(2.0 pontos) Determine todos os valores de (-1-\sqrt{3} i)^{1 / 4}.

Passo 1

O enunciado pede para encontrar todas as raízes quartas de -1 - \sqrt{3}\,i. O conteúdo envolvido é representação de número complexo em forma polar e uso do Teorema de De Moivre para extrair raízes.

Passo 2

Vamos escrever
z = -1 - \sqrt{3}\,i
em forma polar z = r\,\bigl(\cos\theta + i\sin\theta\bigr). Primeiro o módulo:
r = \lvert z\rvert = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2.
Depois o argumento \theta. Como
\tan\theta = \frac{\Im z}{\Re z} = \frac{-\sqrt{3}}{-1} = \sqrt{3},
e \Re z<0,\;\Im z<0, temos \theta = \pi + \tfrac{\pi}{3} = \tfrac{4\pi}{3}. Logo
z = 2\bigl(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\bigr).

Passo 3

Pela fórmula das raízes n-ésimas, as quatro raízes quartas de z são
w_k = r^{1/4}\Bigl[\cos\!\bigl(\tfrac{\theta + 2\pi k}{4}\bigr) + i\,\sin\!\bigl(\tfrac{\theta + 2\pi k}{4}\bigr)\Bigr], \quad k=0,1,2,3.
Aqui r^{1/4} = 2^{1/4} e \theta = \tfrac{4\pi}{3}.

Passo 4

Calculando \phi_k = \tfrac{4\pi/3 + 2\pi k}{4} para k=0,1,2,3:
\phi_0 = \tfrac{4\pi/3}{4} = \tfrac{\pi}{3},\quad \phi_1 = \tfrac{4\pi/3 + 2\pi}{4} = \tfrac{5\pi}{6},\quad \phi_2 = \tfrac{4\pi/3 + 4\pi}{4} = \tfrac{4\pi}{3},\quad \phi_3 = \tfrac{4\pi/3 + 6\pi}{4} = \tfrac{11\pi}{6}.

Passo 5

As raízes ficam
w_k = 2^{1/4}\bigl(\cos\phi_k + i\sin\phi_k\bigr), \quad k=0,1,2,3,
isto é:
w_0 = 2^{1/4}\bigl(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\bigr), \quad w_1 = 2^{1/4}\bigl(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\bigr),
w_2 = 2^{1/4}\bigl(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\bigr), \quad w_3 = 2^{1/4}\bigl(\cos\tfrac{11\pi}{6} + i\sin\tfrac{11\pi}{6}\bigr).

Resposta

2^{1/4}\Bigl(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\Bigr),\quad 2^{1/4}\Bigl(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\Bigr),\quad 2^{1/4}\Bigl(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\Bigr),\quad 2^{1/4}\Bigl(\cos\tfrac{11\pi}{6} + i\sin\tfrac{11\pi}{6}\Bigr).

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