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Forgetting my inverse tangent, polar form of compelx number ASTC

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the polar form of 2i − 1


    Finding polar form is easy r(cosx + isinx)

    call the real part a and imaginary part b

    r = sqrt(a+b)

    theta = arctan (-2) = - 63.43

    This is the wrong angle for theta as it's 116.57 (which is 180 - 64.43), and I guess this if form that quadrant thing

    S | A
    ___|___
    |
    T | C


    okay not the best grid, but can anyone refresh my memory as how to get the correct value of theta?

    Thanks :)
     
  2. jcsd
  3. Feb 17, 2010 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi thomas49th! :smile:

    (have a theta: θ and a square-root: √ :wink:)

    (and you meant √(a2 + b2))


    Yes, your table shows what's positive (sin, tan, cos, or all) …

    (it works because cos*tan = sin, so an even number must be negative, so either 1 or all 3 must be positive :wink:)

    so arctan(-2) can be either top left or bottom right, you can't tell which by looking at it

    you have to go back to the original 2i - 1, which is at (-1,2), so it's obviously top left! :smile:
     
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