solve for z. z^3 + 6√3 - 6i = 0

Answer 1

All of these problems use deMoivre's Theorem. We solve for the base and the angle.

For the n'th root, r = (original r)^(1/n)

theta = theta original/n, (theta original + 2pi)/n, ... (theta original +2pi(n-1))/n

Re-write as z^3 = (6i - 6√3)

Thus, n = 3.

original r = √(6²+(- 6√3)²) = 12

original θ has a sin of 6/12 = 1/2 and a cos of - 6√3/12 = - √3/2, so θ = 150° = 5π/6

Thus, for the solution, r = 12^1/3 or 2^2/33^1/3

θ = (5π/6/3,(5π/6+2π)/3,(5π/6+4π)/3) or (5π/18,17π/18,29π/18)

z = 2^2/33^1/3(cos 5π/18 + i sin 5π/18, cos 17π/18 + i sin 17π/18, cos 29π/18 + i sin 29π/18) or

2^2/33^1/3(cos 50° + i sin 50° , cos 170° + i sin 170° , cos 290° + i sin 290°)

We can also write the solutions as

1.4716 + 1.7538 i, -2.2546 + 0.3976 i, 0.783 + -2.1514 i