Question

For any real $\theta$, $(\cos \theta + i \sin \theta)(\cos \theta - i \sin \theta) =$

• A. $1$
• B. $-1$
• C. $0$
• D. $4i$

Hint:

For any complex number $z = a + bi$, the product $z \cdot \bar{z}$ is always $|z|^2$, which is $a^2 + b^2$. For the unit complex number $e^{i\theta}$, the magnitude is 1, so the product is $1^2 = 1$.

Answer 1

The Correct Option is A

This expression represents the product of a complex number and its conjugate. 1. Algebraic Expansion: Use the difference of squares identity $(a + b)(a - b) = a^2 - b^2$: $$(\cos \theta + i \sin \theta)(\cos \theta - i \sin \theta) = (\cos \theta)^2 - (i \sin \theta)^2$$

2. Applying the property of $i$: Since $i^2 = -1$: $$\cos^2 \theta - (i^2 \sin^2 \theta)$$ $$\cos^2 \theta - (-1 \cdot \sin^2 \theta)$$ $$\cos^2 \theta + \sin^2 \theta$$

3. Using the Pythagorean Identity: In trigonometry, $\cos^2 \theta + \sin^2 \theta = 1$ for any real value of $\theta$. Thus, the product is always equal to 1.