Question

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

• A. $4m\pi$
• B. $\frac{2m\pi }{n(n+1)}$
• C. $\frac{4m\pi }{n(n+1)}$
• D. $\frac{m\pi }{n(n+1)}$

Answer 1

Answer: (C)

$\frac{4m\pi }{n(n+1)}$

Changing the above expression to Eular's form, we get
$e^{i\theta }{e}^{2i\theta }{e}^{3i\theta }...e^{in\theta })=1$
$e^{i(\theta +2\theta +3\theta +...n\theta }=1$
$e^{i\frac{n(n+1)}{2}\theta }=e^{2m\pi }$
Therefore, simplifying we get
$\frac{n(n+1)}{2}\theta =2m\pi$
$\theta =\frac{4m\pi }{n(n+1)}$