For [a C], let [A=z C: Re(a+z) Im(a+z)] and [B=z C: Re(a+z) 0]

Question

For $$a \in \mathbb{C}$$, let $$\mathrm{A}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)\}$$ and $$\mathrm{B}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) 0$$

Accepted Answer

We are given that $a \in \mathbb{C}$ and $z \in \mathbb{C}$. Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \in \mathbb{R}$ We are also given two sets A and B defined as follows : - A is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is greater than the imaginary part of $(\overline{a} + z)$. - B is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is less than the imaginary part of $(\overline{a} + z)$. Statement (S1) says : If the real part and imaginary part of $a$ are both positive, then the set A contains all the real numbers. Statement (S2) says : If the real part and imaginary part of $a$ are both negative, then the set B contains all the real numbers. We need to determine which of these statements are true. Let's evaluate each statement. 1. Statement (S1) : For $z \in A$, $Re(a + \overline{z}) > Im(\overline{a} + z)$ This can be re-written as $x_1 + x_2 > y_2 - y_1$ If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 > 0$, then the condition simplifies to $x_2 > -(x_1 + y_1)$. This indicates that A covers a part of the negative real axis, but not the entire real axis. Therefore, Statement (S1) is false. 2. Statement (S2) : For $z \in B$, $Re(a + \overline{z}) < Im(\overline{a} + z)$ This can be re-written as $x_1 + x_2 < y_2 - y_1$ If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 < 0$, then the condition simplifies to $x_2 < -(x_1 + y_1)$. This indicates that B covers a part of the positive real axis, but not the entire real axis. Therefore, Statement (S2) is false. Therefore, both (S1) and (S2) are false, so the answer is Option A : both are false.

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