Question

If \(\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}\) and \(\sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}\), then \(\cos \theta - i\sin \theta\) is equal to

• A. \(e^{i\theta}\)
• B. \(e^{-i\theta}\)
• C. \(-e^{i\theta}\)
• D. \(-e^{-i\theta}\)

Hint:

Remember the classic exponential identities:
- \( \cos \theta + i\sin \theta = e^{i\theta} \)
- \( \cos \theta - i\sin \theta = e^{-i\theta} \)
These polar-exponential identities are incredibly useful throughout complex analysis.

Answer 1

The Correct Option is B