Definimos z=-\sqrt{2+\sqrt{2}}+i \sqrt{2-\sqrt{2}}.
3- z pode ser escrito na forma exponencial :
a. 2 \mathrm{e}^{\mathrm{i} \frac{7 \pi}{8}}
b. 2 \mathrm{e}^{\mathrm{i} \frac{\pi}{8}}
c. 2 \mathrm{e}^{\mathrm{i} \frac{5 \pi}{8}}
d. 2 \mathrm{e}^{\mathrm{i} \frac{3 \pi}{8}}.

Passo 1

O enunciado define
z = -\sqrt{2+\sqrt{2}} \;+\; i\,\sqrt{2-\sqrt{2}}
e pede para escrevê-lo na forma exponencial re^{i\theta}. Vamos usar as fórmulas de ângulo metade para expressar \sqrt{2\pm\sqrt{2}} em termos de \cos\frac{\pi}{8} e \sin\frac{\pi}{8}, encontrar módulo e argumento de z.

Passo 2

Usando a identidade de ângulo metade para \frac{\pi}{8}=\frac{1}{2}\cdot\frac{\pi}{4}, temos
\cos\frac{\pi}{8} = \sqrt{\frac{1+\cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1+\tfrac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2+\sqrt{2}}}{2},
\sin\frac{\pi}{8} = \sqrt{\frac{1-\cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1-\tfrac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2-\sqrt{2}}}{2}.

Passo 3

Dessa forma
\sqrt{2+\sqrt{2}} = 2\cos\frac{\pi}{8}, \quad \sqrt{2-\sqrt{2}} = 2\sin\frac{\pi}{8},
logo
z = -2\cos\frac{\pi}{8} + i\,2\sin\frac{\pi}{8} = 2\bigl(-\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}\bigr).

Passo 4

Reconhecemos que
-\cos\frac{\pi}{8} + i\sin\frac{\pi}{8} = \cos\Bigl(\pi - \frac{\pi}{8}\Bigr) + i\sin\Bigl(\pi - \frac{\pi}{8}\Bigr) = e^{\,i\bigl(\pi - \frac{\pi}{8}\bigr)} = e^{\,i\frac{7\pi}{8}}.

Passo 5

Portanto
z = 2\,e^{\,i\frac{7\pi}{8}}.

Resposta

a. 2e^{i\frac{7\pi}{8}}

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