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Homework Help: What is the polar form of the complex number 3-4i

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data
    What is the polar form of the complex number 3-4i?


    2. Relevant equations

    z=r*cos(theta)+i*r*sin(theta)

    3. The attempt at a solution
    5(cos(arctan(-4/3))-i*sin(arctan(-4/3)))

    This is what I thought the correct answer would be, but it was a multiple choice quiz and this was not one of the options. The correct answer, according to the quiz, was 5(3/5-i*4/5). I don't see how that is in polar form, since the cos and sin of theta were both evaulated. If that is actually correct, then couldn't you simply make any standard complex number polar form by factoring out r?
     
  2. jcsd
  3. Jan 22, 2010 #2

    Mark44

    Staff: Mentor

    For the angle theta in question, the triangle is a 3-4-5 right triangle. cos(theta) = 3/5 and sin(theta) = -4/5.
     
  4. Jan 22, 2010 #3

    LCKurtz

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    Except for the arithmetic mistake already pointed out, I agree with you. Some authors might call the polar form [itex]r\, cis(\theta)[/itex] or [itex]re^{i\theta}[/itex] which would make the argument even stronger that their "best" answer isn't appropriate.
     
  5. Jan 22, 2010 #4
    I guess my question is, why is it still considered polar form if you carry out the evaluation of cos and sin, thus removing theta? Why wouldn't any polar form then just be r(a/r+ib/r). I guess I don't understand what is technically considered polar form.
     
  6. Jan 22, 2010 #5

    LCKurtz

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    It isn't. The answers to that test question were incorrectly presented.
     
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