Let \(\alpha, \beta\) be the roots of the equation \(\mathrm{x}^{2}-\sqrt{6} \mathrm{x}+3=0\) such that \(\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)\). Let \(\mathrm{a}, \mathrm{b}\) be integers not divisible by 3 and n be a natural number such that \(\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^{n}(a+i b), i=\sqrt{-1}\). Then \(\mathrm{n}+\mathrm{a}+\mathrm{b}\) is equal to

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Complex Numbers

Let

α, β

be the roots of the equation

x^{2} - 6 x + 3 = 0

such that

Im (α) > Im (β)

. Let

a, b

be integers not divisible by 3 and n be a natural number such that

\frac{α ^{99}}{β} + α^{98} = 3^{n} (a + ib), i = - 1

. Then

n + a + b

is equal to

Answer: 49

x^{2} - 6 x + 3 = 0 \Rightarrow x = \frac{6 \pm 6 - 12}{2} = \frac{6}{2} (1 \pm i)

.

Since

Im (α) > Im (β)

α = \frac{6}{2} (1 + i)

β = \frac{6}{2} (1 - i)

.

Write these in polar form:

\frac{6}{2} (1 + i) = \frac{6}{2} \cdot 2 e^{iπ /4} = 3 e^{iπ /4}

, and similarly

β = 3 e^{- iπ /4}

.

So

\frac{α}{β} = e^{iπ /2} = i

\frac{α ^{99}}{β} + α^{98} = α^{98} (\frac{α}{β} + 1) = α^{98} (i + 1)

α^{98} = (3)^{98} e^{i 98 π /4} = 3^{49} e^{i 49 π /2}

. Since

49 π /2 = 24 π + π /2

e^{i 49 π /2} = e^{iπ /2} = i

.

So

α^{98} = 3^{49} i

, and

α^{98} (i + 1) = 3^{49} i (i + 1) = 3^{49} (i^{2} + i) = 3^{49} (- 1 + i)

.

Comparing with

3^{n} (a + ib)

n = 49, a = - 1, b = 1

(neither divisible by

3

n + a + b = 49 - 1 + 1 = 49

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer