If |z| = 1 and ω = (z-1)/(z+1) (where z ≠ -1), then Re(ω) is (A) 0 (B) 1/|z+1|² (C) |z/(z+1)| · 1/|z

Q: Why do we multiply by the conjugate of the denominator? Can't we just substitute $z = x+iy$?

You absolutely can! But in JEE Advanced, time is currency. Using the conjugate property $z\bar{z}=|z|^2$ is the elegant path that avoids messy algebraic expansion. It is about training your intuition to see the structure rather than just grinding through coordinates.

Q: Wait, why is $|z+1|^2$ always real?

Great question! Remember that for any complex number $w$, $w \cdot \bar{w} = |w|^2$. Since $|w|^2$ is defined as the square of the distance from the origin, it must be a non-negative real number. This is a fundamental property that simplifies complex expressions instantly.

Q: Is it possible for $Re(\omega)$ to be non-zero if $z=-1$?

The problem explicitly excludes $z=-1$ because the expression becomes undefined. If $z$ approaches $-1$, the denominator approaches zero, and the magnitude of $\omega$ blows up to infinity. The real part is only defined where the expression is valid.

Question

If $|z| = 1$ and $\omega = \frac{z-1}{z+1}$ (where $z 
eq -1$), then $Re(\omega)$ is

Options:

(A) 0

(B) $\frac{1}{|z+1|^2}$

(C) $\left|\frac{z}{z+1}ight| \cdot \frac{1}{|z+1|^2}$

(D) $\frac{\sqrt{2}}{|z+1|^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the complex plane, a vast, two-dimensional landscape where every point is a number. We are given a constraint: $|z| = 1$.

To a JEE aspirant, this is a command to visualize. It means our point $z$ is tethered to the origin, forever dancing on the circumference of a unit circle.

There is a critical constraint: $z 
eq -1$. Because at $z = -1$, the denominator of our expression $\omega = \frac{z-1}{z+1}$ vanishes, leading to the mathematical abyss of division by zero. We must respect this boundary.

### The Algebraic Dance

We are tasked with finding the real part of $\omega = \frac{z-1}{z+1}$. The most powerful tool in our arsenal is the complex conjugate.

We want to make the denominator real, as this allows us to easily separate the real and imaginary parts of the numerator. We multiply the numerator and the denominator by the conjugate of the denominator, which is $\bar{z}+1$:

$$\omega = \frac{(z-1)(\bar{z}+1)}{(z+1)(\bar{z}+1)}$$

### The Beauty of Simplification

Look at the denominator: $(z+1)(\bar{z}+1)$. This is the product of a complex number and its conjugate, which is simply the square of its modulus: $|z+1|^2$. This is a purely real number.

Now, let us turn our attention to the numerator: $(z-1)(\bar{z}+1)$. Expanding this, we get:

$$z\bar{z} + z - \bar{z} - 1$$

Since $|z|=1$, we know that $z\bar{z} = |z|^2 = 1$. Substituting this into our numerator, we get $1 + z - \bar{z} - 1$. The $1$ and $-1$ cancel out perfectly, leaving us with $z - \bar{z}$.

### The Final Revelation

We are left with the simplified expression:

$$\omega = \frac{z-\bar{z}}{|z+1|^2}$$

Recall the fundamental property of complex numbers: $z - \bar{z} = 2i 	ext{Im}(z)$. This means the numerator is purely imaginary.

Since the denominator $|z+1|^2$ is purely real, the entire expression $\omega$ is purely imaginary. If a complex number is purely imaginary, its real part must be zero.

And there it is—the elegance of the result:

**$Re(\omega) = 0$**

We have navigated the geometry, mastered the algebra, and arrived at the truth. The complex number $\omega$ will always lie on the imaginary axis, no matter where $z$ is on the unit circle.

Animated Solution for Mathematics - Complex Numbers: If $∣ z ∣ = 1$ and $ω = \frac{z - 1}{z + 1}$ (where $z \neq = - 1$ ), then $R e (ω)$ is

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Visualized Solution

Visualizing the Constraint $∣ z ∣ = 1$

The Excluded Point $z \neq = - 1$

Defining $ω$ and the Goal

The Power of the Identity $z \overset{z}{ˉ} = ∣ z ∣^{2}$

Rationalizing the Denominator

Setting up the Multiplication

Simplifying the Denominator

Expanding the Numerator

Substituting $z \overset{z}{ˉ} = 1$

Canceling Terms

Property of $z - \overset{z}{ˉ}$

The Final Form of $ω$

Final Conclusion: $R e (ω) = 0$

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Algebraic Dance

The Beauty of Simplification

The Final Revelation

Similar Questions

$z \neq = 0$ is a complex number

List-I

List-II

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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∣=1 and ω=z+1z−1​ (where z=−1), then Re(ω) is

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Visualized Solution

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

The Excluded Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=−1

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω and the Goal

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