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Graph of a Complex Number (basic concept)

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Sketch [tex]0 \le arg z \le \frac{\pi}{4}[/tex] [tex](z \not= 0)[/tex]


    3. The attempt at a solution

    I know from my book that his is a punctured disk aka deleted neighborhood only because it says so and because it is in the form of [tex]0 < \mid z - z_0 \mid < \epsilon[/tex]. I honestly have no clue how to graph this or even visual how it is a disk with a hole in it. Anyone care to mention or explain anything that will help me understand this concept and eventually be able to graph it. Thank you.
     
  2. jcsd
  3. Sep 11, 2009 #2

    Dick

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    It's not a punctured disk. If you can explain to me what you think arg(z) is, I think I can explain what the graph looks like. If you can explain that clearly, you probably won't need me to explain what the graph looks like.
     
  4. Sep 11, 2009 #3
    honestly I always think of it as the angle in radians you get when you graph a complex number z = x + iy but this may be wrong =[
     
  5. Sep 11, 2009 #4

    Dick

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    No, that's exactly right. So the region you are graphing should look more like a wedge, shouldn't it?
     
  6. Sep 12, 2009 #5
    I do not even know how to begin graphing this so I honestly do not see how it looks like a wedge...sigh.

    All I can visualize is a vector z = x + iy with theta = pi/4 but this is cant be right and 0 must be involved somehow
     
    Last edited: Sep 12, 2009
  7. Sep 12, 2009 #6

    Dick

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    Sigh. The set of points where arg(z)=0 is the positive x-axis, right? The set of points where arg(z)=pi/4 is a ray at 45 degrees to the x-axis, right? What you want are the points whose angles are in between. I would call that a "wedge". What would you call it?
     
  8. Sep 12, 2009 #7
    Wow I feel embarrassed how straight forward it is yet I could not grasp it by myself LOL, and I would call it Sergeant Dick Amazing (lol just kidding) and thank you for clearing this up for me kind sir.
     
  9. Sep 12, 2009 #8

    Dick

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    You'll do better next time, right? Not all 'complex' problems are hard.
     
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