Question

How do you simplify `root8( -16)`?

Answer 1

No root is real. 8 complex roots.

Explanation:

That's not simple.

`z^8 = 16 (cos pi + i sin pi)`

`z = sqrt 2 (cos frac{pi + 2k pi}{8} + i sin frac{pi + 2k pi}{8})`

`k in {0,1,2,3,4,5,6,7}`

`{1,3,5,7,9,11,13,15} * pi/8`