Question

A forma trigonométrica do número complexo $z=\frac{1+i}{i}$ é:

a) $\frac{\sqrt{2}}{2}\left(cos\frac{\pi }{4}-isen\frac{\pi }{4}\right)$
b) $\sqrt{2}\left(cos\frac{5\pi }{4}-isen\frac{5\pi }{4}\right)$
c) $\sqrt{2}\left(cos\frac{7\pi }{4}-isen\frac{7\pi }{4}\right)$
d) $\sqrt{2}\left(cos\frac{\pi }{4}-isen\frac{\pi }{4}\right)$
e) $\sqrt{2}\left(cos\frac{3\pi }{4}-isen\frac{3\pi }{4}\right)$

Answer 1

$\displaystyle \text z= \frac{1+\text i}{\text i} \\\\\ \text z=\frac{(1+\text i)}{\text i}.\frac{\text i}{\text i} \\\\\ \text z=\frac{\text i+\text i ^2}{\text i^2} \\\\\ \text z = \frac{\text i+(-1)}{-1} \\\\ \text z= 1-\text i \\\\\ \underline{\text{na forma trigonom{\'e}trica}}: \\\\ \text z = \text{a + b.i}\\\\ \text z= |\text z|.[ \ \text{cos}(\theta)+\text i.\text{sen}(\theta)\ ] \\\\ \text{onde : }\\\\ |\text z| = \sqrt{(\text a)^2+(\text b)^2} \\\\ \tex{cos}(\theta) = \frac{\text a}{|\text z|}$

$\displaystyle \text{sen}(\theta)=\frac{\text b}{|\text z|}$

$\displaystyle \underline{\text{temos}}: \\\\ \text a = 1 \ ,\text b =-1 \\\\ \text z=\sqrt{1^2+(-1)^2 }\to |\text z| = \sqrt{2} \\\\\ \text{cos}(\theta)=\frac{1}{\sqrt{2}} \to \text{cos}(\theta)=\frac{\sqrt2}{2} \\\\ \text{sen}(\theta)=\frac{-1}{\sqrt 2} \to \text{sen}(\theta)=\frac{-\sqrt2}{2} \\\\\\\ \underline{\text{Sabemos que}}: \\\\ \text{sen}(\frac{\pi}{4}) =\text{cos}(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \\\\ \underline{\text{Para que o seno seja negativo, basta que fa{\c c}amos}}:$

$\displaystyle \theta = 2\pi-\frac{\pi}{4} \to \boxed{\theta = \frac{7\pi}{4}}$

Daí :

$\displaystyle \huge\boxed{\text z= \sqrt2.[\ \text{cos}(\frac{7\pi}{4})+i.\text{sen}(\frac{7\pi}{4})\ ]\ }$  

(você disse que o gabarito é a letra c, mas isso não condiz que o desenvolvimento. Então a questão está anulada)

Faça o teste substituindo os respectivos valores no item c :

$\displaystyle \text z=\sqrt{2}[\ \text{cos}(\frac{7\pi}{4})-\text i.\text{sen}\frac{7\pi}{4} \ ] \\\\\\ \text z = \sqrt{2}.[\ \frac{\sqrt2}{2}-\text i.\frac{(-\sqrt{2})}{2} \ ] \\\\\\ \text z = \frac{\sqrt{2}.\sqrt{2}}{2}+\frac{\text i.\sqrt{2}.\sqrt{2}}{2} \\\\ \text z = 1+\text i \to \text{Diferente do desenvolver de }(\frac{1+\text i}{\text i})$