# How do you simplify #root5( -1)#?

##### 1 Answer

It depends!

#### Explanation:

The function

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

#f^(-1)(x) = root(5)(x)#

This is called the real fifth root.

Note that

So the real fifth root gives us

**Complications**

Note however, that

As a result, when you are dealing primarily with complex numbers you will encounter the principal complex fifth root of

#cos(pi/5) + i sin(pi/5) = 1/4(1+sqrt(5)) + 1/4sqrt(10-2sqrt(5)) i#

This is also denoted