# How do you simplify  root8(-1066)?

Aug 13, 2016

$\sqrt[8]{- 1066}$ is not simplifiable as such, but is expressible in the form $a + b i$:

$\sqrt[8]{- 1066} = \frac{1}{2} \sqrt[8]{1066} \sqrt{2 + \sqrt{2}} + \left(\frac{1}{2} \sqrt[8]{1066} \sqrt{2 - \sqrt{2}}\right) i$

#### Explanation:

First note that $1066 = 2 \cdot 13 \cdot 41$ has no square factors, let alone any higher powers, so this radical does not simplify as such, but we can get it into the form $a + b i$.

I will use:

$\cos \left(\frac{\pi}{8}\right) = \frac{1}{2} \sqrt{2 + \sqrt{2}}$

$\sin \left(\frac{\pi}{8}\right) = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

${\left(\cos \theta + i \sin \theta\right)}^{n} = \left(\cos n \theta + i \sin n \theta\right) \text{ }$ (de Moivre)

So we find:

$\sqrt[8]{- 1066} = {\left(1066 \left(\cos \pi + i \sin \pi\right)\right)}^{\frac{1}{8}}$

$= \sqrt[8]{1066} {\left(\cos \pi + i \sin \pi\right)}^{\frac{1}{8}}$

$= \sqrt[8]{1066} \left(\cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)\right)$

$= \sqrt[8]{1066} \cos \left(\frac{\pi}{8}\right) + \left(\sqrt[8]{1066} \sin \left(\frac{\pi}{8}\right)\right) i$

$= \frac{1}{2} \sqrt[8]{1066} \sqrt{2 + \sqrt{2}} + \left(\frac{1}{2} \sqrt[8]{1066} \sqrt{2 - \sqrt{2}}\right) i$

Note that this is the principal $8$th root, there are $7$ others which may be found by repeatedly multiplying by:

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