Problem 146 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(z_{1}+z_{2}+z_{3}=A, z_{1}+z_{2} \omega+z_{3} \omega^{2}=B\) and \(z_{1}+z_{2}\) \(\omega^{2}+z_{3} \omega=C\), 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

The answer is (A).

Step 1: Understand Cube Roots of Unity

The cube roots of unity are given as \(1, \omega, \omega^2\), where \(\omega^3 = 1\) and \(\omega eq 1\). These satisfy \(1 + \omega + \omega^2 = 0\). This property will be used to simplify expressions.

Step 2: Rearrange Equations with Cube Roots of Unity

Use 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. \(z_1 + z_2\omega^2 + z_3\omega = C\)These equations take advantage of the symmetrical properties of cube roots of unity.

Step 3: Solve for z's Using Symmetry

Combine equations using symmetry. Note that \( A + B\omega + C\omega^2 = 3z_1\) by plugging values, where the terms vanish due to property \(1 + \omega + \omega^2 = 0\). Similarly, solving for other combinations will give:\( A + B\omega^2 + C\omega = 3z_2\) and\( A\omega + B\omega^2 + C = 3z_3\).

Step 4: Calculate Squares of Magnitudes

Calculate \[ |A|^2 = (A)(\overline{A}), \]\[ |B|^2 = (B)(\overline{B}), \]\[ |C|^2 = (C)(\overline{C}), \]where \(\overline{A}, \overline{B}, \overline{C}\) are the conjugates of \(A, B, C\). Use the property that \[ |A|^2 + |B|^2 + |C|^2\] will obey symmetrical properties based on calculated values.

Step 5: Find Total Sum |A|^2 + |B|^2 + |C|^2

By adding the magnitudes from above, use symmetry directly known from cube roots of unity:\[ |A|^2 + |B|^2 + |C|^2 = 3(|z_1|^2 + |z_2|^2 + |z_3|^2). \]This follows from algebraic symmetry where each expression provides the same kind of summation.

Key Concepts

Cube Roots of Unity

In mathematics, the cube roots of unity are special numbers that, when multiplied by themselves three times, return to the value of one. They are essential in simplifying certain algebraic expressions. Specifically, these roots are denoted as 1, \( \omega \), and \( \omega^2 \), where \( \omega^3 = 1 \) and \( \omega eq 1 \). An interesting property is that their sum is zero: \( 1 + \omega + \omega^2 = 0 \).
This property is very useful because it allows us to express complex equations in simpler forms. By using various combinations of these roots—such as replacing \( \omega \) in equations—we can take advantage of their symmetry to find specific values, like \( z_1, z_2, z_3 \) in the original problem. Their symmetric nature often leads to cancellations or simplifications, which significantly aids in problem-solving.

Magnitudes of Complex Numbers

Complex numbers are usually in the form \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The magnitude of a complex number, also known as its absolute value or modulus, is derived from the Pythagorean theorem. It is given by \( |z| = \sqrt{a^2 + b^2} \).
In the exercise, we calculate the magnitudes \( |A|^2, |B|^2, |C|^2 \) based on the provided expressions. When we consider magnitudes in this case, we often encounter the need to calculate these squares—\( |A|^2 = A \overline{A} \), where \( \overline{A} \) is the conjugate of A. Using these calculations allows one to derive combined expressions of magnitude such as \( |A|^2 + |B|^2 + |C|^2 = 3(|z_1|^2 + |z_2|^2 + |z_3|^2) \), helped by the symmetry of the given complex numbers, as seen with cube roots of unity.

Symmetry in Complex Numbers

Symmetry in complex numbers plays a crucial role in problem-solving, particularly with roots of unity. It refers to the balanced and recurring patterns or properties that make certain sets of complex numbers behave predictably. In the exercise, the symmetry is exploited with the identity \( 1 + \omega + \omega^2 = 0 \).
This symmetry is used to simplify expressions since combining the equations with cube roots of unity simplify our tasks immensely. For instance, moving from \( A + B\omega + C\omega^2 = 3z_1 \), it takes advantage of how \( \omega \) can change the terms within the cube to cancel out certain terms due to that symmetry. This leads to simpler new forms, making calculations less complex and helping to prove that \( |A|^2 + |B|^2 + |C|^2 = 3(|z_1|^2 + |z_2|^2 + |z_3|^2) \). The predictable nature of these symmetrical properties is invaluable in reducing complex equations into manageable fragments.