If z and ω are two non-zero complex numbers such that |zω| = 1 and Arg(z) - Arg(ω) = π/2, then z̅ω i

Q: Why can't I just pick values like $z=i$ and $\omega=1$?

You absolutely can! In fact, picking specific values is a brilliant strategy for competitive exams like JEE. If you choose $z=i$ and $\omega=1$, then $|z\omega| = |i| = 1$ and $Arg(z) - Arg(\omega) = \pi/2 - 0 = \pi/2$. Then $\bar{z}\omega = (-i)(1) = -i$. This confirms our result. However, the algebraic method is more robust and helps you understand the underlying structure.

Q: Wait, why is the conjugate $\bar{z}$ written as $|z|e^{-i\theta_1}$?

Great question! The conjugate $\bar{z}$ is the reflection of $z$ across the real axis. If $z = |z|e^{i\theta_1}$, reflecting it across the real axis simply negates the angle. So, the new angle is $-\theta_1$. Thus, $\bar{z} = |z|e^{-i\theta_1}$.

Q: Is there a geometric way to solve this without all the algebra?

Yes! Geometrically, $Arg(z) - Arg(\omega) = \pi/2$ means $z$ is rotated $90^\circ$ counter-clockwise from $\omega$. Multiplying by $\bar{z}$ is equivalent to multiplying by $|z|e^{-i\theta_1}$. This involves both a scaling and a rotation. The algebra is just a formal way of tracking these geometric transformations.

Question

If $z$ and $\omega$ are two non-zero complex numbers such that $|z\omega| = 1$ and $Arg(z) - Arg(\omega) = \pi/2$, then $\bar{z}\omega$ is equal to

Options:

(A) -i

(B) 1

(C) -1

(D) i

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful world of complex numbers. Today, we are not just solving an equation; we are exploring the geometry of the complex plane.

Imagine $z$ and $\omega$ as two vectors, stretching out from the origin like hands on a clock. We do not know their exact lengths, but we know their relationship.

### The Language of Euler

To master this problem, we must speak the language of Euler. Any complex number can be written as $z = |z|e^{i	heta_1}$ and $\omega = |\omega|e^{i	heta_2}$.

This is the most powerful tool in our arsenal. It turns the terrifying prospect of multiplying complex numbers into the simple, elegant addition of exponents.

### Decoding the Conditions

We are given two clues. First, $|z\omega| = 1$. By the properties of the modulus, this means $|z||\omega| = 1$. This is our scaling factor.

Second, $Arg(z) - Arg(\omega) = \frac{\pi}{2}$. This tells us that the angle between these two vectors is exactly $90^\circ$. In our polar notation, this is $	heta_1 - 	heta_2 = \frac{\pi}{2}$.

### The Conjugate Reflection

The question asks for $\bar{z}\omega$. What is $\bar{z}$? It is the mirror image of $z$ across the real axis.

If $z$ is at angle $	heta_1$, then $\bar{z}$ is at angle $-	heta_1$. So, $\bar{z} = |z|e^{-i	heta_1}$.

Now, we multiply:
$$\bar{z}\omega = (|z|e^{-i	heta_1}) \cdot (|\omega|e^{i	heta_2})$$

### The Final Synthesis

Grouping the terms, we get:
$$\bar{z}\omega = (|z||\omega|) \cdot e^{i(	heta_2 - 	heta_1)}$$

We know $|z||\omega| = 1$. And since $	heta_1 - 	heta_2 = \frac{\pi}{2}$, it follows that $	heta_2 - 	heta_1 = -\frac{\pi}{2}$.

Substituting these, we get:
$$\bar{z}\omega = 1 \cdot e^{-i\pi/2}$$

Using Euler's formula:
$$e^{-i\pi/2} = \cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}) = 0 - i = -i$$

And there it is! The complexity melts away, leaving us with the elegant result of **$-i$**. Keep practicing, and you will see this beauty in every problem.

Animated Solution for Mathematics - Complex Numbers: If $z$ and $ω$ are two non-zero complex numbers such that $∣ z ω ∣ = 1$ and $A r g (z) - A r g (ω) = π /2$ , then $\overset{z}{ˉ} ω$ is equal to

Select Answer:

Visualized Solution

The Complex Plane Setup

Exponential Form of Complex Numbers

Analyzing the Magnitude Condition

Analyzing the Argument Condition

The Conjugate of $z$

Setting up the Product $\overset{z}{ˉ} ω$

Grouping Terms

Substituting the Magnitude

Substituting the Argument

Final Evaluation

The Sigma Insight: Argand Plane and Polar Representation

Analyzing the Setup

The Language of Euler

Decoding the Conditions

The Conjugate Reflection

The Final Synthesis

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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; } If z and ω are two non-zero complex numbers such that ∣zω∣=1 and Arg(z)−Arg(ω)=π/2, then zˉω is equal to