Question

How do you simplify `root8(-1066)`?

Answer 1

`root(8)(-1066)` is not simplifiable as such, but is expressible in the form `a+bi`:

`root(8)(-1066) =1/2 root(8)(1066)sqrt(2+sqrt(2))+(1/2 root(8)(1066)sqrt(2-sqrt(2))) i`

Explanation:

First note that `1066 = 2*13*41` has no square factors, let alone any higher powers, so this radical does not simplify as such, but we can get it into the form `a+bi`.

I will use:

`cos(pi/8) = 1/2sqrt(2+sqrt(2))`

`sin(pi/8) = 1/2sqrt(2-sqrt(2))`

`(cos theta + i sin theta)^n = (cos n theta + i sin n theta) " "` (de Moivre)

So we find:

`root(8)(-1066) = (1066(cos pi + i sin pi))^(1/8)`

`=root(8)(1066)(cos pi + i sin pi)^(1/8)`

`=root(8)(1066)(cos (pi/8) + i sin (pi/8))`

`=root(8)(1066)cos(pi/8)+(root(8)(1066)sin(pi/8)) i`

`=1/2 root(8)(1066)sqrt(2+sqrt(2))+(1/2 root(8)(1066)sqrt(2-sqrt(2))) i`

Note that this is the principal `8`th root, there are `7` others which may be found by repeatedly multiplying by:

`cos (pi/4) + i sin (pi/4) = sqrt(2)/2 + (sqrt(2)/2)i`