Let z = cos θ + i sin θ. Then, the value of displaystyle ∑ m=015 Im (z2m-1) at θ = 2° is

Question

Let z = cos θ + i sin θ. Then, the value of displaystyle ∑ m=015 Im (z2m-1) at θ = 2° is (A) (1/ sin 2°) (B) (1/3 sin 2°) (C) (1/2 sin 2°) (D) (1/4 sin 2°)

Tardigrade Admin · Accepted Answer

Given that, z = cosθ + i sin θ = ei θ ∴ displaystyle ∑ m=115 Im(z3m-1)= displaystyle ∑m=115 Im(eiθ)2m-1 = displaystyle ∑m=115 Im ei(2m-1)θ = sin θ + sin 3 θ+ sin 5 θ + ... + sin 29 θ =( sin((θ+29+θ/2) sin((15×2θ/2))/ sin((2θ)2)) =( sin(15θ) sin(15 θ)/ sinθ)=(1/4 sin 2°)

Q. Let $z = cos θ + i sin θ$ . Then, the value of $m = 0 \sum 15 I m (z^{2 m - 1})$ at $θ = 2^{\circ}$ is

$\frac{1}{s i n 2 ^{\circ}}$

$\frac{1}{3 s i n 2 ^{\circ}}$

$\frac{1}{2 s i n 2 ^{\circ}}$

$\frac{1}{4 s i n 2 ^{\circ}}$

Q. Let z=cosθ+isinθ. Then, the value of m=0∑15​Im(z2m−1) at θ=2∘ is

sin2∘1​

3sin2∘1​

2sin2∘1​

4sin2∘1​

Given that, z=cosθ+isinθ=eiθ ∴m=1∑15​Im(z3m−1)=m=1∑15​Im(eiθ)2m−1 =m=1∑15​Imei(2m−1)θ =sinθ+sin3θ+sin5θ+...+sin29θ =sin(22θ​)sin(2θ+29+θ​sin(215×2θ​)​ =sinθsin(15θ)sin(15θ)​=4sin2∘1​

Q. Let $z = cos θ + i sin θ$ . Then, the value of $m = 0 \sum 15 I m (z^{2 m - 1})$ at $θ = 2^{\circ}$ is

$\frac{1}{s i n 2 ^{\circ}}$

$\frac{1}{3 s i n 2 ^{\circ}}$

$\frac{1}{2 s i n 2 ^{\circ}}$

$\frac{1}{4 s i n 2 ^{\circ}}$