Question

For which values of $x$, the complex numbers $\cos x + i \sin 2x$ and $\sin x - i \cos 2x$ are conjugates of each other?

• A. $x = 0$
• B. $x = (n + 1/2)\pi$
• C. No value of $x$
• D. $x = n\pi$

Hint:

When resolving simultaneous trigonometric equations, always verify if their angle periodicities are algebraically compatible.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are complex conjugates of each other if their real parts are equal ($a = c$) and their imaginary parts are opposites ($b = -d$).

Step 2: Detailed Explanation:
Let:
\[ z_1 = \cos x + i \sin 2x \]
\[ z_2 = \sin x - i \cos 2x \]
For $z_1$ and $z_2$ to be conjugates, we require:
1. Real parts to be equal:
\[ \cos x = \sin x \implies \tan x = 1 \implies x = n\pi + \frac{\pi}{4} \quad (\text{for } n \in \mathbb{Z}) \]
2. Imaginary parts to be opposites:
\[ \sin 2x = -(-\cos 2x) \implies \sin 2x = \cos 2x \implies \tan 2x = 1 \implies 2x = m\pi + \frac{\pi}{4} \implies x = \frac{m\pi}{2} + \frac{\pi}{8} \quad (\text{for } m \in \mathbb{Z}) \]
Let us equate the two solutions for $x$ to see if a common solution exists:
\[ n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8} \]
Divide both sides by $\pi$:
\[ n + \frac{1}{4} = \frac{m}{2} + \frac{1}{8} \]
Multiply the entire equation by 8 to clear the denominators:
\[ 8n + 2 = 4m + 1 \]
\[ 8n - 4m = -1 \implies 4(2n - m) = -1 \]
Since $n$ and $m$ are integers, the term $2n - m$ must be an integer.
This implies that the left-hand side, $4(2n - m)$, must be a multiple of 4.
However, the right-hand side is $-1$, which is not a multiple of 4.
This contradiction proves that no integer solution exists for $n$ and $m$.
Thus, there is no value of $x$ for which these two complex numbers are conjugates of each other.

Step 3: Final Answer
The correct option is (C).