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

1 Answer
Mar 5, 2018

Answer:

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)#