Thread: Applying De Moivres theorem..
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Old December 4th, 2008, 07:05 AM
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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
1 - i\sqrt 3 in the form
re^{i\theta }

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



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

\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}
\begin{gathered}so \hfill \\1 - i\sqrt 3 = 2e^{ - i\frac{\pi }{3}} \hfill \\ \end{gathered}

SO BY DE MOIVRES THEOREM

\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...


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

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


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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