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Complex numbers in polar form

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute the 4th roots of -16 in both Cartesian and polar form and plot their positions in the complex plane.


    2. Relevant equations
    z^1/n=(r^1/n)(e^i(theta)/n), (r^1/n)(e^i(theta)/n)(e^i2(pi)/n........


    3. The attempt at a solution
    How do I find the value of r, and theta??
     
  2. jcsd
  3. Aug 19, 2009 #2

    Cyosis

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    Draw z=-16 in the complex plane. The distance from the origin to -16 in the complex plane is r and the angle between the positive real axis and the negative real axis rotating counter clock wise is [itex]\theta[/itex].
     
  4. Aug 19, 2009 #3
    How do I draw -16 in the complex plane, when I don't know r or theta?
     
  5. Aug 19, 2009 #4

    Cyosis

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    Draw the the complex plane and put a dot where -16 is. Then calculate the distance and angle.
     
  6. Aug 19, 2009 #5
    where is -16?
     
  7. Aug 19, 2009 #6

    Cyosis

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    Do you know where -16 is on the line of real numbers?
     
  8. Aug 19, 2009 #7
    are you saying that the argument is zero and that the modulus is 16?
     
  9. Aug 19, 2009 #8

    Cyosis

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    The modulus is 16, but the argument is not 0. If the argument was 0 -16 would be placed on the positive real axis, which it clearly isn't.
     
  10. Aug 19, 2009 #9
    okay so you think the argument in pi
     
  11. Aug 19, 2009 #10
    that's not right
     
  12. Aug 19, 2009 #11
    why are you wasting my time?
     
  13. Aug 19, 2009 #12

    Cyosis

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    Wasting your time? Why would that not be right? You may want to provide some arguments to why this is wrong.

    Either way I can tell you that I am not wrong. Perhaps review the the relevant equation you posted before jumping the gun?
     
    Last edited: Aug 19, 2009
  14. Aug 19, 2009 #13
    Beacause the answer is apparently 2e^i((pi + 2kpi)/4) where K=0,1,2,3

    that's why it wouldn't be right.
     
  15. Aug 19, 2009 #14
    So ...take one of these numbers (say the k=0 one), convert it to Cartesian form, and take its 4th power. You can then check for yourself whether it is right.
     
  16. Aug 19, 2009 #15

    Cyosis

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    It is obvious that every multiple of 2pi added to the original argument will return you to that exact same spot, after all a circle is exactly 2pi radians.
     
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