If $\frac{w - \bar{w}z}{1-z}$ is purely real where $w = \alpha + i\beta, \beta \neq 0$ and $z \neq 1$, then the set of the values of $z$ is Options: (A) $\{z : |z| = 1\}$ (B) $\{z : z = \bar{z}\}$ (C) $\{z : z \neq 1\}$ (D) $\{z : |z| = 1, z \neq 1\}$

### Analyzing the Setup Have you ever looked at a complex number and felt like it was hiding a secret? In the world of JEE Advanced, complex numbers aren't just points on a plane; they are geometric entities waiting to be manipulated. We are given the expression $k = \frac{w - \bar{w}z}{1-z}$ and told it is purely real. Our goal is to find the set of all possible values for $z$. ### The Mirror Symmetry of Real Numbers The problem provides a critical condition: the expression is purely real. In the complex plane, this means the number has no imaginary part, which is equivalent to saying the number is its own reflection across the real axis. If $k$ is real, then $k = \bar{k}$. We set our expression equal to its conjugate: $$\frac{w - \bar{w}z}{1-z} = \overline{\left( \frac{w - \bar{w}z}{1-z} \right)}$$ Applying the properties of conjugates—where the conjugate of a quotient is the quotient of the conjugates—the right side transforms into: $$\frac{\bar{w} - \overline{(\bar{w}z)}}{\overline{1-z}} = \frac{\bar{w} - w\bar{z}}{1-\bar{z}}$$ ### The Algebraic Dance Now we have the equation $\frac{w - \bar{w}z}{1-z} = \frac{\bar{w} - w\bar{z}}{1-\bar{z}}$. To solve this, we cross-multiply to clear the denominators: $$(w - \bar{w}z)(1 - \bar{z}) = (\bar{w} - w\bar{z})(1 - z)$$ Expanding both sides carefully, we obtain: $$w - w\bar{z} - \bar{w}z + \bar{w}z\bar{z} = \bar{w} - \bar{w}z - w\bar{z} + wz\bar{z}$$ Notice that the terms $-w\bar{z}$ and $-\bar{w}z$ appear on both sides. They cancel out perfectly, leaving us with a much cleaner equation: $$w + \bar{w}z\bar{z} = \bar{w} + wz\bar{z}$$ ### The Geometric Revelation Recall that $z\bar{z} = |z|^2$. Substituting this into our equation, we get $w + \bar{w}|z|^2 = \bar{w} + w|z|^2$. Grouping the terms involving $|z|^2$ on one side and the constants on the other yields: $$w - \bar{w} = w|z|^2 - \bar{w}|z|^2$$ $$(w - \bar{w}) = (w - \bar{w})|z|^2$$ $$(w - \bar{w})(1 - |z|^2) = 0$$ Since $w$ is not real, $w - \bar{w} \neq 0$. We can safely divide by $(w - \bar{w})$, which leaves us with $1 - |z|^2 = 0$, or $|z|^2 = 1$. This implies $|z| = 1$. Geometrically, this represents the unit circle centered at the origin. However, we must respect the constraint $z \neq 1$ from the original expression. The final set of values is the **unit circle, excluding the point $z = 1$**.

Why do we reject the case where $w - \bar{w} = 0$?

Great catch! We look at the product $(w - \bar{w})(1 - |z|^2) = 0$. For this to be zero, either $(w - \bar{w}) = 0$ or $(1 - |z|^2) = 0$. The problem explicitly states $w = \alpha + i\beta$ with $\beta \neq 0$. This means $w$ is not a real number. If $w$ were real, $w$ would equal $\bar{w}$. Since $w$ is not real, $w - \bar{w}$ cannot be zero. Therefore, we are mathematically justified in dividing it out.

JEE Advanced 2006

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: 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

Q: Wait, why do we set the expression equal to its conjugate? Is there another way?

That is a fantastic question! Setting $k = \bar{k}$ is the most elegant and direct way to handle the 'purely real' condition. While you could substitute $z = x + iy$ and $w = \alpha + i\beta$, it would lead to a massive algebraic nightmare with real and imaginary parts. Using the conjugate property keeps the math clean and reveals the geometric structure of the solution much faster.

Q: Why is the constraint $z \neq 1$ so important at the end?

It is the 'trap' in the problem! When we solve $(1 - |z|^2) = 0$, we get $|z| = 1$, which is the unit circle. However, the original expression $\frac{w - \bar{w}z}{1-z}$ is undefined if $z = 1$ because the denominator becomes zero. In mathematics, we must always respect the domain of the original expression. Thus, we must exclude the point $z = 1$ from our final set.

Select Answer:

Visualized Solution

Analyze the Given Condition

Given expression: $k = \frac{w - w ˉ z}{1 - z}$ is purely real.
Given: $w = α + i β$ with $β \neq = 0$ .
Constraint: $z \neq = 1$ .

Property of Purely Real Numbers

For any complex number $k$ , if $k \in R$ , then $k = \overset{ˉ}{k}$ .
Applying this to our expression:
$\frac{w - w ˉ z}{1 - z} = \overline{(\frac{w - w ˉ z}{1 - z})}$

Distribute the Conjugate

Using properties of conjugates: $\overline{(\frac{z _{1}}{z _{2}})} = \frac{z ˉ _{1}}{z ˉ _{2}}$ and $\overline{z_{1} \pm z_{2}} = \overset{z}{ˉ}_{1} \pm \overset{z}{ˉ}_{2}$ .
$\frac{w - w ˉ z}{1 - z} = \frac{w ˉ - ( w ˉ z )}{1 - z}$
$\frac{w - w ˉ z}{1 - z} = \frac{w ˉ - w z ˉ}{1 - z ˉ}$

Cross-Multiplication

Cross-multiplying the terms to eliminate fractions:
$(w - \overset{w}{ˉ} z) (1 - \overset{z}{ˉ}) = (\overset{w}{ˉ} - w \overset{z}{ˉ}) (1 - z)$

Expand Both Sides

Expanding the left side: $w - w \overset{z}{ˉ} - \overset{w}{ˉ} z + \overset{w}{ˉ} z \overset{z}{ˉ}$
Expanding the right side: $\overset{w}{ˉ} - \overset{w}{ˉ} z - w \overset{z}{ˉ} + w z \overset{z}{ˉ}$
Equating them:
$w - w \overset{z}{ˉ} - \overset{w}{ˉ} z + \overset{w}{ˉ} z \overset{z}{ˉ} = \overset{w}{ˉ} - \overset{w}{ˉ} z - w \overset{z}{ˉ} + w z \overset{z}{ˉ}$

Cancel Common Terms

Notice the common terms on both sides: $- w \overset{z}{ˉ}$ and $- \overset{w}{ˉ} z$ .
Canceling them out:
$w + \overset{w}{ˉ} z \overset{z}{ˉ} = \overset{w}{ˉ} + w z \overset{z}{ˉ}$

Apply $z \overset{z}{ˉ} = ∣ z ∣^{2}$

Recall the fundamental property of complex numbers: $z \overset{z}{ˉ} = ∣ z ∣^{2}$ .
Substituting this into our simplified equation:
$w + \overset{w}{ˉ} ∣ z ∣^{2} = \overset{w}{ˉ} + w ∣ z ∣^{2}$

Rearrange and Factorize

Bring all terms to one side:
$w - \overset{w}{ˉ} = w ∣ z ∣^{2} - \overset{w}{ˉ} ∣ z ∣^{2}$
Factor out $∣ z ∣^{2}$ on the right side:
$w - \overset{w}{ˉ} = (w - \overset{w}{ˉ}) ∣ z ∣^{2}$
Bring everything to the left to factorize completely:
$(w - \overset{w}{ˉ}) - (w - \overset{w}{ˉ}) ∣ z ∣^{2} = 0$
$(w - \overset{w}{ˉ}) (1 - ∣ z ∣^{2}) = 0$

Analyze the First Factor

We have $(w - \overset{w}{ˉ}) (1 - ∣ z ∣^{2}) = 0$ .
This means either $w - \overset{w}{ˉ} = 0$ or $1 - ∣ z ∣^{2} = 0$ .
Recall the given condition: $w = α + i β$ and $β \neq = 0$ .
Since $β \neq = 0$ , $w$ is not purely real.
Therefore, $w \neq = \overset{w}{ˉ}$ , which means $w - \overset{w}{ˉ} \neq = 0$ .

The Locus of $z$

Since $w - \overset{w}{ˉ} \neq = 0$ , we must have $1 - ∣ z ∣^{2} = 0$ .
This implies $∣ z ∣^{2} = 1$ , so $∣ z ∣ = 1$ .
Geometrically, $∣ z ∣ = 1$ represents a circle centered at the origin with radius $1$ .
But remember the initial constraint: $z \neq = 1$ .

Final Set of Values

The locus is the unit circle $∣ z ∣ = 1$ .
We must exclude the point $z = 1$ to avoid division by zero in the original expression.
The final set of values for $z$ is ${z : ∣ z ∣ = 1, z \neq = 1}$ .

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The Sigma Insight: Conjugate and Modulus

Solution Diagram

Analyzing the Setup

Have you ever looked at a complex number and felt like it was hiding a secret? In the world of JEE Advanced, complex numbers aren't just points on a plane; they are geometric entities waiting to be manipulated.

We are given the expression

k = \frac{w - w ˉ z}{1 - z}

and told it is purely real. Our goal is to find the set of all possible values for

z

The Mirror Symmetry of Real Numbers

The problem provides a critical condition: the expression is purely real. In the complex plane, this means the number has no imaginary part, which is equivalent to saying the number is its own reflection across the real axis.

k

is real, then

k = \overset{ˉ}{k}

. We set our expression equal to its conjugate:

\frac{w - w ˉ z}{1 - z} = \overline{(\frac{w - w ˉ z}{1 - z})}

Applying the properties of conjugates—where the conjugate of a quotient is the quotient of the conjugates—the right side transforms into:

\frac{w ˉ - ( w ˉ z )}{1 - z} = \frac{w ˉ - w z ˉ}{1 - z ˉ}

The Algebraic Dance

Now we have the equation

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

. To solve this, we cross-multiply to clear the denominators:

(w - \overset{w}{ˉ} z) (1 - \overset{z}{ˉ}) = (\overset{w}{ˉ} - w \overset{z}{ˉ}) (1 - z)

Expanding both sides carefully, we obtain:

w - w \overset{z}{ˉ} - \overset{w}{ˉ} z + \overset{w}{ˉ} z \overset{z}{ˉ} = \overset{w}{ˉ} - \overset{w}{ˉ} z - w \overset{z}{ˉ} + w z \overset{z}{ˉ}

Notice that the terms

- w \overset{z}{ˉ}

and

- \overset{w}{ˉ} z

appear on both sides. They cancel out perfectly, leaving us with a much cleaner equation:

w + \overset{w}{ˉ} z \overset{z}{ˉ} = \overset{w}{ˉ} + w z \overset{z}{ˉ}

The Geometric Revelation

Recall that

z \overset{z}{ˉ} = ∣ z ∣^{2}

. Substituting this into our equation, we get

w + \overset{w}{ˉ} ∣ z ∣^{2} = \overset{w}{ˉ} + w ∣ z ∣^{2}

. Grouping the terms involving

∣ z ∣^{2}

on one side and the constants on the other yields:

w - \overset{w}{ˉ} = w ∣ z ∣^{2} - \overset{w}{ˉ} ∣ z ∣^{2}

(w - \overset{w}{ˉ}) = (w - \overset{w}{ˉ}) ∣ z ∣^{2}

(w - \overset{w}{ˉ}) (1 - ∣ z ∣^{2}) = 0

Since

w

is not real, $w - \bar{w} eq 0$. We can safely divide by

(w - \overset{w}{ˉ})

, which leaves us with

1 - ∣ z ∣^{2} = 0

, or

∣ z ∣^{2} = 1

. This implies

∣ z ∣ = 1

Geometrically, this represents the unit circle centered at the origin. However, we must respect the constraint $z eq 1$ from the original expression.

The final set of values is the unit circle, excluding the point $z = 1$ .

Common Traps & FAQs

Wait, why do we set the expression equal to its conjugate? Is there another way?

That is a fantastic question! Setting

k = \overset{ˉ}{k}

is the most elegant and direct way to handle the 'purely real' condition. While you could substitute

z = x + i y

and

w = α + i β

, it would lead to a massive algebraic nightmare with real and imaginary parts. Using the conjugate property keeps the math clean and reveals the geometric structure of the solution much faster.

Why do we reject the case where

w - \overset{w}{ˉ} = 0

Great catch! We look at the product

(w - \overset{w}{ˉ}) (1 - ∣ z ∣^{2}) = 0

. For this to be zero, either

(w - \overset{w}{ˉ}) = 0

(1 - ∣ z ∣^{2}) = 0

. The problem explicitly states

w = α + i β

with

β \neq = 0

. This means

w

is not a real number. If

w

were real,

w

would equal

\overset{w}{ˉ}

. Since

w

is not real,

w - \overset{w}{ˉ}

cannot be zero. Therefore, we are mathematically justified in dividing it out.

Why is the constraint

z \neq = 1

so important at the end?

It is the 'trap' in the problem! When we solve

(1 - ∣ z ∣^{2}) = 0

, we get

∣ z ∣ = 1

, which is the unit circle. However, the original expression

\frac{w - w ˉ z}{1 - z}

is undefined if

z = 1

because the denominator becomes zero. In mathematics, we must always respect the domain of the original expression. Thus, we must exclude the point

z = 1

from our final set.

Animated Solution for Mathematics - Complex Numbers: 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

Select Answer:

Visualized Solution

Analyze the Given Condition

Property of Purely Real Numbers

Distribute the Conjugate

Cross-Multiplication

Expand Both Sides

Cancel Common Terms

Apply $z \overset{z}{ˉ} = ∣ z ∣^{2}$

Rearrange and Factorize

Analyze the First Factor

The Locus of $z$

Final Set of Values

The Sigma Insight: Conjugate and Modulus

Analyzing the Setup

The Mirror Symmetry of Real Numbers

The Algebraic Dance

The Geometric Revelation

Similar Questions

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 $z_{1}, z_{2}$ 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

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

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List-I

List-II

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 $z$ 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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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

If $∣ z - 4/ z ∣ = 2$ , then the maximum value of $∣ Z ∣$ is equal to

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 1−zw−wˉz​ is purely real where w=α+iβ,β=0 and z=1, then the set of the values of z is

Select Answer:

Visualized Solution

Analyze the Given Condition

Property of Purely Real Numbers

Distribute the Conjugate

Cross-Multiplication

Expand Both Sides

Cancel Common Terms

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } zzˉ=∣z∣2

Rearrange and Factorize

Analyze the First Factor

The Locus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Final Set of Values

The Sigma Insight: Conjugate and Modulus

Analyzing the Setup

The Mirror Symmetry of Real Numbers

The Algebraic Dance

The Geometric Revelation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ and z3​ are complex numbers such that ∣z1​∣=∣z2​∣=∣z3​∣=∣1/z1​+1/z2​+1/z3​∣=1, then ∣z1​+z2​+z3​∣ is

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List-I

List-II

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Animated Solution for Mathematics - Complex Numbers: 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

Apply $z \overset{z}{ˉ} = ∣ z ∣^{2}$

The Locus of $z$

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 $z_{1}, z_{2}$ 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

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 $z \neq = 0$ be a complex number such that $∣ z - \frac{1}{z} ∣ = 2$ , then the maximum value 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 $z$ is a complex number of unit modulus and argument $θ$ , then $a r g (\frac{1 + z}{1 + z ˉ})$ equals

If $a > 0$ and $z = \frac{( 1 + i ) ^{2}}{a - i}$ , has magnitude $2/5$ , then $z$ is equal to :

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

If $∣ z - 4/ z ∣ = 2$ , then the maximum value of $∣ Z ∣$ is equal to