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Complex Mapping/Images question

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data
    For the mapping f(z) = [itex]\frac{z - i}{z + i}[/itex], find the image of Im(z) ≥ 0

    3. The attempt at a solution

    So in the z-plane this is clearly everything for y ≥ 0

    I substituted x + iy, and rationalised it to get:

    [itex]u+iv[/itex] = [itex]{\frac {{x}^{2}-1+{y}^{2}}{{x}^{2}+1+{y}^{2}+2\,y}}-{\frac {2\,ix}{{x}
    ^{2}+1+{y}^{2}+2\,y}}
    [/itex]

    The denominator can be simplified to x^2 + (y+1)^2

    The answer says that this is a closed disc of |w| ≤ 1 with an exclusion at w = 1. Which means I should be seeing an answer of the form u^2 + v^2 ≤ 1

    I do not see how they got this.
     
    Last edited: Apr 13, 2012
  2. jcsd
  3. Apr 13, 2012 #2

    Dick

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    Try looking at what happens to the boundary y=0. Can you show that maps into u^2+v^2=1?
     
  4. Apr 13, 2012 #3
    For (x,y)

    I used (0,0) -> (-1,0)
    (1,0) -> (0,-1)
    (-1,0) -> (0,1)

    So one can see that it starts forming a circle, from the answer given I could guess that it is indeed a circle, but since I can't plot (0,-1) since that's not defined, how would I know it is a circle?

    Also it's easy to see that for any values y>0, it maps to a point inside the circle, so it's definitely inside the region.

    Also, is it possible to see what this is without having to plot several points?
     
  5. Apr 13, 2012 #4

    Dick

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    Put y=0 into your u and v expressions. Show u^2+v^2=1. It's pretty easy algebra.
     
  6. Apr 13, 2012 #5
    I do not quite follow. That gives me an expression in x.
     
  7. Apr 13, 2012 #6

    Dick

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    What expression for x? If you did it correctly you should be able to simplify the expression to 1.
     
  8. Apr 13, 2012 #7
    For y = 0:

    u + iv = [itex]{\frac {{x}^{2}-2\,ix-1}{{x}^{2}+1}}[/itex]
     
  9. Apr 13, 2012 #8
    Oh, I see now, when y=0 and x= 0. Thanks!
     
  10. Apr 13, 2012 #9

    Dick

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    Well no, for y=0 and x=anything. ((x^2-1)^2+(2x)^2)/(x^2+1)^2=1. Yes?
     
  11. Apr 13, 2012 #10
    Hmm, yes I see that now after working it out, although I hadn't the faintest idea that would occur by looking at the expressions, how did you figure that out?
     
  12. Apr 13, 2012 #11

    Dick

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    You told me the answer should be u^2+v^2=1. That was my first clue. I know that the function you gave is a Mobius transformation. I know all Mobius transformations map lines to lines or circles. That was a pretty good second clue.
     
  13. Apr 13, 2012 #12
    A Mobius Transformation? Never heard of that, thanks for the tip, I will check it out now.
     
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