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Math Help - Finding polar form

  1. #1
    Member
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    Nov 2008
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    Finding polar form

    This makes no sense to me. I definitely think I'm doing it right.

    Convert 3sqrt(3) - 9i into polar form.

    Every time I do it, I get 6sqrt(3)(1/2 - .....) The second part doesn't work.

    What am I doing wrong?


    EDIT: Bonus Question:
    Prove there is no field F such that R is not a subset of F which is not a subset of C. (R is the real numbers and C is the complex numbers, sorry I don't know how to do the math writing on here well yet.)
    Last edited by Janu42; November 23rd 2008 at 06:51 PM.
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  2. #2
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    Hello, Janu42!

    You never did write it in polar form . . .


    Convert 3\sqrt{3} - 9i into polar form.
    The triangle looks like this:
    Code:
                _
              3√3
          * - - - - *
            * θ     |
              *     |9
             r  *   |
                  * |
                    *

    r \:=\:\sqrt{(3\sqrt{3})^2 + 9^2} \;=\;\sqrt{108} \:=\:6\sqrt{3}

    \cos\theta \:=\:\frac{3\sqrt{3}}{6\sqrt{3}} \:=\:\frac{1}{2} \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{3}

    On the graph, we see that the angle is -\frac{\pi}{3}

    Therefore, the polar form is:

    . . 6\sqrt{3}\bigg[\cos\left(\text{-}\tfrac{\pi}{3}\right) + i\sin\left(\text{-}\tfrac{\pi}{3}\right)\bigg] \;=\;6\sqrt{3}\left(\cos\tfrac{\pi}{3} - i\sin\tfrac{\pi}{3}\right)

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