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Math Help - complex analysis

  1. #1
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    complex analysis

    We know that there is only one linear fractional transformation that maps three given points  z_1, z_2, z_3 to three specified points  w_1, w_2, w_3 . So the mapping  \text{Im} \ z = 0 onto the circle  |w| = 1 is uniquely determined if we choose the points (for example):  z = 0, \ z = 1, \ z = \infty .



    Why do we write  w = e^{i \alpha} \frac{z-z_{0}}{z-z_{1}} . Where did the  e^{i \alpha} come from?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Consider the Mobius transformation  T(z) = \frac{az+b}{cz+d} that maps  Im(z) \geq 0 onto the circle  |w| \leq 1 . Note that  T(z) maps  Im(z) = 0 onto the circle  |w| = 1 .

    Now choose an  \alpha , where  Im(\alpha) > 0 such that  T(\alpha) = 0 or equivalently  \frac{a\alpha+b}{c\alpha+d} = 0 \Rightarrow \alpha = -\frac{b}{a} .

    Note that  \overline{\alpha} is the inverse of  \alpha with respect to the real axis. So  0 and  T(\overline{\alpha}) must be inverses of each other with respect to  |w| = 1 . Hence  T(\overline{\alpha}) = \infty or equivalently  \frac{a\overline{\alpha}+b}{c\overline{\alpha}+d} = \infty \Rightarrow \overline{\alpha} = -\frac{d}{c} .

     T(z) = \frac{az+b}{cz+d} = \frac{a}{c} \; \frac{z+\frac{b}{a}}{z+\frac{d}{c}} = \frac{a}{c} \; \frac{z-\alpha}{z-\overline{\alpha}}

    When  x \in \mathbb{R} \; \; |T(x)| = 1 . Also notice that  |x-\alpha| = |\overline{x-\alpha}| = |x-\overline{\alpha}| \Rightarrow \frac{|x-\alpha|}{|x-\overline{\alpha}|} =1 . Therefore  \left|\frac{a}{c}\right| = 1 .

    By converting  \frac{a}{c} into polar coordinates, the above equality yields  \frac{a}{c} = e^{i\beta} .

    Therefore  T(z) = e^{i\beta}  \frac{z-\alpha}{z-\overline{\alpha}} .



    So notice that only two points were chosen to map the upper-half plane onto the unit disk and certain constraints were given to a third point but no third point was chosen explicitly, therefore it makes sense that there is not a unique solution to do this.

    I hope this answers your question.

    -Chip
    Last edited by chiph588@; January 12th 2009 at 08:14 AM. Reason: forgot to add something
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