Question

Considere os números complexos:

Answer 1

Explicação passo-a-passo:

$\text{cos}~\dfrac{\pi}{6}=\text{cos}~30^{\circ}=\dfrac{\sqrt{3}}{2}$

$\text{sen}~\dfrac{\pi}{6}=\text{sen}~30^{\circ}=\dfrac{1}{2}$

Assim:

$z_1=2\cdot\left(\text{cos}~\dfrac{\pi}{6}+i\cdot\text{sen}~\dfrac{\pi}{6}\right)$

$z_1=2\cdot\left(\dfrac{\sqrt{3}}{2}+i\cdot\dfrac{1}{2}\right)$

$z_1=\sqrt{3}+i$

$\text{cos}~\dfrac{3\pi}{4}=\text{cos}~135^{\circ}=\dfrac{-\sqrt{2}}{2}$

$\text{sen}~\dfrac{3\pi}{4}=\text{sen}~135^{\circ}=\dfrac{\sqrt{2}}{2}$

Assim:

$z_2=4\cdot\left(\text{cos}~\dfrac{3\pi}{4}+i\cdot\text{sen}~\dfrac{3\pi}{4}\right)$

$z_2=4\cdot\left(\dfrac{-\sqrt{2}}{2}+i\cdot\dfrac{\sqrt{2}}{2}\right)$

$z_2=-2\sqrt{2}+2\sqrt{2}\cdot i$

Logo:

$\dfrac{z_2}{z_1}=\dfrac{-2\sqrt{2}+2\sqrt{2}\cdot i}{\sqrt{3}+i}$

$\dfrac{z_2}{z_1}=\dfrac{-2\sqrt{2}+2\sqrt{2}\cdot i}{\sqrt{3}+i}\cdot\dfrac{\sqrt{3}-i}{\sqrt{3}-i}$

$\dfrac{z_2}{z_1}=\dfrac{-2\sqrt{6}+2\sqrt{6}\cdot i+2\sqrt{2}\cdot i-2\sqrt{2}\cdot i^2}{(\sqrt{3})^2-i^2}$

$\dfrac{z_2}{z_1}=\dfrac{2\sqrt{2}-2\sqrt{6}+2\sqrt{6}\cdot i+2\sqrt{2}\cdot i}{3+1}$

$\dfrac{z_2}{z_1}=\dfrac{2\cdot(\sqrt{2}-\sqrt{6}+\sqrt{6}\cdot i+\sqrt{2}\cdot i)}{4}$

$\dfrac{z_2}{z_1}=\dfrac{\sqrt{2}-\sqrt{6}}{2}+\dfrac{(\sqrt{6}+\sqrt{2})\cdot i}{2}$

$\dfrac{z_2}{z_1}=2\cdot\left(\text{cos}~\dfrac{7\pi}{12}+i\cdot\text{sen}~\dfrac{7\pi}{12}\right)$