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dankelly07 · #1 December 4th, 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)