Taking the squared norm of the numerator and denominator separately,
$$
\eqalign{
\left|e^{ 2i\theta}\pm e^{ i\theta}-1\right|^2 &=
\left(e^{ 2i\theta}\pm e^{ i\theta}-1\right)\cdot
\left(e^{-2i\theta}\pm e^{-i\theta}-1\right)\\ &
\matrix{=& 1 & \pm2e^{ i\theta} & -e^{2i\theta} \\\\
& \pm2e^{-i\theta} & +4 & \mp2e^{ i\theta} \\\\
& -e^{2i\theta} & \mp2e^{-i\theta} & +1 }
\\\\ &= 6 - 2\cos 2\theta \pm 2\cos\theta \mp 2\cos\theta
\\\\ &= 6 - 2\cos 2\theta\,.
}
$$
Notice, however, that this no longer depends on the sign,
i.e. it is the same for the numerator and denominator.
But I admit, I like @Raymond's and @Aryabhata's answers much better!