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

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

(B) $\frac{1}{3 \sin 2^{\circ}}$

(C) $\frac{1}{2 \sin 2^{\circ}}$

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Complex Numbers

Let

z = cos θ + i sin θ

. Then the value of

m = 1 \sum 15 Im (z^{2 m - 1})

θ = 2^{\circ}

Answer: (D)

We have

z = e^{i θ}

, so

z^{2 m - 1} = e^{i (2 m - 1) θ}

and

Im (z^{2 m - 1}) = sin ((2 m - 1) θ)

.

Thus:

X = m = 1 \sum 15 sin ((2 m - 1) θ) = sin θ + sin 3 θ + sin 5 θ + \dots + sin 29 θ

Using the product-to-sum formula:

2 (sin θ) X = (1 - cos 2 θ) + (cos 2 θ - cos 4 θ) + \dots + (cos 28 θ - cos 30 θ)

2 (sin θ) X = 1 - cos 30 θ

X = \frac{1 - cos 30 θ}{2 sin θ}

θ = 2°

X = \frac{1 - cos 60°}{2 sin 2°} = \frac{1 - \frac{1}{2}}{2 sin 2°} = \frac{1}{4 sin 2°}

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option