Question

The product of all values of \( \cos(\alpha) + i \sin(\alpha) )^{3/5} \) is equal to

• A. 1
• B. \( \cos\alpha + i \sin\alpha \)
• C. \( \cos 5\alpha + i \sin 5\alpha \)
• D. \( \cos \alpha + i \sin 5\alpha \)

Hint:

De Moivre's theorem is useful for raising complex numbers in polar form to powers.

Answer 1

The Correct Option is C

Step 1: Using De Moivre's Theorem.
De Moivre's theorem states that \( \left( \cos\alpha + i \sin\alpha \right)^n = \cos(n\alpha) + i \sin(n\alpha) \). Thus, the product of all values is \( \cos 5\alpha + i \sin 5\alpha \).

Step 2: Conclusion.
Thus, the correct answer is option (C).

Final Answer: \[ \boxed{\text{(C) } \cos 5\alpha + i \sin 5\alpha} \]