The solutions of the equation $z^2\left(1-z^2\right)=16, z \in \mathbf{C}$, lie on the curve

If $z, \bar{z},-z,-\bar{z}$ forms a rectangle of area $2 \sqrt{3}$ square units, then one such $z$ is

$$ \left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16}= $$