Question related to Complex Analysis

Question

z^6 - 2z^3 - 8 = 0 using Polar Form and De Moivre theorem

Answer 1

This equation is a quadratic in z³...

(z³ - 4)(z³ + 2)

Use De Moivre's theorem to express the roots in polar form...although you really don't need De Moivre's theorem since the cube roots are simply obtained using the 3 cube roots of 1.

Yes...you need to use De Moivre's Theorem on each factor.

However, a short cut is that the 3 solutions of z³=R are R^1/3 times 1, times ω, and times ω².

.....ω and ω² are the 2 complex roots of 1.