Question

Considere os números complexos $Z = \sqrt{2} + i \sqrt{2}$ e $W = 1 + \sqrt{3} i$ . Sendo $m = { | \frac{W ^{6} + {3Z}^{4} + 4i }{Z ^{2} + {W}^{3} + 6 - 2i } | }^{2}$, então $m$ vale :

( Gabarito : 34 )

#Cálculo e explicação

Answer 1

 Olá Emanueli!

 Irei separar a resolução em duas partes: uma para determinar as potências de Z e outra para as potências de W.


PARTE 1:

 Inicialmente, devemos transformar a forma algébrica de Z na forma trigonométrica. Segue,

Módulo de Z:

$\\ \mathsf{\rho^2 = \left ( \sqrt{2} \right )^2 + \left ( \sqrt{2} \right )^2} \\\\ \mathsf{\rho^2 = 2 + 2} \\\\ \mathsf{\rho^2 = 4} \\\\ \boxed{\mathsf{\rho = 2}}$

Argumento de Z:

$\\ \mathsf{\tan \alpha = \frac{\sqrt{2}}{\sqrt{2}}} \\\\ \mathsf{\tan \alpha = 1} \\\\ \boxed{\mathsf{\alpha = \frac{\pi}{4}}}, \ \mathsf{pois \ 0 \leq Z \leq \frac{\pi}{2}.}$
 
 Com isso, temos que:

$\\ \mathsf{Z = \rho \cdot \left ( \cos \alpha + i \cdot \sin \alpha \right )} \\\\ \boxed{\mathsf{Z = 2 \cdot \left ( \cos \frac{\pi}{4} + i \cdot \sin \frac{\pi}{4} \right )}}$


 Isto posto, podemos determinar $\mathsf{Z^2}$ e $\mathsf{Z^4}$ aplicando a 1ª fórmula de MOIVRE. Veja:

$\bullet \quad \mathsf{Z^n = \rho^n \cdot \left [ \cos \left ( n \alpha \right ) + i \cdot \sin \left ( n \alpha \right ) \right ]} \\\\ \mathsf{Z^2 = 2^{2} \cdot \left [ \cos \left ( 2 \cdot \frac{\pi}{4} \right ) + i \cdot \sin \left ( 2 \cdot \frac{\pi}{4} \right ) \right ]} \\\\\\ \mathsf{Z^2 = 4 \cdot \left ( \cos \frac{\pi}{2} + i \cdot \sin \frac{\pi}{2} \right )} \\\\ \mathsf{Z^2 = 4 \cdot \left ( 0 + i \cdot 1 \right )} \\\\ \boxed{\boxed{\mathsf{Z^2 = - 4i}}}$

E,

$\bullet \quad \mathsf{Z^n = \rho^n \cdot \left [ \cos \left ( n \alpha \right ) + i \cdot \sin \left ( n \alpha \right ) \right ]} \\\\ \mathsf{Z^4 = 2^{4} \cdot \left [ \cos \left ( 4 \cdot \frac{\pi}{4} \right ) + i \cdot \sin \left ( 4 \cdot \frac{\pi}{4} \right ) \right ]} \\\\\\ \mathsf{Z^4 = 16 \cdot \left ( \cos \pi + i \cdot \sin \pi \right )} \\\\ \mathsf{Z^2 = 16 \cdot \left ( - 1 + i \cdot 0 \right )} \\\\ \boxed{\boxed{\mathsf{Z^4 = - 16}}}$



 PARTE 2:

 De modo análogo, transformamos a forma algébrica de W na trigonométrica. Vejamos:

Módulo de W:

$\\ \mathsf{\rho^2 = \left(1\right)^2+\left(\sqrt{3}\right)^2} \\\\ \mathsf{\rho^2 = 1 + 3} \\\\ \mathsf{\rho^2 = 4} \\\\ \boxed{\mathsf{\rho = 2}}$

Argumento de W:

$\\ \mathsf{\tan \theta = \frac{\sqrt{3}}{1}} \\\\ \mathsf{\tan \theta = \sqrt{3}} \\\\ \boxed{\mathsf{\theta = \frac{\pi}{3}}}, \ \mathsf{pois \ 0 \leq W \leq \frac{\pi}{2}.}$
 

 Com efeito,

$\\ \mathsf{W = \rho \cdot \left ( \cos \theta + i \cdot \sin \theta \right )} \\\\ \boxed{\mathsf{W = 2 \cdot \left ( \cos \frac{\pi}{3} + i \cdot \sin \frac{\pi}{3} \right )}}$


 Por conseguinte, aplicamos a 1ª fórmula de MOIVRE para encontrar $\mathsf{W^3}$ e $\mathsf{W^6}$.

$\bullet \quad \mathsf{W^n = \rho^n \cdot \left [ \cos \left ( n \theta \right ) + i \cdot \sin \left ( n \theta \right ) \right ]} \\\\ \mathsf{W^3 = 2^{3} \cdot \left [ \cos \left ( 3 \cdot \frac{\pi}{3} \right ) + i \cdot \sin \left ( 3 \cdot \frac{\pi}{3} \right ) \right ]} \\\\\\ \mathsf{W^3 = 8 \cdot \left ( \cos \pi + i \cdot \sin \pi \right )} \\\\ \mathsf{W^3 = 8 \cdot \left ( - 1 + i \cdot 0 \right )} \\\\ \boxed{\boxed{\mathsf{W^3 = - 8}}}$

E,

$\bullet \quad \mathsf{W^n = \rho^n \cdot \left [ \cos \left ( n \theta \right ) + i \cdot \sin \left ( n \theta \right ) \right ]} \\\\ \mathsf{W^6 = 2^{6} \cdot \left [ \cos \left ( 6 \cdot \frac{\pi}{3} \right ) + i \cdot \sin \left ( 6 \cdot \frac{\pi}{3} \right ) \right ]} \\\\\\ \mathsf{W^6 = 64 \cdot \left ( \cos 2\pi + i \cdot \sin 2\pi \right )} \\\\ \mathsf{W^6 = 64 \cdot \left ( 1 + i \cdot 0 \right )} \\\\ \boxed{\boxed{\mathsf{W^6 = 64}}}$


 Por fim, e não menos trabalhoso [risos], temos:

$\\ \displaystyle \mathsf{m = \left | \frac{W^6 + 3Z^4 + 4i}{Z^2 + W^3 + 6 - 2i} \right |^2 = \left | \frac{64 + 3 \cdot (- 16) + 4i}{4i + (- 8) + 6 - 2i} \right |^2} \\\\\\ \mathsf{m = \left | \frac{16 + 4i}{- 2 + 2i} \right |^2 = \left | \frac{8 + 2i}{- 1 + i} \right |^2 = \left | \frac{8 + 2i}{- 1 + i} \cdot \frac{- 1 - i}{- 1 - i}\right |^2} \\\\\\ \mathsf{m = \left | \frac{- 8 - 8i - 2i - 2i^2}{1 - i^2} \right |^2 = \left | \frac{- 6 - 10i}{2} \right |^2 = \left | - 3 - 5i \right |^2}$


 Antes de prosseguir, vale salientar que: dado um número complexo $\mathsf{z_1 = a + bi, \ a, b \in \mathbb{R}}$ qualquer, então representamos e calculamos seu módulo, respectivamente, do seguinte modo:

$\bullet \quad \mathsf{\rho}$

$\bullet \quad \mathsf{|z_1|}$

$\bullet \quad \mathsf{\rho = |z_1| = \sqrt{a^2 + b^2}}$


 Isto posto,

$\\ \displaystyle \mathsf{m = \left | - 3 - 5i \right |^2} \\\\ \mathsf{m = \left ( \sqrt{(- 3)^2 + (- 5)^2} \right )^2} \\\\ \mathsf{m = (- 3)^2 + (- 5)^2} \\\\ \mathsf{m = 9 + 25} \\\\ \boxed{\boxed{\boxed{\boxed{\mathsf{m = 34}}}}}$