Problem 195 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(z\) and \(\omega\) are two non-zero complex numbers such that \(|z \omega|=1\), and \(\operatorname{Arg}(z)-\operatorname{Arg}(\omega)=\frac{\pi}{2}\), then \(\bar{Z} \omega\) is equal to \([\mathbf{2 0 0 3}]\) (A) 1 (B) \(-1\) (C) \(i\) (D) \(-i\)

Step-by-Step Solution

Answer

(C) \(i\)

Step 1: Understand Modulus Condition

The equation \(|z \omega| = 1\) implies that the product of the magnitudes of \(z\) and \(\omega\) is 1. If \(|z| = r\) and \(|\omega| = s\), this means \(r \cdot s = 1\).

Step 2: Understand Argument Condition

The equation \(\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}\) implies that the angle difference between \(z\) and \(\omega\) is \(\frac{\pi}{2}\). This means \(z\) and \(\omega\) are orthogonal in terms of their directions.

Step 3: Rewrite conjugate-product expression

Using the argument condition, relate the complex conjugate: \(\overline{z} = \frac{1}{z}\) due to \(|z|^2 = 1\). Therefore, \(\bar{Z} \omega = \frac{\omega}{z}\).

Step 4: Substitute and Simplify Using Argument

Using \(z = re^{i\theta}\) and \(\omega = se^{i(\theta + \frac{\pi}{2})}\):\[ \overline{z} = \frac{1}{r} e^{-i\theta} \]So, \[ \bar{Z} \omega = \frac{1}{r} e^{-i\theta} \cdot s e^{i(\theta + \frac{\pi}{2})} = \frac{s}{r} (e^{i\frac{\pi}{2}}) = \frac{s}{r} i \].

Step 5: Apply Modulus Condition to Simplify

Given \(r \cdot s = 1\), substitute to get \(\frac{s}{r} = 1\). Thus, \[ \bar{Z} \omega = i \].

Step 6: Conclusion: Identify the Correct Option

The real value of \(\bar{Z} \omega\) based on our simplification is \(i\), matching option C.

Key Concepts

Modulus of Complex Numbers

In the realm of complex numbers, each number can be expressed in the form of \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The modulus of a complex number is essentially a measure of its "size".
To calculate the modulus, denoted as \( |z| \), one uses the formula: \[ |z| = \sqrt{a^2 + b^2} \] Think of it as finding the length of the vector that represents the complex number on the complex plane.

• It has a geometrical interpretation as the distance from the origin to the point \( (a, b) \) on the complex plane.
• It is always a non-negative real number.

In the provided exercise, the condition \(|z \omega| = 1\) shows that the product of the moduli of the two complex numbers \(z\) and \( \omega \) equals 1.
This implies that if one number moves away from the unit distance, the other compensates.Understanding modulus is key in studying properties like magnitude and phase relationships in complex analysis.

Argument of Complex Numbers

The argument of a complex number is all about its direction in the complex plane. Simply put, it tells us the angle that the number makes with the positive real axis.
If you express a complex number \( z = re^{i\theta} \), \( \theta \) is the argument of \( z \), often denoted as \( \operatorname{Arg}(z) \).

• The argument is usually measured in radians and can be any real number.
• For an expression such as \( \, \operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2} \, \), it implies that the two numbers are orthogonal.

In the context of the given exercise, this angle difference corresponds to the numbers being perpendicular to each other if plotted in the plane.
This means knowing about arguments aids in understanding rotations and positions relative to one another.It’s crucial for operations like multiplication and division in polar forms of complex numbers.

Complex Conjugate

The complex conjugate of a number \( z = a + bi \) is \( \overline{z} = a - bi \). The conjugate "flips" the sign of the imaginary part.
Complex conjugates have several interesting properties:

• When a complex number is multiplied by its conjugate, the result is a real number: \( z \cdot \overline{z} = a^2 + b^2 = |z|^2 \).
• The conjugate has the same modulus as the original number, which makes it useful in simplifying expressions.

In the exercise, using \( \overline{z} = \frac{1}{z} \) comes from understanding the normalization process when \(|z|\) is 1. This allows the expression \( \overline{z} \cdot \omega \) to transform into simpler terms, often necessary for separating components into real and imaginary parts.
Understanding conjugates is vital as it often appears in solving equations, simplifying fractions, and representing real-life phenomena in engineering and physics.