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Complex numbers

  1. Sep 24, 2007 #1
    The phase of a complex number is z=re[tex]^{i\theta}[/tex]

    This first example is a simple z=1+i, but where does the r come from for this?
     
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  3. Sep 24, 2007 #2

    bel

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    The "r" is the distance from the origin. Thus, [tex]z=re^{i\psi} [/tex]
    [tex]z= r (cos(\psi)+i sin (\psi) ) [/tex].
    Hence, for [tex]z=1+i[/tex], [tex] r= \sqrt{1^2 + 1^2}= \sqrt{2} [/tex].
     
    Last edited: Sep 24, 2007
  4. Sep 24, 2007 #3
    If it was 1+4i

    Would it be:
    [tex] r= \sqrt{1^2 + 4^2} [/tex]?
     
  5. Sep 24, 2007 #4

    learningphysics

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    yeah, that's right.
     
  6. Sep 24, 2007 #5
    Is a reasonable answer for this phase then:
    /sqrt{2}e ^i(theta) ?
     
  7. Sep 24, 2007 #6
    since its just 'i' and no angle is given
     
  8. Sep 24, 2007 #7

    learningphysics

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    wait... phase usually refers to the angle... what exactly does the question ask you to find?

    from your original post:

    "The phase of a complex number is z=re[tex]^{i\theta}[/tex]"

    that doesn't seem right... did you write this out word for word?
     
  9. Sep 24, 2007 #8
    No, phase is just the angle but there doesn't seem to be any angle so.. yeah...
    exact words are: phase is the value of theta when z is expressed as z=re^i(theta).

    Can you find an angle with just 1+i ?
     
  10. Sep 24, 2007 #9

    learningphysics

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    yes, you can. arctan(1/1). Draw the point on the complex plane... the x-coordinate is the real... y-coordinate is the complex part... what angle does the line from the origin to the point make with the positive x-axis...

    That's the angle you need.
     
  11. Sep 24, 2007 #10

    HallsofIvy

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    No, for any complex number, both r and [itex]\theta[/itex] are both given. Your original question was about r only, that's why nothing was said about [itex]\theta[/itex].

    In the "complex plane" or Argand diagram, any complex number x+ iy can be associated with the point (x,y). [itex]r (cos(\theta)+ i sin(\theta))= re^{i\theta}[/itex] is just that point given in polar coordinates. Since [itex]x^2+ y^2= r^2 cos^2(theta)+ r^2 sin^2(\theta)= r^2[/itex], [itex]r= \sqrt{x^2+ y^2}[/itex]. Since [itex]y/x= r^2 sin^2(\theta)/[r^2 cos^2(\theta)= tan(\theta)[/itex], [itex]\theta= arctan(y/x)[/itex].

    In your first example, 1+i, [itex]r= \sqrt{1^2+ 1^2}= \sqrt{2}[/itex] and [itex]\theta= arctan(1/1)= arctan(1)= \pi/4[/itex].

    In your second example, 1+ 4i, [itex]r= \sqrt{1^2+ 4^2}= \sqrt{17}[/itex] and [itex]\theta= arctan(4/1)= arectan(4)= 1.3 radians approximately.

    Am puzzled by your saying "The phase of a complex number is [itex]z=re^{i\theta}[/itex]. Normally the "phase" is given as an angle. I would have thought the "phase" of the number [itex]x+ iy= r e^{i\theta}[/itex] would be just [itex]\theta[/itex].
     
  12. Sep 24, 2007 #11
    ahhh, 1 and 1 is a 45 degrees.
     
  13. Sep 24, 2007 #12
    when looking at a more complicated one like sqrt[2e] ^ -i(pi)/4
    is the entire exponent my imaginery part?
    and the base is real?
     
  14. Sep 24, 2007 #13
    when theyre like this, all hope of visualizing it is gone.
     
  15. Sep 24, 2007 #14

    learningphysics

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    The e is inside the square root?
     
  16. Sep 24, 2007 #15
    hmm
    the books a little iffy..

    I'll say no its not actually.
     
  17. Sep 24, 2007 #16
    They're very easy to visualize, it's just like a point in the plane described in polar coordinates.

    In your last example, sqrt(2) is the distance from the origin to the point in the plane and -pi/4 is the angle (in radians) starting from the real axis (or the x axis in the analogy).
     
  18. Sep 24, 2007 #17

    learningphysics

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    Then it shouldn't be hard to visualize... r = sqrt(2). theta = -pi/4
     
  19. Sep 24, 2007 #18
    lol yeah but its sqrt[2] multiplied by e and raised to -i(pi)/4 ! :P

    Am I correct in saying the exponent is imaginery?
     
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