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

1 Answer
George C.
Aug 13, 2016

#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#

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