Question

Dados os números complexos Z e W abaixo, encontre: a) ZW; b) Z:W; c) W ao quadrado.

Answer 1

$\Large\boxed{\begin{array}{l}\sf Z=a+bi\\\underline{\rm M\acute odulo~de~um~n\acute umero~complexo}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\rho=\sqrt{a^2+b^2} }}}}\\\underline{\rm Argumento~de~um~n\acute umero~complexo}\\\sf \acute E~o~\hat angulo~\theta~tal~que\\\sf sen(\theta)=\dfrac{a}{\rho}~e~cos(\theta)=\dfrac{b}{\rho}\\\underline{\rm Forma~trigonom\acute etrica~de~um~n\acute umero~complexo}\\\huge\boxed{\boxed{\boxed{\boxed{\sf Z=\rho[cos(\theta)+i~sen(\theta)]}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Operac_{\!\!,}\tilde oes~com~n\acute umeros}\\\underline{\rm complexos~na~forma~trigonom\acute etrica}\\\sf Dados~os~complexos~Z_1=\rho_1[cos(\theta_1)+i~sen(\theta_1)]\\\sf e~ Z_2=\rho_2[cos(\theta_2)+i~sen(\theta_2)]\\\blacksquare~~\underline{\boldsymbol{ Produto}}\\\sf Z_1\cdot Z_2=\rho_1\cdot\rho_2[cos(\theta_1+\theta_2)+i~sen(\theta_1+\theta_2)] \end{array}}$

$\large\boxed{\begin{array}{l}\blacksquare\blacksquare~~\underline{\boldsymbol{ Quociente}}\\\sf\dfrac{Z_1}{Z_2}=\dfrac{\rho_1}{\rho_2}[cos(\theta_1-\theta_2)+i~sen(\theta_1-\theta_2)]\\\blacksquare\blacksquare\blacksquare~~\underline{ \boldsymbol{Pot\hat enciac_{\!\!,}\tilde ao}}\\\sf Z^n=\rho^n[cos(n\theta)+i~sen(n\theta)]\end{array}}$

$\large\boxed{\begin{array}{l}\blacksquare\blacksquare\blacksquare\blacksquare~~\underline{\boldsymbol{ Radiciac_{\!\!,}\tilde ao}}\\\sf W_k=\sqrt[\sf n]{\sf \rho}\bigg[cos\bigg(\dfrac{\theta+2k\pi}{n}\bigg)+i~sen\bigg(\dfrac{\theta+2k\pi}{n}\bigg)\bigg]\end{array}}$

$\Large\boxed{\begin{array}{l}\sf Z=6\bigg[cos\bigg(\dfrac{4\pi}{3}\bigg)+i~sen\bigg(\dfrac{4\pi}{3}\bigg)\bigg]\\\\\sf W=1\bigg[cos\bigg(\dfrac{\pi}{6}\bigg)+i~sen\bigg(\dfrac{\pi}{6}\bigg)\bigg]\end{array}}$

$\large\boxed{\begin{array}{l}\tt a)~\sf Z\cdot W=6\cdot1\bigg[cos\bigg(\dfrac{4\pi}{3}+\dfrac{\pi}{6}\bigg)+i~sen\bigg(\dfrac{4\pi}{3}+\dfrac{\pi}{6}\bigg)\bigg]\\\\\sf Z\cdot W=6\bigg[cos\bigg(\dfrac{8\pi+\pi}{6}\bigg)+i~sen\bigg(\dfrac{8\pi+\pi}{6}\bigg)\bigg]\\\\\sf Z\cdot W=6\bigg[cos\bigg(\dfrac{3\pi}{2}\bigg)+i~sen\bigg(\dfrac{3\pi}{2}\bigg)\bigg]\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf\dfrac{Z}{W}=\dfrac{6}{1}\bigg[cos\bigg(\dfrac{4\pi}{3}-\dfrac{\pi}{6}\bigg)-i~sen\bigg(\dfrac{4\pi}{3}-\dfrac{\pi}{6}\bigg)\bigg]\\\\\sf \dfrac{Z}{W}=6\bigg[cos\bigg(\dfrac{8\pi-\pi}{6}\bigg)+i~sen\bigg(\dfrac{8\pi-\pi}{6}\bigg)\bigg]\\\\\sf \dfrac{Z}{W}=6\bigg[cos\bigg(\dfrac{7\pi}{6}\bigg)+i~sen\bigg(\dfrac{7\pi}{6}\bigg)\bigg]\end{array}}$

$\Large\boxed{\begin{array}{l}\tt c)~\sf w^2=1^2\bigg[cos\bigg(\backslash\!\!\!2\cdot\dfrac{\pi}{\backslash\!\!\!6}\bigg)+i~sen\bigg(\backslash\!\!\!2\cdot\dfrac{\pi}{\backslash\!\!\!6}\bigg)\bigg]\\\sf W^2=cos\bigg(\dfrac{\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{3}\bigg)\end{array}}$