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

Solution

Maths sir · 4 steps

Put a = p + iq and z = x + iy, and both sets turn into half-planes cut by a single line.

1. Write in parts. Let a = p + iq and z = x + iy. Then a + z̄ = (p + x) + i(q - y), so Re(a + z̄) = p + x. Also ā + z = (p + x) + i(y - q), so Im(ā + z) = y - q.

   A: x - y + p + q > 0, B: x - y + p + q < 0

2. Real numbers only. A real number has y = 0. So a real x lies in A only when x > -(p + q), and in B only when x < -(p + q). Each set holds just one side of the point -(p + q) on the real line, never the whole line.
3. Test (S1) and (S2). For (S1), take a = 1 + i: the real number z = -3 gives -3 + 2 = -1, which is not > 0, so -3 ∉ A. For (S2), take a = -1 - i: the real number z = 3 gives 3 - 2 = 1, which is not < 0, so 3 ∉ B.
4. So. No choice of signs for Re(a) and Im(a) can make a half-line cover all real numbers. Both statements are false, which is option (a).

The idea to keep. Turn conditions on Re and Im into an inequality in x and y. A strict linear inequality gives a half-plane, which always leaves out part of any full line.

Answer from the NTA final answer key. Solution by PrepWiser.