Question

Evaluate: \[ \sum_{k=1}^{6} \left[ \sin\left( \frac{2\pi k}{7} \right) - i \cos\left( \frac{2\pi k}{7} \right) \right] \]

• A. \(1\)
• B. \(-i\)
• C. \(i\)
• D. \(-1\)

Hint:

Recognize patterns of complex numbers using Euler's formula: \( e^{i\theta} = \cos\theta + i\sin\theta \), and use roots of unity properties to simplify such sums.

Answer 1

The Correct Option is C

Combine the terms inside the sum: \[ \sum_{k=1}^{6} \left[ \sin\left( \frac{2\pi k}{7} \right) - i \cos\left( \frac{2\pi k}{7} \right) \right] = \sum_{k=1}^{6} \left[ -i \left( \cos\left( \frac{2\pi k}{7} \right) + i \sin\left( \frac{2\pi k}{7} \right) \right) \right] \] Note: \[ \cos\left( \frac{2\pi k}{7} \right) + i \sin\left( \frac{2\pi k}{7} \right) = e^{i \cdot \frac{2\pi k}{7}} \Rightarrow \text{So the sum becomes:} - i \sum_{k=1}^{6} e^{i \cdot \frac{2\pi k}{7}} \] But the full sum of all \( 7 \)th roots of unity is 0: \[ \sum_{k=0}^{6} e^{i \cdot \frac{2\pi k}{7}} = 0 \Rightarrow \sum_{k=1}^{6} e^{i \cdot \frac{2\pi k}{7}} = -1 \] Thus, the total sum: \[ - i \cdot (-1) = i \]