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Math Help - [SOLVED] Mobius Transformation

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Mobius Transformation

    Consider a Mobius transformation

    z \mapsto w =\frac{az + b}{cz + d}

    Determine the constants a, b, c, and d, (all \epsilon \mathbb{C}), such that the

    transformation maps
    0 \rightarrow i, −i \rightarrow 1, and −1 \rightarrow 0.

    How would I work this out?
    Last edited by ronaldo_07; January 21st 2009 at 08:56 PM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    What's mapped to  i ? Looks like you forgot to type it.
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  3. #3
    Member ronaldo_07's Avatar
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    Quote Originally Posted by chiph588@ View Post
    What's mapped to  i ? Looks like you forgot to type it.
    sorry 0 is mapped to i
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    Consider a Mobius transformation

    z \mapsto w =\frac{az + b}{cz + d}

    Determine the constants a, b, c, and d, (all \epsilon \mathbb{C}), such that the

    transformation maps
    0 \rightarrow i, −i \rightarrow 1, and −1 \rightarrow 0.

    How would I work this out?
    The way to do this is to cross products equal to each other.

     (z,0,-i,-1) = (w,i,1,0)

     \frac{z-0}{z-(-i)} \frac{-1-(-i)}{-1-0} = \frac{w-i}{w-1} \frac{0-1}{0-i}

     -\frac{z(-1+i)}{z+i}= -i \frac{w-i}{w-1}

    Solving for  w , we get  w = \frac{-iz-i}{z-1}

     a = -i

     b = -i

     c = 1

     d = -1
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  5. #5
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    Setting the cross products equal, I get

    (z-(-i))/z-(-1) x 0-(-1)/0-(-i) = (w-1)/(w-0) x (i-0)/(i-1)

    (z+i)/(iz+i) = (iw-i)/(iw-w)

    and getting w by itself

    w = (z+1)/(iz-i) a = 1, b = 1, c = i, d = -i
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Multiply numerator and denominator by  -i and you'll get my answer.
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