Find the general value of $\theta$ which satisfies the equation
$\displaystyle (\cos\theta+i\sin\theta)(\cos2\theta+i\sin2\theta)...(\cos n\theta+i\sin n\theta)=1$
My thoughts: Simplest answer is $\theta= 0$
$\displaystyle (\cos \frac{n(n+1)}{2}\theta+i\sin\frac{n(n+1)}{2}\theta)=1$
Now what ?
Since you have $$e^{i\frac{n(n+1)}{2}\theta}=1.$$ Thus equating the real parts and the imaginary parts we get: \begin{align*} \cos \frac{n(n+1)}{2}\theta & =1 \\ \sin \frac{n(n+1)}{2}\theta & =0 \\ \end{align*} This gives that $$n(n+1)\theta=4k\pi, \qquad \text{ where } k \in \mathbb{Z}.$$
