JEE Main 2023 (Online) 11th April Evening ShiftQuestion 1
For $a\in\mathbb{C}$, let $A=\{z\in\mathbb{C}:\operatorname{Re}(a+\bar{z})>\operatorname{Im}(\bar{a}+z)\}$ and $B=\{z\in\mathbb{C}:\operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}$. Then among the two statements:
(S1) : If $\operatorname{Re}(a),\operatorname{Im}(a)>0$, 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,
Complex Numbers
Solution
$\operatorname{Re}(a+\bar z)=\operatorname{Re}a+x$, $\operatorname{Im}(\bar a+z)=-\operatorname{Im}a+y$. For real $z=x$ ($y=0$): $A$ needs $x>-(\operatorname{Re}a+\operatorname{Im}a)$, which fails for very negative $x$. So S1 is false. $B$ needs $x<-(\operatorname{Re}a+\operatorname{Im}a)=$ a positive constant, which fails for large $x$. So S2 is false. Both false: option B.