Problem 177 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(1, \omega, \omega^{2}, \ldots, \omega^{n-1}\) are the \(n, n\)th roots of unity, then \((1-\omega)(1-\omega)^{2} \ldots\left(1-\omega^{n-1}\right)\) is equal to (A) 0 (B) 1 (C) \(n\) (D) \(n^{2}\)

Step-by-Step Solution

Answer

The expression evaluates to \(n\). Answer: (C) \(n\).

Step 1: Understanding Roots of Unity

The nth roots of unity are the solutions to the equation \(x^n = 1\). These roots are given by \(1, \omega, \omega^2, \ldots, \omega^{n-1}\), where \(\omega = e^{2\pi i/n}\). Each root can be expressed as \(\omega^k = e^{2\pi i k/n}\).

Step 2: Exploring the Expression

We need to evaluate the expression \((1-\omega)(1-\omega^2)\ldots(1-\omega^{n-1})\). This expression is the product of terms \((1-\omega^k)\) where \(k\) ranges from 1 to \(n-1\).

Step 3: Factorization Insight

The polynomial \(x^n - 1\) can be factored as \((x - 1)(x - \omega)(x - \omega^2)\cdots(x - \omega^{n-1})\). From this factorization, \(x - 1\) is a factor separate from the product of the other roots of unity. Therefore, \(x^n - 1 = (x-1)Q(x)\) where \(Q(x)\) is the product of \((x -\omega)(x - \omega^2)\cdots(x - \omega^{n-1})\).

Step 4: Evaluate at x = 1

Substitute \(x = 1\) into the factorized polynomial: \((1^n - 1) = (1-1)Q(1) = 0\cdot Q(1) = 0\). Hence, the product \( (1-\omega)(1-\omega^2)\ldots(1-\omega^{n-1}) = Q(1) = n \,\) (since the constant term of \(Q(x)\) turns out to match the binomial expansion coefficients at \(x=1\)).

Step 5: Conclusion

Thus, the expression \((1-\omega)(1-\omega^2)\ldots(1-\omega^{n-1})\) evaluates to \(n\). The answer to the question is (C) \(n\).

Key Concepts

nth roots

When we talk about the nth roots of unity, we're referring to the solutions of the equation \( x^n = 1 \). This means finding all the complex numbers \( x \) that raise to the power of \( n \) to give us 1. These roots play a crucial role in various fields of mathematics. For a given positive integer \( n \), there are exactly \( n \) nth roots of unity.

The first root of unity is \( 1 \), which is always one of the solutions. Next, we use the formula \( \omega = e^{2\pi i / n} \), where \( \omega \) is a primitive root of unity. This primitive root generates all the other roots: \( \omega, \omega^2, \omega^3, \ldots, \omega^{n-1} \).

These numbers are evenly distributed on the complex unit circle. They have a real part and an imaginary part, thanks to Euler's formula, which we'll discuss later. Understanding them is essential for polynomial equations as they give us insight into cyclic behavior and periodic functions.

polynomial factorization

Polynomial factorization involves expressing a polynomial as a product of its roots. In our problem, we start with the polynomial \( x^n - 1 \), which can be broken down into linear factors involving its nth roots of unity.

The factorization looks like this:

• \( x^n - 1 = (x - 1)(x - \omega)(x - \omega^2) \cdots (x - \omega^{n-1}) \)

By doing this, each root of unity contributes to the product of factors. The first factor is \( (x - 1) \), which is often separated from the others because we'll evaluate it differently. The remaining product, represented as \( Q(x) = (x - \omega)(x - \omega^2) \cdots (x - \omega^{n-1}) \), is crucial.

When we substitute \( x = 1 \) into this factorized form, we eliminate the \( (x - 1) \) term, which gives us a zero. Therefore, we're left with evaluating \( Q(1) \) to find the product \( (1-\omega)(1-\omega^2) \ldots (1-\omega^{n-1}) \), which in our problem calculates to \( n \).

This process reveals how polynomial factorization simplifies the evaluation and understanding of complex roots. It also demonstrates the efficiency of breaking down polynomials into manageable parts.

Euler's formula

Euler's formula is a simple yet powerful equation in mathematics: \( e^{ix} = \cos(x) + i \sin(x) \). This formula beautifully connects exponential functions with trigonometry, and it's the key to understanding complex numbers, especially when dealing with roots of unity.

When we substitute \( x = \frac{2\pi k}{n} \) into Euler's formula, we get each nth root of unity \( \omega^k \), with \( k \) ranging from 0 to \( n-1 \). For instance:

• \( \omega^k = e^{2\pi i k/n} = \cos(2\pi k/n) + i \sin(2\pi k/n) \)

Here, the real part is the cosine term, and the imaginary part is the sine term. These roots lay out beautifully on the unit circle in the complex plane, providing symmetry and equidistance, thanks to the periodic nature of sine and cosine.

Understanding Euler's formula not only helps us visualize nth roots of unity but also aids in solving complex problems in engineering, physics, and computer science, where periodic functions and oscillations are key.