Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^{2}+|z|=0$, $z \in C$. Then $4(\alpha^{2}+\beta^{2})$ is equal to

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Complex Numbers

Let

α

and

β

be the sum and the product of all the non-zero solutions of the equation

(\overset{z}{ˉ})^{2} + ∣ z ∣ = 0

z \in C

. Then

4 (α^{2} + β^{2})

is equal to

Answer: (A)

Let

z = x + i y

, so

(\overset{z}{ˉ})^{2} = x^{2} - y^{2} - 2 i x y

and

∣ z ∣ = x^{2} + y^{2}

(\overset{z}{ˉ})^{2} + ∣ z ∣ = 0 \Rightarrow (x^{2} - y^{2} + x^{2} + y^{2}) - 2 i x y = 0.

Imaginary part:

x y = 0

$x = 0$ : $- y^{2} + ∣ y ∣ = 0 \Rightarrow ∣ y ∣ = y^{2} \Rightarrow y = 0, \pm 1$ , giving $z = \pm i$ .
$y = 0$ : $x^{2} + ∣ x ∣ = 0 \Rightarrow x = 0$ .

Non-zero solutions:

i

and

- i

. Sum

α = 0

, product

β = (i) (- i) = 1

4 (α^{2} + β^{2}) = 4 (0 + 1) = 4.

Option A.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option