How do you simplify #root5( -1)#?
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.
So the real fifth root gives us
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