Hi,
can you please tell me how can I fin the roots of x=(-j)^(1/3)

To evaluate this use D'Moivre's theorem:


(R * cis \theta)^n = R^n * cis (\frac {\theta} n)


So:


(-j)^ {1/3} = (1*cis( \frac {3 \pi} 2 + 2 \pi k))^{1/3} = 1^{1/3} * cis( \frac {\pi} 2 + \frac 2 3 k \pi)

where k = 0, 1, or 2

This gives three unique roots, each spaced at 120 degrees around the origin. The first is j (you can verify this by evaluating j^3 and see if it equals -j). Can you now determine the other two?