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[http://www.math.purdue.edu/~bell/MA425/hwk3.pdf HWK 3 problems] | [http://www.math.purdue.edu/~bell/MA425/hwk3.pdf HWK 3 problems] | ||
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| + | Hint for III.9.2 --[[User:Bell|Bell]] | ||
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| + | 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 | ||
| + | point infinity to one. Hence, c is not zero and we may write | ||
| + | |||
| + | <math>\lim_{z\to\infty}L(z)=\frac{a}{c},</math> | ||
| + | |||
| + | and we may conclude that a=c. Divide the numerator and denominator | ||
| + | by c in order to be able to write | ||
| + | |||
| + | <math>L(z)=\frac{z+A}{z+B}.</math> | ||
| + | |||
| + | Now, the two conditions L(i)=i and L(-i)=-i give two equations for | ||
| + | the two unknowns, A and B. | ||
Revision as of 07:45, 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 write
$ \lim_{z\to\infty}L(z)=\frac{a}{c}, $
and we may conclude that a=c. Divide the numerator and denominator 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.
