What is (0-8i)^1/3?

Question

What is (0-8i)^1/3?

 

I think that the answer is -2i^(1/3)

Answer 1

There are 3 cube roots of -8i.

 

In polar form, -8i = 8(cos270° + isin270°).

 

Using DeMoivre's Theorem, one of the cube roots is 8^1/3[cos(270°/3) + isin(270°/3)] 

 

                                                                         = 2(cos90° + isin90°] = 2i

 

Another cube root is 2[cos(90° + 360°/3) + isin(90° + 360°/3)]

 

                         = 2(cos210° + isin210°) = 2(-√3/2 - (1/2)i) = -√3 - i

 

Third cube root = 2[cos(90° + 2(360°/3)) + isin(90° + 2(360°/3))] 

 

                       = 2(cos330° + isin330°) = 2(√3/2 - (1/2)i)

 

                                                            = √3 - i

 

Cube roots of -8i:  2i, -√3 - i, √3 - i