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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 10 of 160

Marks: +1, -0

If

n, K \in \mathbb{N}

n \neq 3K

(\sqrt{3}+i)^{2n}+(\sqrt{3}-i)^{2n}=

$(-1)^n 2^{2n+1}$
$(-1)^{n+1} 2^{2n+1}$
$(-1)^{n+1} 2^{2n}$
$(-1)^{n+1} 2^n$

Solution:

Given,

n, K \in \mathbb{N}

such that

n \neq 3K

Let

z=\sqrt{3}+i

, then

\overline{z}=\sqrt{3}-i

\;|z|=|\overline{z}|=\sqrt{3+1}=2

\; \arg(z)=\tan^{-1}\left(\;\frac{1}{\sqrt{3}}\right)=\;\frac{\pi}{6}

\; \arg(z)=\tan^{-1}\left(-\;\frac{1}{\sqrt{3}}\right)=\;-\frac{\pi}{6}

\; \therefore \;\; z=2\left(\cos \;\frac{\pi}{6}+i \sin \;\frac{\pi}{6}\right)

\; \Rightarrow z^{2n}=2^{2n}\left(\cos \;\frac{2n\pi}{6}+i \sin \;\frac{2n\pi}{6}\right)

\;\; \text{And} \; \overline{z}=2\left(\cos \;\frac{\pi}{6}-i \sin \;\frac{\pi}{6}\right)

\; \Rightarrow \overline{z}^{2n}=2^{2n}\left(\cos \;\frac{2n\pi}{6}-i \sin \;\frac{2n\pi}{6}\right)

\; \therefore z^{2n}+\overline{z}^{2n}=2 \cdot 2^{2n} \cos \left(\;\frac{n\pi}{3}\right)

\; =2^{2n+1} \cos \left(\;\frac{n\pi}{3}\right)

Since,

n \neq 3K \Rightarrow n

is not a multiple of

3\cos\left(\;\frac{n\pi}{3}\right) \neq \cos(k\pi)=\pm 1 = (-1)^{n+1} 2^{2n}

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