Question

NÚMEROS COMPLEXOS

ix^2-2x+√3=0

resposta: S={√2/2+(√6-1)i , -√2/2-(√6/2+1)i}

Answer 1

Olá!

Por Bháskara,

$\\ \displaystyle \mathsf{ix^2 - 2x + \sqrt{3} = 0} \\\\ \mathsf{\Delta = (- 2)^2 - 4 \cdot i \cdot \sqrt{3}} \\\\ \mathsf{\Delta = 4 - 4i\sqrt{3}} \\\\ \mathsf{\Delta = 4 \cdot (1 - i\sqrt{3})} \\\\\\ \mathsf{x = \frac{- (- 2) \pm \sqrt{4 \cdot (1 - i\sqrt{3})}}{2i}} \\\\\\ \mathsf{x = \frac{2 \pm 2 \cdot \sqrt{(1 - i\sqrt{3})}}{2i}} \\\\\\ \mathsf{x = \frac{1 \pm \sqrt{1 - i\sqrt{3}}}{i}}$

$\\ \mathsf{x = \frac{1 \pm \sqrt{1 - i\sqrt{3}}}{i} \cdot \frac{i}{i}} \\\\\\ \mathsf{x = \frac{\left ( 1 \pm \sqrt{1 - i\sqrt{3}} \right ) \cdot i}{i^2}} \\\\\\ \mathsf{x = \frac{\left ( 1 \pm \sqrt{1 - i\sqrt{3}} \right ) \cdot i}{- 1}} \\\\\\ \mathsf{x = \left ( - 1 \pm \sqrt{1 - i\sqrt{3}} \right ) \cdot i}$


 Determinemos as raízes quadradas de $\mathsf{1 - i\sqrt{3}}$ aplicando a 2ª fórmula de Moivre. Segue,

$\boxed{\mathsf{Z_k = \sqrt[n]{\rho} \cdot \left [ \cos \left ( \frac{\theta}{n} + k \cdot \frac{360^o}{n} \right ) + i \cdot \sin \left ( \frac{\theta}{n} + k \cdot \frac{360^o}{n} \right ) \right ], \ k = \left \{ 0, 1 \right \}}}$


 Passando $\mathsf{\(1 - i\sqrt{3}\)}$ para a sua forma trigonométrica, tiramos que:

$\mathsf{1 - \sqrt{3} \cdot i = 2 \cdot \left ( \cos 300^o + i \cdot \sin 300^o \right )}$
 
 Ou seja, $\mathsf{\rho = 2}$ e $\mathsf{\theta = 300^o}$.


 Com efeito,

$\\ \bullet \qquad \mathsf{Z_0 = \sqrt[2]{2} \cdot \left [ \cos \left ( \frac{300^o}{2} + 0 \cdot \frac{360^o}{2} \right ) + i \cdot \sin \left ( \frac{300^o}{2} + 0 \cdot \frac{360^o}{2} \right ) \right ]} \\\\\\ \mathsf{Z_0 = \sqrt{2} \cdot \left ( \cos 150^o + i \cdot \sin 150^o \right )} \\\\\\ \mathsf{Z_0 = \sqrt{2} \cdot \left ( - \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2} \right )} \\\\\\ \boxed{\mathsf{Z_0 = \frac{\sqrt{2}}{2} \cdot \left ( - \sqrt{3} + i \right )}}$

$\\ \bullet \qquad \mathsf{Z_1 = \sqrt[2]{2} \cdot \left [ \cos \left ( \frac{300^o}{2} + 1 \cdot \frac{360^o}{2} \right ) + i \cdot \sin \left ( \frac{300^o}{2} + 1 \cdot \frac{360^o}{2} \right ) \right ]} \\\\\\ \mathsf{Z_1 = \sqrt{2} \cdot \left ( \cos 330^o + i \cdot \sin 330^o \right )} \\\\\\ \mathsf{Z_1 = \sqrt{2} \cdot \left ( \frac{\sqrt{3}}{2} + i \cdot \frac{- 1}{2} \right )} \\\\\\ \boxed{\mathsf{Z_1 = \frac{\sqrt{2}}{2} \cdot \left ( \sqrt{3} - i \right )}}$


 Por fim, substituímos $\mathsf{Z_0}$ e $\mathsf{Z_1}$ em

$\mathsf{x = \left ( - 1 \pm \sqrt{1 - i\sqrt{3}} \right ) \cdot i}$


 Segue,

$\\ \mathsf{x_1 = \left ( - 1 + Z_0 \right ) \cdot i} \\\\\\ \mathsf{x_1 = \left [ - 1 + \frac{\sqrt{2}}{2} \cdot \left ( - \sqrt{3} + i \right ) \right ] \cdot i} \\\\\\ \mathsf{x_1 = - i - \frac{i\sqrt{6}}{2} + \frac{i^2\sqrt{2}}{2}} \\\\\\ \boxed{\boxed{\boxed{\mathsf{x_1 = - \frac{\sqrt{2}}{2} - \left ( 1 + \frac{\sqrt{6}}{2} \right ) \cdot i}}}}$

 E,

$\\ \mathsf{x_2 = \left ( - 1 + Z_1 \right ) \cdot i} \\\\\\ \mathsf{x_2 = \left [ - 1 + \frac{\sqrt{2}}{2} \cdot \left (\sqrt{3} - i \right ) \right ] \cdot i} \\\\\\ \mathsf{x_2 = - i + \frac{i\sqrt{6}}{2} - \frac{i^2\sqrt{2}}{2}} \\\\\\ \boxed{\boxed{\boxed{\mathsf{x_2 = \frac{\sqrt{2}}{2} + \left ( - 1 + \frac{\sqrt{6}}{2} \right ) \cdot i}}}}$