Problem 99 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(z_{1}\) and \(z_{2}(\pm 0)\) are two complex numbers such that \(\left|\frac{z_{1}-z_{2}}{z_{1}+z_{2}}\right|=1\), then (A) \(z_{2}=i k z_{1}, k \in R\) (B) \(z_{2}=k z_{1}, k \in R\) (C) \(z_{2}=z_{1}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (A) \(z_2 = i k z_1, k \in R\).

Step 1: Analyze the Condition

Start by considering the equation \( \left|\frac{z_1 - z_2}{z_1 + z_2}\right| = 1 \). This implies that the modulus of the complex number \( \frac{z_1 - z_2}{z_1 + z_2} \) is 1. A complex number \( \frac{a}{b} \) has a modulus 1 if \( |a| = |b| \). Therefore, \( |z_1 - z_2| = |z_1 + z_2| \).

Step 2: Geometric Interpretation

From the condition \( |z_1 - z_2| = |z_1 + z_2| \), we infer that the complex numbers \( z_1 - z_2 \) and \( z_1 + z_2 \) must be equidistant from the origin on the complex plane. This occurs when one is the negative of a rotation of the other, specifically by \( \frac{\pi}{2} \) or by any odd multiples of \( \frac{\pi}{2} \).

Step 3: Simplifying Using Known Results

If \( z_2 = i k z_1 \), then substitute this in the expression: \( z_1 - z_2 = z_1 - i k z_1 = (1 - i k) z_1 \) and \( z_1 + z_2 = z_1 + i k z_1 = (1 + i k) z_1 \). Simplifying, \( \left|\frac{z_1 - z_2}{z_1 + z_2}\right| = \left|\frac{1 - i k}{1 + i k}\right| \).

Step 4: Calculate Modulus

Calculate the modulus: \( |1 - i k| = \sqrt{1^2 + k^2} = \sqrt{1 + k^2} \) and \( |1 + i k| = \sqrt{1^2 + k^2} = \sqrt{1 + k^2} \). Thus, \( \left|\frac{1 - i k}{1 + i k}\right| = \frac{\sqrt{1 + k^2}}{\sqrt{1 + k^2}} = 1 \), hence satisfying the condition.

Step 5: Conclusion

The condition \( \left|\frac{z_1 - z_2}{z_1 + z_2}\right| = 1 \) holds true when \( z_2 = i k z_1 \) for \( k \) belonging to the real numbers. Refer back to the options, this matches option (A).

Key Concepts

Modulus

The modulus of a complex number is a key concept that helps understand its magnitude. When we talk about a complex number \( z = x + yi \), its modulus is calculated as \( |z| = \sqrt{x^2 + y^2} \). This value represents the distance of the point \((x, y)\) from the origin in the complex plane.

In this exercise, the modulus of the expression \( \left| \frac{z_1 - z_2}{z_1 + z_2} \right| = 1 \) is critical. A modulus of 1 indicates that the expression has a constant value on the unit circle of the complex plane. This arises when the modulus of the numerator equals the modulus of the denominator.

Here, \(|z_1 - z_2| = |z_1 + z_2|\) leads us to understand that there's a special geometric relationship between \(z_1\) and \(z_2\).

Geometric Interpretation

Understanding the geometric interpretation of complex numbers on the plane gives great insights. Complex numbers can be represented as points or vectors on the complex plane, where the real part is on the x-axis and the imaginary part on the y-axis.

For the condition \(|z_1 - z_2| = |z_1 + z_2|\), geometrically, it suggests that \(z_1 - z_2\) and \(z_1 + z_2\) are on opposite sides of \(z_1\), making them symmetrically equidistant from the origin.

This is possible when one of the vectors is a rotation, potentially turned by an angle like \( \frac{\pi}{2} \), of the other. Practically, one number or vector can be achieved by rotating and possibly scaling the other, corroborating with the solution involving \(z_2 = ikz_1\).

Complex Plane

The complex plane, a two-dimensional plane, is essential for visualizing and understanding complex numbers. Each complex number \( z = x + yi \) corresponds to a single point on this plane, where \(x\) is the real part and \(y\) is the imaginary part.

This plane allows for a powerful visualization of operations such as addition, subtraction, and especially multiplication by complex numbers. It illuminates ideas like rotation and reflection, which are core when interpreting the modulus condition \(\left| \frac{z_1 - z_2}{z_1 + z_2} \right| = 1\).

The origin-centered unit circle within the complex plane is crucial here. Since all complexes with a modulus of 1 lie on this circle, it shows how \(z_1\) and \(z_2\) relate through symmetric properties, as deduced in this exercise solution.