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Complex analysis - graphing in complex plane

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Graph the following in the complex plane
    {zϵC: (6+i)z + (6-i)zbar + 5 = 0}

    2. Relevant equations

    z=x+iy
    zbar=x-iy

    3. The attempt at a solution

    Substituting the equations gives
    2(6x-y) + 5 = 0
    => y = 6x + (5/2)

    But that's a line in R^2. The imaginary parts canceled. The question asks to graph it in the complex plane. So what will it look like?
     
  2. jcsd
  3. Sep 20, 2009 #2
    I was told that it was all complex numbers of form x + (6x + 5/2)i but i don't understand how that's derived since I got only a line in R^2
     
  4. Sep 20, 2009 #3
    Identify x+iy with the point (x,y) in the plane.

    x+iy = (x,y)

    Your answer is the same as the suggested answer. You found {(x,y) : y=6x +5/2}. The suggested answer is the same thing:
    {x+iy : y=6x + 5/2}, i.e. {x + i(6x+5/2) : x is real}
     
  5. Sep 20, 2009 #4
    Thanks, I get the part about substituting x for y. But what does that look like in the complex plane in terms of where it crosses the real and imaginary axes?
     
  6. Sep 20, 2009 #5
    It's the line y=6x+5/2, just like in elementary algebra. The real axis is the x-axis, and the imaginary axis is the y-axis.
     
  7. Sep 20, 2009 #6
    but then isn't it an Argand diagram with coordinate axes of y and x. isn't that somewhat different than if the coordinate axes were Im and Re? (it is just somewhat confusing since in graphing it like in elementary algebra, the i is implicit otherwise it is as if it's in R^2 !!! strange...)
     
    Last edited: Sep 20, 2009
  8. Sep 20, 2009 #7
    No difference. That's what graphing in the complex plane means.
     
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