Question

Given \(\varepsilon = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right)\), find the value of \[ \prod_{k=0}^{n-1} \left( \varepsilon^2k - 2\varepsilon k \cos \theta + 1 \right) \]

• A. \(2(1 - \cos n\theta)\)
• B. \(2(1 + \cos n\theta)\)
• C. \((1 - \cos n\theta)^2\)
• D. \(1 + \cos^2 n\theta\)

Answer 1

The Correct Option is A

This is a standard identity from the product over roots of unity in complex numbers. Using De Moivre’s theorem and factorization techniques from trigonometric identities, we derive: \[ \prod_{k=0}^{n-1} \left( \varepsilon^2k - 2\varepsilon k \cos \theta + 1 \right) = 2(1 - \cos n\theta) \] (Full derivation involves higher algebra from complex roots of unity.)