Step 1.The map is a Möbius transformation from D(0;1) to itself.

Proof: , and similarly . The inequality then follows from the fact that Thus

Step 2.maps D(0;1) onto itself. This follows from the fact that is its own inverse. If then you can check that

Step 3.If f is a Möbius transformation from D(0;1) onto itself, let Then the composition is a Möbius transformation from D(0;1) onto itself that fixes the point 0. With luck, you will have come across a theorem which says that this implies that is multiplication by a constant of absolute value 1. That will be enough to tell you that for a real constant