If ( cos θ+i sin θ)( cos 2 θ+i sin 2 θ) ldots . .( cos n θ+i sin n θ)=1, then the value of θ is, m ∈ N

Question

If ( cos θ+i sin θ)( cos 2 θ+i sin 2 θ) ldots . .( cos n θ+i sin n θ)=1, then the value of θ is, m ∈ N (A) (2 m π/n(n+1)) (B) 4 m π  (C) (4 m π/ n ( n +1)) (D) (m π/n(n+1))

Tardigrade Admin · Accepted Answer

Writing in euler's form, we get e i θ e 2 i θ e 3 i θ ldots e in θ=1 e i (θ+2 θ+3 θ+ ldots n θ=1 ei (n(n+1)/2) θ=e2 m π Therefore ( n ( n +1)/2) θ=2 m π θ=(4 m π/n(n+1))

Q. If $(cos θ + i sin θ) (cos 2 θ + i sin 2 θ) \dots .. (cos n θ + i sin n θ) = 1$ , then the value of $θ$ is, $m \in N$

$\frac{2 mπ}{n ( n + 1 )}$

$4 m π$

$\frac{4 mπ}{n ( n + 1 )}$

$\frac{mπ}{n ( n + 1 )}$

Writing in euler's form, we get
$e^{i θ} e^{2 i θ} e^{3 i θ} \dots e^{in θ} = 1$
$e^{i (θ + 2 θ + 3 θ + \dots n θ} = 1$
$e^{i \frac{n ( n + 1 )}{2} θ} = e^{2 mπ}$
Therefore
$\frac{n ( n + 1 )}{2} θ = 2 mπ$
$θ = \frac{4 mπ}{n ( n + 1 )}$

Q. If (cosθ+isinθ)(cos2θ+isin2θ)…..(cosnθ+isinnθ)=1, then the value of θ is, m∈N

n(n+1)2mπ​

4mπ

n(n+1)4mπ​

n(n+1)mπ​