Q:

Help on a complex number equality problem

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 ?

A:

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 
$$n(n+1)\theta=4k\pi, \qquad \text{ where } k \in \mathbb{Z}.$$

