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If $n$ is a positive integer, then $(1+i \sqrt{3})^n+(1-i \sqrt{3})^n$ is equal to
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2770 Upvotes
Verified Answer
The correct answer is:
$2^{n+1} \cos \frac{n \pi}{3}$
We have,
$$
\begin{aligned}
& (1+\sqrt{3} i)^n-(1-\sqrt{3} i)^n \\
& =2^n\left[\left(\cos \frac{\pi}{3}+2 \sin \frac{\pi}{3}\right)^n+\left(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right)^n\right] \\
& \quad\left[\because 1 \pm \sqrt{3} i=2\left(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}\right)\right] \\
& =2^n\left[\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3}+\cos \frac{n \pi}{3}-i \sin \frac{n \pi}{3}\right] \\
& =2^{n+1} \cos \frac{n \pi}{3}
\end{aligned}
$$
$$
\begin{aligned}
& (1+\sqrt{3} i)^n-(1-\sqrt{3} i)^n \\
& =2^n\left[\left(\cos \frac{\pi}{3}+2 \sin \frac{\pi}{3}\right)^n+\left(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right)^n\right] \\
& \quad\left[\because 1 \pm \sqrt{3} i=2\left(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}\right)\right] \\
& =2^n\left[\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3}+\cos \frac{n \pi}{3}-i \sin \frac{n \pi}{3}\right] \\
& =2^{n+1} \cos \frac{n \pi}{3}
\end{aligned}
$$
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