Find the general value of $\theta$

Question

Find the general value of $\theta$ which satisfies the equation $(\cos\theta+i\sin\theta)(\cos3\theta+i\sin3\theta)\dots \{\cos(2n-1)\theta+i\sin(2n-1)\theta\}=1$.

My attempt:

$(\cos\theta+i\sin\theta)(\cos3\theta+i\sin3\theta)\dots \{\cos(2n-1)\theta+i\sin(2n-1)\theta\}$

$=\cos\{\theta+3\theta+\dots+(2n-1)\theta\}+i\sin\{\theta+3\theta+\dots+(2n-1)\theta\}$

$=\cos(n^2\theta)+i\sin(n^2\theta)$

$\therefore \cos(n^2\theta)+i\sin(n^2\theta)=1$

Equating real and imaginary parts, we get,

$\cos(n^2\theta)=1$ and $\sin(n^2\theta)=0$

$\therefore n^2\theta=2k\pi$ and $n^2\theta=k\pi$

How to proceed further? Does $\theta$ have to simultaneously satisfy both the equations or only $n^2\theta=2k\pi?$

Answer 1

Note that the equation is same: $$e^{i\theta(1+3+\cdots+2n-1)}=e^{i2\pi k}$$