Problem 24 - A Complete Resource Book in Mathematics for JEE Main

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 \((1-a_1)(1-a_2)...(1-a_{n-1})\) is \(n\). Choose (B) \(n\).

Step 1: Understanding Roots of Unity

The nth roots of unity are the solutions to the equation \(x^n = 1\). These roots are of the form \(a_k = e^{2 \pi i k / n}\) for \(k = 0, 1, 2, ..., n-1\), where \(a_0 = 1\). Thus, \(1, a_1, a_2, ..., a_{n-1}\) represent all roots of \(x^n = 1\).

Step 2: Roots of Unity Polynomial

The polynomial whose roots are the nth roots of unity is \(x^n - 1\). This can be rewritten as \((x-1)(x-a_1)(x-a_2)...(x-a_{n-1}) = 0\). Hence, \((x-a_1)(x-a_2)...(x-a_{n-1}) = \frac{x^n - 1}{x-1}\).

Step 3: Evaluating at x=1

Plug \(x = 1\) into the expression \((x-a_1)(x-a_2)...(x-a_{n-1}) = \frac{x^n - 1}{x-1}\) to find the product \((1-a_1)(1-a_2)...(1-a_{n-1})\).

Step 4: Simplify the Expression

Substitute \(x=1\) leading to: \(\frac{1^n - 1}{1 - 1} = \frac{0}{0}\), which is undefined. However, this simplifies to evaluating the limit as \(x\) approaches 1: \(\lim\_{x \to 1} \frac{x^n - 1}{x - 1} = n\), using the derivative of \(x^n - 1\) at \(x = 1\) which is \(nx^{n-1}\). By L'Hôpital's rule, this limit is \(n\).

Step 5: Conclusion

The value of \((1-a_1)(1-a_2)...(1-a_{n-1})\) is \(n\). Therefore, option (B) \(n\) is the correct answer.

Key Concepts

Complex Numbers

Complex numbers are numbers that include the imaginary unit, usually denoted as \(i\), where \(i^2 = -1\). A complex number is often expressed in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. Complex numbers are crucial in finding roots of unity, as these roots are expressed in terms of complex numbers.
When dealing with nth roots of unity, you encounter expressions like \(e^{2\pi i k / n}\), where \(k\) is an integer. This notation, derived from Euler's formula, represents the complex numbers forming a unit circle in the complex plane. These values have a magnitude of 1 and equally divide the circle, meaning they lie on the circumference of a circle of radius one centered at the origin of the plane.

Polynomials

Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. When exploring the roots of unity, we encounter the polynomial \(x^n - 1\), which is significant as its roots are precisely the nth roots of unity.
The polynomial \(x^n - 1 = (x - 1)(x - a_1)(x - a_2)...(x - a_{n-1})\) reveals that each factor corresponds to one of the roots of unity. This polynomial structure aids in understanding how roots of unity relate to one another and can be used to simplify complex computations. The factor \((x - a_k)\) means each root \(a_k\) is a solution that, when substituted for \(x\), will make the polynomial equal zero. These polynomials form the foundation for many algebraic operations, including finding limits and solving equations.

Limits

Limits are a fundamental concept in calculus, defining the value that a function approaches as the input approaches a point. In the exercise, after substituting \(x = 1\) into the polynomial derived from the nth roots of unity, we faced an undefined expression \(\frac{0}{0}\). This scenario is typical and necessitates finding a limit.
To solve this, we use the limit \(\lim_{x \to 1} \frac{x^n - 1}{x - 1}\), which simplifies using methods like L'Hôpital's rule. The principle of a limit helps navigate undefined expressions and ensures accurate calculations of values an expression approaches. It verifies that complex operations such as polynomial division and simplification result in a well-defined answer, preventing indeterminate forms.

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool in calculus for resolving indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When we are confronted with these forms while calculating limits, L'Hôpital's Rule allows us to simplify them by differentiating the numerator and denominator separately.
In this exercise, after substituting \(x = 1\) in \(\frac{x^n - 1}{x - 1}\), the expression becomes \(\frac{0}{0}\). Applying L'Hôpital's rule involves computing the derivative of both the numerator \(x^n - 1\) and the denominator \(x - 1\), resulting in \(\lim_{x \to 1} \frac{d}{dx}(x^n - 1) / \frac{d}{dx}(x - 1) = \lim_{x \to 1} \frac{nx^{n-1}}{1}\).
This process yields \(n\), providing us with the correct solution. This method clarifies potential pitfalls when direct substitution leads to uncertainties and ensures precise outcomes.