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

1 Answer

Vinícius Ferraz

Jun 14, 2017

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#

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