Determine z_{1}, z_{2} e \frac{z_{1}}{z_{2}} na forma a+b i, sendo
z_{1}=2\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \text { e } z_{2}=4\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right)

Passo 1

O enunciado dá
z_{1}=2\bigl(\cos\tfrac{\pi}{8}+i\sin\tfrac{\pi}{8}\bigr), \quad z_{2}=4\bigl(\cos\tfrac{3\pi}{8}+i\sin\tfrac{3\pi}{8}\bigr)
e pede z_{1}, z_{2} e \frac{z_{1}}{z_{2}} na forma a+bi. Vamos usar fórmulas de ângulo metade e propriedades da forma trigonométrica de números complexos.

Passo 2

Usamos as fórmulas de ângulo metade:
\cos\tfrac{\pi}{8} =\sqrt{\frac{1+\cos\tfrac{\pi}{4}}{2}} =\sqrt{\frac{1+\tfrac{\sqrt{2}}{2}}{2}} =\frac{\sqrt{2+\sqrt{2}}}{2},
\sin\tfrac{\pi}{8} =\sqrt{\frac{1-\cos\tfrac{\pi}{4}}{2}} =\sqrt{\frac{1-\tfrac{\sqrt{2}}{2}}{2}} =\frac{\sqrt{2-\sqrt{2}}}{2}.

Passo 3

Substituindo em z_{1}:
z_{1} =2\Bigl(\frac{\sqrt{2+\sqrt{2}}}{2} +i\,\frac{\sqrt{2-\sqrt{2}}}{2}\Bigr) =\sqrt{2+\sqrt{2}}+i\,\sqrt{2-\sqrt{2}}.

Passo 4

Para \cos\tfrac{3\pi}{8} e \sin\tfrac{3\pi}{8} usamos \cos\tfrac{3\pi}{8}=\sin\tfrac{\pi}{8} e \sin\tfrac{3\pi}{8}=\cos\tfrac{\pi}{8}. Logo
\cos\tfrac{3\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}, \quad \sin\tfrac{3\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}.

Passo 5

Substituindo em z_{2}:
z_{2} =4\Bigl(\frac{\sqrt{2-\sqrt{2}}}{2} +i\,\frac{\sqrt{2+\sqrt{2}}}{2}\Bigr) =2\,\sqrt{2-\sqrt{2}}+i\,2\,\sqrt{2+\sqrt{2}}.

Passo 6

Para o quociente usamos forma polar: \lvert z_{1}\rvert=2, \arg z_{1}=\tfrac{\pi}{8}; \lvert z_{2}\rvert=4, \arg z_{2}=\tfrac{3\pi}{8}.
Então
\frac{z_{1}}{z_{2}} =\frac{2}{4}\,\cis\Bigl(\tfrac{\pi}{8}-\tfrac{3\pi}{8}\Bigr) =\frac12\cis\bigl(-\tfrac{\pi}{4}\bigr) =\frac12\bigl(\cos(-\tfrac{\pi}{4})+i\sin(-\tfrac{\pi}{4})\bigr).
Como \cos(-\tfrac{\pi}{4})=\tfrac{\sqrt{2}}{2} e \sin(-\tfrac{\pi}{4})=-\tfrac{\sqrt{2}}{2}, resulta
\frac{z_{1}}{z_{2}} =\frac12\Bigl(\frac{\sqrt{2}}{2}-i\,\frac{\sqrt{2}}{2}\Bigr) =\frac{\sqrt{2}}{4}-i\,\frac{\sqrt{2}}{4}.

Resposta

z_{1}=\sqrt{2+\sqrt{2}}+i\,\sqrt{2-\sqrt{2}} z_{2}=2\,\sqrt{2-\sqrt{2}}+i\,2\,\sqrt{2+\sqrt{2}} \displaystyle \frac{z_{1}}{z_{2}}=\frac{\sqrt{2}}{4}-i\,\frac{\sqrt{2}}{4}

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