphic to PSL(2,R) — or equivalently, this is the group of conformal automorphisms of the upper half-plane.) The group Aut(int(D^2)) is isomorphic to SL(2,R) / {I,-I}, since the elements of G = Aut(int(D^2)) = {z |—> (az+b)/(cz+d) | a, b, c, d in R with ad-bc = 1} multiply just as do the 2x2 matrices of SL(2,R), and any element (az+b)/(cz+d) is the same mapping when (a,b,c,d) is changed to (-a,-b,-c,-d), and only then. Now given the setup with D below: Since elements of G carries circles to circles, we may find an element of G that takes D to a disk centered at the origin. This takes the other relevant circles to circles, but these new circles all have the same radius. Without loss of generality we assume the center of D is at the origin. In this conformally equivalent picture to the original one, either a) Copies of the first additional disk C below (i.e., having the same radius as C) either fit consecutively around D and return exactly to C after finitely many new disks are placed as below, or b) not. And a) will be true or not independent of where the first C is placed, by circular symmetry. –Dan
On Dec 18, 2015, at 6:56 AM, Erich Friedman <erichfriedman68@gmail.com> wrote:
back in undergraduate complex analysis, i learned the following theorem, whose proof is easy enough using conformal mappings. does this theorem have a name?
Let D be a disk entirely inside the unit circle. Consider all collections of non-overlapping disks C so that each member of C is:
a) inside the unit circle and tangent to the unit circle, b) doesn't overlap D but is tangent to D, and c) tangent to exactly two other members of C. (Thus the collection C forms a tangent ring around D.)
For a given D, the collection is either empty or infinite.