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(C) Given $|x|=1$ and $\operatorname{Arg}(x)=2\alpha$,we have $x=e^{i 2\alpha}$.Given $|y|=1$ and $\operatorname{Arg}(y)=3\beta$,we have $y=e^{i 3\bet

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(C) Given $|x|=1$ and $\operatorname{Arg}(x)=2\alpha$,we have $x=e^{i 2\alpha}$.Given $|y|=1$ and $\operatorname{Arg}(y)=3\beta$,we have $y=e^{i 3\beta}$.Then $x^6 y^4 = (e^{i 2\alpha})^6 (e^{i 3\beta})^4 = e^{i 12\alpha} e^{i 12\beta} = e^{i 12(\alpha+\beta)}$.Given $\alpha+\beta = \frac{\pi}{36}$,we substitute this into the expression:$x^6 y^4 = e^{i 12(\frac{\pi}{36})} = e^{i \frac{\pi}{3}}$.Now,$x^6 y^4 + \frac{1}{x^6 y^4} = e^{i \frac{\pi}{3}} + e^{-i \frac{\pi}{3}}$.Using Euler's formula $e^{i 	heta} + e^{-i 	heta} = 2 \cos 	heta$,we get:$2 \cos(\frac{\pi}{3}) = 2 	imes \frac{1}{2} = 1$.

$x$ and $y$ are two complex numbers such that $|x|=|y|=1$. If $\operatorname{Arg}(x)=2 \alpha$,$\operatorname{Arg}(y)=3 \beta$,and $\alpha+\beta=\frac{\pi}{36}$,then $x^6 y^4+\frac{1}{x^6 y^4}=$

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