If alpha, beta are the roots of the equation | vedclass

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(B) Given equation is $x^2-2x+2=0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get: $x = \frac{2 \pm \sqrt{4-8}}{2} = \frac

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$-2^{1011}$

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(B) Given equation is $x^2-2x+2=0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get: $x = \frac{2 \pm \sqrt{4-8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i$. Let $\alpha = 1+i$ and $\beta = 1-i$. Convert to polar form: $\alpha = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = \sqrt{2} e^{i\pi/4}$. $\beta = \sqrt{2}(\cos \frac{\pi}{4} - i \sin \frac{\pi}{4}) = \sqrt{2} e^{-i\pi/4}$. Then,$\alpha^{2020} = (\sqrt{2})^{2020} e^{i(2020\pi/4)} = 2^{1010} e^{i(505\pi)}$. Since $e^{i(505\pi)} = \cos(505\pi) + i \sin(505\pi) = -1 + 0 = -1$,$\alpha^{2020} = -2^{1010}$. Similarly,$\beta^{2020} = (\sqrt{2})^{2020} e^{-i(505\pi)} = 2^{1010} (-1) = -2^{1010}$. Therefore,$\alpha^{2020} + \beta^{2020} = -2^{1010} - 2^{1010} = -2 \cdot 2^{1010} = -2^{1011}$.

If $\alpha, \beta$ are the roots of the equation $x^2-2x+2=0$,then $\alpha^{2020}+\beta^{2020}=$

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