Solution
Maana a=x1+iy1 aur z=x+iy.
Yahan aˉ=x1−iy1 aur zˉ=x−iy hoga.
Set A ke liye condition:
Re(a+zˉ)=Re(x1+iy1+x−iy)=x1+x
Im(aˉ+z)=Im(x1−iy1+x+iy)=y−y1
Condition A: x_1 + x > |y - y_1|
Statement (S1) ke liye:
Sawal kehta hai z ek "real number" hai, matlab y=0.
Condition ban jayegi: x_1 + x > |-y_1| \implies x_1 + x > |y_1|.
Chunki x_1, y_1 > 0, toh ∣y1∣=y1.
Condition: x > y_1 - x_1.
Yeh condition hamesha sach nahi hai sabhi real numbers x ke liye (sirf y1−x1 se bade x ke liye sahi hai).
Isliye (S1) galat (False) hai.
Set B ke liye condition:
Re(a+zˉ)=x1+x
Im(aˉ+z)=y−y1
Condition B: x_1 + x < y - y_1
Statement (S2) ke liye:
Yahan bhi z real hai, toh y=0.
Condition ban jayegi: x_1 + x < -y_1 \implies x < -y_1 - x_1.
Yeh condition bhi sabhi real numbers x ke liye sahi nahi hai.
Isliye (S2) bhi galat (False) hai.
Correct Option: (2) both are false
