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

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Answer 1 · 2017-06-14T15:34:34

No root is real. 8 complex roots.

That's not simple.

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

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

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

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

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