Mathematics · Complex Numbers

JEE Main 2024 — 4 April, Shift 1 — Question 15

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 $4\left(\alpha^{2}+\beta^{2}\right)$ is equal to :

Option A:
6
Option B:
4
Correct
Option C:
8
Option D:
2

Answer: B

Step-by-step solution

$\mathrm{z}=\mathrm{x}+\mathrm{iy}$

$\bar{z}=x-i y$

$\bar{z}^{2}=x^{2}-y^{2}-2 i x y$

$\Rightarrow x^{2}-y^{2}-2 i x y+\sqrt{x^{2}+y^{2}}=0$

$\Rightarrow \mathrm{x}=0 \quad$ or $\quad \mathrm{y}=0$

$-y^{2}+|y|=0 \quad x^{2}+|x|=0$

$|y|=|y|^{2} \quad \Rightarrow x=0$

$y=0, \pm 1$

$\Rightarrow \mathrm{i},-\mathrm{i}$

$\Rightarrow \alpha=\mathrm{i}-\mathrm{i}=0$

are roots

$\beta=\mathrm{i}(-\mathrm{i})=1$

$4(0+1)=4$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Complex Numbers
Topic: Properties of Complex Numbers