N°8 Equazioni biquadratiche

$z^{4} + 3 (1 + i) z^{2} + 9 i = 0$

$z^{2} = \frac{- 3 (1 + i) \pm \sqrt{9 {(1 + i)}^{2} - 36 i}}{2} = \frac{- 3 (1 + i) \pm 3 \sqrt{1 + 2 i + i^{2} - 4 i}}{2} = \frac{- 3 (1 + i) \pm 3 \sqrt{1 - 2 i + i^{2}}}{2} = \frac{- 3 (1 + i) \pm 3 \sqrt{{(1 - i)}^{2}}}{2} = \frac{- 3 (1 + i) \pm 3 (1 - i)}{2}$

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

z_{a 0,1}^{2} = - 3 i \to ∣ - 3 i ∣ = 3

\arg (- 3 i) = \frac{3}{2} π


$z_{a 0} = \sqrt{3} \cdot [\cos (\frac{\frac{3}{2} π + 2 π \cdot 0}{2}) + i \cdot \sin (\frac{\frac{3}{2} π + 2 π \cdot 0}{2})] = \sqrt{3} \cdot [\cos (\frac{3}{4} π) + i \cdot \sin (\frac{3}{4} π)] = \sqrt{3} \cdot (- \frac{\sqrt{2}}{2} + i \cdot \frac{\sqrt{2}}{2}) = - \frac{\sqrt{6}}{2} + i \frac{\sqrt{6}}{2}$
$z_{a 1} = \sqrt{3} \cdot [\cos (\frac{\frac{3}{2} π + 2 π \cdot 1}{2}) + i \cdot \sin (\frac{\frac{3}{2} π + 2 π \cdot 1}{2})] = \sqrt{3} \cdot [\cos (\frac{7}{4} π) + i \cdot \sin (\frac{7}{4} π)] = \sqrt{3} \cdot (\frac{\sqrt{2}}{2} - i \cdot \frac{\sqrt{2}}{2}) = \frac{\sqrt{6}}{2} - i \frac{\sqrt{6}}{2}$

z_{b 0,1}^{2} = - 3 \to ∣ - 3 ∣ = 3

\arg (- 3) = π


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