Question

If \(1, \alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_{n-1}\) are n roots of the equation, \( x^n = 1 \) then the value of \( (1-\alpha_1)(1-\alpha_2)(1-\alpha_3)\ldots(1-\alpha_{n-1}) \) is

• A. n
• B. n-1
• C. n-2
• D. n-3

Hint:

This is a standard result related to roots of unity. The polynomial \( \Phi_n(x) = \frac{x^n-1}{x-1} \) is called the nth cyclotomic polynomial, and its roots are the primitive nth roots of unity (for prime n). The value of this polynomial at \(x=1\) gives the desired product, which is always \(n\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The equation \(x^n = 1\) defines the \(n^{th}\) roots of unity. The question asks for the product of terms of the form \( (1 - \alpha_k) \), where the \( \alpha_k \) are the roots of unity excluding 1 itself.
Step 2: Key Formula or Approach:
The polynomial \( P(x) = x^n - 1 \) has the \(n\) roots of unity as its roots. Therefore, it can be factored as:
\[ x^n - 1 = (x-1)(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1}) \] We can also use the formula for the sum of a geometric series to write:
\[ x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \ldots + x + 1) \] By comparing these two factorizations, we can deduce an expression for the product involving the roots.
Step 3: Detailed Explanation:
From the two factorizations of \( x^n - 1 \), we can equate the parts that do not contain the \( (x-1) \) factor:
\[ (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1}) = x^{n-1} + x^{n-2} + \ldots + x + 1 \] We are asked to find the value of the expression \( (1-\alpha_1)(1-\alpha_2)\ldots(1-\alpha_{n-1}) \).
This can be obtained by substituting \( x = 1 \) into the polynomial identity above.
Let \( Q(x) = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1}) \).
We need to find \( Q(1) \).
\[ Q(1) = 1^{n-1} + 1^{n-2} + \ldots + 1^1 + 1 \] The sum on the right-hand side has \(n\) terms, and each term is equal to 1.
\[ Q(1) = \underbrace{1 + 1 + \ldots + 1}_{n \text{ terms}} = n \] Step 4: Final Answer:
The value of the expression is \( n \).