Applying De Moivres theorem..

dankelly07 · #1 $Old$ 12-04-2008, 06:46 AM

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)

mr fantastic · #2 $Old$ 12-04-2008, 07:05 AM

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).