If z_n = (1 + i sqrt{2})^n, n in Z, then frac | vedclass

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(C) Given $z_n = (1 + i \sqrt{2})^n$. Then $z_4 = (1 + i \sqrt{2})^4$ and $\bar{z}_5 = (1 - i \sqrt{2})^5$. Consider the product $z_4 \bar{z}_5 = (1 +

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$9$

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(C) Given $z_n = (1 + i \sqrt{2})^n$. Then $z_4 = (1 + i \sqrt{2})^4$ and $\bar{z}_5 = (1 - i \sqrt{2})^5$. Consider the product $z_4 \bar{z}_5 = (1 + i \sqrt{2})^4 (1 - i \sqrt{2})^5$. We can rewrite this as $z_4 \bar{z}_5 = [(1 + i \sqrt{2})(1 - i \sqrt{2})]^4 (1 - i \sqrt{2})$. Since $(1 + i \sqrt{2})(1 - i \sqrt{2}) = 1^2 + (\sqrt{2})^2 = 1 + 2 = 3$, we have: $z_4 \bar{z}_5 = 3^4 (1 - i \sqrt{2}) = 81(1 - i \sqrt{2}) = 81 - 81i \sqrt{2}$. The real part is $\operatorname{Re}(z_4 \bar{z}_5) = 81$. Therefore, $\frac{1}{9} \operatorname{Re}(z_4 \bar{z}_5) = \frac{81}{9} = 9$.

If $z_n = (1 + i \sqrt{2})^n$, $n \in Z$, then $\frac{1}{9} \operatorname{Re}(z_4 \bar{z}_5) = $

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