# 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
$
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)

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
$
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).