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Math Help - extremely confused.. please help

  1. #1
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    extremely confused.. please help

    let f(x) = (ax + b) / (cx + d) , x [does not equal] -d/c

    a. Show that f is one to one if ad-b[doesnt equal] 0.

    b. Suppose that ad-bc[doesnt equal] 0. Find f^-1.
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  2. #2
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    Quote Originally Posted by runner07 View Post
    let f(x) = (ax + b) / (cx + d) , x [does not equal] -d/c
    This map has a special name to it called the "Mobius Transformation".

    We need to show if f(x_1)=f(x_2)\implies x_1=x_2.
    Thus,
    \frac{ax_1+b}{cx_1+d} = \frac{ax_2+b}{cx_2+d}
    Thus,
    (ax_1+b)(cx_2+d)=(ax_2+b)(cx_1+d)
    Thus,
    acx_1x_2+bcx_2+adx_1+bd = acx_1x_2+bcx_1+adx_2+bd
    Thus,
    (ad-bc)x_1 = (ad-bc)x_2
    You can cancel for ad\not = bc
    Thus,
    x_1=x_2.
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  3. #3
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    how does that show ad-b[doesnt equal] 0?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by runner07 View Post
    how does that show ad-b[doesnt equal] 0?
    1. You have a misprint, it should be ad-bc != 0

    2. You need this condition so that the brackets (ad-bc) can be cancelled
    in ImPerfectHacker's post, as otherwise x_1 and x_2 can have any values
    (not necessarily equal) and still satisfy:

    (ad-bc)x_1 = (ad-bc)x_2

    as this would be 0 x_1 = 0 x_2 if ad-bc=0

    RonL
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  5. #5
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    thank you!
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  6. #6
    Super Member Rebesques's Avatar
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    If you don't want to do all that work, there's always the "cheating" way.

    Notice that ad-bc is also the determinant of the matrix <br />
\left(\begin{array}{cc}<br />
a  \ b \\ c   \ d <br />
\end{array}\right)<br />
. This suggests f and that matrix share some nice properties; Especially, if s=ad-bc is not zero, the inverse matrix <br />
\left(\begin{array}{cc}<br />
d/s  \ -b/s \\ -c/s    \ \ a/s <br />
\end{array}\right)<br />
exists and defines the wanted mapping f^{-1}=\frac{(d/s)x-b/s}{(-c/s)x+a/s}.
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  7. #7
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    I never did conformal mapping.
    But this transformation produces some beautiful pictures, look for pictures here.
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  8. #8
    Super Member Rebesques's Avatar
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    "Mobius Transformation"

    By the way, did you foresee how you can construct some symmetric (S_4), alternating (A_5) or dihedral (D_4) groups with the help of Mobius transformations?
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