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If $z_n=(1+i \sqrt{2})^n, n \in Z$, then $\frac{1}{9} \operatorname{Re}\left(z_4 \bar{z}_5\right)=$
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Verified Answer
The correct answer is:
9
We have,
$$
\begin{aligned}
z_n & =(1+\sqrt{2} i)^n \\
z_4 & =(1+\sqrt{2} i)^4 \\
\bar{z}_5 & =(1-\sqrt{2} i)^5 \\
z_4 \bar{z}_5 & =(1+\sqrt{2} i)^4(1-\sqrt{2} i)^5 \\
z_4 \bar{z}_5 & =(1+\sqrt{2} i)^4(1-\sqrt{2} i)^4(1-\sqrt{2} i) \\
& =(1+2)^4(1-\sqrt{2} i) \\
z_4 \bar{z}_5 & =3^4(1-\sqrt{2} i) \\
\frac{1}{9}\left(\operatorname{Re} z_4 \bar{z}_5\right) & =\frac{1}{9} \times 3^4=9
\end{aligned}
$$
$$
\begin{aligned}
z_n & =(1+\sqrt{2} i)^n \\
z_4 & =(1+\sqrt{2} i)^4 \\
\bar{z}_5 & =(1-\sqrt{2} i)^5 \\
z_4 \bar{z}_5 & =(1+\sqrt{2} i)^4(1-\sqrt{2} i)^5 \\
z_4 \bar{z}_5 & =(1+\sqrt{2} i)^4(1-\sqrt{2} i)^4(1-\sqrt{2} i) \\
& =(1+2)^4(1-\sqrt{2} i) \\
z_4 \bar{z}_5 & =3^4(1-\sqrt{2} i) \\
\frac{1}{9}\left(\operatorname{Re} z_4 \bar{z}_5\right) & =\frac{1}{9} \times 3^4=9
\end{aligned}
$$
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