Question

For \(a∈ C, let A = {z∈C : Re (a + z ) > Im ( a + z)}\) and \(B = {z∈C : Re (a + z ) < Im ( a + z)}.\) 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

• A. only (S2) is true
• B. only (S1) is true
• C. both are true
• D. both are false

Answer 1

The Correct Option is D

We are tasked with analyzing the given inequalities for the statements \( S1 \) and \( S2 \) involving complex numbers \( a = x_1 + iy_1 \) and \( z = x + iy \).

Step 1: Analyze the Inequality 

The inequality is given as: \[ \text{Re}(a + z) > \text{Im}(a + z). \]

Expanding both sides: \[ \text{Re}(a + z) = x_1 + x, \quad \text{Im}(a + z) = -y_1 + y. \]

The inequality becomes: \[ x_1 + x > -y_1 + y. \]

Step 2: Check Statement \( S1 \)

For \( S1 \), we are given: \[ x_1 = 2, \, y_1 = 10, \, x = -12, \, y = 0. \]

Substitute these values into the inequality: \[ x_1 + x > -y_1 + y \implies 2 - 12 > -(10) + 0 \implies -10 > -10. \]

This inequality is not valid. Hence, \( S1 \) is false.

Step 3: Check Statement \( S2 \)

For \( S2 \), the inequality is: \[ \text{Re}(a + z) < \text{Im}(a + z). \]

Expanding: \[ x_1 + x < -y_1 + y. \]

For \( S2 \), we are given: \[ x_1 = -2, \, y_1 = -10, \, x = 12, \, y = 0. \]

Substitute these values: \[ x_1 + x < -y_1 + y \implies -2 + 12 < -(-10) + 0 \implies 10 < 10. \]

This inequality is not valid. Hence, \( S2 \) is false.

Final Answer:

Both \( S1 \) and \( S2 \) are false.