# Math Help - linear fractional transformations 2

1. ## 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.

2. 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 $T$ and $S$ be Mobius transformations on the extended complex plane with $T(z_k) = S(z_k)$ for $k=1,2,3$ and $z_k \in \mathbb{C}_{\infty}$.
Remember that the Mobius transformations form a group under function composition on the Riemann sphere. Therefore $T\circ S^{-1}$ is a Mobius transformation with $T\circ S^{-1} (z_k) = z_k$ for $k=1,2,3$. However, Mobius transformations have at most two fixes points other than the identity. Therefore, $T\circ S^{-1}$ is the identity Mobius transformation on $\mathbb{C}_{\infty}$ and so $S=T$.