Let $\alpha, \beta$ be the roots of the equation $x^{2}-a x-b=0$ with $\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)$. Let $P_{n}=\alpha^{n}-\beta^{n}$. If $P_{3}=-5 \sqrt{7} i$, $P_{4}=-3 \sqrt{7} i$, $P_{5}=11 \sqrt{7} i$ and $P_{6}=45 \sqrt{7} i$, then $\left|\alpha^{4}+\beta^{4}\right|$ is equal to ______.

Complex Numbers — JEE Main · 2025 · 23 Jan 2025 · Shift 2 Question with Solution

Let

α, β

be the roots of the equation

x^{2} - a x - b = 0

with

Im (α) < Im (β)

. Let

P_{n} = α^{n} - β^{n}

. If

P_{3} = - 57 i

P_{4} = - 37 i

P_{5} = 117 i

and

P_{6} = 457 i

, then

α^{4} + β^{4}

is equal to ______.

Answer: 31

Since

α, β

are roots of

x^{2} - a x - b = 0

, they satisfy

α^{2} = a α + b

and

β^{2} = a β + b

. Multiplying by

α^{n - 2}

and

β^{n - 2}

and subtracting gives the recurrence

P_{n} = a P_{n - 1} + b P_{n - 2}

Using

P_{6} = a P_{5} + b P_{4}

457 i = a (117 i) + b (- 37 i) \Rightarrow 11 a - 3 b = 45 (i)

Using

P_{5} = a P_{4} + b P_{3}

117 i = a (- 37 i) + b (- 57 i) \Rightarrow - 3 a - 5 b = 11 (ii)

Solving (i) and (ii) simultaneously gives

a = 3

b = - 4

.

Since

α + β = a = 3

and

α β = - b = 4

, the discriminant

a^{2} + 4 b = 9 - 16 = - 7 < 0

, so

α, β

are complex conjugates:

α = \frac{3 - i 7}{2}, β = \frac{3 + i 7}{2}

Then

α^{2} = \frac{9 - 6 i 7 - 7}{4} = \frac{2 - 6 i 7}{4} = \frac{1 - 3 i 7}{2}

α^{4} = (\frac{1 - 3 i 7}{2})^{2} = \frac{1 - 6 i 7 - 63}{4} = \frac{- 62 - 6 i 7}{4} = \frac{- 31 - 3 i 7}{2}

Since

β = \overset{α}{ˉ}

β^{4} = \overline{α^{4}} = \frac{- 31 + 3 i 7}{2}

, so

α^{4} + β^{4} = \frac{- 31 - 3 i 7 - 31 + 3 i 7}{2} = - 31

∣ α^{4} + β^{4} ∣ = 31

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Integer answer