Question

The set of all values of $\theta$ such that $\frac{1-i\cos\theta}{1+2i\sin\theta}$ is purely imaginary is

• A. $\{n\pi + (-1)^n \frac{\pi}{4}, n \in \mathbb{Z}\}$
• B. $\{\frac{n\pi}{2} + (-1)^n \frac{\pi}{4}, n \in \mathbb{Z}\}$
• C. $\{n\pi + (-1)^n \frac{\pi}{2}, n \in \mathbb{Z}\}$
• D. $\{2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z}\}$

Hint:

To check if a complex number is purely real or purely imaginary, the condition is often simpler using conjugates. A number $z$ is purely imaginary if $z = - \bar{z}$ (and $z \neq 0$). For this problem, setting $\text{Re}(z)=0$ after rationalization is the most direct method. Be familiar with general solutions of trigonometric equations like $\sin x = a, \cos x = a, \tan x = a$.

Answer 1

The Correct Option is B

Step 1: Simplify the Complex Number: Let \( z = \frac{1-i\cos\theta}{1+2i\sin\theta} \). Multiply numerator and denominator by the conjugate of the denominator \( 1 - 2i\sin\theta \): \[ z = \frac{(1-i\cos\theta)(1-2i\sin\theta)}{(1+2i\sin\theta)(1-2i\sin\theta)} = \frac{1 - 2i\sin\theta - i\cos\theta + 2i^2\sin\theta\cos\theta}{1 + 4\sin^2\theta} \] \[ z = \frac{(1 - 2\sin\theta\cos\theta) - i(2\sin\theta + \cos\theta)}{1 + 4\sin^2\theta} \]
Step 2: Condition for Purely Imaginary: A complex number is purely imaginary if its Real part is zero. \[ \text{Re}(z) = \frac{1 - 2\sin\theta\cos\theta}{1 + 4\sin^2\theta} = 0 \] \[ 1 - 2\sin\theta\cos\theta = 0 \] \[ \sin(2\theta) = 1 \]
Step 3: Solve for \( \theta \): \[ 2\theta = 2n\pi + \frac{\pi}{2} \] \[ \theta = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} \] This represents the set \( \{ \dots, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}, \dots \} \). Option (B) in the image roughly corresponds to this form (or implies the general solution for a related sine equation). The correct mathematical set is \( \{ n\pi + \pi/4 \} \).