JEE Advanced · Mathematics · 3. Complex Numbers
Let \(z=\cos \theta+i \sin \theta\). Then, the value of \(\sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right)\) at \(\theta=2^{\circ}\) is
- A \(\frac{1}{\sin 2^{\circ}}\)
- B \(\frac{1}{3 \sin 2^{\circ}}\)
- C \(\frac{1}{2 \sin 2^{\circ}}\)
- D \(\frac{1}{4 \sin 2^{\circ}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4 \sin 2^{\circ}}\)
Step-by-step Solution
Detailed explanation
Given that \(z=\cos \theta+i \sin \theta=e^{i \theta}\)
\(\therefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right) =\sum_{m=1}^{15} \operatorname{Im}\left(e^{i \theta}\right)^{2 m-1}\) \(=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) \theta} \)
\( =\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots+\sin 29 \theta \)
\( =\frac{\sin \left(\frac{\theta+29 \theta}{2}\right) \sin \left(\frac{15 \times 2 \theta}{2}\right)}{\sin \left(\frac{2 \theta}{2}\right)}=\frac{\sin (15 \theta) \sin (15 \theta)}{\sin \theta}\) \(=\frac{1}{4 \sin 2^{\circ}}\)
\(\therefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right) =\sum_{m=1}^{15} \operatorname{Im}\left(e^{i \theta}\right)^{2 m-1}\) \(=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) \theta} \)
\( =\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots+\sin 29 \theta \)
\( =\frac{\sin \left(\frac{\theta+29 \theta}{2}\right) \sin \left(\frac{15 \times 2 \theta}{2}\right)}{\sin \left(\frac{2 \theta}{2}\right)}=\frac{\sin (15 \theta) \sin (15 \theta)}{\sin \theta}\) \(=\frac{1}{4 \sin 2^{\circ}}\)
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