Homework 3
Hint for III.9.2 --Steve Bell
Suppose $ L(z)=\frac{az+b}{cz+d} $ is an 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.
Hint for the last problem --Steve Bell
$ \left|\frac{z-a}{1-\bar a z}\right|<1 $
if and only if
$ \left|\frac{z-a}{1-\bar a z}\right|^2<1 $
if and only if
$ |z-a|^2<|1-\bar a z|^2 $
if and only if
$ (z-a)\overline{(z-a)}<(1-\bar a z)\overline{(1-\bar a z)} $
if and only if
$ |z|^2-\bar a z-a\bar z +|a|^2<1-\bar a z-a\bar z-|a|^2|z|^2 $
and then show that this inequality is true if |a|<1 and |z|<1.
