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Math Help - Show f is a conformal mapping

  1. #1
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    Show f is a conformal mapping

    My working: A basis for the tangent space (at any point) is (1,0), (0,1) so I need only show it for this . df(1,0)=df/dx and df(0,1)=df/dy Is this the right approach? Thanks
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  2. #2
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    Re: Show f is a conformal mapping

    deleted.
    Last edited by xxp9; May 10th 2013 at 06:49 AM.
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  3. #3
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    Re: Show f is a conformal mapping

    yes I was right. A parametrization is conformal if its first fundamental form is of the form g(du^2+dv^2). So just compute its first fundamental form.
    Let A=1+x^2+y^2. The coordinate vectors are:
    \frac{\partial}{\partial x}=\frac{(2A-4x^2, -4xy, -4x)}{A^2}
    \frac{\partial}{\partial y}=\frac{(-4xy, 2A-4y^2, -4y)}{A^2}
    It is easy to verify that \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial x} \rangle=\langle \frac{\partial}{\partial y}, \frac{\partial}{\partial y} \rangle
    And \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \rangle=0
    Thanks from Plato13
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    Re: Show f is a conformal mapping

    I was hoping to solve by using that theorem that a map is conformal if at each point p there is a basis e,e' such that the df(e).df(e')=lambda x e.e' where lamda is the scalar factor. I've just realised I ought to use polar co-ordinates x=rcost, y=rsint. That will simplify things right?
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  5. #5
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    Re: Show f is a conformal mapping

    Your theorem is nothing but the statement of "the first fundamental form is of the form \lambda(dx^2+dy^2)"
    Note that in my post, \frac{\partial}{\partial x} = df(e_1), and \frac{\partial}{\partial y} = df(e_2), given e_1=(1,0) and e_2=(0,1).
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  6. #6
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    Re: Show f is a conformal mapping

    thanks for you help and patience
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