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Math Help - Complex number trig conversion.

  1. #1
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    Complex number trig conversion.

    question asks:

    (1-root 3 i)^6

    I can do this but the confusion is the trig part here.
    we know that tan theta = b/a

    thus tan theta = - root 3
    thus theta = - pi/3

    But in the solution they subtract this from 2pi to get 5pi/3 as theta.
    Why?
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  2. #2
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    In what quadrants is \tan \theta negative?

    Is there a difference between \displaystyle \frac{-\pi}{3} and \displaystyle \frac{5\pi}{3} ?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    In what quadrants is \tan \theta negative?

    Is there a difference between \displaystyle \frac{-\pi}{3} and \displaystyle \frac{5\pi}{3} ?
    Right, they are the same thing.
    okay so one last thing. I got the answer as:

    2^6 = r
    so its r(cos n theta + i sin n theta)
    I get:

    64(cos 2pi + i sin 2pi) or if you use 5pi/3 its 64(cos 10pi + i sin 10pi)

    but sin 2pi is 0, so the answer should just be 64. The solution shows its 64+64i, is that a mistake?
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  4. #4
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    The correct answer is 64, so you are correct.
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