Usando a fórmula de De Moivre, calculem as potências:
(1+\sqrt{3} i)^{4}

Passo 1

O enunciado pede para calcular
(1+\sqrt{3}\,i)^{4}
usando a fórmula de De Moivre.
Para isso, vamos:
  1. Escrever 1+\sqrt{3}\,i em forma polar r(\cos\theta + i\sin\theta).
  2. Aplicar a fórmula \bigl(r(\cos\theta + i\sin\theta)\bigr)^{n} = r^{n}\bigl(\cos(n\theta)+i\sin(n\theta)\bigr).

Passo 2

Calculamos o módulo r e o argumento \theta de z=1+\sqrt{3}\,i:
r = |z| = \sqrt{1^{2}+(\sqrt{3})^{2}} = \sqrt{1+3} = 2,
\theta = \arg(z) = \arctan\!\bigl(\tfrac{\sqrt{3}}{1}\bigr) = \tfrac{\pi}{3}.
Logo
1+\sqrt{3}\,i = 2\bigl(\cos(\tfrac{\pi}{3}) + i\sin(\tfrac{\pi}{3})\bigr).

Passo 3

Aplicando De Moivre com n=4, obtemos
(1+\sqrt{3}\,i)^{4} = \bigl[2(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3})\bigr]^{4} = 2^{4}\bigl(\cos(4\cdot\tfrac{\pi}{3}) + i\sin(4\cdot\tfrac{\pi}{3})\bigr) =16\bigl(\cos(\tfrac{4\pi}{3}) + i\sin(\tfrac{4\pi}{3})\bigr).

Passo 4

No círculo trigonométrico temos
\cos\bigl(\tfrac{4\pi}{3}\bigr) = -\tfrac12, \quad \sin\bigl(\tfrac{4\pi}{3}\bigr) = -\tfrac{\sqrt{3}}{2}.
Substituindo,
(1+\sqrt{3}\,i)^{4} =16\Bigl(-\tfrac12 + i\bigl(-\tfrac{\sqrt{3}}{2}\bigr)\Bigr) = -8 - 8\sqrt{3}\,i.

Resposta

(1+\sqrt{3}\,i)^{4} = -8 - 8\sqrt{3}\,i.

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