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Math Help - Comlex numbers

  1. #1
    fgg
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    Comlex numbers

    Hi.


    I wonder if anyone know how to solve this math problem?


    z^9 = 1 + i * square root of 3 , answer on the form a + bi


    Should I solve it like this:
    z^9= 2^9(cos (pi/3 *9) + isin (pi/3 *9)) = -512


    or like this:
    r^9(cos 9u + isin9u) = 2(cos 60 + isin 60)
    n=0 zo=2^(1/9)(cos 60/9 + isin 60/9) = 2^(1/9) *(cos( 60/9) + isin (60/9))
    n=1
    .
    .
    .
    n=9

    Which is the right way to solve this problem?
    Tank you.
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  2. #2
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    Quote Originally Posted by fgg View Post
    Hi.


    I wonder if anyone know how to solve this math problem?


    z^9 = 1 + i * square root of 3 , answer on the form a + bi


    Should I solve it like this:
    z^9= 2^9(cos (pi/3 *9) + isin (pi/3 *9)) = -512


    or like this:
    r^9(cos 9u + isin9u) = 2(cos 60 + isin 60)
    n=0 zo=2^(1/9)(cos 60/9 + isin 60/9) = 2^(1/9) *(cos( 60/9) + isin (60/9))
    n=1
    .
    .
    .
    n=9

    Which is the right way to solve this problem?
    Tank you.
    Does giving one root solve a problem with 9?

    \displaystyle w_k=r^{1/n}\left[\cos\left(\frac{\theta+2\pi k}{n}\right)+i\sin\left(\frac{\theta+2\pi k}{n}\right)\right]

    k\in\mathbb{Z}, \ 0\leq k\leq 8, \ \theta=\frac{\pi}{3}, \ n=9, \ r=2
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  3. #3
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    \displaystyle z^9 = 1 + i\sqrt{3}.


    \displaystyle |z^9| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2.


    \displaystyle \arg{(z^9)} = \arctan{\left(\frac{\sqrt{3}}{1}\right)} = \arctan{(\sqrt{3})} = \frac{\pi}{3}.


    So \displaystyle z^9 = 2e^{i\frac{\pi}{3}}

    \displaystyle z = \left(2e^{i\frac{\pi}{3}}\right)^{\frac{1}{9}}

    \displaystyle z = \sqrt[9]{2}\,e^{i\frac{\pi}{27}}.


    This is the first of the nine square roots. The rest are evenly spaced around a circle, so have the same magnitude and are all separated by an angle of \displaystyle \frac{2\pi}{9}.

    So the roots are \displaystyle \sqrt[9]{2}\,e^{i\frac{\pi}{27}}, \sqrt[9]{2}\,e^{i\left(\frac{\pi}{27} + \frac{2\pi}{9}\right)}, \sqrt[9]{2}\,e^{i\left(\frac{\pi}{27} + \frac{4\pi}{9}\right)}, \dots

    Evaluate the rest of the roots, then convert them back to Cartesians.
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  4. #4
    fgg
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    Thank you very much for your help!
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