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For $a \in C$, let $A =\{z \in C: \operatorname{Re}( a +\overline{ z }) > \operatorname{Im}(\bar{a}+z)\}$ and $B=\{z \in C: \operatorname{Re}(a+\bar{z}) < \operatorname{Im}(\bar{a}+z)\}$. Then among the two statements :
$(S 1)$ : 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,
Only $(S1)$ is true
both are false
Only $(S2)$ is true
Both are true
Solution
Let $a=x_1+i y_1 z=x+i y$
Now $\operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)$
$\therefore x _1+ x >- y _1+ y$
$x _1=2, y _1=10, x =-12, y =0$
Given inequality is not valid for these values.
$S 1$ is false.
Now $\operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)$
$x _1+ x < – y _1+ y$
$x _1=-2, y _1=-10, x =12, y =0$
Given inequality is not valid for these values.
$S2$ is false.
