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Math Help - mobius transform question

  1. #1
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    mobius transform question


    D^+=\left \{z:|z|<1,Im \left \{ z \right \}>0  \right \}
    so it represents the northen hemisphere of a circle with radius 1.
    f(z)=\frac{2z-i}{2-iz}
    i need to find what is the picture of f(d^+)=?

    i tried to solve it like this:
    my area is bounded by a line and a curve.
    i want to see what each one transforms to
    curve points:
    f(1)=1
    f(-1)=-1
    f(i)=i/3
    so it will look like this
    http://i46.tinypic.com/3129rg4.gif

    line points:
    f(1)=1
    f(-1)=-1
    f(0)=-i/2
    so it looks like this
    http://i49.tinypic.com/dnnh2u.jpg

    and when i try and see where the inside goes :
    f(i/2)=0
    so the answer represents their intersection area
    (i mistakenly marked only a part of the interssection marked it should be the whole intersection)
    http://i49.tinypic.com/14bi2qr.gif

    but my prof thinks otherwise
    http://i48.tinypic.com/2jff52r.jpg

    who is worng here? why?
    so its in their intersections
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    D^+=\left \{z:|z|<1,Im \left \{ z \right \}>0 \right \}
    so it represents the northen hemisphere of a circle with radius 1.
    f(z)=\frac{2z-i}{2-iz}
    i need to find what is the picture of f(d^+)=?

    i tried to solve it like this:
    my area is bounded by a line and a curve.
    i want to see what each one transforms to
    curve points:
    f(1)=1
    f(-1)=-1
    f(i)=i/3
    so it will look like this
    http://i46.tinypic.com/3129rg4.gif

    line points:
    f(1)=1
    f(-1)=-1
    f(0)=-i/2
    so it looks like this
    http://i49.tinypic.com/dnnh2u.jpg

    and when i try and see where the inside goes :
    f(i/2)=0
    so the answer represents their intersection area
    (i mistakenly marked only a part of the interssection marked it should be the whole intersection)
    http://i49.tinypic.com/14bi2qr.gif

    but my prof thinks otherwise
    http://i48.tinypic.com/2jff52r.jpg

    who is worng here? why?
    so its in their intersections

    Put z=x+iy\,,\,\,so\,\,\,z\in\ D^+\Longrightarrow x^2+y^2<1\,,\,\,y>0 , and from here:

    \frac{2z-i}{2-iz}=\frac{5x}{x^2+(y+2)^2}+\frac{2\left[x^2+(y+2)\left(y-\frac{1}{2}\right)\right]}{x^2+(y+2)^2}\,i

    It's easy to see that the imaginary part above can be negative, for example take z=0.1+0.1i\in D^+, so the image cannot be what your prof says since that image has all its imaginary part positive...

    Tonio
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  3. #3
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    thanks
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