# How do you simplify (-1)^(-1/8)?

##### 2 Answers
Dec 28, 2015

Deleted answer.

Dec 28, 2015

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

#### Explanation:

We can use the half angle formulas to find $\cos \left(\frac{\pi}{8}\right)$ and $\sin \left(\frac{\pi}{8}\right)$, which we use later...

${\cos}^{2} \left(\frac{\theta}{2}\right) = \frac{1}{2} \left(1 + \cos \left(\theta\right)\right)$

${\sin}^{2} \left(\frac{\theta}{2}\right) = \frac{1}{2} \left(1 - \cos \left(\theta\right)\right)$

Let $\theta = \frac{\pi}{4}$.

Then $\cos \left(\theta\right) = \frac{\sqrt{2}}{2}$ and we find:

${\cos}^{2} \left(\frac{\pi}{8}\right) = \frac{1}{2} \left(1 + \frac{\sqrt{2}}{2}\right) = \frac{2 + \sqrt{2}}{4}$

${\sin}^{2} \left(\frac{\pi}{8}\right) = \frac{1}{2} \left(1 - \frac{\sqrt{2}}{2}\right) = \frac{2 - \sqrt{2}}{4}$

So, since $\frac{\pi}{8}$ is in Q1 we want the positive square roots:

$\cos \left(\frac{\pi}{8}\right) = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\sin \left(\frac{\pi}{8}\right) = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

Note that: ${e}^{i \pi} = - 1$

So ${\left(- 1\right)}^{\frac{1}{8}} = {\left({e}^{i \pi}\right)}^{\frac{1}{8}} = {e}^{i \frac{\pi}{8}} = \cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)$

There are $7$ other $8$th roots of $- 1$, but this is the principal one.

So we find:

${\left(- 1\right)}^{- \frac{1}{8}}$

$= \frac{1}{- 1} ^ \left(\frac{1}{8}\right)$

$= \frac{1}{\sqrt[8]{- 1}}$

$= \frac{1}{\cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)}$

=(cos(pi/8) - i sin(pi/8))/((cos(pi/8) - i sin(pi/8))(cos(pi/8) + i sin (pi/8))

$= \frac{\cos \left(\frac{\pi}{8}\right) - i \sin \left(\frac{\pi}{8}\right)}{{\cos}^{2} \left(\frac{\pi}{8}\right) + {\sin}^{2} \left(\frac{\pi}{8}\right)}$

$= \cos \left(\frac{\pi}{8}\right) - i \sin \left(\frac{\pi}{8}\right)$

$= \frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{2} i$