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Now, the two conditions L(i)=i and L(-i)=-i give two equations for | Now, the two conditions L(i)=i and L(-i)=-i give two equations for | ||
the two unknowns, A and B. | the two unknowns, A and B. | ||
| + | |||
| + | Hint for the last problem --[[User:Bell|Bell]] | ||
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| + | <math>\left|\frac{z-a}{1-\bar a z}\right|<1</math> | ||
| + | |||
| + | if and only if | ||
| + | |||
| + | <math>\left|\frac{z-a}{1-\bar a z}\right|^2<1</math> | ||
| + | |||
| + | if and only if | ||
| + | |||
| + | <math>|z-a|^2<|1-\bar a z|^2</math> | ||
| + | |||
| + | if and only if | ||
| + | |||
| + | <math>(z-a)\overline{(z-a)}<(1-\bar a z)\overline{(1-\bar a z)}</math> | ||
| + | |||
| + | if and only if | ||
| + | |||
| + | <math>|z|^2-\bar a z-a\bar z +|a|^2<1-\bar a z-a\bar z-|a|^2|z|^2</math> | ||
| + | |||
| + | and then show that this inequality is true if |a|<1 and |z|<1. | ||
Revision as of 10:56, 17 September 2009
Homework 3
Hint for III.9.2 --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 --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.
