Let z1 and z2 be two complex numbers such that z̄1 = iz̄2 and arg(z1/z̄2) = π. Then
(A) arg z2 = π/4

Q: Wait, why does $\arg(\bar{z}) = -\arg(z)$? I always get confused with the sign.

Think of the complex plane as a map. If $z$ is at an angle $\theta$ above the real axis, its conjugate $\bar{z}$ is its mirror image below the real axis. Therefore, the angle of $\bar{z}$ is simply the negative of the angle of $z$. It is a perfect symmetry! Just remember: reflection across the real axis flips the sign of the argument.

Q: Why can't we just use the modulus? The problem doesn't give us any information about the lengths.

That is a brilliant observation! You are absolutely right. The problem provides no information about the modulus (the length of the vectors), which is a huge hint that the answer must be independent of the lengths. We are working entirely in the domain of angles (arguments). When you see a problem like this, focus solely on the rotational properties.

Q: Why is $\arg(-i) = -\frac{\pi}{2}$? Can't it be $\frac{3\pi}{2}$?

In the context of the principal argument, we usually restrict the range to $(-\pi, \pi]$. The complex number $-i$ sits on the negative imaginary axis. Moving clockwise from the positive real axis, we reach it at $-\frac{\pi}{2}$. While $\frac{3\pi}{2}$ is coterminal, using $-\frac{\pi}{2}$ is standard and keeps our algebraic calculations clean and consistent.

Question

Let $z_1$ and $z_2$ be two complex numbers such that $\bar{z}_1 = i\bar{z}_2$ and $\arg\left(\frac{z_1}{\bar{z}_2}\right) = \pi$. Then

Options:

(A) $\arg z_2 = \frac{\pi}{4}$

(B) $\arg z_2 = -\frac{3\pi}{4}$

(C) $\arg z_1 = \frac{\pi}{4}$

(D) $\arg z_1 = -\frac{3\pi}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are choreographing a dance between two complex numbers, $z_1$ and $z_2$.

In the JEE Advanced arena, complex numbers are often presented as abstract algebraic entities, but I want you to see them as vectors—arrows pointing from the origin into the complex plane. When we manipulate these numbers, we are essentially rotating and reflecting these vectors.

### Decoding the Conjugate

We begin with the given condition: $\bar{z}_1 = i\bar{z}_2$. At first glance, this looks like a static equation.

To understand the relationship between $z_1$ and $z_2$ directly, we apply the conjugate operation to both sides. Using the property $\overline{z \cdot w} = \bar{z} \cdot \bar{w}$, we obtain:

$$\overline{(\bar{z}_1)} = \overline{(i\bar{z}_2)}$$

Since the conjugate of a conjugate returns the original number, and the conjugate of $i$ is $-i$, we arrive at the elegant relationship:

$$z_1 = -iz_2$$

This tells us that $z_1$ is simply $z_2$ rotated by $-90^\circ$ (or $-\frac{\pi}{2}$ radians).

### The Argument Transformation

Now, let us shift our focus to the second condition: $\arg\left(\frac{z_1}{\bar{z}_2}\right) = \pi$. We utilize the property that the argument of a quotient is the difference of the arguments:

$$\arg(z_1) - \arg(\bar{z}_2) = \pi$$

Recalling that $\arg(\bar{z}_2) = -\arg(z_2)$, we substitute this into our equation:

$$\arg(z_1) - (-\arg(z_2)) = \pi \Rightarrow \arg(z_1) + \arg(z_2) = \pi$$

This is our second key equation. We have successfully translated a complex division into a simple linear sum of angles.

### The System of Equations

We are now in the home stretch. We have the following system:

1. From $z_1 = -iz_2$, we know $\arg(z_1) = \arg(-i) + \arg(z_2) = \arg(z_2) - \frac{\pi}{2}$.
2. From our second derivation: $\arg(z_1) + \arg(z_2) = \pi$.

Substituting the first into the second:

$$\left(\arg(z_2) - \frac{\pi}{2}\right) + \arg(z_2) = \pi$$

$$2\arg(z_2) = \frac{3\pi}{2} \Rightarrow \arg(z_2) = \frac{3\pi}{4}$$

Finally, we find $\arg(z_1)$ by substituting back:

$$\arg(z_1) = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$$

### Conclusion: The Elegance of Symmetry

We found that $z_2$ points into the second quadrant at $\frac{3\pi}{4}$, and $z_1$ points into the first quadrant at $\frac{\pi}{4}$.

The beauty of this problem lies not in the calculation, but in the realization that complex numbers are just vectors waiting to be understood. Keep this perspective, and no problem will ever be too daunting. You have mastered the dance!

Animated Solution for Mathematics - Complex Numbers: Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\overset{z}{ˉ}_{1} = i \overset{z}{ˉ}_{2}$ and $ar g (\frac{z _{1}}{z ˉ _{2}}) = π$ . Then

Select Answer:

Visualized Solution

Problem Statement

Conjugating the First Condition

Simplifying the Relation

Applying Argument on Both Sides

Evaluating $ar g (- i)$

Analyzing the Second Condition

Property of Conjugate Argument

Combining the Two Equations

Solving for $ar g (z_{2})$

Solving for $ar g (z_{1})$

Final Conclusion

The Sigma Insight: Argand Plane and Polar Representation

Analyzing the Setup

Decoding the Conjugate

The Argument Transformation

The System of Equations

Conclusion

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z1​ and z2​ be two complex numbers such that zˉ1​=izˉ2​ and arg(zˉ2​z1​​)=π. Then