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Complex numbers

  1. May 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Lets find the set of points z in the complex plane which satisfy the condition:

    a)|z-a|=r , r>0 where a is fixed point from the same plane, and r is positive real number.

    b)|z-a|=|z-b|, [itex]a \neq b[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I don't know how to solve this at all. Is z something to do with z=a+bi ? is a=Re(z) ? ?
     
  2. jcsd
  3. May 11, 2008 #2

    malawi_glenn

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    yes, you can start with:

    z=x+yi, where x is coordinate on real axis and y coordinate on imaginary axis.

    Then you have to know how |z'| is defined (z' is an arbitrary complex number)
     
  4. May 11, 2008 #3

    tiny-tim

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    Hi Physicsissuef! :smile:

    Nooo … a is another "complex number", exactly like z. :smile:

    (and so is b)
     
  5. May 11, 2008 #4

    malawi_glenn

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    hehe I must have missread your post Physicsissuef :P
     
  6. May 12, 2008 #5

    Gib Z

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    Take a geometric interpretation of the problem; a is some point on the Argand Plane, |z-a| is the distance between the satisfying points z and the point a, and this distance must be exactly r units. Does this remind you of any shape =] ?

    We can use a geometric approach for the second one as well: "Find all points where the distance from point a is equal to the distance from point b". Good luck.
     
  7. May 12, 2008 #6
    Do you have some diagram which shows |z-a|?
    Lets say z=x+yi and a=c+di then we can represent z and a as points.

    z(x,y)
    a(c,d)

    So |z-a|=[tex]\sqrt{(x-c)^2+(y-d)^2}[/tex], right?
     
  8. May 12, 2008 #7

    Gib Z

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    Well that definitely is true. By doing that, we change back to Cartesian co-ordinates, and now that weve done that : You know that expression is equal to r. Square both sides. Does this look familiar =] ?
     
  9. May 12, 2008 #8

    dx

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    Yes thats right. Now you have an expression for |z-a|, and you know its supposed to be equal to r. If you equate |z-a| to r, you get an equation that must be satisfied by (x,y). Ofcourse, theres more than one (x,y) that satisfies this equation, so you have to figure out how to obtain the whole set of solutions. A good idea is to look at what this equation means geometrically. |z-a| is the distance from z to a in the complex plane. What is the locus of the set of z that are a distance r from a?
     
  10. May 12, 2008 #9
    Ohhh.... Circular is [tex]r^2=(x-p)^2+(y-q)^2[/tex]

    I also got the same...
    [tex]|z-a|=r=\sqrt{(x-c)^2+(y-d)^2}[/tex]

    [tex]r^2=(x-c)^2+(y-d)^2[/tex]

    So it is equation of circular right?
     
  11. May 12, 2008 #10

    Gib Z

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    Yes, it is a circle. Now just the easy questions of what its radius and center is.
     
  12. May 12, 2008 #11
    And what about the second one?
    b)|z-a|=|z-b|,
    [itex]
    a \neq b
    [/itex]

    ?
     
  13. May 12, 2008 #12

    Gib Z

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    Come on, you don't need to be baby stepped through both =] Again, think of them as distances from points a and b. Give it a good try yourself.
     
  14. May 12, 2008 #13
    z(x,y)
    a(c,d)
    b(e,f)
    [tex]|z-a|=\sqrt{(x-c)^2+(y-d)^2}[/tex]
    [tex] |z-b|=\sqrt{(x-e)^2+(y-f)^2}[/tex]
    [tex]\sqrt{(x-c)^2+(y-d)^2}=\sqrt{(x-e)^2+(y-f)^2}
    [/tex]

    Equal distance from one point?

    For example. AB=AD
     
  15. May 12, 2008 #14

    Gib Z

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    Well, yes...But you don't need to set it up like that! From the point of Euclidean geometry, can you tell me the locus of a point equidistant from two points?
     
  16. May 12, 2008 #15

    tiny-tim

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    Hi Physicsissuef! :smile:

    Assuming you're just being slow and careful, and wanting to prove things from the axiomatic definition of the complex plane rather than from the analogy with ordinary plane geometry …

    When you have an equation with a square root on one side (or both sides, of course), just square the equation!
    You'll find lots of things then cancel. :smile:
     
  17. May 12, 2008 #16
    [tex]x=\frac{x_1+x_2}{2}[/tex]

    [tex]y=\frac{y_1+y_2}{2}[/tex]

    But isn't mine correect?
     
  18. May 12, 2008 #17

    Gib Z

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    Yes, that is ONE point that satisfies it =] Ok, connect the two points, and draw the perpendicular bisector, which will go through the midpoint ( the point you just mentioned). Draw some triangles, can you see that every point on this line will be equidistant from the two points as well =] ?

    EDIT: O yes, as tiny-tim said, yours is correct, but it might be easier to understand it this way..
     
  19. May 12, 2008 #18
    You mean like a symmetrical line ? It is same like mine...
     
  20. May 12, 2008 #19

    Gib Z

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    It is called the perpendicular bisector. Find the equation of that line, you should be set.
     
  21. May 12, 2008 #20
    Ok, thank you very much for the help. Thanks for all of you, who helped me...
     
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