# use DeMoivre's theorem to simplify ?

## How do you use DeMoivre's theorem to simplify ${\left(- {\left(3\right)}^{0.5} + i\right)}^{\frac{9}{2}}$

Jan 17, 2018

${\left(- \sqrt{3} + i\right)}^{\frac{9}{2}} = 16 - 16 i$

#### Explanation:

DeMoivre's theorem states that if a complex number in polar form is $a + i b = r \left(\cos \theta + i \sin \theta\right)$, then

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

Now we can write ${\left(- {\left(3\right)}^{0.5} + i\right)}^{\frac{9}{2}}$ as

${\left(- \sqrt{3} + i\right)}^{\frac{9}{2}}$

= ${\left\{2 \left(- \frac{\sqrt{3}}{2} + i \frac{1}{2}\right)\right\}}^{\frac{9}{2}}$

= ${\left\{2 \left(\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right)\right\}}^{\frac{9}{2}}$

= ${2}^{\frac{9}{2}} \left(\cos \left(\frac{5 \pi}{6} \times \frac{9}{2}\right) + i \sin \left(\frac{5 \pi}{6} \times \frac{9}{2}\right)\right\}$

= $16 \sqrt{2} \left(\cos \left(\frac{15 \pi}{4}\right) + i \sin \left(\frac{15 \pi}{4}\right)\right)$

= $16 \sqrt{2} \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

= $16 \sqrt{2} \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)$

= $16 - 16 i$