Prove or Disprove that $\left|\frac{e^{2i\theta} -2e^{i\theta} - 1}{e^{2i\theta} + 2e^{i\theta} -1}\right| = 1$ Prove or disprove that 
$$\left|\frac{e^{2i\theta} -2e^{i\theta} - 1}{e^{2i\theta} + 2e^{i\theta} -1}\right| = 1$$
This is a step in an attempt to solve a much larger problem, thus I'm fairly sure it's true but not absolutely sure.  It looks like it should be simple but it's resisted all my attempts so far.
 A: Divide by $e^{i\theta}$ the numerator and denominator :
$$\left|\frac{e^{2i\theta} -2e^{i\theta} - 1}{e^{2i\theta} + 2e^{i\theta} -1}\right|=\left|\frac{e^{i\theta} -2 - e^{-i\theta}}{e^{i\theta} +2 - e^{-i\theta}}\right|$$
Think at the complex conjuguate of the numerator and conclude!
A: 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!
A: This is true.
$$|z - \frac{1}{z} -2 | = |z - \frac{1}{z} + 2|$$
where $z = e^{i\theta}$, since $\Re(z - \frac{1}{z}) = 0$.
Geometrically, $z - \frac{1}{z}$ lies on the $y$-axis (perpendicular bisector of $(2,0)$ and $(-2,0)$).
