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Math Help - how to show this is a conformal map

  1. #1
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    how to show this is a conformal map

     f(z)=\frac{z}{(z-1)^{2}} maps disc |z|<1 to  C\ \{x+yi: x\leq -\frac{1}{4}, y=0 \}
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  2. #2
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    Quote Originally Posted by szpengchao View Post
     f(z)=\frac{z}{(z-1)^{2}} maps disc |z|<1 to  C\ \{x+yi: x\leq -\frac{1}{4}, y=0 \}
    Let f_1 be the map z\mapsto \tfrac{1+z}{1-z}.
    Let f_2 be the map z\mapsto z+1.
    Let f_3 be the map z\mapsto -\tfrac{1}{2}z.
    Let f_4 be the map z\mapsto z^2.
    Let f_5 be the map z\mapsto z+1.

    Then, z\mapsto \tfrac{z}{(z-1)^2} is f_5\circ f_4\circ f_3 \circ f_2 \circ f_1.

    1)The conformal map f_1 maps the disk |z|<1 to the half-plane \Re (z) > 0.
    2)The conformal map f_2 maps the half-plane \Re(z) > 0 to \Re(z) > 1.
    3)The conformal map f_3 maps the half-plane \Re(z) > 1 to  \Re(z) < - \tfrac{1}{2}.
    4)The problem is that f_4 is not conformal on this half-plane.
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