Question

Oi !
Coloque esses números em sua forma algébrica!

Z1 =
$\frac{ \sqrt{2 - i} }{ \sqrt{2} + 2i }$


Z2 =
$\frac{6 + 2i}{ - 1 + 3i}$
​

Answer 1

Explicação passo-a-passo:

a)

$\sf z_1=\dfrac{\sqrt{2}-i}{\sqrt{2}+2i}$

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

$\sf z_1=\dfrac{2-i\sqrt{2}-2i\sqrt{2}+2i^2}{(\sqrt{2})^2-(2i)^2}$

$\sf z_1=\dfrac{2-i\sqrt{2}-2i\sqrt{2}+2\cdot(-1)}{2-4i^2}$

$\sf z_1=\dfrac{2-i\sqrt{2}-2i\sqrt{2}-2}{2-4\cdot(-1)}$

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

$\sf z_1=\dfrac{-3\sqrt{2}\cdot i}{6}$

$\sf \red{z_1=\dfrac{-\sqrt{2}\cdot i}{2}}$

b)

$\sf z_2=\dfrac{6+2i}{-1+3i}$

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

$\sf z_2=\dfrac{-6-2i-18i-6i^2}{(-1)^2-(3i)^2}$

$\sf z_2=\dfrac{-6-2i-18i-6\cdot(-1)}{1-9i^2}$

$\sf z_2=\dfrac{-6-2i-18i+6}{1-9\cdot(-1)}$

$\sf z_2=\dfrac{-6+6-2i-18i-6i^2}{1+9}$

$\sf z_2=\dfrac{-20i}{10}$

$\sf \red{z_2=-2i}$