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- Thread starter StefanM
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- Thread starter
- #1

- Jan 29, 2012

- 661

- Thread starter
- #3

- Jan 29, 2012

- 661

If you prefer this method, the result is: there exists a Möbius transformation $w$ such that $w(z_1)=w_1,w(z_2)=w_2,w(z_3)=w_3$ . This transformation is defined by $\dfrac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\dfrac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ . As you say, $\dfrac{2(w-i)}{(w+1)(1-i)}=\dfrac{2z}{z+1}$ . Now, you can easily express $w=\ldots$ as a function of $z$ .I saw an example f(z)=(z-z1)(z2-z3)/(z-z3)(z2-z1)....and I've used that and I got 2z/(z+1) and for f(w) I got 2(w-i)/(w+1)(1-i)...then by computing f(z) o f^(-1)(w)=mobius transformation.....but I don't know how to get f^(-1)(w).