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Linear Fractional Transformation (Mobius Transformation) from circle to line

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

    find linear fractional transformation that carries circle |z|=1 onto the line Re((1+i)w)=0

    2. Relevant equations

    linear fractional transformation is of the form az+b/cz+b where ad-bc≠0

    3. The attempt at a solution

    Re((1+i)w)=0 means that the line is just the y axis, but then I am unsure.. I think I might have to pick points and figure out the transformation that maps the points where i want them to go?
     
  2. jcsd
  3. Apr 14, 2012 #2
    Yes, you do need to pick points!

    Now the question is, how many points and how do you strategically choose them so the LFT you solve for is unique?
     
  4. Apr 14, 2012 #3
    I would say 3 points.. But geometrically does the circle basically just unravel and sit on that line?
     
  5. Apr 14, 2012 #4
    If that's the case I would have to pick a point that maps to the origin and then pick another that maps to infinity? And then I'm not sure about the third...
     
  6. Apr 14, 2012 #5
    You can say that, yes. Every poit on your circle will map to a point on the line.


    The way to tackle these types of problems is to map two symmetric points to two symmetric points, and then to map a boundary point to a boundary point. In this case your are sending a circle to a line. You need to choose two points that are symmetric to the circle (one inside of the circle, one outside) and send them to two points that are symmetric to the line you want to map to. Lastly, you want to map a point on the circle to a point on the line. This will force the new LFT to map that circle to that line.

    You can pick ANY two symmetric points to the circle mapping to ANY two other symmetric points to the line as well as any point on the circle to any point on the line; the answer will be the same. Because of this, try to pick easy values to work with to make your life a bit easier.
     
  7. Apr 14, 2012 #6
    For this problem since I am mapping from |z|=1 there is no inside and outside of the circle right? It is just the boundary that is being mapped.. So I would have to pick points from the boundary right?
     
  8. Apr 14, 2012 #7
    I think I could map 1 to infinity and map -1 to 0 but after that i'm not sure..
     
  9. Apr 14, 2012 #8
    Sorry for my late response, I was at work all day.

    You have it backwards! Your circle is the unit disk. However, your two points of 1 and -1 are NOT symmetric to the unit circle, they are symmetric to the y-axis. I think you just mixed up the terms, you had the right idea.

    Okay, so let's say your LFT is T(z). Set [itex]T(0) = -1 \text{ and } T(\infty) = 1[/itex]. Now you need one more point! Just take any point on the unit circle and map it to any point on the y axis! Solve for your LFT and you are done!

    You can check yourself by picking any point on the unit disk, say, r. Then take T(r) and you will see that it will have no real part.
     
  10. Apr 15, 2012 #9
    but I didn't think 0 and ∞ were on the circle |z|=1
     
  11. Apr 15, 2012 #10
    They're not! Maybe I'm assuming too much, is your only method available to constructing LFTs to just map three points on a circle to three points on a circle? (remember line and circle are the same in this context) if that's the case, pick any three points on the circle and map them to any three points on the line.

    I was suggesting a somewhat easier method. Have you gone over symmetric points yet? You map two symmetric points to two symmetric points, and then you map one point on the circle to one point on the line.

    Can you tell me which way you are supposed to go about this?
     
  12. Apr 15, 2012 #11
    I don't see anything in the chapter about symmetric points... so i'm assuming just 3 points to 3 points on the line.. but it seems like there would be more than one mapping... the answer the book gives is λ(1-i)[itex]\frac{z+1}{z-1}[/itex] so i'm just trying to understand it
     
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