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Equation of a line in complex plane

  1. May 12, 2013 #1
    All the derivation of the equation of line on complex plane uses the fact that (x,y) \in R^2 can be identified with x+iy \in C.

    Thus, they begin with ax+by+c = 0 then re-write x = (z+\bar{z})/2 and y = (z-\bar{z})/(2i), and substitute it into real plane line equation to get it in complex form.

    What I don't quite understand is, since (x,y) is identified with (x,iy),

    don't we need to write ax+by+c=0 into ax+biy+c=0 before we proceed with the substitution? Why don't we need to do such thing?

    Thank you.
     
  2. jcsd
  3. May 12, 2013 #2

    tiny-tim

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    hi 4everphysics! :smile:
    (x,iy) is not in ℂ :wink:

    ℂ is a set whose elements are single items (traditionally called "z")

    ℂ is not a direct product of two sets, with elements that are ordered pairs
     
  4. May 13, 2013 #3

    HallsofIvy

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    In other words, we are not "identifying (x, y) with (x, iy)". We are identifying the pair of real numbers, (x, y) with the single complex number x+ iy.
     
  5. May 13, 2013 #4

    Bacle2

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    Still, one thing that I think is good to take into account is that a complex line is 2-dimensional as a real object. Notice that the 1st complex projective space is a 2-sphere, but 1st real projective space is a circle.
     
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