Question

If \( x_k = \cos \left( \frac{2\pi k}{n} \right) + i \sin \left( \frac{2\pi k}{n} \right) \), then \[ \sum_{k=1}^{n} x_k = ? \]

• A. 1
• B. -1
• C. 0
• D. None of these

Hint:

The sum of the \( n \)-th roots of unity is always zero. This can be derived from their geometric interpretation or using properties of polynomials.

Answer 1

The Correct Option is C

Step 1: Recognize the given expression.
The expression \( x_k = \cos \left( \frac{2\pi k}{n} \right) + i \sin \left( \frac{2\pi k}{n} \right) \) is the general form of a complex number on the unit circle in the complex plane, and it represents the \( n \)-th roots of unity. These roots of unity are the values that satisfy the equation: \[ z^n = 1, \] where \( z \) is a complex number.

Step 2: Use the sum of the \( n \)-th roots of unity.
The sum of all the \( n \)-th roots of unity is known to be zero. This result can be derived from the geometric interpretation of these roots on the unit circle, or from the fact that they are the solutions to the equation \( z^n = 1 \), and the sum of all roots of any polynomial equation is zero (for a polynomial with no constant term). Specifically, \[ \sum_{k=1}^{n} \left( \cos \left( \frac{2\pi k}{n} \right) + i \sin \left( \frac{2\pi k}{n} \right) \right) = 0. \]

Step 3: Conclusion.
Thus, \[ \sum_{k=1}^{n} x_k = 0. \] The correct answer is (c) 0.