Problem 145 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(z_{1}+z_{2}+z_{3}=A, z_{1}+z_{2} \omega+z_{3} \omega^{2}=B\) and \(z_{l}+z_{2}\) \(\omega^{2}+z_{3} \omega=C\), where \(1, \omega, \omega^{2}\) are the three cube roots of unity, then \(|A|^{2}+|B|^{2}+|C|^{2}=\) (A) \(3\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2}\right)\) (B) \(2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2}\right)\) (C) \(\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2}\right)\) (D) None of these

Step-by-Step Solution

Answer

(A) \(3\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2}\right)\).

Step 1: Understanding Cube Roots of Unity

The cube roots of unity are \(1, \omega, \omega^2\), where \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\). These roots are used in expressions for symmetry and simplification in calculations.

Step 2: Express the Given Equations

We have:1. \(z_1 + z_2 + z_3 = A\)2. \(z_1 + z_2 \omega + z_3 \omega^2 = B\)3. \(z_1 + z_2 \omega^2 + z_3 \omega = C\)

Step 3: Calculate \(|A|^2\)

Since \(A = z_1 + z_2 + z_3\),\(|A|^2 = (z_1 + z_2 + z_3)\overline{(z_1 + z_2 + z_3)} = |z_1|^2 + |z_2|^2 + |z_3|^2 + 2\Re(z_1\overline{z_2} + z_1\overline{z_3} + z_2\overline{z_3})\).

Step 4: Calculate \(|B|^2\)

For \(B = z_1 + z_2 \omega + z_3 \omega^2\), \(|B|^2 = (z_1 + z_2\omega + z_3\omega^2)\overline{(z_1 + z_2\omega + z_3\omega^2)}\).Using \(\omega\overline{\omega} = 1\), expand this expression and simplify.

Step 5: Calculate \(|C|^2\)

For \(C = z_1 + z_2 \omega^2 + z_3 \omega\),\(|C|^2 = (z_1 + z_2\omega^2 + z_3\omega)\overline{(z_1 + z_2\omega^2 + z_3\omega)}\).Using similar expansion methods as in \(|B|^2\), simplify this expression.

Step 6: Evaluate \(|A|^2 + |B|^2 + |C|^2\)

After expanding and simplifying \(|A|^2\), \(|B|^2\), and \(|C|^2\), combine the results. Observe the symmetry due to cube roots of unity, resulting in terms involving \(|z_1|^2, |z_2|^2, |z_3|^2\) only. The interference terms related to \(\omega\) cancel by symmetry.

Step 7: Simplify to Find the Final Expression

The expressions simplify to \(|A|^2 + |B|^2 + |C|^2 = 3(|z_1|^2 + |z_2|^2 + |z_3|^2)\), as interference terms cancel. Thus, the answer aligns with choice \(\text{(A)}\).

Key Concepts

Cube Roots of Unity

When dealing with complex numbers, the cube roots of unity are fundamental. These roots are the solutions to the equation \(x^3 - 1 = 0\). The cube roots of unity are explicitly given as \(1, \omega, \omega^2\), where \(\omega\) is a complex number and satisfies the following properties:

• \(\omega^3 = 1\) which indicates that after three multiplications, \(\omega\) returns to 1.
• \(1 + \omega + \omega^2 = 0\), a critical property that simplifies many algebraic expressions.

The role of \(\omega\) and its powers, \(\omega^2\), is significant in expressing problems with cyclical symmetry, much like the roots of unity in trigonometric and exponential forms. In the problem, \(\omega\) enables the distribution of the complex numbers symmetrically, hence reducing complexity and revealing patterns.

Modulus of Complex Numbers

The modulus of a complex number \(z = a + bi\) is the distance of \(z\) from the origin in the complex plane. This modulus is calculated using the formula \(|z| = \sqrt{a^2 + b^2}\).

In our problem, calculating the modulus \(|A|^2, |B|^2, |C|^2\) involves finding each complex expression's magnitude and observing their relationships. These relate to the sum of squares and cross-product terms, which arise from multiplying a complex number by its complex conjugate.

• For example, \(|A|^2 = (z_1 + z_2 + z_3)\overline{(z_1 + z_2 + z_3)}\), expands into individual modulus squares and cross terms.
• The technique involves expanding the expression using properties of complex-conjugate multiplication.

Understanding the modulus helps interpret the geometry and symmetry present with complex numbers, as it provides a scalar measure of their size.

Symmetry in Complex Expressions

Symmetry is a powerful tool when dealing with complex numbers, especially in expressions involving cube roots of unity. In the given problem, symmetry manifests through expressions for \(A, B,\) and \(C\) as they are derived from sums of complex numbers weighted by roots of unity.

Here's how symmetry plays a role:

• When you compute \(|B|^2\) or \(|C|^2\), using \(\omega \cdot \overline{\omega} = 1\) helps simplify expressions by ensuring that cross product terms involving \(\omega\) cancel out.
• The cyclical symmetry due to \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\) leads to many interference terms canceling as seen in \(|A|^2 + |B|^2 + |C|^2 = 3(|z_1|^2 + |z_2|^2 + |z_3|^2)\).

Recognizing these patterns helps simplify the algebraic manipulations, allowing simplification from complex conjugates and reducing expressions elegantly by leveraging the inherent balance in the complex plane dictated by these roots.