N°5 Equazioni biquadratiche

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

$z^{2} = \frac{- (1 - 3 i) \pm \sqrt{{(1 - 3 i)}^{2} + 12 i}}{2} = \frac{- (1 - 3 i) \pm \sqrt{1 - 6 i + {9 i}^{2} + 12 i}}{2} = \frac{- (1 - 3 i) \pm \sqrt{1 + 6 i + {9 i}^{2}}}{2} = \frac{- (1 - 3 i) \pm \sqrt{{(1 + 3 i)}^{2}}}{2} = \frac{- (1 - 3 i) \pm (1 + 3 i)}{2}$

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

z_{a 0,1}^{2} = 3 i \to ∣ 3 i ∣ = 3

\arg (3 i) = \frac{1}{2} π


$z_{a 0} = \sqrt{3} \cdot [\cos (\frac{\frac{1}{2} π + 2 π \cdot 0}{2}) + i \cdot \sin (\frac{\frac{1}{2} π + 2 π \cdot 0}{2})] = \sqrt{3} \cdot [\cos (\frac{1}{4} π) + i \cdot \sin (\frac{1}{4} π)] = \sqrt{3} \cdot (\frac{\sqrt{2}}{2} + i \cdot \frac{\sqrt{2}}{2}) = \frac{\sqrt{6}}{2} + i \frac{\sqrt{6}}{2}$
$z_{a 1} = \sqrt{3} \cdot [\cos (\frac{\frac{1}{2} π + 2 π \cdot 1}{2}) + i \cdot \sin (\frac{\frac{1}{2} π + 2 π \cdot 1}{2})] = \sqrt{3} \cdot [\cos (\frac{5}{4} π) + i \cdot \sin (\frac{5}{4} π)] = \sqrt{3} \cdot (- \frac{\sqrt{2}}{2} - i \cdot \frac{\sqrt{2}}{2}) = - \frac{\sqrt{6}}{2} - i \frac{\sqrt{6}}{2}$

z_{b 0,1}^{2} = - 1 \to ∣ - 1 ∣ = 1

\arg (- 1) = π


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