How do you simplify #root8( -16)#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 1 Answer VinÃcius 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# Answer link Related questions How do you simplify radical expressions? How do you simplify radical expressions with fractions? How do you simplify radical expressions with variables? What are radical expressions? How do you simplify #root{3}{-125}#? How do you write # ""^4sqrt(zw)# as a rational exponent? How do you simplify # ""^5sqrt(96)# How do you write # ""^9sqrt(y^3)# as a rational exponent? How do you simplify #sqrt(75a^12b^3c^5)#? How do you simplify #sqrt(50)-sqrt(2)#? See all questions in Simplification of Radical Expressions Impact of this question 1236 views around the world You can reuse this answer Creative Commons License