Question

What is the value of `root5 -1`?

Answer 1

see below

Explanation:

`root(5)(-1)=(-1)^(1/5)`

Since,
`e^(ipi),e^(3ipi),e^(5ipi),e^(-ipi),e^(-3ipi)=-1`

`=>root(5)(-1)=(e^(ipi))^(1/5)=e^((ipi)/5)=cos(pi/5)+i*sin(pi/5)`

Similarly,

`=>root(5)(-1)=e^(ipi/5),e^(3ipi/5),e^(ipi),e^(-ipi/5),e^(-3ipi/5)`

`=>root(5)(-1)=cos(pi/5)+i*sin(pi/5), cos((3pi)/5)+i*sin((3pi)/5), cos(pi)+i*sin(pi), cos(-pi/5)+i*sin(-pi/5), cos((-3pi)/5)+i*sin((-3pi)/5)`

`=>root(5)(-1)=(0.80902+-0.58779i), (-1), (-0.30902+-0.95106i)`

Answer 2

`root5(-1) = -1`

Explanation:

Consider other roots of `1`

`sqrt1 = 1" "rArr1^2 =1`

`root3(-1) = -1" "rArr (-1)^3 = (-1)(-1)(-1) = -1`

`root5(-1) = -1`

`[(-1)^5 = (-1)(-1)(-1)(-1)(-1)=-1]`

If the radicand is negative, the root must have been a negative number, raised to an odd power.

Answer 3

It depends...

Explanation:

The expression `root(5)(-1)` can mean the real fifth root or the principal primitive complex fifth root of `-1`.

As a real valued function of reals, `x^5` is strictly monotonically increasing and continuous, with domain and range the whole of `RR`. As a result, the real fifth root of any real number is uniquely defined. In the case of `-1`, we find:

`(-1)^5 = -1`

and hence:

`root(5)(-1) = -1`

As a complex valued function of complex numbers, `x^5` is many to one, with exactly `5` solutions to `x^5 = -1`, namely:

`e^(pi/5i) = 1/4(1+sqrt(5))+1/4sqrt(10-2sqrt(5))i`

`e^((3pi)/5i) = 1/4(1-sqrt(5))+1/4sqrt(10+2sqrt(5))i`

`e^(pii) = -1`

`e^((7pi)/5i) = 1/4(1-sqrt(5))-1/4sqrt(10+2sqrt(5))i`

`e^((9pi)/5i) = 1/4(1+sqrt(5))-1/4sqrt(10-2sqrt(5))i`

The first of these can be considered the principal complex fifth root of `-1`.