there is a unique mobius transformation taking three points to three points, proof?Watch

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Thread starter 14 years ago
#1
Show that, if z1, z2, z3 and w1, w2, w3 are two triples of
distinct points in C, there exists a unique Mobius transformation that
takes zj to wj [j = 1; 2; 3].

Errr, I can't find my notes on this, and I know there's a really snazzy way to do it. Help would be much appreciated.

Lex!
0
14 years ago
#2
Define a transform T(z) = (z-z1)(z2-z3)/(z-z3)(z2-z1), so that T(z1) = 0, T(z2) = 1, T(z3) = infinity
And a transform W(w) = (w-w1)(w2-w3)/(w-w3)(w2-w1), then W(w1) = 0, W(w2) = 1, W(w3) = infinity

Then we have the required transform X = (W^-1)T

To prove it is unique, assume it isnt, so that X(z_i) = Y(z_i) for i=1,2,3, then you can define a transform which has 3 fixed points, thus is the identity which contradicts X =/= Y.
0
14 years ago
#3
let A(z)=(z-z1)(z2-z3)/(z-z3)(z2-z1)
then A(z) is a mob transform
let B(w) be s.t B(w)=(w-w1)(w2-w3)/(w-w3)(w2-w1)
then B(w) is a mob transform.
then
T(z)=B^-1A(z) is a mobius transformation with T(zi)=wi.
for uniqueness need the fact a mob trans can only have 2 fixed points unless
t(z)=z for all z.
assume T(z) and S(z) are mob trans s.t T(zi)=S(zi)=wi i=1,2,3
then S^-1.T(zi)=zi i=1,2,3
ie S^-1T has 3 fixed points => S^-1T(z)=z for all z
ie S=T.
to show the 2 fixed point "fact"
consider T(z)=az+b/cz+d with ab-cd<>0
if c=0 then t(z)=z is a linear equation so only 1 solution if T(z) <> z for all z.
If c<>0 then t(z)=z is a quadratic with at most 2 solutions.
0
14 years ago
#4
(Original post by JamesF)
To prove it is unique, assume it isnt, so that X(z_i) = Y(z_i) for i=1,2,3, then you can define a transform which has 3 fixed points, thus is the identity which contradicts X =/= Y.
I don't see the logic here - though I don't dispute the result.

Why need it be the identity?

It's easy to check only the identity fixes 0,1, and infinity, [as this gives equations in a,b,c,d] and then one can get uniqueness for a general triple from this.
0
14 years ago
#5
I have no idea how many times i've posted this now but i'll try again.

If there exists two different transforms T(z) and S(z) which are equal at 3 distinct points, then TS^-1 has 3 fixed points - so it must be the identity, therefore T = S.
You can show that a transform with 3 fixed points must be the identity by considering the order of the polynomial you get for the fixed point transform;
T(z) = z = (az+b)/(cz+d)
is a quadratic and linear if c = 0. You only get more than 2 solutions if T(z) is identically z.
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