QJEEM23Ma686c: JEE Main 2023 Mathematics - Class 11 JEEM23Ma686c Solution: For a ∈ C , let A = { z ∈ C : Re ( a + z ¯ ) > Im ( a ¯ + z ) } …

We are given that $a \in \mathbb{C}$ and $z \in \mathbb{C}$ .

Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \in \mathbb{R}$

We are also given two sets A and B defined as follows :

- A is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is greater than the imaginary part of $(\overline{a} + z)$ .

- B is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is less than the imaginary part of $(\overline{a} + z)$ .

Statement (S1) says : If the real part and imaginary part of $a$ are both positive, then the set A contains all the real numbers.

Statement (S2) says : If the real part and imaginary part of $a$ are both negative, then the set B contains all the real numbers.

We need to determine which of these statements are true.

Let's evaluate each statement.

1. Statement (S1) : For $z \in A$ ,

Re(a + \overline{z}) > Im(\overline{a} + z)

This can be re-written as x_1 + x_2 > y_2 - y_1

If we consider only real z (i.e. $y_2 = 0$ ) and given that x_1, y_1 > 0 , then the condition simplifies to x_2 > -(x_1 + y_1) .

This indicates that A covers a part of the negative real axis, but not the entire real axis. Therefore, Statement (S1) is false.

2. Statement (S2) : For $z \in B$ ,

Re(a + \overline{z}) < Im(\overline{a} + z)

This can be re-written as x_1 + x_2 < y_2 - y_1

If we consider only real z (i.e. $y_2 = 0$ ) and given that x_1, y_1 < 0 , then the condition simplifies to x_2 < -(x_1 + y_1) .

This indicates that B covers a part of the positive real axis, but not the entire real axis. Therefore, Statement (S2) is false.

Therefore, both (S1) and (S2) are false, so the answer is Option A : both are false.