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Thread: how to show this is a conformal map

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

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

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

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