$$\text{Im} \left(z^n + \dfrac 1{z^n}\right) = 0 , n \in \mathbb Z$$
The correct solution is:
$\cos (nθ) + i \sin (nθ) + \cos (nθ) – i \sin (nθ) = 2 \cos (nθ) $, imaginary part of which is $0.$
Doubt:
why is it not $-\cos nθ$ when we convert the second term of the equation $\left(\dfrac1{z^n}\right)$ to the modulus argument form?
If $z=e^{i\theta}=\cos\theta+i(\sin\theta)$, then $z^n=e^{in\theta}=\cos n\theta+i\sin n\theta$. Furthermore $z^{-n}=e^{-in\theta}=\cos(- n\theta)+i\sin(-n\theta)$. Now, $\cos$ is an even function which means that $\cos(-x)=\cos x$ for all $x$ while $\sin$ is an odd function, which means that $\sin(-x)=-\sin x$ for all $x$. Consequently $\cos(-n\theta)+i\sin(-n\theta)=\cos(n\theta)-i\sin(n\theta)$.