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Someone explain continuity principle

  1. Aug 29, 2012 #1
  2. jcsd
  3. Aug 29, 2012 #2
    You can show this algebraically. Let's take our circle with radius 1.
    Then the red circle has center at (0,0) and has radius 1. The equation for such a circle is
    The blue circle has center at (5,0) and has radius 1. The equation is

    We can now find the points in the intersection of these two circles. We know from the first equation that


    Substituting that in the second equation gets us

    [tex](x-5)^2 + (1 -x^2 )=1[/tex]

    This is an equation that can easily be solved. we get x=5/2. We substitute that in the first equation and get

    and thus

    [tex]y=\pm i\sqrt{21}/2[/tex]

    So the points of intersection are [itex](5/2,i\sqrt{21}/2)[/itex] and [itex](5/2,-i\sqrt{21}/2)[/itex].
  4. Aug 30, 2012 #3


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    But the y values are imaginary numbers while the numbers defining the coordinate system must be real numbers- so to say the circles "intersect" there is generalizing "intersect" a heck of a lot!
  5. Aug 30, 2012 #4
    is it possible to plot the circles with y-axis having the imaginary part and x axis having the real part(on the complex plane)?
  6. Aug 30, 2012 #5


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    [itex]e^{R\theta}[/itex] gives a circle with center at 0 and radius R in the complex plane. You cannot plot an equation like y= f(x) with y and x complex numbers because you would have to have real and complex axes for both x and y- and that requires 4 dimensions.
  7. Sep 2, 2012 #6
    It is my understanding that the intersection does exist, just not in the euclidian plane. So it can be said that the circles intersect without changing the meaning of intersection
  8. Sep 17, 2012 #7


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    this principle in its simplest form says that the equations X^2 = t always have two solutions no matter what t is. if you believe that, then you must also believe the original assertion, as micromass showed.
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