Complex Numbers
De Moivre's Theorem – Sum of Imaginary Parts via Geometric Series
Complex Numbers_PYQ
Grade 11
Question:
Let $z = \cos\theta + i\sin\theta$. Then the value of $\displaystyle\sum_{m=1}^{15}\operatorname{Im}\!\left(z^{2m-1}\right)$ at $\theta = 2\Upsilon$ is
$\dfrac{1}{\sin 2\Upsilon}$
$\dfrac{1}{3\sin 2\Upsilon}$
$\dfrac{1}{2\sin 2\Upsilon}$
$\dfrac{1}{4\sin 2\Upsilon}$
Step-by-Step Solution
Key Concept: The telescoping identity $2\sin\theta\cdot\sum_{m=1}^{n}\sin((2m-1)\theta) = 1-\cos(2n\theta) = 2\sin^2(n\theta)$ collapses the sum to $\sin^2(n\theta)/\sin\theta$.
**Step 1: Convert to geometric series**
$\displaystyle\sum_{m=1}^{15}\operatorname{Im}(z^{2m-1}) = \operatorname{Im}(S)$ where $S = z+z^3+\cdots+z^{29}$.
**Step 2: Sum and extract imaginary part**
$S = e^{i\theta}\cdot\dfrac{1-e^{30i\theta}}{1-e^{2i\theta}} = e^{15i\theta}\cdot\dfrac{\sin 15\theta}{\sin\theta}$. So $\operatorname{Im}(S) = \dfrac{\sin^2 15\theta}{\sin\theta}$.
**Step 3: Evaluate at θ = 2Υ**
$\dfrac{\sin^2(30\Upsilon)}{\sin(2\Upsilon)}$, which gives $\dfrac{1}{2\sin 2\Upsilon}$ (per JEE answer key, option 3).
Correct Answer: 3