Problem 10 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(\omega_{1}\) and \(\omega_{2}\) are complex cube roots of unity, then prove that \(\omega_{1}^{4}+\omega_{2}^{4}=-\frac{1}{\omega_{1} \omega_{2}}\).

Step-by-Step Solution

Answer

The result is verified: \(\omega_1^4 + \omega_2^4 = -\frac{1}{\omega_1 \omega_2} = -1\).

Step 1: Understanding Cube Roots of Unity

Complex cube roots of unity are the solutions to the equation \(x^3 = 1\). They are given by \(1, \omega_1, \omega_2\), where \(\omega_1 = e^{2\pi i / 3}\) and \(\omega_2 = e^{-2\pi i / 3}\). These roots satisfy the relations \(\omega_1^3 = 1\), \(\omega_2^3 = 1\), and \(\omega_1 \cdot \omega_2 = 1\).

Step 2: Expressing Powers of Roots

Calculate higher powers of \(\omega_1\) and \(\omega_2\). Since \(\omega_1^3 = 1\), we know \(\omega_1^4 = \omega_1\) and similarly \(\omega_2^4 = \omega_2\). Thus, \(\omega_1^4 + \omega_2^4 = \omega_1 + \omega_2\).

Step 3: Using the Sum of Roots Property

For cube roots of unity, the sum \(1 + \omega_1 + \omega_2 = 0\) holds. Thus, rearranging gives \(\omega_1 + \omega_2 = -1\). Therefore, \(\omega_1^4 + \omega_2^4 = \omega_1 + \omega_2 = -1\).

Step 4: Verify the Identity with Product Relation

We need to show that \(-1 = -\frac{1}{\omega_1 \omega_2}\). From cube roots of unity, \(\omega_1 \omega_2 = 1\). Thus, \(-\frac{1}{\omega_1 \omega_2} = -1\), which matches \(\omega_1^4 + \omega_2^4\).

Key Concepts

Cube Roots of Unity

To truly grasp the concept of cube roots of unity, consider that they are the solutions to the equation \(x^3 = 1\). This means these are the complex numbers that, when raised to the power of three, yield 1. There are always three cube roots for any number, but in the case of unity (1), the roots are well-defined as \(1\), \(\omega_1\), and \(\omega_2\). Here’s how they work:

• \(\omega_1 = e^{2\pi i / 3}\): A complex number representing a 120-degree rotation on the complex plane.
• \(\omega_2 = e^{-2\pi i / 3}\): Similarly, it represents a 240-degree rotation.

These roots are deeply connected through the relation \(\omega_1^3 = \omega_2^3 = 1\), and interestingly enough, the product \(\omega_1 \cdot \omega_2 = 1\), meaning multiplying them together returns us to unity. These properties form the basis for manipulating any powers of cube roots.

Powers of Complex Numbers

When dealing with the powers of complex numbers, particularly powers of cube roots of unity, there are specific properties that simplify computations. For example, since \(\omega_1^3 = 1\), higher powers like \(\omega_1^4\) can be simplified:

• \(\omega_1^4 = \omega_1 \cdot \omega_1^3 = \omega_1\)
• Similarly, \(\omega_2^4 = \omega_2\)

Thus, the expression \(\omega_1^4 + \omega_2^4\) reduces to the simpler calculation of \(\omega_1 + \omega_2\). Knowing how these numbers cycle through their powers makes it easier to solve or verify identities and formulae involving them.

Properties of Roots of Unity

The properties of cube roots of unity reveal fascinating and useful relationships. One such property is their sum:

• \(1 + \omega_1 + \omega_2 = 0\)

This means that the sum of all cube roots of unity is always zero. From this equation, we can infer that:

• \(\omega_1 + \omega_2 = -1\)

Furthermore, using the product of \(\omega_1 \cdot \omega_2 = 1\), we can conclude:

• \(-\frac{1}{\omega_1 \omega_2} = -1\)

These properties allow for converting complex identities into simpler equations, facilitating problem-solving in exercises like the one detailed above, where the goal was to confirm that \(\omega_1^4 + \omega_2^4 = -\frac{1}{\omega_1 \omega_2}\).