Problem 175 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(1, a_{1}, a_{2}, \ldots, a_{n-1}\) are the \(n, n\)th roots of unity, then \(\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right) \ldots\left(1-a_{n-1}\right)=\) (A) \(n+1\) (B) \(n\) (C) \(n-1\) (D) None of these.

Step-by-Step Solution

Answer

The product is \(0\), so the answer is (D) None of these.

Step 1: Define the problem

We are given that \(1, a_1, a_2, \ldots, a_{n-1}\) are the \(n\)th roots of unity. This means these are the solutions of the equation \(x^n - 1 = 0\). We need to find the product of \((1 - a_1)(1 - a_2)\ldots(1 - a_{n-1})\).

Step 2: Understanding n-th roots of unity

The \(n\)th roots of unity are given by \(1, \omega, \omega^2, \ldots, \omega^{n-1}\), where \(\omega = e^{2\pi i/n}\) is a primitive \(n\)th root of unity. These roots satisfy the property \(\sum_{k=0}^{n-1} \omega^k = 0\).

Step 3: Consider the polynomial

The \(n\)th roots of unity are the roots of the polynomial \(x^n - 1 = 0\). This can be factored as \((x-1)(x-a_1)(x-a_2)\ldots(x-a_{n-1}) = x^n - 1\).

Step 4: Substitute x = 1

Substitute \(x = 1\) in the factored form of the polynomial. This gives \((1-1)(1-a_1)(1-a_2)\ldots(1-a_{n-1}) = 1^n - 1 = 0\).

Step 5: Solve for product

Since \((1-1) = 0\), the product \((1-a_1)(1-a_2)\ldots(1-a_{n-1}) = 0\) as one of the factors (\((1 - 1)\)) in the product is zero.

Key Concepts

n-th roots

The concept of the "n-th roots" is pivotal in understanding many algebraic problems, especially in the field of complex numbers. The "n-th roots of unity" represent special solutions to a polynomial that are inherently structured and symmetrical. These roots are solutions to the equation \(x^n = 1\), meaning if you raise any of these roots to the power of \(n\), the result will be 1.
However, not all these roots are 1; most of them are complex numbers known as "primitive roots of unity."

• The first root is always 1, because any number to the power of 0 is 1.


• The remaining roots, \( \omega, \omega^2, \ldots, \omega^{n-1} \), are complex numbers evenly distributed along the circumference of the unit circle in the complex plane.


• The primitive \( n \)-th root of unity is given by \( \omega = e^{2\pi i / n} \).

These roots together satisfy the crucial identity \( \sum_{k=0}^{n-1} \omega^k = 0 \), which is instrumental in simplifying and solving polynomials like \( x^n - 1 = 0 \). Each root has a geometric interpretation and a particular angle, which repeats every \( n \)th step.Understanding how these roots work allows you to solve many complex problems in both theoretical and applied mathematics.

polynomial equation

A "polynomial equation" is a foundational concept in mathematics and forms the basis for much of algebra. At its core, a polynomial equation is an expression made up of variables and coefficients, involving terms only up to a given power. Specifically, a polynomial equation in one variable \( x \) looks like \( a_n x^n + a_{n-1} x^{n-1} + \, \ldots \, + a_1 x + a_0 = 0 \).
Each term is a product of a constant and a variable to a non-negative integer power.In the exercise discussed, the polynomial is \( x^n - 1 = 0 \). This specific form is significant because:

• It represents an equation where \( x^n \) is set equal to 1, seeking those \( x \) values (the roots) that satisfy this relationship.


• The polynomial \( x^n - 1 = 0 \) can be factored into \((x - 1)(x - a_1)\ldots(x - a_{n-1}) = 0\), where the \( a_i \)'s are the non-trivial \( n \)-th roots of unity, excluding the root 1.


• This factorization helps to locate all solutions (roots) of the equation.

Each root makes the polynomial equal to zero when substituted for \( x \). Therefore, these roots represent the solutions of the equation and are crucial in solving the products and sums of complex polynomials.

complex numbers

"Complex numbers" extend the traditional concept of numbers by introducing an imaginary unit \( i \), where \( i^2 = -1 \). This is a groundbreaking concept because it allows every polynomial equation to have solutions, ensuring all roots can be accounted for.

• A complex number has the form \( a + bi \), where \( a \) is the real part, and \( bi \) is the imaginary part.


• These numbers can be plotted on the complex plane, with the real part on the x-axis and the imaginary part on the y-axis.


• In calculations, the magnitude (\(|z|\), where \(z = a + bi\)), is found as \( \sqrt{a^2 + b^2} \), representing a distance from the origin.

The exercise shows this concept by involving \( n \)-th roots of unity, where most roots are complex numbers. These roots are situated on the unit circle (radius 1) in the complex plane and are equidistant, forming regular polygons.
The depth and beauty of complex numbers are not just in solving equations, but in their ability to represent periodic phenomena naturally, which is central to physics, engineering, and many other fields.