Question

I need it ASAP. Solve it please!! {z}^{3} - 8i = 0​

Answer 1

Step-by-step explanation:

one of the cube roots (there are 3 of them) is

81/3[cos(270°/3) + isin(270°/3)] = 2(cos90° + isin90°) = 2i

The 3 cube roots are equally spaced around the circle centered at (0,0) with radius 81/3 = 2

360° / 3 = 120°

Another cube root of -8i is

2[cos(90° + 120°) + isin(90° + 120°)] = 2[cos(210°) + isin(210°)] = 2(-√3/2 + i(-1/2)) = -√3 - i

The third cube root of -8i is

2[cos(90°+2(120°)) + isin(90°+2(120°)) = 2[cos330° + isin(330°)] = 2[√3/2 +i(-1/2)] = √3 - i

So, the cube roots of -8i are 2i, -√3 - i, and √3 - i