Find the three cube roots of −343i. (Give your answers in the form a + bi, where a and b are real numbers. Enter your answers as a comma-separated list.)

Answer 1

Every number has 3 cube roots.

The first cube root of -343i is 7i.

The other 2 cube roots are 7i times the complex cube roots of 1 which are (-1±i√3)/2.

I will leave it to you to do the necessary multiplications.

BTW: the reason for this answer is that x³ - R³ factors as (x-R) (x²+Rx+R²) and the second factor factors as R times each of the 2 complex roots of 1 as stated above.

Answer 2

z = (-343i)^1/3 = [343(cos(3π/2 + 2πn) + isin(3π/2 + 2πn))]^1/3 = 7[cos(π/2 + 2πn/3) + isin(π/2 + 2πn/3)]

n = 0, z₁ = 7(cosπ/2 + isinπ/2) = 7i;

n = 1, z₂ = 7[cos(π/2 + 2π/3) + isin(π/2 + 2π/3)] = -7√3/2 - 7/2i

n = 2; z₃ = 7[cos(π/2 + 4π/3) + isin(π/2 + 4π/3)] = 7/2 - 7√3/2i