Question

Determine o módulo, argumento e a fórmula trigonometrica dos números complexos abaixo e, em seguida, represente-os no plano de argand Gauss :​

Answer 1

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

$\large\boxed{\begin{array}{l}\rm a)~\sf Z=1+\sqrt{3}i\\\sf \rho=\sqrt{1^2+(\sqrt{3})^2}\\\sf \rho=\sqrt{1+3}\\\sf \rho=\sqrt{4}\\\sf \rho=2\\\begin{cases}\sf cos(\theta)=\dfrac{1}{2}\\\\\sf sen(\theta)=\dfrac{\sqrt{3}}{2}\end{cases}\!\!\!\!\!\implies\sf \theta=\dfrac{\pi}{3}\\\\\sf Z=2\cdot\bigg[cos\bigg(\dfrac{\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{3}\bigg)\bigg]\\\\\tt gr\acute afico:\boldsymbol{vide\,anexo\,1}\end{array}}$

$\large\boxed{\begin{array}{l}\rm b)~\sf Z=7-7i\\\sf \rho=\sqrt{7^2+(-7)^2}\\\sf \rho=\sqrt{49+49}\\\sf \rho=\sqrt{49\cdot2}\\\sf \rho=7\sqrt{2}\\\begin{cases}\sf cos(\theta)=\dfrac{7}{7\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\sf sen(\theta)=-\dfrac{7}{7\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\end{cases}\!\!\!\!\longrightarrow\sf \theta=\dfrac{7\pi}{4}\\\\\sf Z=7\sqrt{2}\bigg[cos\bigg(\dfrac{7\pi}{4}\bigg)+i~sen\bigg(\dfrac{7\pi}{4}\bigg)\bigg]\\\\\tt gr\acute afico:\boldsymbol{vide\,anexo\,2}\end{array}}$

$\large\boxed{\begin{array}{l}\rm c)~\sf z=\sqrt{3}-i\\\sf \rho=\sqrt{(\sqrt{3})^2+1^2}\\\sf \rho=\sqrt{3+1}\\\sf \rho=\sqrt{4}\\\sf \rho=2\\\begin{cases}\sf cos(\theta)=\dfrac{\sqrt{3}}{2}\\\\\sf sen(\theta)=-\dfrac{1}{2}\end{cases}\!\!\!\!\longrightarrow\sf\theta=\dfrac{11\pi}{6}\\\\\sf z=2\cdot\bigg[cos\bigg(\dfrac{11\pi}{6}\bigg)+i~sen\bigg(\dfrac{11\pi}{6}\bigg)\bigg]\\\\\tt gr\acute afico:\boldsymbol{vide\,anexo\,3}\end{array}}$

$\large\boxed{\begin{array}{l}\sf\rm d)~\sf z=-5+0i\\\sf \rho=\sqrt{(-5)^2}\\\sf \rho=|-5|\\\sf\rho=5\\\begin{cases}\sf cos(\theta)=\dfrac{-5}{5}=-1\\\\\sf sen(\theta)=\dfrac{0}{5}=0\end{cases}\sf\longrightarrow \theta=\pi\\\\\sf\theta=5\cdot[cos(\pi)+i~sen(\pi)]\\\tt gr\acute afico:\boldsymbol{vide\,anexo\,4}\end{array}}$

$\large\boxed{\begin{array}{l}\rm e)~\sf z=-1+i\\\sf \rho=\sqrt{(-1)^2+1^2}\\\sf\rho=\sqrt{1+1}\\\sf \rho=\sqrt{2}\\\begin{cases}\sf cos(\theta)=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\\\\\sf sen(\theta)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\end{cases}\sf\!\!\!\!\!\longrightarrow\theta=\dfrac{3\pi}{4}\\\\\sf z=\sqrt{2}\bigg[cos\bigg(\dfrac{3\pi}{4}\bigg)+i~sen\bigg(\dfrac{3\pi}{4}\bigg)\bigg]\\\\\tt gr\acute afico:\boldsymbol{vide\,anexo\,5}\end{array}}$

$\large\boxed{\begin{array}{l}\rm f)~\sf z=1-i\\\sf \rho=\sqrt{1^2+(-1)^2}\\\sf \rho=\sqrt{1+1}\\\sf\rho=\sqrt{2}\\\begin{cases}\sf c os(\theta)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\\\sf sen(\theta)=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\end{cases}\sf\!\!\!\!\!\longrightarrow\theta=\dfrac{7\pi}{4}\\\\\sf z=\sqrt{2}\cdot\bigg[cos\bigg(\dfrac{7\pi}{4}\bigg)+i~sen\bigg(\dfrac{7\pi}{4}\bigg)\bigg]\\\\\tt gr\acute afico:\boldsymbol{vide\,anexo\,5}\end{array}}$