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: Understanding the Concept
A complex number $z = x+iy$ is purely imaginary if its real part is zero ($x=0$) and its imaginary part is non-zero ($y \neq 0$).
Step 2: Key Formula or Approach
To find the real part of the given complex number, we first need to write it in the standard form $x+iy$. This is done by multiplying the numerator and denominator by the conjugate of the denominator.
Step 3: Detailed Explanation
Let $z = \frac{1-i\cos\theta}{1+2i\sin\theta}$.
Multiply the numerator and denominator by the conjugate of the denominator, which is $1-2i\sin\theta$.
\[ z = \frac{1-i\cos\theta}{1+2i\sin\theta} \times \frac{1-2i\sin\theta}{1-2i\sin\theta} \] The denominator becomes:
\[ (1+2i\sin\theta)(1-2i\sin\theta) = 1^2 - (2i\sin\theta)^2 = 1 - 4i^2\sin^2\theta = 1 + 4\sin^2\theta \] The numerator becomes:
\[ (1-i\cos\theta)(1-2i\sin\theta) = 1 - 2i\sin\theta - i\cos\theta + 2i^2\cos\theta\sin\theta \] \[ = 1 - i(2\sin\theta+\cos\theta) - 2\sin\theta\cos\theta \] \[ = (1 - 2\sin\theta\cos\theta) - i(2\sin\theta+\cos\theta) \] So, the complex number is: \[ z = \frac{(1 - 2\sin\theta\cos\theta) - i(2\sin\theta+\cos\theta)}{1 + 4\sin^2\theta} \] \[ z = \frac{1 - \sin(2\theta)}{1 + 4\sin^2\theta} - i \frac{2\sin\theta+\cos\theta}{1 + 4\sin^2\theta} \] For $z$ to be purely imaginary, its real part must be zero.
\[ \text{Re}(z) = \frac{1 - \sin(2\theta)}{1 + 4\sin^2\theta} = 0 \] Since $1+4\sin^2\theta$ is always positive (it's between 1 and 5), the numerator must be zero.
\[ 1 - \sin(2\theta) = 0 \implies \sin(2\theta) = 1 \] The general solution for $\sin(x)=1$ is $x = 2k\pi + \frac{\pi}{2}$, where $k \in \mathbb{Z}$. So, $2\theta = 2k\pi + \frac{\pi}{2}$.
\[ \theta = k\pi + \frac{\pi}{4}, \quad k \in \mathbb{Z} \] We must also ensure the imaginary part is non-zero. For $\theta = k\pi + \pi/4$, $\sin\theta$ and $\cos\theta$ are $\pm 1/\sqrt{2}$. The imaginary part's numerator is $-(2\sin\theta+\cos\theta)$ which is never zero for these values.
The solution set is $\{ k\pi + \frac{\pi}{4} \mid k \in \mathbb{Z} \}$.
Let's check the options. The provided correct option is (B). Let's see if it is equivalent to our solution.
Let's test option (B): $\theta = \frac{n\pi}{2} + (-1)^n \frac{\pi}{4}$.
If $n$ is even, let $n=2k$: $\theta = \frac{2k\pi}{2} + (-1)^{2k}\frac{\pi}{4} = k\pi + \frac{\pi}{4}$. If $n$ is odd, let $n=2k+1$: $\theta = \frac{(2k+1)\pi}{2} + (-1)^{2k+1}\frac{\pi}{4} = k\pi + \frac{\pi}{2} - \frac{\pi}{4} = k\pi + \frac{\pi}{4}$.
In both cases, we get $\theta = k\pi + \frac{\pi}{4}$. So option (B) is an equivalent representation of the solution set.
Step 4: Final Answer
The set of all values of $\theta$ is given by $\theta = k\pi + \frac{\pi}{4}$ for any integer $k$. This is represented by the expression in option (B).