For a ∈ C, let A = {z ∈ C : Re(a + z̄) Im(ā + z)} and B = {z ∈ C : Re(a + z̄)

Q: Wait, why do we assume $y_2 = 0$ for the real numbers? Isn't $z$ a complex number?

That is a fantastic question! You are absolutely right that $z$ is a complex number, but the question specifically asks if the set contains *all* real numbers. By definition, a complex number $z = x_2 + iy_2$ is a real number if and only if its imaginary part $y_2$ is zero. So, to test if the set contains all real numbers, we must restrict our focus to the subset of the complex plane where $y_2 = 0$.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! While physics problems often rely on conservation laws, this is a pure mathematics problem involving the geometry of the complex plane. Here, our 'conservation' is the logical consistency of the inequalities. We are essentially mapping the regions defined by the inequalities onto the real axis.

Q: Is it possible for the set to contain all real numbers if the inequality was different?

Yes, absolutely! If the inequality had simplified to something like $0 > -1$, which is always true, then the set would indeed contain all real numbers. The fact that we ended up with a condition like $x_2 > -k$ (a ray) instead of a tautology is exactly why the statements are false. Keep looking for those boundary conditions!

Question

For $a \in \mathbb{C}$, let $A = \{z \in \mathbb{C} : ext{Re}(a + \bar{z}) > ext{Im}(\bar{a} + z)\}$ and $B = \{z \in \mathbb{C} : ext{Re}(a + \bar{z}) < ext{Im}(\bar{a} + z)\}$. Then among the two statements: (S1): If $ ext{Re}(a), ext{Im}(a) > 0$, then the set $A$ contains all the real numbers. (S2): If $ ext{Re}(a), ext{Im}(a) < 0$, then the set $B$ contains all the real numbers. Options: (A) Only (S2) is true (B) only (S1) is true (C) Both are true (D) Both are false

Sigma Percentile · Accepted Answer

### Analyzing the Setup The notation $z \in \mathbb{C}$ and $a \in \mathbb{C}$ can feel like a fog. Let us clear it by using the Cartesian form. We define our constant $a = x_1 + iy_1$ and our variable $z = x_2 + iy_2$, where $x_1, y_1, x_2, y_2$ are all real numbers. This is our anchor. The conjugate of $a$ is $\bar{a} = x_1 - iy_1$, and the conjugate of $z$ is $\bar{z} = x_2 - iy_2$. By simply flipping the sign of the imaginary part, we have unlocked the key to the entire problem. ### The Algebraic Heavy Lifting Now, let us evaluate the expressions inside the inequalities. For the left side, we have: $$a + \bar{z} = (x_1 + iy_1) + (x_2 - iy_2) = (x_1 + x_2) + i(y_1 - y_2)$$ Therefore, the real part is simply $ ext{Re}(a + \bar{z}) = x_1 + x_2$. For the right side, we have: $$\bar{a} + z = (x_1 - iy_1) + (x_2 + iy_2) = (x_1 + x_2) + i(y_2 - y_1)$$ Thus, the imaginary part is $ ext{Im}(\bar{a} + z) = y_2 - y_1$. We have successfully translated the complex expressions into simple real-valued components. ### The Boundary Line Now, let us look at the condition for set $A$: $x_1 + x_2 > y_2 - y_1$. Rearranging this to isolate the variables of $z$ on one side, we get: $$x_2 - y_2 > -(x_1 + y_1)$$ This is the equation of a half-plane in the complex plane. Similarly, for set $B$, the condition is $x_2 - y_2 < -(x_1 + y_1)$. These two sets are simply the two sides of the line $x_2 - y_2 = -(x_1 + y_1)$. ### Testing the Real Axis The question asks if these sets contain *all* real numbers. A complex number $z$ is real if and only if its imaginary part $y_2 = 0$. Let us substitute $y_2 = 0$ into our inequalities. For set $A$, the condition becomes $x_2 > -(x_1 + y_1)$. For set $B$, it becomes $x_2 < -(x_1 + y_1)$. Now, look at statement (S1): if $ ext{Re}(a) > 0$ and $ ext{Im}(a) > 0$, then $x_1 > 0$ and $y_1 > 0$. This means $-(x_1 + y_1)$ is a negative number. Let us call it $-k$ where $k > 0$. The condition for set $A$ becomes $x_2 > -k$. This is the interval $(-k, \infty)$, which is just a ray, not the entire real line. Thus, **(S1) is false**. Similarly, for statement (S2), if $ ext{Re}(a) < 0$ and $ ext{Im}(a) < 0$, then $x_1 < 0$ and $y_1 < 0$. This means $-(x_1 + y_1)$ is a positive number, say $k$. The condition for set $B$ becomes $x_2 < k$. This is the interval $(-\infty, k)$, which is also just a ray. Thus, **(S2) is also false**. We have arrived at our destination: **both statements are false**.

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Visualized Solution

The Complex Sets $A$ and $B$

Cartesian Representation

Complex Conjugates

Evaluating $Re (a + \overset{z}{ˉ})$

Evaluating $Im (\overset{a}{ˉ} + z)$

Inequality for Set $A$

Inequality for Set $B$

Intersection with Real Numbers

Evaluating Statement (S1)

Evaluating Statement (S2)

Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Algebraic Heavy Lifting

The Boundary Line

Testing the Real Axis

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