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Thread: linear mapping

  1. #1
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    linear mapping

    Consider the inversion f(z)=1/z. What happens to the line y=2 under this mapping?

    In such a question the way to solve them is:
    1/z=1/(x+iy)=(x-iy)/(x^2+y^2)
    so u= x/(x^2+y^2) and v=(-y)/(x^2+y^2)
    so x= u/(u^2+v^2) y=-v/(u^2+v^2)

    u^2 + v^2 +v/2=0 by complete the square
    => u^2+(v+1/4)^2=1/16

    so the center is(0,1/4) ad my radius = 1/root(16)?

    Thanks
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  2. #2
    MHF Contributor

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    Quote Originally Posted by NaNa View Post
    Consider the inversion f(z)=1/z. What happens to the line y=2 under this mapping?

    In such a question the way to solve them is:
    1/z=1/(x+iy)=(x-iy)/(x^2+y^2)
    so u= x/(x^2+y^2) and v=(-y)/(x^2+y^2)
    so x= u/(u^2+v^2) y=-v/(u^2+v^2)

    u^2 + v^2 +v/2=0 by complete the square
    => u^2+(v+1/4)^2=1/16

    so the center is(0,1/4) ad my radius = 1/root(16)?

    Thanks
    You might want to say that the radius is 1/4 rather than 1/root(16)!
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  3. #3
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    yes!
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  4. #4
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    so is right!?
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