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Math Help - 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
    <br />
1 - i\sqrt 3 <br />
in the form
     <br />
re^{i\theta } <br />

    and then express <br />
(1 - i\sqrt 3 )^{10} <br />
    in the form a + ib



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

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

    SO BY DE MOIVRES THEOREM

    <br />
\begin{gathered}<br />
  \left( {1 - i\sqrt 3 } \right)^{10}  \hfill \\<br />
   = \left( {2e^{ - i\frac{\pi }<br />
{3}} } \right)^{10}  \hfill \\<br />
   = 2^{10} e^{ - 10i\frac{\pi }<br />
{3}}  \hfill \\ <br />
\end{gathered} <br />

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


    <br />
 = 2^{10} e^{4i\pi  - 10i\frac{\pi }<br />
{3}} <br />

    <br />
2^{10} e^{2i\frac{\pi }<br />
{3}} <br />


    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
    <br />
1 - i\sqrt 3 <br />
in the form
     <br />
re^{i\theta } <br />

    and then express <br />
(1 - i\sqrt 3 )^{10} <br />
    in the form a + ib



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

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

    SO BY DE MOIVRES THEOREM

    <br />
\begin{gathered}<br />
\left( {1 - i\sqrt 3 } \right)^{10} \hfill \\<br />
= \left( {2e^{ - i\frac{\pi }<br />
{3}} } \right)^{10} \hfill \\<br />
= 2^{10} e^{ - 10i\frac{\pi }<br />
{3}} \hfill \\ <br />
\end{gathered} <br />

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


    <br />
= 2^{10} e^{4i\pi - 10i\frac{\pi }<br />
{3}} <br />

    <br />
2^{10} e^{2i\frac{\pi }<br />
{3}} <br />


    Then you plug into the formula r(cos(theta)+isin(theta)
    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 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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