Let $$\alpha,\beta$$ be the roots of the equation $$x^{2}-ax-b=0$$ with $$Im(\alpha) < Im(\beta)$$. Let $$P_{n}=\alpha^{n}-\beta^{n}$$. If $$P_{3}=-5\sqrt{7}i,P_{4}-3\sqrt{7}i,P_{5}=11\sqrt{7}i\text{ and }P_{}=45\sqrt{7}i$$.then $$|\alpha^{4}+\beta^{4}|$$ is equal to

We have $$\alpha, \beta$$ as roots of $$x^2 - ax - b = 0$$. Given $$P_n = \alpha^n - \beta^n$$, with $$P_3 = -5\sqrt{7}i$$, $$P_4 = -3\sqrt{7}i$$, $$P_5 = 11\sqrt{7}i$$, $$P_6 = 45\sqrt{7}i$$.

Since $$\alpha, \beta$$ satisfy $$x^2 = ax + b$$, the sequence $$P_n$$ satisfies the recurrence relation

$$P_n = aP_{n-1} + bP_{n-2}$$

Substituting $$n=5$$ gives

$$P_5 = aP_4 + bP_3$$

Therefore,

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

$$11 = -3a - 5b$$ ... (i)

For $$n=6$$,

$$P_6 = aP_5 + bP_4$$

Thus,

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

$$45 = 11a - 3b$$ ... (ii)

From (i): $$3a + 5b = -11$$ ... (i')

From (ii): $$11a - 3b = 45$$ ... (ii')

Multiply (i') by 3: $$9a + 15b = -33$$

Multiply (ii') by 5: $$55a - 15b = 225$$

Adding: $$64a = 192 \Rightarrow a = 3$$

From (i'): $$9 + 5b = -11 \Rightarrow b = -4$$

We know: $$\alpha + \beta = a = 3$$ and $$\alpha\beta = -b = 4$$

$$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 9 - 8 = 1$$

$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = 1 - 2(16) = 1 - 32 = -31$$

$$|\alpha^4 + \beta^4| = |-31| = 31$$

The answer is 31.