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Math Help - Complex Analysis -- functions on a connected set

  1. #1
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    Complex Analysis -- functions on a connected set

    Suppose that U is a simply-connected open domain in C and assume
    that f,g : U \rightarrow U are one-to-one and onto maps which are holomorphic mappings with the property that f' and g' are non-zero for all points of
    U. Prove that if f(z_i) = g(z_i) for i = 1, 2 and z_1 \neq z_2 then f = g.

    I am not sure where to start on this one. I know that if f and g are equal on a set which is not discrete then they are the same, but in this case the set is just two elements. I'm not sure what piece of the puzzle I am missing -- any help would be appreciated.
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  2. #2
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    Re: Complex Analysis -- functions on a connected set

    Let H be the upper half-space. We know that automorphisms of H are of the form

    f(z)=\frac{az+b}{cz+d} \qquad a,b,c,d\in \mathbb{R} \qquad ad-bc=1

    then (working with f\circ g^{-1}=h) assume h is such a function with two fixed points. Without loss of generality assume h(i)=i then h(z)=\frac{az+b}{-bz+a}. Let w\in H be the other fixed point then

    w=h(w)=\frac{aw+b}{-bw+a}

    and this leads to, assuming b\neq 0, w=i or w=-i so we have b=0 or h(z)=z.

    Now, since automorphisms of \mathbb{C} are affine transformations the result in this case is obvious.

    Conclude by the Riemann mapping theorem.
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