JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $S = \{z = i(z^2 + \text{Re}(z))\}$ . Then, $\sum_{z \in S} |z|^2$ is equal to:

Choose the correct answer:

Explanation

Solution:

Let $z = x + iy$ , then $\bar{z} = x - iy$ .

Substituting in the equation: $x - iy = i(x^2 - y^2 + 2ixy + x)$

$\Rightarrow x - iy = -2xy + i(x^2 - y^2 + x)$

Comparing real and imaginary parts:

$x = -2xy$ and $-y = x^2 - y^2 + x$

$\Rightarrow x(1 + 2y) = 0$ and $(x+y)(x-y) + (x+y) = 0$

$\Rightarrow (x = 0 \text{ or } y = -1/2)$ and $(x+y)(x-y+1) = 0$

Case 1: If $x = 0$ , then $y = 0, 1$ .
Case 2: If $y = -1/2$ , then $(x - 1/2)(x + 1/2 + 1) = 0 \Rightarrow x = 1/2, -3/2$ .

The set $S = \{0 + 0i, 0 + i, \frac{1}{2} - \frac{i}{2}, -\frac{3}{2} - \frac{i}{2}\}$ .

$\sum_{z \in S} |z|^2 = (0^2+0^2) + (0^2+1^2) + (\frac{1}{4} + \frac{1}{4}) + (\frac{9}{4} + \frac{1}{4}) = 0 + 1 + \frac{1}{2} + \frac{10}{4} = 4$ .

Correct Option: (1) 4

Choose the correct answer:

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