Complex Numbers
Complex Number in Iota Form
Complex Numbers_PYQ
Grade 11
Question:
A value of $\theta$ for which $\dfrac{2 + 3i\sin\theta}{1 - 2i\sin\theta}$ is purely imaginary, is
$\dfrac{\pi}{3}$
$\dfrac{\pi}{6}$
$\sin^{-1}\!\left(\dfrac{\sqrt{3}}{4}\right)$
$\sin^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right)$
Step-by-Step Solution
Key Concept: The real part of $\frac{p + qi}{r + si}$ after rationalisation equals $\frac{pr + qs}{r^2 + s^2}$; set this to zero to enforce purely imaginary.
**Step 1: Rationalise and extract real part**
Multiplying by conjugate $1 + 2i\sin\theta$:
$$\text{Re} = \dfrac{2 - 6\sin^2\theta}{1 + 4\sin^2\theta}.$$
**Step 2: Set real part to 0**
$2 - 6\sin^2\theta = 0 \Rightarrow \sin^2\theta = \dfrac{1}{3} \Rightarrow \sin\theta = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = \sin^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right)$.
Correct Answer: 4