Question

Dados os números complexos z=6 (cos 30° + i sen 30°) e w=2 (cos 180° + i sen 180°), calculando Z/W obtemos:

A) 3(cos 30° + i sen 30°)
B) 3( cos 330° + i sen 330°)
C) 3(cos 90° + i sen 90°)
D) 3(cos 210° + i sen 210°)

Gente, por favor, eu preciso de resposta fundamentada!!!

Answer 1

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/39113180

                                                   

$\boxed{\sf{\underline{M\acute{o}dulo~de~um~n\acute{u}mero~complexo~z=a+bi}}}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf{\rho=\sqrt{a^2+b^2}}}}}}}$

$\boxed{\sf{\underline{Argumento~de~um~n\acute{u}mero~complexo}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{cos(\theta)=\dfrac{a}{\rho}~~~sen(\theta)=\dfrac{b}{\rho}}}}}}$

$\boxed{\sf{\underline{Forma~trigonom\acute{e}trica~de~um~n\acute{u}mero~complexo}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{Z=\rho\left[cos(\theta)+i~sen(\theta)\right]}}}}}$

$\large\boxed{\begin{array}{l}\sf Divis\tilde ao~de~n\acute umeros~complexos\\\sf na~forma~trigonom\acute etrica\end{array}}\\\\\large\boxed{\boxed{\boxed{\boxed{\sf\dfrac{z_1}{z_2}=\dfrac{\rho_1}{\rho_2}[cos(\theta_1-\theta_2)+i~sen(\theta_1-\theta_2)]}}}}$

$\boxed{\begin{array}{l}\sf z=6[cos(30^\circ)+i~sen(30^\circ)]\\\sf w=2[cos(180^\circ)+i~sen(180^\circ)]\\\sf \dfrac{z}{w}=\dfrac{6}{2}[cos(30^\circ-180^\circ)+i~sen(30^\circ-180^\circ)]\\\\\sf\dfrac{z}{w}=3[cos(-150^\circ)+i~sen(-150^\circ)]\\\sf\dfrac{z}{w}=3[cos(150^\circ)-i~sen(150^\circ)]\\\sf por~gentileza~revise~o~gabarito \end{array}}$