# How do you simplify root5( -1)?

Mar 5, 2018

It depends!

#### Explanation:

The function $f \left(x\right) = {x}^{5}$ is continuous and strictly monotonically increasing from $\mathbb{R}$ onto $\mathbb{R}$.

As a result it is one to one with a well defined inverse that is also continuous, strictly monotonically increasing and one to one from $\mathbb{R}$ onto $\mathbb{R}$:

${f}^{- 1} \left(x\right) = \sqrt[5]{x}$

This is called the real fifth root.

Note that ${\left(- 1\right)}^{5} = - 1$

So the real fifth root gives us $\sqrt[5]{- 1} = - 1$

Complications

Note however, that $f \left(x\right) = {x}^{5}$ is not one to one as a complex valued function of complex numbers.

As a result, when you are dealing primarily with complex numbers you will encounter the principal complex fifth root of $- 1$, which is not $- 1$ but:

$\cos \left(\frac{\pi}{5}\right) + i \sin \left(\frac{\pi}{5}\right) = \frac{1}{4} \left(1 + \sqrt{5}\right) + \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} i$

This is also denoted $\sqrt[5]{- 1}$