Can any one explain me how to solve z^3=8-√3i by the "roots of complex numbers" method?

Question

Very urgent due to a test TOMORROW!

Answer 1

The best way to deal with roots of complex numbers is in polar form
8-sqrt3 i = sqrt(67) < Arctg ( -(sqrt3)/8)=-0.2132 +2kpi rad Its cubic roots are

r=67^1/6 and angles -0.2132/3 +2/3 kpi with k = 0,1,2

So you get three roots and if you wish you can go to binomial form

x= r cos fi y=r sin fi

Answer 2

well this is what u do............car+shop= CALCLATOR