Let $\mathrm{p}, \mathrm{q} \in \mathbb{R}$ and $(1-\sqrt{3} \mathrm{i})^{200}=2^{199}(\mathrm{p}+\mathrm{iq})$, $\mathrm{i}=\sqrt{-1}$. Then $\mathrm{p}+\mathrm{q}+\mathrm{q}^{2}$ and $\mathrm{p}-\mathrm{q}+\mathrm{q}^{2}$ are roots of the equation.

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Complex Numbers

Let

p, q \in R

and

(1 - 3 i)^{200} = 2^{199} (p + iq)

i = - 1

. Then

p + q + q^{2}

and

p - q + q^{2}

are roots of the equation.

Answer: (B)

Write

1 - 3 i = 2 e^{- iπ /3}

, so

(1 - 3 i)^{200} = 2^{200} e^{- i 200 π /3} = 2^{200} e^{- i 2 π /3} = 2^{200} (- \frac{1}{2} - \frac{3}{2} i)

. Dividing by

2^{199}

gives

p + i q = - 1 - 3 i

, so

p = - 1

and

q = - 3

. Then

p + q + q^{2} = 2 - 3

and

p - q + q^{2} = 2 + 3

, whose sum is

4

and product is

1

. The required equation is

x^{2} - 4 x + 1 = 0.

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option