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

$\begin{array}{l}\alpha ,\beta are roots of x^2+\left(2i-1\right)=0\Rightarrow x^2=1-2i\\ \Rightarrow {\mathrm{\alpha }}^2=1-2\mathrm{i}\text{ and }{\mathrm{\beta }}^2=1-2\mathrm{i}\\ \Rightarrow {\mathrm{\alpha }}^2={\mathrm{\beta }}^2=1-2\mathrm{i}\\ \Rightarrow {\mathrm{\alpha }}^8={\mathrm{\beta }}^8\end{array}$
Hence $\left|{\mathrm{\alpha }}^8+{\mathrm{\beta }}^8\right|=2{\left|{\mathrm{\alpha }}^2\right|}^4=2(\sqrt{1+4}{)}^4=50$

Explanation:

Given: α and β be the roots of x² + (2i - 1) = 0

→ x² + (2i - 1) = 0

→ x² = 1 - 2i

→ α² = 1 - 2i and β² = 1 - 2i

→ α² = β²

→ (α²)⁴ = (β²)⁴

→ α⁸ = β⁸

∴ α⁸ + β⁸ = 2α⁸

∴ α⁸ + β⁸ = 2(α²)⁴

→ α⁸ + β⁸ = 2(1 - 2i)⁴

→ |α⁸ + β⁸| = 2|1 - 2i|⁴

→ |α⁸ + β⁸| = 2 × ( √(1² + (-2)²) )⁴

→ |α⁸ + β⁸| = 2 × (√5)⁴

→ |α⁸ + β⁸| = 2(25) = 50