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Deduce the locus of points of the following equation

  1. Feb 7, 2006 #1
    Hi, I would like to know if there is a quick way to deducing the locus of points of the following, where z is a complex number.

    (i) |z-1| + |z+3| = 4
    (ii) |z+1| - |z-4| = 2

    I know that the first one is an ellipse and the second one is a hyperbola (if I remember correctly anyway) since there is a plus in the first one and a minus in the second one.

    I'm looking for a general method or observation which I can use to quickly deduce whatt he locus is(assuming that such a method exists). For example |z - (a+bi)| < c is a circle with centre (a,b) and radius c.

    Any help would be good thanks.
     
  2. jcsd
  3. Feb 7, 2006 #2

    Galileo

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    Interpret the expression geometrically. You should know that the distance between two complex numbers z and w is |z-w|.

    So the locus of |z-a|=r is the set of points whose distance from a is r, hence a circle of radius r with center a.

    Can you interpret |z-1|+|z+3|=4 and |z+1|-|z-4|=2 in this way?
     
  4. Feb 7, 2006 #3
    For the first one |z-1| + |z+3| = 4. The key points are -3 and 1 on the real axis. For a point z_0 which satisfies the equation, the sum of the distances between z_0 and each of -3 + 0i and 1 + 0i is 4. From that I can really only get a rough sketch. I am trying to find out if I can get the lengths of the major and minor axes just from the equation.
     
  5. Feb 8, 2006 #4

    Galileo

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    Ok, I thought you'd geometrically wanted to see why it was an ellipse, and that's is what you described just now.
    Well, the foci are -3 and 1. The major axis is the line through these points and the center is (-3+1)/2=-1. I guess you can find the length of the semi-minor axis by considering a point on the intersection of the ellipse and the semi-minor axis. The distances from this point to both foci is equal, hence 2, and you can find it by Pythagoras.
     
  6. Feb 8, 2006 #5

    Galileo

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    By the way, if the the major axis lies on the x-axis you can simply translate the ellipse with the center at the origin. Then the equation is in standard form: [itex]x^2/a^2+y^2/b^2=1[/itex] and the length of the semi-axes are a and b.
     
  7. Feb 8, 2006 #6

    benorin

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    The [geometric] definition of an ellipse is the locus of points the sum of whose distances from two fixed points (called foci) is a positive constant: in terms of complex numbers, this is

    [tex]|z-z_1|+|z-z_2|=k[/tex], where z1 and z2 are the foci and k>0.

    This is easy to grasp mentally when considering the two tacks, string, and pencil construction of an ellipse: you fix both ends of a string in place with the tacks (leave some slack when you do so) and draw with the pencil so that the string is kept taunt by the pencil tip throughout tracing out of the ellipse.
     
  8. Feb 8, 2006 #7
    Thanks for the help guys.
     
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