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(B) Let $a = x_1 + i y_1$ and $z = x + i y$, where $x, y, x_1, y_1 \in \mathbb{R}$.For set $A$, the condition is $\operatorname{Re}(a + \bar{z}) > \op

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Both are false

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(B) Let $a = x_1 + i y_1$ and $z = x + i y$, where $x, y, x_1, y_1 \in \mathbb{R}$.For set $A$, the condition is $\operatorname{Re}(a + \bar{z}) > \operatorname{Im}(\bar{a} + z)$.$\operatorname{Re}(x_1 + i y_1 + x - i y) > \operatorname{Im}(x_1 - i y_1 + x + i y)$$x_1 + x > -y_1 + y \implies y If $z$ is a real number, then $y = 0$. The condition becomes $0  -(x_1 + y_1)$. This is not true for all $x \in \mathbb{R}$ (e.g., choose $x$ very small). Thus, $(S1)$ is false.For set $B$, the condition is $\operatorname{Re}(a + \bar{z}) $x_1 + x  x + x_1 + y_1$.If $z$ is a real number, then $y = 0$. The condition becomes $0 > x + x_1 + y_1$, which implies $x Therefore, both statements are false.

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