Thread: linear fractional transformations 2

Show that there is only one linear fractional transformation that maps three given distinct points Z1, Z2, Z3 in the extended Z plane onto three specified distinct points W1, W2, W3 in the extended W plane.

Suggestion: Let T and S be two such linear transformations. Then, after pointing out why S^(-1)[T(Zk)] = Zk (K=1,2,3), use the fact that a composition of two linear fractional transformations is a linear fractional transformation and that every linear fractional transformation has at most two fixed points in the extended plane. Use this to show that S^(-1)[T(Z)]=Z for all z. This show that T(z)=S(z) for all z.

Originally Posted by mndi1105

Show that there is only one linear fractional transformation that maps three given distinct points Z1, Z2, Z3 in the extended Z plane onto three specified distinct points W1, W2, W3 in the extended W plane.

Suggestion: Let T and S be two such linear transformations. Then, after pointing out why S^(-1)[T(Zk)] = Zk (K=1,2,3), use the fact that a composition of two linear fractional transformations is a linear fractional transformation and that every linear fractional transformation has at most two fixed points in the extended plane. Use this to show that S^(-1)[T(Z)]=Z for all z. This show that T(z)=S(z) for all z.

Let and be Mobius transformations on the extended complex plane with for and .
Remember that the Mobius transformations form a group under function composition on the Riemann sphere. Therefore is a Mobius transformation with for . However, Mobius transformations have at most two fixes points other than the identity. Therefore, is the identity Mobius transformation on and so .