If for $\mathrm{z}=\alpha+\mathrm{i} \beta,|\mathrm{z}+2|=\mathrm{z}+4(1+\mathrm{i})$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation

(A) $\mathrm{x}^{2}+7 \mathrm{x}+12=0$

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Complex Numbers

If for

z = α + i β, ∣ z + 2∣ = z + 4 (1 + i)

, then

α + β

and

α β

are the roots of the equation

Answer: (A)

Given

∣ z + 2∣ = z + 4 (1 + i)

, where

z = α + i β

.

LHS:

∣ z + 2∣ = ∣ (α + 2) + i β ∣ = (α + 2)^{2} + β^{2}

(real number).

RHS:

z + 4 (1 + i) = (α + 4) + i (β + 4)

.

For equality, the RHS must be real, so

β + 4 = 0

, giving

β = - 4

.

Then:

(α + 2)^{2} + 16 = α + 4

.

Squaring:

(α + 2)^{2} + 16 = (α + 4)^{2}

α^{2} + 4 α + 4 + 16 = α^{2} + 8 α + 16

4 α = 4

, so

α = 1

.

Thus

α + β = 1 + (- 4) = - 3

and

α β = 1 \times (- 4) = - 4

.

The quadratic with roots

- 3

and

- 4

x^{2} - (- 3 - 4) x + (- 3) (- 4) = x^{2} + 7 x + 12 = 0

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option