Let z= θ+i θ . Then, the value of ∑_ m=1 ^ 15 Im (z^ 2 m-1 ) at θ=2^ is

Correct answer: D. 1 4 2^ Given that z= θ+i θ=e^ i θ aligned ∑_ m=1 ^ 15 Im(z^ 2 m-1 ) & =∑_ m=1 ^ 15 Im(e^ i θ )^ 2 m-1 =∑_ m=1 ^ 15 Im e^ i(2 m-1) θ & = θ+ 3 θ+ 5 θ+ + 29 θ & =( ((θ+29 θ)/(2)) ((15 × 2 θ)/(2)))/( ((2 θ)/(2)))=( (15 θ) (15 θ))/( θ)= 1 4 2^ ^ aligned

JEE Advanced 2009Paper 1

Exam: JEE Advanced
Year: 2009
Paper: Paper 1
Subject: Mathematics
Chapter: Complex Number

JEE Advanced 2009 · Paper 1 · Mathematics · Complex Number

Let z= θ+i θ . Then, the value of ∑_ m=1 ^ 15 Im (z^ 2 m-1 ) at...

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

Options

A $\frac{1}{\sin 2^{\circ}}$
B $\frac{1}{3 \sin 2^{\circ}}$
C $\frac{1}{2 \sin 2^{\circ}}$
D $\frac{1}{4 \sin 2^{\circ}}$

Correct answer

D. $\frac{1}{4 \sin 2^{\circ}}$

Step-by-step solution

Given that

z=\cos \theta+i \sin \theta=e^{i \theta}

\begin{aligned} \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+\sin3 \theta+\sin5 \theta+\ldots+\sin29 \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}}} \end{aligned}

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