If z and ω are two complex numbers such that |zω| = 1 and arg(z) - arg(ω) = 3π/2, then arg((1-2z̄ω)/

Q: Why can't we just solve for $z$ and $\omega$ separately?

Because the problem doesn't give us enough constraints to find $z$ and $\omega$ individually. We only know their product's magnitude and the difference of their arguments. This is a classic JEE trick: the problem is designed to be solved by manipulating the expression as a whole, not by finding the individual components.

Q: Wait, why is $\arg(z) - \arg(\omega) = 3\pi/2$ equivalent to $\bar{z}\omega = i$?

Great observation! Remember that $\bar{z} = re^{-i\theta}$. When you multiply $\bar{z}$ by $\omega = \frac{1}{r}e^{i\phi}$, you get $e^{i(\phi - \theta)}$. Since $\theta - \phi = 3\pi/2$, then $\phi - \theta = -3\pi/2$. And $e^{-i(3\pi/2)} = e^{i\pi/2} = i$. It's all about the properties of exponents!

Q: How do I know which quadrant the final complex number is in?

Once you simplify the expression to $-\frac{1}{2} - \frac{1}{2}i$, look at the real and imaginary parts. Both are negative. In the complex plane, a negative real part and a negative imaginary part place the point squarely in the third quadrant. From there, use the standard formula for the principal argument in the third quadrant: $-\pi + \tan^{-1}(|y/x|)$.

Question

If $z$ and $\omega$ are two complex numbers such that $|z\omega| = 1$ and $\arg(z) - \arg(\omega) = \frac{3\pi}{2}$, then $\arg\left(\frac{1-2\bar{z}\omega}{1+3\bar{z}\omega}\right)$ is : (Here $\arg(z)$ denotes the principal argument of complex number $z$)

Options:

(A) $\frac{\pi}{4}$

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

(C) $-\frac{\pi}{4}$

(D) $\frac{3\pi}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the complex plane, looking at two mysterious numbers, $z$ and $\omega$. At first glance, they seem like independent entities, but the problem provides two beautiful constraints: $|z\omega| = 1$ and $\arg(z) - \arg(\omega) = 3\pi/2$.

The first condition, $|z\omega| = 1$, tells us that the product of their lengths is unity. If $z$ is a vector stretching far from the origin, $\omega$ must be a tiny vector tucked inside the unit circle to compensate.

The second condition, $\arg(z) - \arg(\omega) = 3\pi/2$, describes their orientation. They are separated by a massive angular gap of $270^\circ$. This is the setup for a beautiful simplification.

### The Great Collapse

We do not need to find $z$ or $\omega$ individually. Instead, let us examine the product $\bar{z}\omega$.

Using Euler's form, let $z = re^{i	heta}$. Then, its conjugate is $\bar{z} = re^{-i	heta}$. Let $\omega = \frac{1}{r}e^{i\phi}$.

When we multiply $\bar{z}$ and $\omega$, the $r$ and $1/r$ terms cancel out perfectly:

$$\bar{z}\omega = \left(re^{-i	heta}ight) \left(\frac{1}{r}e^{i\phi}ight) = e^{i(\phi - 	heta)}$$

Since $	heta - \phi = 3\pi/2$, it follows that $\phi - 	heta = -3\pi/2$. Thus, $\bar{z}\omega = e^{-i(3\pi/2)}$.

An angle of $-3\pi/2$ is equivalent to $\pi/2$ in the positive direction. Since $e^{i\pi/2} = i$, the entire expression has collapsed into the imaginary unit:

$$\bar{z}\omega = i$$

### The Algebraic Journey

Now, the target expression $\frac{1-2\bar{z}\omega}{1+3\bar{z}\omega}$ simplifies significantly by substituting $\bar{z}\omega = i$:

$$\frac{1-2i}{1+3i}$$

To simplify this, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, $1-3i$:

$$\frac{(1-2i)(1-3i)}{(1+3i)(1-3i)} = \frac{1 - 3i - 2i + 6i^2}{1^2 + 3^2}$$

Since $i^2 = -1$, the numerator becomes $1 - 5i - 6 = -5 - 5i$. The denominator becomes $1 + 9 = 10$.

$$\frac{-5 - 5i}{10} = -\frac{1}{2} - \frac{1}{2}i$$

### The Final Destination

We have arrived at the point $(-\frac{1}{2}, -\frac{1}{2})$ on the complex plane. Both the real and imaginary parts are negative, placing us firmly in the third quadrant.

The acute angle $\alpha$ is given by:

$$\alpha = 	an^{-1}\left( \left| \frac{-1/2}{-1/2} ight| ight) = 	an^{-1}(1) = \frac{\pi}{4}$$

In the third quadrant, the principal argument is $-\pi + \alpha = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$.

We have navigated the complexity and found the final answer: **$-3\pi/4$**. The complexity of the initial expression was merely a mask for this elegant geometric truth.

Animated Solution for Mathematics - Complex Numbers: If $z$ and $ω$ are two complex numbers such that $∣ z ω ∣ = 1$ and $ar g (z) - ar g (ω) = \frac{3 π}{2}$ , then $ar g (\frac{1 - 2 z ˉ ω}{1 + 3 z ˉ ω})$ is : (Here $ar g (z)$ denotes the principal argument of complex number $z$ )

Select Answer:

Visualized Solution

Analyze Magnitudes $∣ z ω ∣ = 1$

Relate Arguments $ar g (z) - ar g (ω) = \frac{3 π}{2}$

Exponential Form of $z$ and $ω$

Calculate Product $\overset{z}{ˉ} ω$

Simplify $e^{- i \frac{3 π}{2}}$

Substitute into the Target Expression

Rationalize the Denominator

Final Simplified Complex Number

Calculate the Principal Argument

The Sigma Insight: Argand Plane and Polar Representation

Analyzing the Setup

The Great Collapse

The Algebraic Journey

The Final Destination

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Select Answer:

Visualized Solution

Analyze Magnitudes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣zω∣=1

Relate Arguments .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } arg(z)−arg(ω)=23π​

Exponential Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

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