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Math Help - Complex Numbers: Cartesian to Polar

  1. #1
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    Complex Numbers: Cartesian to Polar

    Hi there,

    If I have a complex number z = 1 + root(3)i

    Then changing it to polar form is z = r cis(11pi/6)

    But my book does mention that the angle must lie between –pi and pi.

    Using logic from circular trigonometry 11pi/6 = -pi/6

    Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

    What does this restriction on the angle represent?
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  2. #2
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    Most likely just restricting the domain of the Cosine function.
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  3. #3
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    Quote Originally Posted by Bushy View Post
    Hi there,

    If I have a complex number z = 1 + root(3)i

    Then changing it to polar form is z = r cis(11pi/6)

    But my book does mention that the angle must lie between –pi and pi.

    Using logic from circular trigonometry 11pi/6 = -pi/6

    Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

    What does this restriction on the angle represent?
    How could \theta = \frac{11\pi}{6} when z is clearly in the first quadrant?


    Remember that r = \sqrt{1^2 + (\sqrt{3})^2} = 2

    and

    \theta = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}.


    Therefore z = 2\,\textrm{cis}\,\frac{\pi}{3}.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    How could \theta = \frac{11\pi}{6} when z is clearly in the first quadrant?

    Sorry I meant to write z = 1-root(3)i

    Then....


    changing it to polar form is z = r cis(11pi/6)

    But my book does mention that the angle must lie between –pi and pi.

    Using logic from circular trigonometry 11pi/6 = -pi/6

    Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

    What does this restriction on the angle represent?
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  5. #5
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    Quote Originally Posted by Bushy View Post
    Sorry I meant to write z = 1-root(3)i

    Then....


    changing it to polar form is z = r cis(11pi/6)

    But my book does mention that the angle must lie between –pi and pi.

    Using logic from circular trigonometry 11pi/6 = -pi/6

    Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

    What does this restriction on the angle represent?
    The reason there is a restriction on the argument is because an INFINITE number of angles actually fit.

    Namely -\frac{\pi}{6} + 2\pi n, where n is an integer representing the number of times you have gone around the unit circle.

    In order to avoid confusion, one had to define the "Principle Argument".

    And it was chosen that the restriction would be -\pi < \theta \leq \pi.

    There is no reason why 0 \leq \theta < 2\pi could not have been chosen, but it wasn't, so we have to deal with what we have been given to use.


    Anyway, to answer your question, there is a restriction on \theta so that you only get one value of \theta instead of a family of values.
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