Let alpha and beta be the roots of the equati | vedclass

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(A) Given the equation $x^{2} + (2i - 1) = 0$,we have $x^{2} = 1 - 2i$.Since $\alpha$ and $\beta$ are the roots,$\alpha^{2} = 1 - 2i$ and $\beta^{2} =

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$50$

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(A) Given the equation $x^{2} + (2i - 1) = 0$,we have $x^{2} = 1 - 2i$.Since $\alpha$ and $\beta$ are the roots,$\alpha^{2} = 1 - 2i$ and $\beta^{2} = 1 - 2i$.Thus,$\alpha^{8} = (\alpha^{2})^{4} = (1 - 2i)^{4}$ and $\beta^{8} = (\beta^{2})^{4} = (1 - 2i)^{4}$.Therefore,$\alpha^{8} = \beta^{8}$.We need to find $|\alpha^{8} + \beta^{8}| = |2\alpha^{8}| = 2|\alpha^{8}| = 2|\alpha^{2}|^{4}$.Calculate the modulus $|\alpha^{2}| = |1 - 2i| = \sqrt{1^{2} + (-2)^{2}} = \sqrt{1 + 4} = \sqrt{5}$.Then,$|\alpha^{8} + \beta^{8}| = 2(\sqrt{5})^{4} = 2(5^{2}) = 2(25) = 50$.

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2} + (2i - 1) = 0$. Then,the value of $|\alpha^{8} + \beta^{8}|$ is equal to

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