Question 1032727
{{{x^3 = -i = cos(3*pi/2 + 2k*pi) + i sin(3*pi/2 +2k*pi)}}}

==> {{{x = (-i)^(1/3) = (cos(3*pi/2 + 2k*pi) +  isin(3*pi/2 +2k*pi))^(1/3)}}} for k = 0, 1, 2, 3, 4,...

Choose k = 0, 1, and 2 to find the three distinct cube roots, and apply de Moivre's theorem


==>  {{{x = cos(pi/2 + 2k*pi/3) +  isin(pi/2 +2k*pi/3)}}}.
==> {{{x[1] = cos(pi/2) +  isin(pi/2) = i}}},

{{{x[2] = cos(7pi/6) +  isin(7pi/6) = -sqrt(3)/2 - i/2}}}, and 

{{{x[3] = cos(11pi/6) +  isin(11pi/2) = sqrt(3)/2 - i/2}}}.