If n, K in N such that n neq 3K,then (sqrt{3} | vedclass

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(C) Let $z = \sqrt{3} + i$. We can write $z$ in polar form as $z = 2(\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}))$. Then $z^{2n} = 2^{2n}(\cos(\frac{n

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

Vedclass · Answer

(C) Let $z = \sqrt{3} + i$. We can write $z$ in polar form as $z = 2(\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}))$. Then $z^{2n} = 2^{2n}(\cos(\frac{n\pi}{3}) + i \sin(\frac{n\pi}{3}))$. Similarly,$(\sqrt{3}-i)^{2n} = 2^{2n}(\cos(\frac{n\pi}{3}) - i \sin(\frac{n\pi}{3}))$. Adding these,we get $2^{2n} 	imes 2 \cos(\frac{n\pi}{3}) = 2^{2n+1} \cos(\frac{n\pi}{3})$. Since $n 
eq 3K$,$n$ is not a multiple of $3$. If $n = 3K \pm 1$,then $\cos(\frac{n\pi}{3}) = \cos(\frac{(3K \pm 1)\pi}{3}) = \cos(K\pi \pm \frac{\pi}{3}) = (-1)^K \cos(\frac{\pi}{3}) = (-1)^K 	imes \frac{1}{2}$. This does not lead to a simple power of $2$ unless we re-evaluate the expression. Actually,for $n$ not a multiple of $3$,the expression simplifies based on the specific values. Given the options,the correct form is $(-1)^{n+1} 2^{n+1}$ is not listed,but evaluating for $n=1$: $(\sqrt{3}+i)^2 + (\sqrt{3}-i)^2 = (3-1+2\sqrt{3}i) + (3-1-2\sqrt{3}i) = 4$. For $n=1$,option $B$ gives $(-1)^2 2^{2(1)+1} = 8$ (Incorrect). Wait,let's re-check: $(\sqrt{3}+i)^2 = 2+2\sqrt{3}i$. $(\sqrt{3}-i)^2 = 2-2\sqrt{3}i$. Sum is $4$. Option $C$ for $n=1$: $(-1)^2 2^{2(1)} = 4$. This matches. Thus,the correct option is $C$.

If $n, K \in N$ such that $n \neq 3K$,then $(\sqrt{3}+i)^{2n} + (\sqrt{3}-i)^{2n} = $

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