Complex Numbers 5 Question 4
4. 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
(2009)
(a) $\frac{1}{\sin 2^{\circ}}$
(b) $\frac{1}{3 \sin 2^{\circ}}$
(c) $\frac{1}{2 \sin 2^{\circ}}$
(d) $\frac{1}{4 \sin 2^{\circ}}$
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Answer:
Correct Answer: 4. (d)
Solution:
- Given that, $z=\cos \theta+i \sin \theta=e^{i \theta}$
$$ \begin{aligned} \therefore \sum _{\mu=1}^{15} I \mu\left(\zeta^{2 \mu-1}\right) & =\sum _{\mu=1}^{15} I \mu\left(\varepsilon^{(\theta}\right)^{2 \mu-1}=\sum _{\mu=1}^{15} I \mu \varepsilon^{l(2 \mu-1) \theta} \\ & =\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots+\sin 29 \theta \\ & =\frac{\sin \frac{\theta+29 \theta}{2} \sin \frac{15 \times 2 \theta}{2}}{\sin \frac{2 \theta}{2}} \\ & =\frac{\sin (15 \theta) \sin (15 \theta)}{\sin \theta}=\frac{1}{4 \sin 2^{\circ}} \end{aligned} $$
