Mobius transformations f

StefanM

New member
I want to understand how to make a Mobius Transformation.If someone can help me with an example that will be great.
Let's say we have f(0) = i, f(1) = 1, f(−1) = −1 for instance ...how should I proceed in finding one?Thank you

Fernando Revilla

Well-known member
MHB Math Helper
Write $f(z)=\dfrac{az+b}{cz+d}$ , then $f(0)=i\Leftrightarrow b=di\;,\;\ldots$ etc , and solve the system on the unknowns $a,b,c,d$ .

StefanM

New member
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).

Fernando Revilla

Well-known member
MHB Math Helper
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).
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 $z$ .