When n=8,(sqrt{3}+i)^n+(sqrt{3}-i)^n= | vedclass

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(A) Given that,$n=8$. We need to evaluate $(\sqrt{3}+i)^8+(\sqrt{3}-i)^8$. First,convert the complex numbers to polar form: $\sqrt{3}+i = 2(\cos \frac

Vedclass · Accepted Answer

$-256$

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(A) Given that,$n=8$. We need to evaluate $(\sqrt{3}+i)^8+(\sqrt{3}-i)^8$. First,convert the complex numbers to polar form: $\sqrt{3}+i = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2e^{i\pi/6}$ and $\sqrt{3}-i = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6}) = 2e^{-i\pi/6}$. Then,$(\sqrt{3}+i)^8 = (2e^{i\pi/6})^8 = 2^8 e^{i8\pi/6} = 256 e^{i4\pi/3}$. And $(\sqrt{3}-i)^8 = (2e^{-i\pi/6})^8 = 2^8 e^{-i8\pi/6} = 256 e^{-i4\pi/3}$. Summing these: $256(e^{i4\pi/3} + e^{-i4\pi/3}) = 256(2 \cos \frac{4\pi}{3})$. Since $\cos \frac{4\pi}{3} = \cos(240^\circ) = -\frac{1}{2}$,the expression becomes $256 	imes 2 	imes (-\frac{1}{2}) = -256$.

When $n=8$,$(\sqrt{3}+i)^n+(\sqrt{3}-i)^n=$

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