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Pretty dumb question involving complex numbers

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm asked to describe geometrically the set of points in the complex plane describing some equations. I got them all right except this one:

    |z+1| + |z-1| = 8

    2. Relevant equations

    |z| = sqrt( x2 + y2 )

    3. The attempt at a solution

    Well, I know that an equation of the |z+1| = 8 type would be a circle centered at (-1,0) with a radius of 8. Such a circle has the following equation:

    (x+1)2 + y2 = 82

    I started from there to write my equation for |z+1| + |z-1| = 8 using x's and y's.


    (x+1)2 + y2 + (x-1)2 + y2 = 82

    x2 - 2x + 1 + x2 +2x +1 + 2y2 = 82

    2x2 + 2y2 = 14


    I end up with the equation of a circle, but the solution manual says the solution is an ellipse
    of foci (-1,0), (1,0), semi-major axis = 4. What did I do wrong?
     
  2. jcsd
  3. Mar 10, 2012 #2

    SammyS

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    Start with the basics.

    [itex]\left|z\right|=\sqrt{z\bar{z}}=\sqrt{(x+iy)(x-iy)}=\sqrt{x^2+y^2}[/itex]

    [itex]|z+1|=\sqrt{(x+1)^2+y^2}[/itex]

    [itex]|z-1|=\underline{\ \ ?\ \ }[/itex]
     
  4. Mar 10, 2012 #3
    |z-1| = sqrt ((x-1)2 + y2 ), no?
     
  5. Mar 10, 2012 #4
    Well essentially you have two points given to you that are the foci of the ellipse.

    What is the equation telling you in terms of length and the two points?
     
  6. Mar 10, 2012 #5
    Well when I did the problem I didn't intuitively notice that the equation would give me an ellipse, that's why I developped it the way I did. I'm trying to figure out what I did wrong, since I got the equation of a circle.

    Why isn't |z+1| + |z-1| = 8

    equal to

    sqrt [(x+1)2 + y2] + sqrt[(x-1)2 + y2] = 8

    (x+1)2 + y2 + (x-1)2 + y2 = 82

    ??
     
  7. Mar 11, 2012 #6
    Exactly! You forgot the square root. That's what you did wrong.
    If you notice, putting in the square root makes it look like the locus of points having the sum of distances from two fixed points equal to constant.
    And that is what the definition of an ellipse is!
     
  8. Mar 11, 2012 #7
    I didn't forget it, I simply started at the step right after it in my initial post. Notice the squared 8, I squared the equation to remove the square roots. If you simplify the (x+1)2 and (x-1)2 terms, you end up with 2x2 + 2, so the equation becomes

    2x2 + 2y2 = 82 - 2

    x2 + y2 = 7
     
  9. Mar 11, 2012 #8

    SammyS

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    (a+b)2 ≠ a + b !!!!
     
  10. Mar 11, 2012 #9
    Oh god I'm an idiot, thanks everyone. I guess I shouldn't be doing maths at 1:30am when sleep deprived :P
     
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