What are the values of all the cube roots (Z = -4v3 - 4i)?

Question

I have to remember to convert -4√3 – 4i) in to polar form

Answer 1

z= -4sqr3 -4i

find 3 cubic roots

rectangular coordinates are (-4sqr3, 4i) = (2,270) in polar form = (2,7pi/6)

z^(1/3) =(-4sqr3-4i)^(1/3)

=8^1/3(cos(210/3) +isin(210/3) = 2(cos70+isin70)

one cube root of z = about .684+1.879i

= about .7 +1.9i

another is 2(cos10+isin10)

=about 2(.985+.174i)

= 2+.3i

a 3rd is 2(cis130)

= about -1.3 +1.5i

check the 1st cube root:

z^(1/3) = a +bi

z = (a+bi)^3 = a^3 +3a^2bi -3ab^2 -b^3

z= (a^3-ab^2) +(a^2b-b^3)i

a^3-3ab^2 =-4sqr3

3a^2b -b^3 = -4

(.684)^3 -3(.684)(1.879)^2

=.32-7.245

=- 6.9

=-4sqr3 =-6.9

Answer 2

r =(((-4)(3)^(1/2))^2  + ((-4)^2))^(1/2) = 8

θ = arctan (-4/(-4(3)^(1/2) = 7∏/6.

roots = 2 (cos (7∏/18 + 12n∏/18) = i sin(7∏/18 + 12n∏/18), n = 0, 1, 2

Answer 3

z = 8 cis (2πk-5π/6) for k = 1,2,3.

z^1/3 = 2 cis (2πk/3 - 5π/18) for k = 1,2,3.

      = 2 cos(π/18) + 2i sin(π/18)

       or 2 cos(7π/18) + 2i sin(7π/18)

       or 2 cos(13π/18) + 2i sin(13π/18)