"Let $\alpha, \beta$ be the roots of the equation $x^2-\mathrm{ax}-\mathrm{b}=0$ with $\operatorname{Im}(\alpha)"

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We begin with the equations for the roots:

$\alpha + \beta = \mathrm{a}$

$\alpha \beta = -\mathrm{b}$

Given:

$ \mathrm{P}_6 = \mathrm{aP}_5 + \mathrm{bP}_4 $

$ \mathrm{P}_5 = \mathrm{aP}_4 + \mathrm{bP}_3 $

Using the given values:

For $\mathrm{P}_6$:

$ 45 \sqrt{7} i = \mathrm{a} \times 11 \sqrt{7} i + \mathrm{b}(-3 \sqrt{7}) i $

Simplifying, we obtain:

$ 45 = 11 \mathrm{a} - 3 \mathrm{b} \quad \text{(Equation 1)} $

For $\mathrm{P}_5$:

$ 11 \sqrt{7} i = \mathrm{a}(-3 \sqrt{7} i) + \mathrm{b}(-5 \sqrt{7} i) $

Simplifying, we obtain:

$ 11 = -3 \mathrm{a} - 5 \mathrm{b} \quad \text{(Equation 2)} $

Solving these linear equations, we find:

$ \mathrm{a} = 3 $

$ \mathrm{b} = -4 $

Now, we calculate $\left|\alpha^4 + \beta^4\right|$ using the relation:

$ \left|\alpha^4 + \beta^4\right| = \sqrt{(\alpha^4 - \beta^4)^2 + 4 (\alpha^4 \beta^4)} $

From $\mathrm{b} = -4$, we know:

$ \alpha \beta = -\mathrm{b} = 4 \quad \Rightarrow \alpha^4 \beta^4 = (\alpha \beta)^4 = 4^4 = 256 $

Substitute into the relation:

$ \left|\alpha^4 + \beta^4\right| = \sqrt{(-63) + 1024} $

$ = \sqrt{961} = 31 $

Thus, $\left|\alpha^4 + \beta^4\right|$ is equal to 31.

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