Question

Escreva na forma trigonométrica o resultado da operação:

Answer 1

$\boxed{\begin{array}{l}\sf\dfrac{1}{(1+i\sqrt{3})}\cdot\dfrac{(1-i\sqrt{3})}{(1-i\sqrt{3})}\\\\\sf\dfrac{1-i\sqrt{3}}{1^2-(i\sqrt{3})^2}\\\\\sf\dfrac{1-i\sqrt{3}}{1-3i^2}=\dfrac{1-i\sqrt{3}}{1-3\cdot(-1)}\\\\\sf\dfrac{1-i\sqrt{3}}{1+3}=\dfrac{1}{4}-\dfrac{\sqrt{3}}{4}i\end{array}}$

$\boxed{\begin{array}{l}\sf z=\dfrac{1}{4}-\dfrac{\sqrt{3}}{4}i\\\\\sf \rho=\sqrt{\bigg(\dfrac{1}{4}\bigg)^2+\bigg(\dfrac{\sqrt{3}}{4}\bigg)^2}\\\\\sf\rho=\sqrt{\dfrac{1}{16}+\dfrac{3}{16}}=\sqrt{\dfrac{4}{16}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\\\sf cos(\theta)=\dfrac{\frac{1}{4}}{\frac{1}{2}}=\dfrac{1}{4}\cdot2=\dfrac{1}{2}\\\\\sf sen(\theta)=\dfrac{-\frac{\sqrt{3}}{4}}{\frac{1}{2}}\\\\\sf sen(\theta)=-\dfrac{\sqrt{3}}{4}\cdot 2=-\dfrac{\sqrt{3}}{2}\\\sf\theta=\dfrac{5\pi}{3}\\\huge\boxed{\boxed{\boxed{\boxed{\sf z=\dfrac{1}{2}\bigg[cos\bigg(\dfrac{5\pi}{3}\bigg)+i~sen\bigg(\dfrac{5\pi}{3}\bigg)\bigg]}}}} \end{array}}$