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Math Help - linear fractional transformations 2

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by mndi1105 View Post
    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.
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