N°9 Equazioni biquadratiche

$z^{6} + 2 (1 - i) z^{3} - 4 i = 0$

$z^{3} = \frac{- 2 (1 - i) \pm \sqrt{{4 (1 - i)}^{2} + 16 i}}{2} = \frac{- 2 (1 - i) \pm \sqrt{4 - 8 i + {4 i}^{2} + 12 i}}{2} = \frac{- 2 (1 - i) \pm \sqrt{4 + 4 i + i^{2}}}{2} = \frac{- 2 (1 - i) \pm \sqrt{{(2 + 2 i)}^{2}}}{2} = \frac{- 2 (1 - i) \pm (2 + 2 i)}{2}$

${\begin{matrix} z_{a 0,1,2}^{3} = \frac{1}{2} \cdot (- 2 + 2 i + 2 + 2 i) = 2 i \\ z_{b 0,1,2}^{3} = \frac{1}{2} \cdot (- 2 + 2 i - 2 - 2 i) = - 2 \end{matrix}$

z_{a 0,1,3}^{3} = 2 i \to ∣ 2 i ∣ = 2

\arg (2 i) = \frac{π}{2}


$z_{a 0} = \sqrt[3]{2} \cdot [\cos (\frac{\frac{π}{2} + 2 π \cdot 0}{3}) + i \cdot \sin (\frac{\frac{π}{2} + 2 π \cdot 0}{3})] = \sqrt[3]{2} [\cos (\frac{1}{6} π) + i \cdot \sin (\frac{1}{6} π)] = \sqrt[3]{2} \cdot (\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2})$
$z_{a 1} = \sqrt[3]{2} \cdot [\cos (\frac{\frac{1}{2} π + 2 π \cdot 1}{3}) + i \cdot \sin (\frac{\frac{1}{2} π + 2 π \cdot 1}{3})] = \sqrt[3]{2} \cdot [\cos (\frac{5}{6} π) + i \cdot \sin (\frac{5}{6} π)] = \sqrt[3]{2} \cdot (- \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2})$ $z_{a 2} = \sqrt[3]{2} \cdot [\cos (\frac{\frac{1}{2} π + 2 π \cdot 2}{3}) + i \cdot \sin (\frac{\frac{1}{2} π + 2 π \cdot 2}{3})] = \sqrt[3]{2} \cdot [\cos (\frac{3}{2} π) + i \cdot \sin (\frac{3}{2} π)] = - i \cdot \sqrt[3]{2}$

z_{b 0,1,2}^{3} = - 2 \to ∣ - 2 ∣ = 2

\arg (- 2) = π


$z_{b 0} = \sqrt[3]{2} \cdot [\cos (\frac{π + 2 π \cdot 0}{3}) + i \cdot \sin (\frac{π + 2 π \cdot 0}{3})] = \sqrt[3]{2} \cdot [\cos (\frac{1}{3} π) + i \cdot \sin (\frac{1}{3} π)] = \sqrt[3]{2} \cdot (\frac{1}{2} + i \frac{\sqrt{3}}{2})$
$z_{b 1} = \sqrt[3]{2} \cdot [\cos (\frac{π + 2 π \cdot 1}{3}) + i \cdot \sin (\frac{π + 2 π \cdot 1}{3})] = \sqrt[3]{2} \cdot [\cos (π) + i \cdot \sin (π)] = - \sqrt[3]{2}$ $z_{b 2} = \sqrt[3]{2} \cdot [\cos (\frac{π + 2 π \cdot 2}{3}) + i \cdot \sin (\frac{π + 2 π \cdot 2}{3})] = \sqrt[3]{2} \cdot [\cos (\frac{5}{3} π) + i \cdot \sin (\frac{5}{3} π)] = \sqrt[3]{2} \cdot (\frac{1}{2} - i \frac{\sqrt{3}}{2})$