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Describe the closure of the set with formulas

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data
    -∏<arg(z)<∏ (z≠0)



    2. Relevant equations
    arg(z) is the angle from y=0


    3. The attempt at a solution

    Arg(z) spans the entire graph since -pi to pi is the full 360 degrees so I put:
    -∏<arg(z)<∏ -->

    0<arg(z)<2∏+k∏, (k ε Z) -->

    arg(z) [itex]\subset[/itex] R -->

    arg(z) = R: all real numbers

    but I don't know if that describes the closure since last time when describing the closure it would be Cl(E)={zεC :-∏<arg(z)<∏}

    Would what I did be showing the closure with formulas? Also it's an open set, so I thought it would be an interior as oppose to a closure.
     
  2. jcsd
  3. Jan 27, 2014 #2

    Dick

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    arg(z) isn't all real numbers. It can't be equal to π. Which number in (-π,π) is equivalent to π as an argument? Try and describe your set as a subset of the complex numbers.
     
  4. Jan 27, 2014 #3
    180 is equivalent to pi. Wouldn't it only not be able to be: z≠0+2k*pi? Why can't it equal pi?
     
  5. Jan 27, 2014 #4

    Dick

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    Because there is no number in (-π,π) that differs from π by a multiple of 2π. Notice the 'less than' signs, not 'less than or equal to'.
     
  6. Jan 27, 2014 #5
    Why would it need to differ from pi by a multiple of 2pi? 2π+π=3π or π and 2π-π=π as well. They are the same thing
     
  7. Jan 27, 2014 #6

    Dick

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    You aren't getting me. -pi and pi aren't in the interval -pi<arg(z)<pi. What point in that interval could be equal to pi if you add 2*pi*k?
     
  8. Jan 27, 2014 #7
    I see what you're saying. It would be all real numbers excluding any number differing from pi by 2pi*k right?
     
  9. Jan 27, 2014 #8

    Dick

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    Yes! So what portion of the complex plane is excluded? Can you describe it in words that don't involve 'arg'?
     
  10. Jan 27, 2014 #9
    Cl(E)={zεC : z=C[STRIKE]ε[/STRIKE]∏+2k∏, k ε Z}

    The closure is equivalent to the entire plane excluding z=pi+2k*pi where k is in integer.
     
  11. Jan 27, 2014 #10

    Dick

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    No, there's a big difference between z=pi and arg(z)=pi. Do you know what arg is? And we haven't even gotten to the closure Cl(E) yet, I'm still trying to get you describe the set E.
     
  12. Jan 27, 2014 #11
    arg(z) is the angle from the real axis to z. So arg(z) inplace of z in the closure? I just started this class this week, so sorry about that.
     
  13. Jan 27, 2014 #12

    Dick

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    That's ok, take it one step at a time. What complex numbers have an angle from the real axis of pi?
     
  14. Jan 27, 2014 #13
    We haven't really covered what arg(z) is equivalent too. I know it's the angle, but we never went over how to find the z value within arg(z). I looked it up and it says it's equivalent to atan (imag(z)/real(z)), but I've never heard of atan, so I'm lost there too. Can't you specify the closure with arg(z) instead of specifically using z?
     
  15. Jan 27, 2014 #14

    Dick

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    You are making this much more complicated than it has to be. Look at http://mathworld.wolfram.com/ArgandDiagram.html Concentrate on the picture. Not the words. What complex numbers have arg(z)=pi?
     
  16. Jan 27, 2014 #15
    From the real axis to pi wouldn't it just be 0+ik to 180+ik where k is any real number?
     
  17. Jan 27, 2014 #16

    Dick

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    Look, 0+ik where k is real is the imaginary axis (x=0). 180+ik is line parallel to the imaginary axis 180 units to the right of it (x=180). I don't think that's what you mean. THINK about the what the symbols you are writing mean, ok? Look at the Argand diagram again.
     
  18. Jan 27, 2014 #17
    From the real axis to the imaginary axis it would be from 0 to 0+90i or 0+pi/2 i? and from the imaginary axis back down to the real axis would it be pi or 180 + 0i?
     
  19. Jan 27, 2014 #18

    Dick

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    Angles aren't imaginary. One definition of arg(z) is the angle measured counterclockwise from the POSITIVE part of the real axis. As the picture was supposed to show. To go from that to the POSITIVE part of the imaginary axis is an angle of +pi/2, counterclockwise. What happens if you rotate by another +pi/2 in the same direction. Where do you wind up? And what is the arg of those numbers?
     
  20. Jan 27, 2014 #19
    pi/2 brings the angle to the imaginary axis and another pi/2 you're asking? It would bring it back down to the real axis at +pi. I don't understand what you're saying about the arg of those numbers. Isn't it just 0 to pi/2 to get to the imaginary axis and another pi/2 to get back down to the positive part of the real axis?
     
  21. Jan 27, 2014 #20

    Dick

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    If you rotate in opposite directions, yes. But you aren't getting the arg thing. Look, if I take a point on the positive real axis and rotate by pi, where do I wind up?
     
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