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.

2. Originally Posted by runner07
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,
$x_1=x_2$.

3. how does that show ad-b[doesnt equal] 0?

4. Originally Posted by runner07
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:

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

RonL

5. thank you!

6. 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 $
\left(\begin{array}{cc}
a \ b \\ c \ d
\end{array}\right)
$
. This suggests f and that matrix share some nice properties; Especially, if s=ad-bc is not zero, the inverse matrix $
\left(\begin{array}{cc}
d/s \ -b/s \\ -c/s \ \ a/s
\end{array}\right)
$
exists and defines the wanted mapping $f^{-1}=\frac{(d/s)x-b/s}{(-c/s)x+a/s}$.

7. I never did conformal mapping.
But this transformation produces some beautiful pictures, look for pictures here.

8. "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?