(sqrt{3}+i)^{10}+(sqrt{3}-i)^{10}= | vedclass

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(B) Let $z = \sqrt{3} + i$. Converting to polar form,$r = \sqrt{(\sqrt{3})^2 + 1^2} = 2$. The argument $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac

Vedclass · Accepted Answer

$1024$

Vedclass · Answer

(B) Let $z = \sqrt{3} + i$. Converting to polar form,$r = \sqrt{(\sqrt{3})^2 + 1^2} = 2$. The argument $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.So,$z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.By De Moivre's Theorem,$z^{10} = 2^{10}(\cos \frac{10\pi}{6} + i \sin \frac{10\pi}{6}) = 1024(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$.Since $\cos \frac{5\pi}{3} = \frac{1}{2}$ and $\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$,we have $z^{10} = 1024(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 512 - 512i\sqrt{3}$.Similarly,for $\bar{z} = \sqrt{3} - i$,$\bar{z}^{10} = 1024(\cos(-\frac{5\pi}{3}) + i \sin(-\frac{5\pi}{3})) = 1024(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 512 + 512i\sqrt{3}$.Adding these,$(\sqrt{3}+i)^{10} + (\sqrt{3}-i)^{10} = (512 - 512i\sqrt{3}) + (512 + 512i\sqrt{3}) = 1024$.

$(\sqrt{3}+i)^{10}+(\sqrt{3}-i)^{10}=$

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