Prove or Disprove that $\left|\frac{e^{2i\theta} -2e^{i\theta} - 1}{e^{2i\theta} + 2e^{i\theta} -1}\right| = 1$

Solution 1:

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 $(-2,0)$).

Solution 2:

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!

Solution 3:

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!