Results 1 to 2 of 2

Thread: complex analysis

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    156

    complex analysis

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



    Why do we write $\displaystyle w = e^{i \alpha} \frac{z-z_{0}}{z-z_{1}} $. Where did the $\displaystyle e^{i \alpha} $ come from?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Consider the Mobius transformation $\displaystyle T(z) = \frac{az+b}{cz+d} $ that maps $\displaystyle Im(z) \geq 0 $ onto the circle $\displaystyle |w| \leq 1 $. Note that $\displaystyle T(z) $ maps $\displaystyle Im(z) = 0 $ onto the circle $\displaystyle |w| = 1 $.

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

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

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

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

    Therefore $\displaystyle 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@; Jan 12th 2009 at 08:14 AM. Reason: forgot to add something
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 4th 2011, 05:30 AM
  2. Replies: 6
    Last Post: Sep 13th 2011, 07:16 AM
  3. Replies: 1
    Last Post: Oct 2nd 2010, 01:54 PM
  4. Replies: 12
    Last Post: Jun 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: Mar 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum