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Image of two homographic transformations (Möbius transformations)

  1. Jan 2, 2014 #1
    The problem statement, all variables and given/known data.

    1) Let ##T## be the transformation over the extended complex plane that sends the poins ##0,i,-i## to the points ##0,1,\infty##. Prove that the image of the circle centered at the origin and of radius ##1## under this transformation is the line ##\{re(z)=1\}##.

    2) For ##\alpha \in \mathbb C## such that ##|\alpha|\neq 1##, prove that the homographic transformation ##T(z)=\dfrac{z-\alpha}{-\overline\alpha z+1}## sends the circle ##\{|z|=1\}## to itself and sends ##\alpha## to ##0##.


    The attempt at a solution.

    1) Let ##T(z)=\dfrac{az+b}{cz+d}##. Then, ##T(0)=0##, so ##b=0##. ##T(i)=1## and from this condition we get that ##ai=ci+d##, so ##a=c-di##. From the last condition we get that ##c(-i)+d=0##, so ##d=ci##. Finally, ##T(z)=\dfrac{2z}{z+i}##.

    Sorry if this is obvious but how do I show that for any ##z \in \{|z|=1\}##, ##T(z)=w## with ##w## of the form ##w=1+bi##, ##b \in \mathbb R##?.

    2) I have the same problem, if ##z \in \{|z|=1\}##, I want to show that ##T(z)=w## such that if ##w=a+bi##, then ##a^2+b^2=1##.

    Edition:

    I could solve 2), what I did was ##{T(z)}^2=T(z) \overline {T(z)}## and I could prove that the last expression equals to ##1##.

    I would appreciate some help for 1).
     
    Last edited: Jan 2, 2014
  2. jcsd
  3. Jan 3, 2014 #2

    Dick

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    Science Advisor
    Homework Helper

    For 1) if you know Mobius transformations will map circles to either lines or circles then if you just compute T(1), T(i) and T(-i) and notice they are all on the line Re(z)=1 that will do it. If you want to be more explicit I'd multiply numerator and denominator of 2z/(z+i) by z* (remembering zz*=1 since |z|=1) and then use that z=a+bi where a^2+b^2=1 and convert it to rectangular form.
     
  4. Jan 3, 2014 #3
    Thanks Dick!
     
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