# How do you use DeMoivre's theorem to simplify (-1/2+sqrt3/2i)^3?

How do you use DeMoivre's theorem to simplify (-1/2+sqrt3/2i)^3?

Answer 1

The answer is $= 1$

#### Explanation:

Let $z = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$

We have to write $z$ in trigonometric form in order to apply DeMoivre's theorem

$z = r \left(\cos \theta + i \sin \theta\right)$

$r = 1$

$\cos \theta = - \frac{1}{2}$

$\sin \theta = \frac{\sqrt{3}}{2}$

So, $\theta$ is located in the 2nd quadrant

$\theta = 2 \frac{\pi}{3}$

Therefore, $z = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$

We need ${z}^{3}$

We use DeMoivre 's theorem,
${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

${z}^{3} = {\left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)}^{3}$

$= \cos \left(2 \frac{\pi}{3} \cdot 3\right) + i \sin \left(2 \frac{\pi}{3} \cdot 3\right)$

$= \cos 2 \pi + i \sin 2 \pi$

$= 1$

${\left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)}^{3} = 1$

#### Explanation:

According to DeMoivre's theorem, if $a + i b = r \cos \theta + i r \sin \theta$

then ${\left(a + i b\right)}^{n} = {r}^{n} \left(\cos n \theta + i \sin n \theta\right)$

Writing $- \frac{1}{2} + \frac{\sqrt{3}}{2} i$ in trigonometric form

$- \frac{1}{2} + \frac{\sqrt{3}}{2} i = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$

Hence ${\left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)}^{3} = {1}^{3} \left(\cos \left(3 \times \frac{2 \pi}{3}\right) + i \sin \left(3 \times \frac{2 \pi}{3}\right)\right)$

= $\cos \left(2 \pi\right) + i \sin \left(2 \pi\right)$

= $1 + i 0$

= $1$

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