# How do you use demoivre's theorem to simplify 4(1-sqrt3i)^3?

May 24, 2017

$- 32$

#### Explanation:

DeMoivre's theorem for exponents says that any complex number $z$ can be written as $r \left(\cos \theta + i \sin \theta\right)$, or $r \textcolor{w h i t e}{\text{." "cis}} \left(\theta\right)$ for short.

It continues to say that when raising an imaginary number to a certain power, the result is:

${z}^{n} = {r}^{n} \textcolor{w h i t e}{\text{.""cis}} \left(n \theta\right)$

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To simplify this, let's first calculate ${\left(1 - i \sqrt{3}\right)}^{3}$.

We find $r$ by using Pythagoras' theorem:

$r = \sqrt{{1}^{2} + {\sqrt{3}}^{2}} = \sqrt{4} = 2$

We find $\theta$ by taking the inverse tangent of (Im(z))/(Re(z).

$\theta = {\tan}^{-} 1 \left(\frac{- \sqrt{3}}{1}\right) = - \frac{\pi}{3}$

So $\left(1 - i \sqrt{3}\right) = 2 \text{cis} \left(- \frac{\pi}{3}\right)$. Therefore:

${\left(1 - i \sqrt{3}\right)}^{3} = {2}^{3} \text{cis"(3(-pi/3)) = 8"cis} \left(\pi\right)$

Finally, we expand $\text{cis} \left(\pi\right)$.

$8 \text{cis} \left(\pi\right) = 8 \cos \pi + 8 i \sin \pi = 8 \left(- 1\right) + 8 i \left(0\right) = - 8$

And don't forget to multiply by 4!

$- 8 \cdot 4 = - 32$

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So $4 {\left(1 - i \sqrt{3}\right)}^{3} = - 32$.