Axiomatic Geometry - Mathematics 360 - Spring 2009

For Homework 16

1. The group of hyperbolic transformations can be defined as: H = the set of all Mobius transformations of the form T(z) = (az + b)/(cz + d) such that a and d are complex conjugates of each other, b and c are complex conjugates of each other, and T has determinant 1. We won't prove here that this definition is equivalent to the one given in class or in the book (see problem 3 on page 84). Instead, just prove that this set of Mobius transformations is a group. (Hint: use matrices.)
2. Let T be an elliptic Mobius transformation that fixes points p and q. Let C be a cline. Prove that the following are equivalent (i.e., each implies the others). (Hint: Using the notation of Chapter 5, first prove this for the transformation R = STS^(-1). Then prove it for T.)
   1. C is a Steiner circle of the second kind for p and q.
   2. p and q are symmetric with respect to C.
   3. T(C) = C.
3. Let T be a hyperbolic Mobius transformation that fixes points p and q. Let C be a cline. Prove that the following are equivalent.
   1. C is a Steiner circle of the first kind for p and q.
   2. T(C) = C.
4. Let T be a parabolic Mobius transformation with fixed point p. Let C be a cline. Prove or disprove each of the following.
   1. If  T(C) = C, then p is on C.
   2. If p is on C, then T(C) = C.
5. Let T be a loxodromic Mobius transformation that is neither elliptic nor hyperbolic. Prove that T does not take any cline to itself.
6. Let T be in H. Prove that if T has exactly one fixed point, then that fixed point must be on the unit circle S^1. Hint: use the above.
7. Prove that every transformation in H is of one of the following types:
   1. A parabolic Mobius transformation with fixed point p on S^1.
   2. A hyperbolic Mobius transformation with fixed points p and q on S^1.
   3. An elliptic Mobius transformation with fixed points p and q symmetric with respect to (but not on) S^1.