For \(a \in \mathbb{C}\), let \(A=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z})>\operatorname{Im}(\bar{a}+z)\}\) and \(B=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) 0\), then the set A contains all the real numbers (S2): If \(\operatorname{Re}(a), \operatorname{Im}(a)<0\), then the set B contains all the real numbers,

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Complex Numbers

For

a \in C

, let

A = {z \in C : Re (a + \overset{z}{ˉ}) > Im (\overset{a}{ˉ} + z)}

and

B = {z \in C : Re (a + \overset{z}{ˉ}) < Im (\overset{a}{ˉ} + z)}

. Then among the two statements:
(S1): If

Re (a), Im (a) > 0

, then the set A contains all the real numbers
(S2): If

Re (a), Im (a) < 0

, then the set B contains all the real numbers,

Answer: (B)

Let

a = x_{1} + i y_{1}

and

z = x + i y

.

For set A:

Re (a + \overset{z}{ˉ}) > Im (\overset{a}{ˉ} + z)

x_{1} + x > - y_{1} + y ⟹ x_{1} + y_{1} > y - x

For real

z

(i.e.,

y = 0

x_{1} + y_{1} > - x

This is NOT satisfied for all real

x

. Counterexample:

x_{1} = 2, y_{1} = 10, x = - 12

gives

12 > 12

, which is false. So (S1) is false.

For set B:

Re (a + \overset{z}{ˉ}) < Im (\overset{a}{ˉ} + z)

x_{1} + x < - y_{1} + y

For real

z

(

y = 0

x_{1} + x < - y_{1}

This is NOT satisfied for all real

x

. Counterexample:

x_{1} = - 2, y_{1} = - 10, x = 12

gives

10 < 10

, which is false. So (S2) is false.

Both statements are false.

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option