How do you find cube roots of imaginary numbers?

Answer 1

For a pure imaginary number: {i = sqrt(-1)} times a real coefficient {r}, you have i*r. The cube_root(i*r) = cube_root(i)*cube_root(r), so find the cube root of r in the normal way, then we just need to find the cube root of i. For any cubic function (which has a polynomial, in which the highest term is x3) will always have 3 roots. There are 3 values, which when cubed will equal the imaginary number i:

• -i will do it: (-i)3 = (-i)2 * (-i) = -1 * (-i) = i
• The other two are complex: sqrt(3)/2 + i/2 and -sqrt(3)/2 + i/2

If you cube either of the two complex binomials by multiplying out, you will end up with 0 + i as the answer in both cases.

Note: the possible roots for any cubic are: 3 real roots, or 1 real root and 2 complex root, or 1 pure imaginary root, and 2 complex roots.

For your original question, if you want to stay in the pure imaginary domain, then you can use: Cube_root(i*r) = -i * cube_root(r) to find an answer.