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Image under a mobius transformation

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the Mobius transformation which carries the points [tex]0,1,-i[/tex] to the points [tex]-1,0,\infty[/tex] respectively. Find the image of the domain [tex]\{z:x<0,-x+y<t\}[/tex] under this mobius transformation.


    2. Relevant equations



    3. The attempt at a solution
    Let [tex]T(z)=\frac{az+b}{cz+d}[/tex].
    Then [tex]T(0)=-1\Longrightarrow \frac{b}{d} \iff b=-d[/tex]

    [tex]T(1)=0\Longrightarrow \frac{a+b}{c+d}=0\Longrightarrow a+b=0 \iff a=-b=d[/tex]

    [tex]T(-i)=\infty\Longrightarrow \frac{-ia+b}{-ci+d}\iff d-ci=0\iff c=\frac{d}{i}=bi=-ai[/tex]

    Now we have [tex]T(z)=\frac{az-a}{a-aiz}=\frac{z-1}{1-zi}[/tex]

    So I now have to find the image under this map which is where I'm a bit stumped. Would it help to find where the intersections of the boundary of the domain with the axes are mapped to?
     
  2. jcsd
  3. Mar 26, 2014 #2
    Yes, the first part seems to be correct. :smile:
     
    Last edited: Mar 26, 2014
  4. Mar 26, 2014 #3

    HallsofIvy

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    Now let z= x+iy where x> 0 and x+ y< t for some fixed real number, t. For example, if t= 1, that would be the part of the complex plane to the right of the imaginary axis and below x+ y= 1. Some points on the boundary of that set would be 1, i, 2- i, etc. Determining what the transformation does to boundary points should help you see what the image is.
     
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