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(C) The given equation is $x^2-2x+4=0$.Using the quadratic formula,the roots are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm i\sqrt{3}$.In polar form,$1

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$2^{13}$

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(C) The given equation is $x^2-2x+4=0$.Using the quadratic formula,the roots are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm i\sqrt{3}$.In polar form,$1 + i\sqrt{3} = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$ and $1 - i\sqrt{3} = 2e^{-i\pi/3}$.Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$.Then $\alpha^{12} = (2e^{i\pi/3})^{12} = 2^{12} e^{i4\pi} = 2^{12}(1) = 2^{12}$.Similarly,$\beta^{12} = (2e^{-i\pi/3})^{12} = 2^{12} e^{-i4\pi} = 2^{12}(1) = 2^{12}$.Therefore,$\alpha^{12} + \beta^{12} = 2^{12} + 2^{12} = 2 	imes 2^{12} = 2^{13}$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^{12}+\beta^{12}=$

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