Solveeit Logo
Login

Question

Mathematics Question on Complex Numbers and Quadratic Equations

Let z=cos⁡θ+isin⁡θz = \cos \theta + i \sin \theta. Then, the value of ∑m=015Im(z2m−1)\displaystyle \sum _{m=0}^{15} Im (z^{2m-1}) at θ=2∘\theta = 2^\circ is

A

1sin⁡2∘\frac{1}{\sin 2^\circ}

B

13sin⁡2∘\frac{1}{3 \sin 2^\circ}

C

12sin⁡2∘\frac{1}{2 \sin 2^\circ}

D

14sin⁡2∘\frac{1}{4 \sin 2^\circ}

Answer

14sin⁡2∘\frac{1}{4 \sin 2^\circ}

Explanation

Solution

Given that, z=cos⁡θ+isin⁡θ=eiθz = \cos\theta + i \sin \theta = e^i \theta
∴           ∑m=115im(z3m−1)=∑m=115im(e(iθ)2m−1\therefore \, \, \, \, \, \, \, \, \, \, \, \displaystyle \sum _{m=1}^{15} im(z^{3m-1})=\displaystyle \sum_{m=1}^{15} im(e^(i\theta)^{2m-1}
            =∑m=115im ei(2m−1)θ\, \, \, \, \, \, \, \, \, \, \, \, = \displaystyle \sum_{m=1}^{15} im \, e^i{^{(2m-1)\theta}}
=sin⁡θ+sin⁡3θ+sin⁡5θ+...+sin⁡29θ= \sin \theta + \sin 3 \theta+ \sin 5 \theta + ... + \sin 29 \theta
=sin⁡(θ+29+θ2sin⁡(15×2θ2)sin⁡(2θ2)=\frac{\sin\bigg(\frac{\theta+29+\theta}{2}\sin\bigg(\frac{15\times2\theta}{2}\bigg)}{\sin\bigg(\frac{2\theta}{2}\bigg)}
=sin⁡(15θ)sin⁡(15θ)sin⁡θ=14sin⁡2∘=\frac{\sin(15\theta) \sin(15 \theta)}{\sin\theta}=\frac{1}{4 \sin 2^{\circ}}