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Thread: Applying De Moivres theorem..

  1. #1
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    Applying De Moivres theorem..

    CAn anyone help? I can get to the point where de moivres theorem comes in but I dont understand how to do this...

    So the question is;
    Express
    $\displaystyle
    1 - i\sqrt 3
    $ in the form
    $\displaystyle
    re^{i\theta }
    $

    and then express $\displaystyle
    (1 - i\sqrt 3 )^{10}
    $
    in the form a + ib



    SO i can do this part
    $\displaystyle
    \begin{gathered}
    = (\sqrt {1^2 } + ( - \sqrt 3 )^2 \hfill \\
    = \sqrt 4 \hfill \\
    = 2 \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    \hfill \\
    \theta = \tan ^{ - 1} \left( {\frac{y}
    {x}} \right) \hfill \\
    = \tan ^{ - 1} \theta \left( { - \frac{{\sqrt 3 }}
    {1}} \right) \hfill \\
    = - \frac{\pi }
    {3} \hfill \\
    \end{gathered}
    $
    $\displaystyle
    \begin{gathered}
    so \hfill \\
    1 - i\sqrt 3 = 2e^{ - i\frac{\pi }
    {3}} \hfill \\
    \end{gathered}
    $

    SO BY DE MOIVRES THEOREM

    $\displaystyle
    \begin{gathered}
    \left( {1 - i\sqrt 3 } \right)^{10} \hfill \\
    = \left( {2e^{ - i\frac{\pi }
    {3}} } \right)^{10} \hfill \\
    = 2^{10} e^{ - 10i\frac{\pi }
    {3}} \hfill \\
    \end{gathered}
    $

    This is where I can get up to, but what needs to happen here??
    in my notes this happens..but I dont undertand it...


    $\displaystyle
    = 2^{10} e^{4i\pi - 10i\frac{\pi }
    {3}}
    $

    $\displaystyle
    2^{10} e^{2i\frac{\pi }
    {3}}
    $


    Then you plug into the formula r(cos(theta)+isin(theta)
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  2. #2
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    Quote Originally Posted by dankelly07 View Post
    CAn anyone help? I can get to the point where de moivres theorem comes in but I dont understand how to do this...

    So the question is;
    Express
    $\displaystyle
    1 - i\sqrt 3
    $ in the form
    $\displaystyle
    re^{i\theta }
    $

    and then express $\displaystyle
    (1 - i\sqrt 3 )^{10}
    $
    in the form a + ib



    SO i can do this part
    $\displaystyle
    \begin{gathered}
    = (\sqrt {1^2 } + ( - \sqrt 3 )^2 \hfill \\
    = \sqrt 4 \hfill \\
    = 2 \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    \hfill \\
    \theta = \tan ^{ - 1} \left( {\frac{y}
    {x}} \right) \hfill \\
    = \tan ^{ - 1} \theta \left( { - \frac{{\sqrt 3 }}
    {1}} \right) \hfill \\
    = - \frac{\pi }
    {3} \hfill \\
    \end{gathered}
    $
    $\displaystyle
    \begin{gathered}
    so \hfill \\
    1 - i\sqrt 3 = 2e^{ - i\frac{\pi }
    {3}} \hfill \\
    \end{gathered}
    $

    SO BY DE MOIVRES THEOREM

    $\displaystyle
    \begin{gathered}
    \left( {1 - i\sqrt 3 } \right)^{10} \hfill \\
    = \left( {2e^{ - i\frac{\pi }
    {3}} } \right)^{10} \hfill \\
    = 2^{10} e^{ - 10i\frac{\pi }
    {3}} \hfill \\
    \end{gathered}
    $

    This is where I can get up to, but what needs to happen here??
    in my notes this happens..but I dont undertand it...


    $\displaystyle
    = 2^{10} e^{4i\pi - 10i\frac{\pi }
    {3}}
    $

    $\displaystyle
    2^{10} e^{2i\frac{\pi }
    {3}}
    $


    Then you plug into the formula r(cos(theta)+isin(theta)
    $\displaystyle e^{-\frac{10 \pi}{3} i} = e^{-\frac{12 \pi}{3} i} \cdot e^{\frac{2 \pi}{3} i} = e^{-4 \pi i} \cdot e^{\frac{2 \pi}{3} i} = 1 \cdot e^{\frac{2 \pi}{3} i} = e^{\frac{2 \pi}{3} i}$.

    You're expected to know that $\displaystyle e^{2 n \pi i} = 1$ for all integer values of n (convert this into cis form and it's obvious).
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