Question

Escrevendo o complexo
$z = \frac{ 1 - i}{1 - \sqrt{3}i }$
calcule os valores do módulo e do argumento:
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Answer 1

Resposta:

$z=\frac{1-i}{1-\sqrt{3}i} .\frac{1+\sqrt{3}i}{1+\sqrt{3}i}=\frac{(1-i)(1+\sqrt{3}i)}{1^{2}-(\sqrt{3}i)^{2}}=\frac{1+\sqrt{3}i-i-\sqrt{3}i^{2}}{1-3(-1)}=\frac{1+\sqrt{3}-(1-\sqrt{3})i}{4}\\\\z=\frac{1+\sqrt{3}}{4}-\frac{1-\sqrt{3}}{4}i$

O módulo de z:

$|z|^2=(\frac{1+\sqrt{3}}{4})^{2}+(\frac{-1+\sqrt{3}}{4})^2\\\\|z|^2=\frac{1+2\sqrt{3}+(\sqrt{3})^{2}}{4^{2}}+\frac{1-2\sqrt{3}+(\sqrt{3})^{2}}{4^{2}}\\\\|z|^2=\frac{8}{16}=\frac{1}{2}\\\\|z|=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}.\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}\approx0,7071$

Argumento de z:

$cos\alpha=\frac{\frac{1+\sqrt{3}}{4}}{\frac{1}{\sqrt{2}}}={\frac{1+\sqrt{3}}{4}.\sqrt{2}\approx0,9659$

$\alpha=arcos(0,9659)=15^{o}$

Obs. quadrado da diferença:

(a²-b²)=(a-b)(a+b)