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If $|x+i y|=\sqrt{x^2+y^2}$, then $\left|(1-\sqrt{3} i)^9+(\sqrt{3}+i)^9\right|=$
Options:
Solution:
2525 Upvotes
Verified Answer
The correct answer is:
$2^{\frac{19}{2}}$
Given $|\mathrm{x}+\mathrm{iy}|=\sqrt{\mathrm{x}^2+\mathrm{y}^2}$
Take $\left|(1-\sqrt{3} i)^9+(\sqrt{3}+\mathrm{i})^9\right|$
$$
\begin{aligned}
& \left|\left(2 \cos \frac{\pi}{3}-\mathrm{i} \sin \frac{\pi}{3}\right)^9+\left(2\left(\cos \frac{\pi}{6}+\mathrm{i} \sin \left(\frac{\pi}{6}\right)\right)\right)^9\right| \\
& =(2)^9|-1-\mathrm{i}|=(2)^9 \times \sqrt{1+1} \\
& =2^{9+\frac{1}{2}}=2^{\frac{19}{2}}
\end{aligned}
$$
So, option (d) is correct.
Take $\left|(1-\sqrt{3} i)^9+(\sqrt{3}+\mathrm{i})^9\right|$
$$
\begin{aligned}
& \left|\left(2 \cos \frac{\pi}{3}-\mathrm{i} \sin \frac{\pi}{3}\right)^9+\left(2\left(\cos \frac{\pi}{6}+\mathrm{i} \sin \left(\frac{\pi}{6}\right)\right)\right)^9\right| \\
& =(2)^9|-1-\mathrm{i}|=(2)^9 \times \sqrt{1+1} \\
& =2^{9+\frac{1}{2}}=2^{\frac{19}{2}}
\end{aligned}
$$
So, option (d) is correct.
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