(sqrt{3}+i)^7+(sqrt{3}-i)^7 is equal to (in s | vedclass

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(C) Let $z = \sqrt{3}+i$. Converting to polar form: $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$. $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{\p

Vedclass · Accepted Answer

$-128$

Vedclass · Answer

(C) Let $z = \sqrt{3}+i$. Converting to polar form: $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$. $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$. So,$z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$. Then $\bar{z} = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6})$. Using De Moivre's theorem: $z^7 + \bar{z}^7 = 2^7(\cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6}) + 2^7(\cos \frac{7\pi}{6} - i \sin \frac{7\pi}{6})$. $= 2^7 	imes 2 \cos \frac{7\pi}{6} = 2^8 \cos(\pi + \frac{\pi}{6})$. $= -2^8 \cos \frac{\pi}{6} = -256 	imes \frac{\sqrt{3}}{2} = -128 \sqrt{3}$.

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

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