Calcule as seguintes raízes complexas:
(a) \sqrt[3]{-1};
(b) \sqrt[4]{-1+i \sqrt{3}};
(c) (1+i \sqrt{3})^{\frac{3}{2}};
(d) \sqrt[3]{8 i}.

Passo 1

O enunciado pede para calcular várias raízes e potências fracionárias de números complexos. O conteúdo central é a representação na forma polar z = r\,e^{i\theta} (ou r\cis\theta), que permite usar as fórmulas gerais das raízes
z^{1/n} = r^{1/n}\cis\Bigl(\frac{\theta+2\pi k}{n}\Bigr),\quad k=0,1,\dots,n-1
e da potência fracionária
z^{p/q} = r^{p/q}\cis\Bigl(\frac{p}{q}(\theta+2\pi k)\Bigr),\quad k=0,1,\dots,q-1.
Vamos em cada item passar o número para a forma polar e aplicar essas fórmulas.

Passo 2

Item (a): \sqrt[3]{-1}.
Escrevemos -1 = 1\cis\pi. Então as três raízes são
\sqrt[3]{-1} = 1^{1/3}\cis\!\Bigl(\frac{\pi+2\pi k}{3}\Bigr) = \cis\!\Bigl(\frac{\pi+2\pi k}{3}\Bigr),\quad k=0,1,2.
Para k=0,1,2 obtemos os valores
\cis\!\Bigl(\tfrac{\pi}{3}\Bigr)=\tfrac12 + \tfrac{\sqrt{3}}2\,i,\quad \cis\!\bigl(\pi\bigr)=-1,\quad \cis\!\Bigl(\tfrac{5\pi}{3}\Bigr)=\tfrac12 - \tfrac{\sqrt{3}}2\,i.

Passo 3

Item (b): \sqrt[4]{-1 + i\sqrt{3}}.
Primeiro, -1 + i\sqrt{3} tem módulo \bigl|{-1 + i\sqrt{3}}\bigr| = 2
e argumento \arg(-1 + i\sqrt{3}) = \tfrac{2\pi}{3}.
Logo
-1 + i\sqrt{3} = 2\cis\!\Bigl(\tfrac{2\pi}{3}\Bigr),
e as quatro raízes são
\sqrt[4]{2}\,\cis\!\Bigl(\frac{\tfrac{2\pi}{3} + 2\pi k}{4}\Bigr) = 2^{1/4}\cis\!\Bigl(\tfrac{\pi}{6} + \tfrac{\pi k}{2}\Bigr), \quad k=0,1,2,3.

Passo 4

Item (c): (1 + i\sqrt{3})^{3/2}.
Escrevemos 1 + i\sqrt{3} = 2\cis\!\bigl(\tfrac{\pi}{3}\bigr).
Usando a fórmula da potência fracionária com p=3, q=2,
(1 + i\sqrt{3})^{3/2} = 2^{3/2}\cis\!\Bigl(\tfrac{3}{2}\bigl(\tfrac{\pi}{3} + 2\pi k\bigr)\Bigr) = 2\sqrt{2}\,\cis\!\bigl(\tfrac{\pi}{2} + 3\pi k\bigr), \quad k=0,1.
Para k=0 e k=1 isso dá
2\sqrt{2}\cis\!\bigl(\tfrac{\pi}{2}\bigr)=2\sqrt{2}\,i, \quad 2\sqrt{2}\cis\!\bigl(\tfrac{\pi}{2}+3\pi\bigr) =2\sqrt{2}\cis\!\bigl(\tfrac{7\pi}{2}\bigr) =2\sqrt{2}\cis\!\bigl(\tfrac{3\pi}{2}\bigr) =-2\sqrt{2}\,i.

Passo 5

Item (d): \sqrt[3]{8\,i}.
Escrevemos 8\,i = 8\cis\!\bigl(\tfrac{\pi}{2}\bigr).
As três raízes são
8^{1/3}\cis\!\Bigl(\frac{\tfrac{\pi}{2} + 2\pi k}{3}\Bigr) = 2\cis\!\Bigl(\tfrac{\pi}{6} + \tfrac{2\pi k}{3}\Bigr), \quad k=0,1,2.
Isso corresponde aos argumentos \pi/6, 5\pi/6 e 9\pi/6=3\pi/2.

Resposta

(a) \sqrt[3]{-1} = \bigl\{\tfrac12 + \tfrac{\sqrt{3}}2\,i,\;-1,\;\tfrac12 - \tfrac{\sqrt{3}}2\,i\bigr\}.
(b) \sqrt[4]{-1 + i\sqrt{3}} = \bigl\{2^{1/4}\cis(\tfrac{\pi}{6} + \tfrac{\pi k}{2})\mid k=0,1,2,3\bigr\}.
(c) (1 + i\sqrt{3})^{3/2} = \{\,2\sqrt{2}\,i,\;-2\sqrt{2}\,i\}.
(d) \sqrt[3]{8\,i} = \bigl\{2\cis(\tfrac{\pi}{6} + \tfrac{2\pi k}{3})\mid k=0,1,2\bigr\}.

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