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Suppose <math>L(z)=\frac{az+b}{cz+d}</math> is a LFT such that | Suppose <math>L(z)=\frac{az+b}{cz+d}</math> is a LFT such that | ||
L maps the point at infinity to 1. If c=0, L cannot map the | 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 | + | point infinity to one. Hence, c is not zero and we may compute |
<math>\lim_{z\to\infty}L(z)=\frac{a}{c},</math> | <math>\lim_{z\to\infty}L(z)=\frac{a}{c},</math> | ||
| − | and we | + | and we conclude that a=c. Divide the numerator and denominator |
| − | by c in order to be able to write | + | in the formula for L by c in order to be able to write |
<math>L(z)=\frac{z+A}{z+B}.</math> | <math>L(z)=\frac{z+A}{z+B}.</math> | ||
Revision as of 06:47, 16 September 2009
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.
