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Construct a one to one map

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Using the properties of the exponential map, construct a one to one mapping of the strip S to the sector C:
    [itex]S=\{-\sqrt{2}x-t<y<-\sqrt{2}x\} , C=\{-\frac{\pi}{6}<arg(w)<\frac{\pi}{3}\}[/itex], where [itex]t[/itex] is a fixed positive real number. Here we let [itex]z=x+iy[/itex].


    2. Relevant equations



    3. The attempt at a solution
    The strip [itex]S[/itex] makes an angle of [itex]arctan(\sqrt{-2})[/itex] with the x/real axis so I would start by rotating the strip by this angle in the opposite direction to make the strip parallel to the axis. Is this a good start?
     
  2. jcsd
  3. Feb 28, 2014 #2

    tiny-tim

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    yes :smile:
     
  4. Mar 2, 2014 #3
    Ok so multiplying by [itex]e^{i\theta}[/itex] where [itex]\theta = arctan(\sqrt{-2})[/itex] I get a new strip [itex]S*=\{(-2\sqrt{x}-t)e^{i arctan(x)}<y<-2\sqrt{x}e^{i arctan(x)}\}[/itex].
    Any suggestions on where to go from here?
     
  5. Mar 2, 2014 #4

    PeroK

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    I must admit, I can't see the point of rotating the strip. Here's the process I would use:

    1) Map the line y = -√2x to the line θ = -π/6.

    1a) For x ≥ 0, map to r ≥ 1.
    1b) For x < 0, map to 0 < r < 1.

    2) Map the line y = -√2x - t to the line θ = π/3.

    1a) For x ≥ 0, map to r ≥ 1.
    1b) For x < 0, map to 0 < r < 1.

    3) For any line y = -√2x - a (0 < a < t), map to the line θ = -π/6 + (a/t)π/2.

    This third step is the key. Can you produce the mapping for this?

    This question is a beaut or a brute depending on your point of view!
     
    Last edited: Mar 2, 2014
  6. Mar 2, 2014 #5
    Thanks Perok. The questions states "using the properties of the exponential map" which suggests to me the rotation is wanted.
    How would you go about formally writing the maps 1) and 2) you mention?
     
  7. Mar 2, 2014 #6

    PeroK

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    As you seem to be really stuck:

    [tex]e^xe^{-\frac{i\pi}{6}}[/tex]

    Would map (1-1) a boundary line of the strip to a boundary line of the segment.

    You'll need to modify this to map the interior of the strip to the interior of the segment.
     
  8. Mar 3, 2014 #7
    I'm definately having a hard time visualising these types of maps. I'll see if I grasp what your are saying Perok:
    You are suggesting the map [tex]z\mapsto e^{z}e^{-\frac{\pi}{6}}[/tex]
    So [tex]-\sqrt{2}x-t\mapsto e^{-\sqrt{2}x-t}e^{-\frac{\pi}{6}}[/tex]
    and [tex]-\sqrt{2}x\mapsto e^{-\sqrt{2}x}e^{-\frac{\pi}{6}}[/tex]
     
  9. Mar 3, 2014 #8

    PeroK

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    Maybe you're not following. First, forget about equations and formulas. That's what I did. So, how do you map a strip to a segment? In terms of visualisation:

    You can think of the strip as a set of parallel lines. And the segment is a set of rays. So, I thought of mapping each line to a ray.

    So, first, how can you map a single line to a ray? One problem, of course, is that the line goes to infinity in two directions, where the ray goes to zero in one direction and infinity in the other.

    The solution was to map the middle of the line (x = 0) to a point 1 unit along the ray (r = 1). Then map one half of the line (x > 0) to the infinite bit of the ray (r > 1); and map the -ve half of the line (x < 0) to the ray r < 1.

    Can you visualise that?

    One good way to achieve this is using the real exp function (hence the hint in your question), since exp maps (-∞, ∞) to (0, ∞).

    The next step was think how to map every line in the strip to a different ray.

    I thought: map one boundary line to one boundary ray, then proportionally map every line to a ray. Working your way across the strip from t to 0, while working your way across the segment from -π/6 to π/3. (Although in the mapping we want, we're only mapping the interior, not the boundaries as well.)

    At this point, I guess, algebra has to come in and we have to start writing down formulas.

    So, leaving the formula out for now, can you visualise this?
     
  10. Mar 3, 2014 #9
    Yes Im with you on your intuition. The problem is I've only seen the exponential map used for horizontal or vertical strips and this is diagonal so how do I know what angles the boundaries make with the axis after using the exponential map?
     
  11. Mar 3, 2014 #10

    PeroK

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    Okay, so now we need some equations:

    First, let's redefine the strip S as a set of parallel lines

    [tex]S = \lbrace L_a: 0 < a < t \rbrace \ where \ L_a = \lbrace x + iy: -∞ < x < ∞ ; y = -√2x - a \rbrace[/tex]
    And the rays are easy:
    [tex]C = \lbrace R_θ: - \frac{\pi}{6} < θ < \frac{\pi}{3} \rbrace \ where \ R_θ = \lbrace r e^{iθ}: 0 < r < ∞ \rbrace [/tex]
    Finally, we need to map a to θ proportionally:
    [tex]\theta = - \frac{\pi}{6} + \frac{a}{t} \frac{\pi}{2}[/tex]
    And, recall that we are mapping x to r by:
    [tex]r = e^x[/tex]
    Can you put all that together into one mapping from S to C?
     
  12. Mar 3, 2014 #11
    [tex]e^{x+i(\frac{a\pi}{2t}-\frac{\pi}{6})}[/tex]
     
  13. Mar 3, 2014 #12

    PeroK

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    Yes, that'll do. Where a = -√2x - y.
     
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