it seems like i'm completely missing something here.
I need to prove that if
$$\ z_1 \times z_2 \not= -1 \\ |z_1| = |z_2| = 1$$
then
$$\ \frac{z_1+z_2}{1+z_1z_2} \in \mathbb R$$
so if
$$\ z_1 = r_1(\cos\alpha + i \sin \alpha) \\ z_2 = r_2(\cos\beta + i \sin \beta) \\ z_1 \times z_2 = r_1 r_2 (\cos (\alpha+\beta) + i \sin (\alpha + \beta)) $$
and because $\ z_1 z_2 \not = -1 $ we can conclude that $\ \alpha + \beta \not = \pi $ and also $\ r_1 = r_2 = 1 $
but if we pick two values $\ \alpha = \frac{\pi}{2} $ and $\ \beta = \frac{\pi}{3} , \alpha + \beta = \frac{5\pi}{6} \not = \pi$ and we know that $\ r = 1 $
we can see that
$$\ \frac{z_1 + z_2}{1+ z_1z_2} = \frac{1(\cos\frac{\pi}{2} + i \sin \frac{\pi}{2}) + 1(\cos\frac{\pi}{3} + i \sin \frac{\pi}{3})}{1+ 1(\cos \frac{5\pi}{6}+ i \sin \frac{5\pi}{6})} \not \in \mathbb R$$
Hint:
You need only to show that for $q = \ \frac{z_1+z_2}{1+z_1z_2}$ holds $\bar q = q$.
Use $\bar z_i = \frac{1}{z_i}$ ($i = 1,2$).