# Thread: Applying De Moivres theorem..

1. ## 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)

2. Originally Posted by dankelly07
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).