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Mobius transformation sharing fixed point commute

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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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, ...