Homework 3
Hint for III.9.2 --Bell
Suppose $ L(z)=\frac{az+b}{cz+d} $ is a LFT such that L maps the point at infinity to 1. If c=0, L cannot map the point infinity to one. Hence, c is not zero and we may compute
$ \lim_{z\to\infty}L(z)=\frac{a}{c}, $
and we conclude that a=c. Divide the numerator and denominator in the formula for L by c in order to be able to write
$ L(z)=\frac{z+A}{z+B}. $
Now, the two conditions L(i)=i and L(-i)=-i give two equations for the two unknowns, A and B.
