Question

achar o resultado desta questão.Se alguém puder ajudar.​

Answer 1

Podemos fazer de duas formas

1ª forma :  Racionalizando

$\displaystyle[\frac{1-\text i}{1+\text i}]^{2011} \to [\frac{(1-\text i)}{(1+\text i)}.\frac{(1-\text i)}{(1-\text i)}]^{2011} \\\\\\\ [\frac{1-2\text i+\text i^2}{1^2-(\text i)^2}}]^{2011} \to [\frac{1-2\text i-1}{1-(-1)}}]^{2011} \\\\\\ \ [\frac{-2\text i}{2}}]^{2011} \to [-\text i]^{2011} \to -\text i \\\\ \text{Portanto}: \\\\ \huge\boxed{(\frac{1-\text i}{1+\text i})^{2011}=- \text i \ }\checkmark$

2ª forma : Colocando na forma trigonométrica.

$\displaystyle \text z= \text a+\text {b.i}\\\\ \underline{\text{forma trigonom{\'e}trica}}: \\\\ \text z=|\text z_1|[\text{cos }\theta + \text {i.sen }\theta] \to \text z = |\text z|\text{Cis }\theta \\\\ \text{onde}: \\\\ |\text z| = \sqrt{\text a^2+\text b^2} \\\\ \text{cos }\theta = \frac{\text a}{|\text z|} \ \ ;\ \ \text{sen } \theta = \frac{\text b}{|\text z|}$

Propriedades que vamos usar :

$1) \ \displaystyle \text{Sejam} \ \text z_1 \ \text e \ \text z_2 \ \text{ n{\'u}meros : complexos}\\\\ \frac{|\text z_1|\text{Cis }\theta }{|\text z_2|\text{Cis }\alpha} = \frac{|\text z_1|}{|\text z_2|}.\text{Cis}(\theta -\alpha ) \\\\\\\\ 2) \ \text z^{\text k } = |\text z|^{\text k}.\text{Cis (k.}\theta )$

numerador:

$\displaystyle \text z_1 = 1-\text i \\\\ |\text z_1| = \sqrt{1^2+(-1)^2} =\sqrt{2 } \\\\\ \text{cos }\theta = \frac{1}{\sqrt{2}} \to \text{cos }\theta =\frac{\sqrt{2}}{2} \\\\\\ \text{sen }\theta = \frac{-1}{\sqrt{2}} \to \text{sen }\theta = \frac{-\sqrt{2}}{2}$

Estamos lidando com o ângulo de 45º, porém no 4º quadrante pois o cosseno deu positivo e seno negativo, então :

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

Daí :

$\displaystyle \text z_ 1= \sqrt{2}.\text{Cis }(\frac{7\pi}{4})$

denominador:

$\displaystyle \text z_2 = 1+\text i \\\\ |\text z_2| = \sqrt{1^2+1^2} =\sqrt{2 } \\\\\ \text{cos }\alpha = \frac{1}{\sqrt{2}} \to \text{cos }\alpha =\frac{\sqrt{2}}{2} \\\\\\ \text{sen }\alpha = \frac{-1}{\sqrt{2}} \to \text{sen }\alpha = \frac{\sqrt{2}}{2} \\\\ \text{Da{\'i}} : \\\\ \theta = \frac{\pi}{4} \\\\ \text{Ent{\~a}o} : \\\\\ \text z_2=\sqrt{2}.\text{Cis}(\frac{\pi}{4})$

Temos :

$\displaystyle [\frac{\displaystyle \sqrt{2}.\text{Cis}(\frac{7\pi}{4})}{\displaystyle \sqrt{2}.\text{Cis}(\frac{\pi}{4})}] ^{2011} \to [\text{Cis}(\frac{7\pi}{4}-\frac{\pi}{4})]^{2011} \\\\\ [\text{Cis}(\frac{3\pi}{2})]^{2011} \to [-\text i]^{2011} \\\\ (-1)^{2011}.[(\text i)^{4}]^{502}.(\text i)^3 \\\\ \huge\boxed{-\text i\ }\checkmark$