Question

Dentre os números complexos a seguir, aquele que NÃO é uma das raízes quartas do número dado por z = −8+ 8√3i é:

(A) 1 −√3i

(B) −1 + √3i

(C) −√3 − i

(D) 1 + √3i

(E) √3+ i

Answer 1

As raízes enésimas de um complexo Z é dado da seguinte forma :

$\displaystyle \sqrt[\text n]{\text Z} = \sqrt[\text n]{|\text Z|}.\text{cis}(\frac{\theta + 2\text k.\pi}{\text n}) \ , \ \text{com k }:\{0,1,2,\ ...\ , \text n -1\}$

Analisando as raízes quartas do complexo dado :

$\text Z = -8+\text i.8.\sqrt{3}$

deixando na forma trigonométrica :

$\displaystyle \text Z =16(\frac{-1}{2}+\frac{\text i.\sqrt{3}}{2}) \\\\\\ \text Z = 16.\text{cis}(\frac{2\pi}{3}) \\\\ \underline{\text {tirando a raiz quarta }}: \\\\ \sqrt[4]{\text Z} = \sqrt[4]{16}.\text{cis}[\ \frac{1}{4}(\frac{2\pi}{3}+2\text k.\pi )\ ] \\\\\\ \sqrt[4]{\text Z} = 2.\text{cis}[\ \frac{\pi}{6}+\frac{\text k.\pi}{2}\ ]$

Substituindo os valores de K = 0,1,2,3.

$\displaystyle \text k = 0 \to \text Z_1 =2.\text{cis}(\frac{\pi}{6}+0) \to 1+\text i.\sqrt{3} \\\\\\ \text k = 1 \to \text Z_2 = 2\text{cis}(\frac{\pi}{6}+\frac{1.\pi}{2}) \to 2\text{cis}(\frac{2\pi}{3}) \to -1+\text i.\sqrt{3} \\\\\\ \text k= 2 \to \text Z_3 = 2.\text{cis}(\frac{\pi}{6}+\frac{2\pi}{2}) \to 2\text{cis}(\frac{7\pi}{6}) \to -\sqrt{3}-\text i \\\\\\ \text k = 3 \to \text Z_4 = 2\text{cis}(\frac{\pi}{6}+\frac{3\pi}{2}) \to 2\text{cis}(\frac{5\pi}{3}) \to 1-\text i\sqrt{3}$

letra E