Mobius transformation sharing fixed point commute
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Let T be a Mobius transformation, T doesn't equal to identity. Show that a Mobius transformation S commutes with T if S and T have the same fixed points.
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Definition:
Mobius transformation can be defined as
T(z) = a(z+b)/(z+c), where a, b, and c are some complex numbers.
Let us parametrize the other Mobius transformation as
S(z) = A(z+B)/(z+C), where A, B, and C are some other complex numbers.
We are to prove that if T and S have the same fixed points, then T(S(z)) = S(T(z))
Let us now see what are the fixed points.
For T, if z is a fixed point, then
z = T(z) = a(z+b)/(z+c).
We can solve this by multiplying the equation by the denominator and getting the quadratic equation:
z^2 +(c-a)z - ab = 0.
There are two ways to proceed now, depending on whether you are familiar with Vieta formulae.
If you have not studied Vieta formulae, ...
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