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Regions of the complex plane - finding the locus

  1. Apr 28, 2012 #1
    Find the locus defined by [tex]|z-2|-|z+2|=3[/tex]

    The given example rewrites the left hand side as:

    [tex]\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2}[/tex]

    1) When they rewrite it as that, they square it and square root it right? Why isn't it squaring the whole expression and rooting it (isn't that the rule?)
    i.e.
    [tex]\sqrt{((x-2)+y)^2}-\sqrt{((x+2)+y)^2}[/tex]

    2) Why does the 'i' disappear?

    Then they move the second expression to the right hand side, square both sides and complete the square for bot sides, square both sides again and they end up with the locus being a hyperbola with asymptotes at y=±√(7)x/3 and x≤-(3/2)

    They don't explain the reason for taking these steps (terse book) - is it just standard procedure for these type of questions? To simplify + complete the square until you get a familiar looking equation?
     
  2. jcsd
  3. Apr 28, 2012 #2


    You should read a little about "complex numbers", "absolute value of complex numbers"

    or "module of a complex number" .It's pretty easy and this will clear out your doubts.

    DonAntonio
     
  4. Apr 28, 2012 #3
    ^ Ok, will do
     
  5. Apr 28, 2012 #4
    From what I understand, the modulus of a complex number is its distance from the origin, [tex]|z| = \sqrt{(x^2+y^2}[/tex], so...for [tex]|z-2| = |(x-2)+yi|[/tex], the modulus bars just denote the distance, which justifies the next part (second equation in my original post)

    Why did they still write the i though? Sorry, I forgot to show that they also wrote [tex]|x+yi-2|-|x+yi+2|=3[/tex]

    And I still don't understand how they figured how all the steps to 'reduce' the equation into one where they could find the locus

    By the way, when I asked why the 'i' disappeared, it was because they'd shown that [tex]|z-2||/tex] was equivalent to |(x-2)+yi| - if they
     
  6. Apr 28, 2012 #5


    But if you read the WHOLE chapter on complex numbers, then it becomes clear that when taking the module one squares the real and

    imaginary parts, sum them and take the square root of it. Not "i" involved...Of course, this is due to the fact that we can

    identify the complex number with real points on the plane xy (or [itex]\mathbb R^2[/itex] , if you will), so the module is, as you

    mention, just the distance from the origin of a point on the plane = a complex number.

    DonAntonio
     
  7. Apr 28, 2012 #6
    So when they change |x+yi-2| to √((x-2)^2+y^2), they're only looking at the points on the plane......if |x+yi-2| already denotes distance, why write 'yi'?Why not just 'y'?

    And I'm just being pedantic for this...the stuff that I can't find any good resources on is finding the locus defined by it
     
  8. Apr 28, 2012 #7

    As you wrote in your first post, and after we understood (??) how things work, you have [tex]\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2}=3[/tex] so passing one of the roots to the other side and squaring one gets [tex](x-2)^2+y^2=9+(x+2)^2+y^2+6\sqrt{(x+2)^2+y^2}[/tex]

    Now some algebraic hokus-pokus and one gets [tex]-8x-9=6\sqrt{(x+2)^2+y^2}\Longrightarrow 64x^2+144x+81=36(x+2)^2+36y^2\Longrightarrow[/tex][tex]28x^2-36y^2-63=0\Longleftrightarrow \frac{x^2}{\left(3/2\right)^2}-\frac{y^2}{\left(\sqrt{7}/2\right)^2}=1[/tex] and so this is a hyperbola

    DonAntonio

    Pd. Check the above calculations. It'd not be that strange if there's some mistake.
     
  9. Apr 28, 2012 #8

    Office_Shredder

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    It's not just pedantics, it's important to understand what the notation is saying so you don't get confused later.

    Let's consider the case when x=1 and y=1.
    [tex] |1+1i-2| = |-1+i| = \sqrt{(-1)^2+1^2} = \sqrt{2}[/tex]
    If we just didn't write in yi and wrote in simply y instead, then we are taking the absolute value of a different number
    [tex] |1+1-2| = |0| = 0[/tex]

    It might seem obvious when the variables are left in that x and y are referring to the real and imaginary parts of a complex number, but if you don't include the i then y is not referring to the imaginary part of a complex number anymore
     
  10. Apr 29, 2012 #9
    @DonAntonio: How do you know to pass one square root to the other side and square? Basically I want to know how you know which algebraic 'hocus-pocus' to do to get find the locus . My book doesn't explain any of these, it just shows the steps without explaining why they took took the steps I did (described in my initial post, and it seems to be the steps you've taken as well)

    @Office_Shredder: So the point of the i there is to just indicate that you're referring to the imaginary part, and that goes when you take the modulus you're taking the magnitude and it's obvious that it's two different parts?
     
  11. Apr 29, 2012 #10

    Mentallic

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    Moving one square root to the other side wasn't necessary. What we want to do is get rid of all the square roots, and that involves squaring of course. But when two surds are present, you'll need to square things twice.

    For example, if we have something of the form

    [tex]\sqrt{a}+\sqrt{b}=c[/tex]

    Then squaring will yield

    [tex]a+2\sqrt{ab}+b=c^2[/tex]

    And as you can see, we still have a surd to deal with. Even if we moved one or both of the terms to the other side at the start and squared, we'd still have surds to deal with.

    Anyway, from this point we will move everything that is not a square root to one side, leaving the square root alone

    [tex]2\sqrt{ab}=c^2-a-b[/tex]

    Now we can square again and finally be rid of the surd.
     
  12. Apr 29, 2012 #11
    ^ So the goal is to get rid of the square roots, and once that's done, you should have some equation for a locus?
     
  13. Apr 29, 2012 #12

    Mentallic

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    Yes, but let's be clear about this. The form [tex]\sqrt{a}+\sqrt{b}=c[/tex] is already an equation of the locus that describes the original question, but we're just trying to convert it into a form that makes more sense to us.
    And that's what DonAntonio has done.
     
  14. May 2, 2012 #13
    Can the forms be any sort of random shape on the plane? Or is it just circles, ellipses, hyperbolas and such?
     
  15. May 2, 2012 #14

    chiro

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    It will depend on the equation you use. Different equations lead to different geometric representations and it's not a specific thing to answer unless you specify the form of the identity you are using (i.e. blah1 = blah2).
     
  16. May 2, 2012 #15
    So they're just like...'normal' equations then? (in the sense that you can have any sort shape)
     
  17. May 3, 2012 #16

    chiro

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    Yeah pretty much. You could end up having an explicit form or an explicit form. Implicit forms in general allow you to have very complex and general kinds of structures, but if you are only dealing with functions involving x and y (even implicit), the output will still be a curve although it may not technically be a function per se.

    You will end up with a line that takes on a trajectory of some sort and if you only have the moduli of complex numbers and functions involving either only a constant functions of x,y or both on the RHS side, then you will always get a line that is deformed in some way in your normal 2D space.

    If you wanted to find the one dimensional form of the line, you would need to parameterize the line to getting x(t) and y(t) where t is your parameter (often from -infinity to +infinity) but this is far from easy for the general case.
     
  18. May 3, 2012 #17
    ^ thanks, I'm still trying to trying to understand what the difference between a parametric equation and a regular (algebraic?) equation is, parameter and variable....will probably be my next question if I can't figure it out
     
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