Question

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

Answer 1

It depends!

Explanation:

The function `f(x) = x^5` is continuous and strictly monotonically increasing from `RR` onto `RR`.

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 `RR` onto `RR`:

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

This is called the real fifth root.

Note that `(-1)^5 = -1`

So the real fifth root gives us `root(5)(-1) = -1`

Complications

Note however, that `f(x) = 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(pi/5) + i sin(pi/5) = 1/4(1+sqrt(5)) + 1/4sqrt(10-2sqrt(5)) i`

This is also denoted `root(5)(-1)`