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Set of points z=x+iy

  1. May 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the set of points z=x+iy from the complex plane which satisfy this conditions:

    a)Re(z)[itex]\leq[/itex]0

    b)|z|=3

    2. Relevant equations



    3. The attempt at a solution

    I've solved the first one...

    Re(z)=x
    so x[itex]\leq[/itex]0

    x[itex]\in[/itex](-infinity,0]

    But for the second one? Should I find x, y or should I find z?
     
  2. jcsd
  3. May 12, 2008 #2

    Gib Z

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    B) is just a circle of radius 3, so it would be the same values that satisfy x^2 + y^2 = 9. Although I don't think you can simplify the set of points much further than |z| = 3 anyway lol.
     
  4. May 12, 2008 #3

    Redbelly98

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    (b) has the same form as a problem you've solved already:

    |z-a|=r
     
  5. May 12, 2008 #4
    So what will be the set of points? :smile:

    (-infinity,+infinity) ?
     
  6. May 12, 2008 #5

    Redbelly98

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    The set of points is what Gib Z said it is.
     
  7. May 12, 2008 #6
    But in the task some is asked for some set of z points...

    Is that set [itex]x^2+y^2=9[/tex]?
     
  8. May 12, 2008 #7

    Defennder

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    Yes that is correct.
     
  9. May 12, 2008 #8
    And what about Re(z)[itex]\geq[/itex]0 , Im(z)[itex]\geq[/itex]0, |z|[itex]\leq[/itex]2 ?

    for Re(z), x [itex]\in (0,+\propto)[/itex]

    for Im(z) y [itex]\in (0,+\propto)[/itex]

    and for |z|[itex]\leq[/itex]2

    [itex]x^2+y^2[/itex]=4

    What to do next?
     
    Last edited: May 12, 2008
  10. May 12, 2008 #9

    HallsofIvy

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    Yes, which means the set of points is the right half plane (including the line y-axis).

    Yes, which means the set of points is the upper half plane (including the x-axis)

    No, you have forgotten the "<". The set of points is the disk with center at (0,0) and radius 2, including the boundary.

    To do what?
     
  11. May 12, 2008 #10
    What are the set of points z for that conditions?

    Is it x,y [itex]\in (0,2)[/itex]?

    Since |z|[itex]\leq[/itex]4
     
  12. May 12, 2008 #11

    Redbelly98

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    Physicsissuef,

    You're asking a lot of similar questions that people are helping you with, but you don't really seem to understand the answers that people are providing or helping you find.

    If you're satisfied with the earlier solution
    [itex]
    x^2+y^2=9
    [/itex]

    then what is troubling you about
    [itex]
    x^2+y^2 \leq 4
    [/itex]

    Can you say what format your teacher or professor is expecting the solution to have?
     
  13. May 12, 2008 #12
    Are the solutions:

    x,y [itex]
    \in (0,2)
    [/itex]

    ?
    I understand the answers... Just I need confirmation, that's alll....
     
  14. May 12, 2008 #13

    HallsofIvy

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    Very well then! Every 'answer' you have given is wrong. And since you have been told what the correct answers are, and say that you understand them, that mysifies me!
     
  15. May 12, 2008 #14
    :smile: Ok. I don't understand something.

    Can somebody please exactly give me what is the set of z points?

    Is it [itex]x^2+y^2 \leq 4[/itex] ?

    Is it x [itex]
    \in (0,+\propto)
    [/itex]

    Is it
    [itex]
    \in (0,+\propto)
    [/itex]
     
    Last edited: May 12, 2008
  16. May 12, 2008 #15

    dx

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    None of those are correct. Are you aware of the interpretation of complex numbers as points in the complex plane?
     
  17. May 12, 2008 #16

    dx

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    For part a, the set of points is the left half of the complex plane, including the imaginary axis. For part b, the set of points is a circle of radius 3 around the origin.

    Do you understand why?
     
  18. May 13, 2008 #17
    So I should present the results in circular, right?
     
  19. May 13, 2008 #18

    Redbelly98

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    I don't understand this question.
     
  20. May 13, 2008 #19
    I asked my professor and he told me that I should draw a circle and present the results...
     
  21. May 13, 2008 #20

    dx

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    Yes, the solution to part b is a circle in the complex plane. But the question is, do you understand why it is a circle?
     
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