Trigonometry - Evaluating cube root

Question

Hi peeps, how ya'll doing?

I'm having a hard time to understand the following questions, would someone please explain them to me?

Thank you so much in advance!

1. Evaluate the cube root of z = 27cis(240°). Then raise them to the cube.
2. Evaluate [ 5√3 (√3/2 + i/2) ]¹⁰
3. Find z1/z2 in in polar form. Z1 = 21cis(135°) &  Z2 = 3cis(75°)

Answer 1

1. z = 27cis(240°) = 27cis(240° + 360°n); z^1/3 = 3(cis(80° + 120°n);

n = 0, z₁^1/3 = 3cis80°; n = 1, z₂^1/3 = 3cis200°; n = 2, z₃^1/3 = 3cis320°. This is 3 values of cube root of z.

(z₁^1/3)³ = 27cis240°); (z₂^1/3)³ = 27cis(600°) = 27cis(240°); (z₃^1/3)³ = 27cis(960°) = 27cis(240°)

2. [ 5√3 (√3/2 + i/2) ]¹⁰= 5¹⁰·3⁵[cis(30°)]¹⁰ = 5¹⁰3⁵cis300°= 5¹⁰3⁵(1/2 - i√3/2)

3. z_1/z₂ = 7cis(60°)