The modulus of the complex number left(frac{2 | vedclass

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(A) Let $z = \left(\frac{2+i \sqrt{5}}{2-i \sqrt{5}}ight)^{10} + \left(\frac{2-i \sqrt{5}}{2+i \sqrt{5}}ight)^{10}$.Let $2 = r \cos 	heta$ and $\

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$2 \cos \left(20 \cos ^{-1}\left(\frac{2}{3}\right)\right)$

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(A) Let $z = \left(\frac{2+i \sqrt{5}}{2-i \sqrt{5}}ight)^{10} + \left(\frac{2-i \sqrt{5}}{2+i \sqrt{5}}ight)^{10}$.Let $2 = r \cos 	heta$ and $\sqrt{5} = r \sin 	heta$. Then $r = \sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4+5} = 3$.So,$\cos 	heta = \frac{2}{3}$ and $\sin 	heta = \frac{\sqrt{5}}{3}$.The expression becomes $z = \left(\frac{\cos 	heta + i \sin 	heta}{\cos 	heta - i \sin 	heta}ight)^{10} + \left(\frac{\cos 	heta - i \sin 	heta}{\cos 	heta + i \sin 	heta}ight)^{10}$.Using De Moivre's theorem,$\frac{\cos 	heta + i \sin 	heta}{\cos 	heta - i \sin 	heta} = e^{i	heta} / e^{-i	heta} = e^{i2	heta} = \cos 2	heta + i \sin 2	heta$.Thus,$z = (e^{i2	heta})^{10} + (e^{-i2	heta})^{10} = e^{i20	heta} + e^{-i20	heta} = 2 \cos(20	heta)$.Since $\cos 	heta = \frac{2}{3}$,we have $	heta = \cos^{-1}(\frac{2}{3})$.Therefore,$|z| = |2 \cos(20	heta)| = 2 \cos(20 \cos^{-1}(\frac{2}{3}))$,as $20	heta$ lies in the first quadrant where cosine is positive.

The modulus of the complex number $\left(\frac{2+i \sqrt{5}}{2-i \sqrt{5}}\right)^{10}+\left(\frac{2-i \sqrt{5}}{2+i \sqrt{5}}\right)^{10}$ is

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