Question

How do you find the three cube roots of `-2-2i \sqrt{3}`?

Answer 1

Let `z=re^{i theta}` be the three roots.

`z^3=-2-2isqrt{3}=4(-1/2-sqrt{3}/2i)`

by rewriting in exponential form,

`Rightarrow r^3e^{i(3theta)}=4e^{i({4pi}/3+2npi)}`,

where `n` is any integer,

`Rightarrow {(r^3=4 Rightarrow r=root(3){4}),(3theta={4pi}/3+2npi Rightarrow theta={4pi}/9+{2npi}/3):}`

Hence, the three roots are

`z={ root(3){4}e^{i{4pi}/9},root(3){4}e^{i{10pi}/9},root(3){4}e^{i{16pi}/9}}`.

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I hope that this was helpful.