<s>[INST] Proving the following number is real Let $z_i$ be complex numbers such that $|z_i| = 1$ .
Prove that :
$$ z\, :=\, \frac{z_1+z_2+z_3 +z_1z_2+z_2z_3+z_1z_3}{1+z_1z_2z_3} \in \mathbb{R} $$
This problem was featured on my son's final exam today, I tried helping him with this, but I guess I got a bit rusty myself.
Any ideas ?
 [/INST] If $|z|=1$, then $z+\frac1z=z+\bar{z}=2\mathrm{Re}(z)$. Therefore, the following are both real:
$$
\begin{align}
\small\left(z_1+\frac1{z_1}\right)\left(z_2+\frac1{z_2}\right)\left(z_3+\frac1{z_3}\right)
&\small=\frac{z_1^2z_2^2z_3^2+z_1^2z_2^2+z_2^2z_3^2+z_3^2z_1^2+z_1^2+z_2^2+z_3^2+1}{z_1z_2z_3}\tag{1}\\
z_1z_2z_3+\frac1{z_1z_2z_3}
&=\frac{z_1^2z_2^2z_3^2+1}{z_1z_2z_3}\tag{2}
\end{align}
$$
The difference of $(1)$ and $(2)$ is also real:
$$
\frac{z_1^2z_2^2+z_2^2z_3^2+z_3^2z_1^2+z_1^2+z_2^2+z_3^2}{z_1z_2z_3}\tag{3}
$$
The ratio of $(3)$ and $(2)$ is also real:
$$
\frac{z_1^2z_2^2+z_2^2z_3^2+z_3^2z_1^2+z_1^2+z_2^2+z_3^2}{z_1^2z_2^2z_3^2+1}\tag{4}
$$
For any $w_1$, $w_2$, and $w_3$ whose absolute values are $1$, we can find $z_1$, $z_2$, and $z_3$ whose absolute values are $1$ so that $w_1=z_1^2$, $w_2=z_2^2$, and $w_3=z_3^2$. Plugging into $(4)$, we get that for any $w_1$, $w_2$, and $w_3$ whose absolute values are $1$, the following is real:
$$
\frac{w_1+w_2+w_3+w_1w_2+w_2w_3+w_3w_1}{1+w_1w_2w_3}\tag{5}
$$
</s>