JEE Main 2015

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: A complex number $z$ is said to be unimodular if $∣ z ∣ = 1$ . Suppose $z_{1}$ and $z_{2}$ are complex numbers such that $\frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}}$ is unimodular and $z_{2}$ is not unimodular. Then the point $z_{1}$ lies on a

Q: Is there a geometric way to see why the cross terms cancel?

Yes! The symmetry in the expression $\frac{z_1 - 2z_2}{2 - z_1\bar{z}_2}$ is not accidental. It is a form of a Mobius transformation. The cancellation happens because the expression is essentially 'self-conjugate' in a way that preserves the modulus. It is a beautiful piece of algebraic harmony.

Select Answer:

Progressive Hints

Hint 1

Think about what it means to be 'unimodular' geometrically. It is not just an algebraic constraint; it is a statement about distance from the origin. If you see a modulus expression equal to 1, visualize a unit circle. How does that help you simplify the expression?

Hint 2

When you have an equation involving moduli, like

∣ A ∣ = ∣ B ∣

, squaring both sides is your best friend. But do not just square it blindly! Use the golden rule of complex numbers:

∣ z ∣^{2} = z \overset{z}{ˉ}

. This is the bridge between the world of absolute values and the world of standard algebra.

Hint 3

Start by writing the condition as

∣ z_{1} - 2 z_{2} ∣ = ∣2 - z_{1} \overset{z}{ˉ}_{2} ∣

. Square both sides to get

∣ z_{1} - 2 z_{2} ∣^{2} = ∣2 - z_{1} \overset{z}{ˉ}_{2} ∣^{2}

. Now, expand both sides using the identity

∣ z ∣^{2} = z \overset{z}{ˉ}

. Watch carefully as the cross-terms emerge—you will be surprised at how much cancels out!

Visualized Solution

Understanding Unimodular

A complex number $z$ is unimodular if $∣ z ∣ = 1$ .
Geometrically, it lies on a unit circle centered at the origin.
Given: $z_{2}$ is not unimodular, so $∣ z_{2} ∣ \neq = 1$ .

The Modulus Condition

Let $w = \frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}}$ .
We are given that $w$ is unimodular.
Therefore, $\frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}} = 1$ .

Cross-Multiplication and Squaring

Cross-multiplying: $∣ z_{1} - 2 z_{2} ∣ = ∣2 - z_{1} \overset{z}{ˉ}_{2} ∣$
Squaring both sides: $∣ z_{1} - 2 z_{2} ∣^{2} = ∣2 - z_{1} \overset{z}{ˉ}_{2} ∣^{2}$

Property of Modulus Squared

Recall the fundamental property: $∣ z ∣^{2} = z \overset{z}{ˉ}$
This converts modulus into standard complex multiplication.

Applying the Property

Applying to LHS: $(z_{1} - 2 z_{2}) (\overline{z_{1} - 2 z_{2}})$
Applying to RHS: $(2 - z_{1} \overset{z}{ˉ}_{2}) (\overline{2 - z_{1} \overset{z}{ˉ}_{2}})$
Distributing the bar: $(z_{1} - 2 z_{2}) (\overset{z}{ˉ}_{1} - 2 \overset{z}{ˉ}_{2}) = (2 - z_{1} \overset{z}{ˉ}_{2}) (2 - \overset{z}{ˉ}_{1} z_{2})$

Expanding Both Sides

LHS: $z_{1} \overset{z}{ˉ}_{1} - 2 z_{1} \overset{z}{ˉ}_{2} - 2 \overset{z}{ˉ}_{1} z_{2} + 4 z_{2} \overset{z}{ˉ}_{2}$
RHS: $4 - 2 z_{1} \overset{z}{ˉ}_{2} - 2 \overset{z}{ˉ}_{1} z_{2} + z_{1} \overset{z}{ˉ}_{1} z_{2} \overset{z}{ˉ}_{2}$
Using $z \overset{z}{ˉ} = ∣ z ∣^{2}$ :
LHS: $∣ z_{1} ∣^{2} - 2 z_{1} \overset{z}{ˉ}_{2} - 2 \overset{z}{ˉ}_{1} z_{2} + 4∣ z_{2} ∣^{2}$
RHS: $4 - 2 z_{1} \overset{z}{ˉ}_{2} - 2 \overset{z}{ˉ}_{1} z_{2} + ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2}$

Canceling Common Terms

Notice the terms $- 2 z_{1} \overset{z}{ˉ}_{2}$ and $- 2 \overset{z}{ˉ}_{1} z_{2}$ appear on both sides.
Canceling them out leaves:
$∣ z_{1} ∣^{2} + 4∣ z_{2} ∣^{2} = 4 + ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2}$

Factorization

Bring all terms to one side:
$∣ z_{1} ∣^{2} - 4 - ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2} + 4∣ z_{2} ∣^{2} = 0$
Grouping terms:
$∣ z_{1} ∣^{2} (1 - ∣ z_{2} ∣^{2}) - 4 (1 - ∣ z_{2} ∣^{2}) = 0$
Factoring out $(1 - ∣ z_{2} ∣^{2})$ :
$(∣ z_{1} ∣^{2} - 4) (1 - ∣ z_{2} ∣^{2}) = 0$

Applying the Constraint

We have $(∣ z_{1} ∣^{2} - 4) (1 - ∣ z_{2} ∣^{2}) = 0$ .
Recall the initial condition: $z_{2}$ is not unimodular.
Therefore, $∣ z_{2} ∣ \neq = 1 ⟹ 1 - ∣ z_{2} ∣^{2} \neq = 0$ .

The Final Locus

Since the second bracket is non-zero, the first must be zero:
$∣ z_{1} ∣^{2} - 4 = 0 ⟹ ∣ z_{1} ∣^{2} = 4$
Taking the square root: $∣ z_{1} ∣ = 2$
This represents a circle of radius 2 centered at the origin.

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The Sigma Insight: Conjugate and Modulus

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler of the complex plane! Today, we are going to unravel a problem that might look like a tangled mess of variables, but is actually a beautiful, symmetrical dance of complex numbers.

Imagine you are standing on the Argand plane. You have a point

z_{1}

that is moving, and a point

z_{2}

that is fixed but "not unimodular."

We are given the relationship:

w = \frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}}

This expression

w

is unimodular, which means the distance of this complex number

w

from the origin is exactly

1

. We start by writing

∣ w ∣ = 1

The Magic of Conjugates

Now, the magic begins. We cross-multiply to get

∣ z_{1} - 2 z_{2} ∣ = ∣2 - z_{1} \overset{z}{ˉ}_{2} ∣

Squaring both sides is our next logical step, leading us to:

∣ z_{1} - 2 z_{2} ∣^{2} = ∣2 - z_{1} \overset{z}{ˉ}_{2} ∣^{2}

Here is where the "Golden Key" of complex numbers comes in: the identity

∣ z ∣^{2} = z \overset{z}{ˉ}

. By applying this to both sides, we transform the modulus into a product of the number and its conjugate.

As we expand the expression:

(z_{1} - 2 z_{2}) (\overset{z}{ˉ}_{1} - 2 \overset{z}{ˉ}_{2}) = (2 - z_{1} \overset{z}{ˉ}_{2}) (2 - \overset{z}{ˉ}_{1} z_{2})

Something miraculous happens. The cross-terms, those pesky

- 2 z_{1} \overset{z}{ˉ}_{2}

and

- 2 \overset{z}{ˉ}_{1} z_{2}

, appear on both sides of the equation! They cancel out perfectly, leaving us with a clean, elegant relationship:

∣ z_{1} ∣^{2} + 4∣ z_{2} ∣^{2} = 4 + ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2}

The Final Revelation

We rearrange this to factorize it into:

(∣ z_{1} ∣^{2} - 4) (1 - ∣ z_{2} ∣^{2}) = 0

Since the problem explicitly tells us that

z_{2}

is not unimodular, we know that

1 - ∣ z_{2} ∣^{2}

cannot be zero. Therefore, the only way for this product to be zero is if

∣ z_{1} ∣^{2} - 4 = 0

This simplifies to:

∣ z_{1} ∣ = 2

And there it is! The locus of

z_{1}

is a circle of radius 2 centered at the origin. You have just navigated through the algebra to find a geometric truth.

Common Traps & FAQs

Why do we square both sides? Can we not just solve it as is?

Great question! Dealing with moduli directly is like trying to solve an equation with square roots—it is messy and prone to errors. Squaring both sides allows us to use the identity

∣ z ∣^{2} = z \overset{z}{ˉ}

, which transforms the modulus into simple multiplication. It turns a 'geometry' problem into an 'algebra' problem, which is much easier to handle.

Wait, why did we reject the case where

1 - ∣ z_{2} ∣^{2} = 0

Ah, look closely at the problem statement! It explicitly says '

z_{2}

is not unimodular.' This means

∣ z_{2} ∣ \neq = 1

. If

∣ z_{2} ∣

were 1, then

1 - ∣ z_{2} ∣^{2}

would be 0, and the entire equation would be satisfied regardless of

z_{1}

. By excluding that case, we are forced to look at the only other possibility:

∣ z_{1} ∣^{2} - 4 = 0

. Always respect the constraints!

Is there a geometric way to see why the cross terms cancel?

Yes! The symmetry in the expression

\frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}}

is not accidental. It is a form of a Mobius transformation. The cancellation happens because the expression is essentially 'self-conjugate' in a way that preserves the modulus. It is a beautiful piece of algebraic harmony.

Select Answer:

Visualized Solution

Understanding Unimodular

The Modulus Condition

Cross-Multiplication and Squaring

Property of Modulus Squared

Applying the Property

Expanding Both Sides

Canceling Common Terms

Factorization

Applying the Constraint

The Final Locus

The Sigma Insight: Conjugate and Modulus

Analyzing the Setup

The Magic of Conjugates

The Final Revelation

Similar Questions

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Let $z$ and $ω$ be two complex numbers such that $∣ z ∣ \leq 1, ∣ ω ∣ \leq 1$ and $∣ z + iω ∣ = ∣ z - i \overset{ω}{ˉ} ∣ = 2$ then $z$ equals

If $z_{1}$ and $z_{2}$ are two complex numbers such taht $∣ z_{1} ∣ < 1 < ∣ z_{2} ∣$ then prove that $∣ \frac{1 - z _{1} z ˉ _{2}}{z _{1} - z _{2}} ∣ < 1$ .

If $z$ is a complex number of unit modulus and argument $θ$ , then $a r g (\frac{1 + z}{1 + z ˉ})$ equals

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If $z_{1}, z_{2}$ and $z_{3}$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = ∣1/ z_{1} + 1/ z_{2} + 1/ z_{3} ∣ = 1$ , then $∣ z_{1} + z_{2} + z_{3} ∣$ is

Let a complex number $z, ∣ z ∣ \neq = 1$ , satisfy $lo g_{\frac{1}{2}} (\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}}) \leq 2$ . Then, the largest value of $∣ z ∣$ is equal to

Let $z$ be complex number satisfying $∣ z ∣^{3} + 2 z^{2} + 4 \overset{z}{ˉ} - 8 = 0$ , where $\overset{z}{ˉ}$ denotes the complex conjugate of $z$ . Let the imaginary part of $z$ be nonzero. Match each entry in List-I to the correct entries in List-II.

List-I

List-II

Select Answer:

Visualized Solution

Understanding Unimodular

The Modulus Condition

Cross-Multiplication and Squaring

Property of Modulus Squared

Applying the Property

Expanding Both Sides

Canceling Common Terms

Factorization

Applying the Constraint

The Final Locus

The Sigma Insight: Conjugate and Modulus

Analyzing the Setup

The Magic of Conjugates

The Final Revelation

Similar Questions

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List-I

List-II

All the points in the set $S = {\frac{α + i}{α - i} : α \in R} (i = - 1)$ lie on a

Let $z$ and $ω$ be two complex numbers such that $∣ z ∣ \leq 1, ∣ ω ∣ \leq 1$ and $∣ z + iω ∣ = ∣ z - i \overset{ω}{ˉ} ∣ = 2$ then $z$ equals

If $z_{1}$ and $z_{2}$ are two complex numbers such taht $∣ z_{1} ∣ < 1 < ∣ z_{2} ∣$ then prove that $∣ \frac{1 - z _{1} z ˉ _{2}}{z _{1} - z _{2}} ∣ < 1$ .

If $z$ is a complex number of unit modulus and argument $θ$ , then $a r g (\frac{1 + z}{1 + z ˉ})$ equals

If $z \neq = 0$ be a complex number such that $∣ z - \frac{1}{z} ∣ = 2$ , then the maximum value of $∣ z ∣$ is:

If $\frac{w - w ˉ z}{1 - z}$ is purely real where $w = α + i β, β \neq = 0$ and $z \neq = 1$ , then the set of the values of $z$ is

If $z = x + i y, x y \neq = 0$ , satisfies the equation $z^{2} + i \overset{z}{ˉ} = 0$ , then $∣ z ∣^{2}$ is equal to :

If $z_{1}, z_{2}$ and $z_{3}$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = ∣1/ z_{1} + 1/ z_{2} + 1/ z_{3} ∣ = 1$ , then $∣ z_{1} + z_{2} + z_{3} ∣$ is

Let a complex number $z, ∣ z ∣ \neq = 1$ , satisfy $lo g_{\frac{1}{2}} (\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}}) \leq 2$ . Then, the largest value of $∣ z ∣$ is equal to