If z1, z2 are two distinct complex number such that |(z1-2z2)/(1/2-z1z̄2)| = 2, then (1) z1 lies on

Q: Wait, why do we square both sides? Can't we just solve it with the modulus directly?

That is a great question! While you could technically work with the modulus, it is essentially a square root. Squaring both sides is the standard, most efficient way to eliminate the radical and unlock the algebraic potential of the $|z|^2 = z\bar{z}$ identity. It transforms a geometric problem into a clean algebraic one.

Q: What happens to the conjugate of the denominator? Why does $\bar{z}_1z_2$ appear?

Remember the property $\overline{ab} = \bar{a}\bar{b}$ and $\overline{a-b} = \bar{a}-\bar{b}$. When you take the conjugate of $(\frac{1}{2} - z_1\bar{z}_2)$, the $\frac{1}{2}$ stays $\frac{1}{2}$ because it is real, and the conjugate of $z_1\bar{z}_2$ becomes $\bar{z}_1\bar{\bar{z}}_2$, which simplifies to $\bar{z}_1z_2$. It is just a direct application of conjugate rules!

Q: How do we know to factor the final expression like that? Is there a trick?

It is not so much a trick as it is pattern recognition. Once you have expanded and cancelled the cross terms, you are left with terms involving $|z_1|^2$ and $|z_2|^2$. By grouping them, you will see a common binomial factor emerge. It is a classic technique in complex number problems to look for these factorable structures.

Question

If $z_1, z_2$ are two distinct complex number such that $\left| \frac{z_1-2z_2}{\frac{1}{2}-z_1\bar{z}_2} \right| = 2$, then

Options:

(A) $z_1$ lies on a circle of radius $\frac{1}{2}$ and $z_2$ lies on a circle of radius 1 .

(B) both $z_1$ and $z_2$ lie on the same circle.

(C) either $z_1$ lies on a circle of radius $\frac{1}{2}$ or $z_2$ lies on a circle of radius 1 .

(D) either $z_1$ lies on a circle of radius 1 or $z_2$ lies on a circle of radius $\frac{1}{2}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow travelers of the mathematical landscape. Today, we are going to unravel a problem that, at first glance, might seem like a tangled mess of variables and bars.

We are looking at the equation:

$$ \left| \frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2} \right| = 2 $$

It is a classic JEE Advanced challenge that tests not just your algebraic stamina, but your ability to see the geometric truth hidden behind the symbols. We are working on the Argand plane, where every complex number is a point, and our goal is to find the locus that $z_1$ and $z_2$ trace out.

### Breaking the Fraction

The first thing we notice is the fraction inside the modulus. We utilize the property that the modulus of a quotient is the quotient of the moduli, $| \frac{a}{b} | = \frac{|a|}{|b|}$.

By applying this, we rewrite our equation as:

$$ |z_1 - 2z_2| = 2 \left| \frac{1}{2} - z_1\bar{z}_2 \right| $$

Suddenly, the fraction is gone, and we have a much more manageable linear relationship. This is the first victory in our journey.

### The Power of Squaring

To handle the modulus, we square both sides. This is a classic move in complex numbers that transforms the equation into:

$$ |z_1 - 2z_2|^2 = 4 \left| \frac{1}{2} - z_1\bar{z}_2 \right|^2 $$

Remember, when you square the $2$ on the right, it becomes a $4$. It is a small detail, but in the heat of an exam, it is where the most brilliant minds often stumble.

### The Identity that Changes Everything

Here is where the magic happens. We invoke the most powerful identity in our toolkit: $|z|^2 = z\bar{z}$.

This allows us to replace the squared modulus with the product of the complex number and its conjugate:

$$ (z_1 - 2z_2)(\bar{z}_1 - 2\bar{z}_2) = 4 \left( \frac{1}{2} - z_1\bar{z}_2 \right) \left( \frac{1}{2} - \bar{z}_1z_2 \right) $$

It looks long, but we are simply breaking down the complex variables into their fundamental components.

### The Expansion and the Elegant Cancellation

Now, we distribute. On the left, we expand $(z_1 - 2z_2)(\bar{z}_1 - 2\bar{z}_2)$ to get:

$$ |z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1z_2 + 4|z_2|^2 $$

On the right, we keep the $4$ outside and expand the bracket to get:

$$ 4 \left( \frac{1}{4} - \frac{1}{2}\bar{z}_1z_2 - \frac{1}{2}z_1\bar{z}_2 + |z_1|^2|z_2|^2 \right) $$

When we distribute the $4$ into the right side, the cross terms $-2z_1\bar{z}_2$ and $-2\bar{z}_1z_2$ appear on both sides of the equation. They cancel out perfectly!

### The Final Revelation

We are left with the simplified expression:

$$ |z_1|^2 + 4|z_2|^2 = 1 + 4|z_1|^2|z_2|^2 $$

By rearranging this, we get:

$$ |z_1|^2 - 1 + 4|z_2|^2 - 4|z_1|^2|z_2|^2 = 0 $$

Factoring by grouping, we arrive at:

$$ (|z_1|^2 - 1)(1 - 4|z_2|^2) = 0 $$

This tells us that either **$|z_1| = 1$** or **$|z_2| = \frac{1}{2}$**. In geometric terms, this means either $z_1$ lies on a circle of radius $1$, or $z_2$ lies on a circle of radius $\frac{1}{2}$.

Animated Solution for Mathematics - Complex Numbers: If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

Select Answer:

Visualized Solution

Analyze the Given Equation

Apply Modulus Properties

Square Both Sides

Use the Identity $∣ z ∣^{2} = z \overset{z}{ˉ}$

Distribute the Conjugate

Expand the Left Hand Side

Expand the Right Hand Side

Simplify the Equation

Rearrange and Group Terms

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Breaking the Fraction

The Power of Squaring

The Identity that Changes Everything

The Expansion and the Elegant Cancellation

The Final Revelation

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