Key Concepts and Formulas

• Complex Number Representation: A complex number $z$ can be represented as $z = x + iy$, where $x = \operatorname{Re}(z)$ is the real part and $y = \operatorname{Im}(z)$ is the imaginary part.
• Complex Conjugate: The complex conjugate of $z = x + iy$ is $\bar{z} = x - iy$.
• Real and Imaginary Parts of Sums: For complex numbers $z_1$ and $z_2$, $\operatorname{Re}(z_1 + z_2) = \operatorname{Re}(z_1) + \operatorname{Re}(z_2)$ and $\operatorname{Im}(z_1 + z_2) = \operatorname{Im}(z_1) + \operatorname{Im}(z_2)$.

Step 1: Express $a$ and $z$ in rectangular form and find their conjugates

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 need to express the given conditions in terms of these real variables. First, we find the conjugates of $a$ and $z$: $\bar{a} = x_1 - iy_1$ $\bar{z} = x_2 - iy_2$

Step 2: Calculate $\operatorname{Re}(a+\bar{z})$ and $\operatorname{Im}(\bar{a}+z)$

Now we calculate $a + \bar{z}$: $a + \bar{z} = (x_1 + iy_1) + (x_2 - iy_2) = (x_1 + x_2) + i(y_1 - y_2)$ Therefore, $\operatorname{Re}(a + \bar{z}) = x_1 + x_2$

Next, we calculate $\bar{a} + z$: $\bar{a} + z = (x_1 - iy_1) + (x_2 + iy_2) = (x_1 + x_2) + i(-y_1 + y_2)$ Therefore, $\operatorname{Im}(\bar{a} + z) = -y_1 + y_2 = y_2 - y_1$

Step 3: Define the conditions for sets A and B

The set A is defined by $\operatorname{Re}(a + \bar{z}) > \operatorname{Im}(\bar{a} + z)$. Substituting our expressions, we get: $x_1 + x_2 > y_2 - y_1$ $x_1 + x_2 + y_1 - y_2 > 0 \quad (*)$

The set B is defined by $\operatorname{Re}(a + \bar{z}) < \operatorname{Im}(\bar{a} + z)$. Substituting our expressions, we get: $x_1 + x_2 < y_2 - y_1$ $x_1 + x_2 + y_1 - y_2 < 0 \quad (**)$

Step 4: Analyze Statement (S1)

Statement (S1): If $\operatorname{Re}(a), \operatorname{Im}(a) > 0$ (i.e., $x_1 > 0$ and $y_1 > 0$), then the set A contains all the real numbers. If $z$ is a real number, then $z = x_2 + i(0)$, so $y_2 = 0$. Substituting $y_2 = 0$ into the inequality for set A $(*)$, we get: $x_1 + x_2 + y_1 - 0 > 0$ $x_1 + x_2 + y_1 > 0$ $x_2 > -x_1 - y_1$ Since $x_1 > 0$ and $y_1 > 0$, we have $x_1 + y_1 > 0$, so $-x_1 - y_1 < 0$.

The statement claims that all real numbers $x_2$ satisfy $x_2 > -x_1 - y_1$. However, this is false. For example, if $x_2 = -x_1 - y_1 - 1$, then $x_2 < -x_1 - y_1$, which contradicts the condition for $z$ to be in set A. Therefore, Statement (S1) is false.

Step 5: Analyze Statement (S2)

Statement (S2): If $\operatorname{Re}(a), \operatorname{Im}(a) < 0$ (i.e., $x_1 < 0$ and $y_1 < 0$), then the set B contains all the real numbers. If $z$ is a real number, then $z = x_2 + i(0)$, so $y_2 = 0$. Substituting $y_2 = 0$ into the inequality for set B $(**)$, we get: $x_1 + x_2 + y_1 - 0 < 0$ $x_1 + x_2 + y_1 < 0$ $x_2 < -x_1 - y_1$ Since $x_1 < 0$ and $y_1 < 0$, we have $x_1 + y_1 < 0$, so $-x_1 - y_1 > 0$.

The statement claims that all real numbers $x_2$ satisfy $x_2 < -x_1 - y_1$. However, this is false. For example, if $x_2 = -x_1 - y_1 + 1$, then $x_2 > -x_1 - y_1$, which contradicts the condition for $z$ to be in set B. Therefore, Statement (S2) is false.

Common Mistakes & Tips

• Be careful with the signs when manipulating inequalities. A common mistake is to forget to flip the inequality sign when multiplying or dividing by a negative number.
• When a statement claims that a set contains "all real numbers," test the condition with extreme values (very large positive and very large negative numbers). If the condition fails for even one real number, the statement is false.
• Understanding the rectangular form of complex numbers ($z=x+iy$) is essential for solving problems involving real and imaginary parts.

Summary

Both statements (S1) and (S2) are false because in each case, assuming $z$ is real leads to a condition of the form $x_2 > \text{constant}$ or $x_2 < \text{constant}$, which cannot be true for all real numbers $x_2$. Therefore, the correct option is (A).

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).