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 ?

Since $|z_i|=1\iff z_i\overline{z_i}=1$, one has$$\begin{align}\overline{\left(\frac{z_1+z_2+z_3+z_1z_2+z_2z_3+z_3z_1}{1+z_1z_2z_3}\right)}&=\frac{\overline{z_1+z_2+z_3+z_1z_2+z_2z_3+z_3z_1}}{\overline{1+z_1z_2z_3}}\\&=\frac{\overline{z_1}+\overline{z_2}+\overline{z_3}+\overline{z_1z_2}+\overline{z_2z_3}+\overline{z_3z_1}}{1+\overline{z_1z_2z_3}}\\&=\frac{\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_1z_2}+\frac{1}{z_2z_3}+\frac{1}{z_3z_1}}{1+\frac{1}{z_1z_2z_3}}\\&=\frac{z_2z_3+z_3z_1+z_1z_2+z_3+z_1+z_2}{z_1z_2z_3+1}\\&=\frac{z_1+z_2+z_3+z_1z_2+z_2z_3+z_3z_1}{1+z_1z_2z_3}\end{align}$$

By $\,z_i \bar z_i = 1,\,$ $\ {z\!+\!1}\,=\,\frac{(1+z_1)(1+z_2)(1+z_3)}{1+z_1z_2z_3}\, =\, \frac{z_1z_2z_3}{z_1z_2z_3}\frac{(\bar z_1+1)(\bar z_2+1)(\bar z_3+1)}{\bar z_1\bar z_2\bar z_3+\,1} = \overline{z\!+\!1},\,$ so $\ z = \bar z $

Remark $\ $ The innate symmetry can be seen nicely by polynomial reciprocation (reversal). Consider the (Vieta) generating function of the elementary symmetric functions $\,e_i$ of the $\,z_i$

$$ g(x,z_i)\, =\, (x-z_1)\cdots (x-z_n)\, =\, x^n - e_1 x^{n-1} + e_2 x^{n-2} -\cdots + (-1)^n e_n $$

Conjugation has the effect of reciprocating the $\,z_i,\,$ i.e $\,z_i\to \color{#c0f}{z_i^{-1}}.\,$ The effect of this on each factor $\,x - z_i\,$ is closely related to reciprocating $\,x,\,$ i.e. $\,x\to \color{#0a0}{x^{-1}},\,$ namely

$$\begin{align} z_i (x-\color{#c0f}{z_i^{-1}})\, & =\,z_i x - 1\, =\, -x\,(\color{#0a0}{x^{-1}}-z_i)\\ z_if(x,\,\color{#c0f}{z_i^{-1}})\, &=\, z_ix-1 \, =\, - xf(\color{#0a0}{\color{#0a0}{x^{-1}}},\,z_i)\end{align} $$

Applied to each linear factor $\,f\,$ in $\,g\,$ this yields

$$ z_1\!\cdots z_n\, g(x, \color{#c0f}{z_i^{-1}})\ =\ (-x)^n g(\color{#0a0}{x^{-1}}, z_i)$$

The RHS is up to sign the reciprocal polynomial of $\,g(x),\,$ i.e. that obtained by reversing the order of its coefficients. Comparing their coefficients of $\,x^k$ we get a $\,\rm\color{#c00}{{reflection}}\,$ formula

$$ \color{#c00}{e_{\large n}\,\overline{e_{\large k}}}\, =\, e_{\large n}\, e_{\large k}(z_i^{-1})\, =\,\color{#c00}{e_{\large n-k}} \quad $$

$${\rm e.g.}\quad \color{#c00}{e_3 \bar e_1 } \, =\, z_1 z_2 z_3 \left[\frac{1}{z_1} + \frac{1}{z_2}+\frac{1}{z_3}\right]\, =\, z_2z_3 + z_3 z_1 + z_1z_2\, =\, \color{#c00}{e_2}$$

Such reflections are evident in the calculations in mathlove's answer, viz.

$$\bar z\, =\, \frac{\bar e_1+\bar e_2}{\bar e_0+\bar e_3}\, =\, \frac{e_3}{e_3}\,\frac{\bar e_1+\bar e_2}{\bar e_0+\bar e_3}\, =\, \frac{\color{#c00}{ e_3 \bar e_1}+ e_3 \bar e_2}{e_3\bar e_0 + e_3\bar e_3}\, =\, \frac{\color{#c00}{e_2} +e_1}{e_3+e_0}\, =\, z$$

As always: $ $ exploit innate symmetry!

First, note that if $|z|=1$, then $z=\cos\theta+i\sin\theta$ for some $\theta$ and $1+z=1+\cos\theta+i\sin\theta=2\cos\frac{\theta}{2}e^{\frac{i\theta}{2}}$. Adding 1 to the original expression in the question, we have \begin{align*} \frac{(1+z_1)(1+z_2)(1+z_3)}{1+z_1z_2z_3}&=\frac{8\cos^2\frac{\theta_1}{2}\cos^2\frac{\theta_2}{2}\cos^2\frac{\theta_3}{2}e^{\frac{i(\theta_1+\theta_2+\theta_3)}{2}}}{2\cos^2\frac{\theta_1+\theta_2+\theta_3}{2}e^{\frac{i(\theta_1+\theta_2+\theta_3)}{2}}}\\ &=\frac{4\cos^2\frac{\theta_1}{2}\cos\frac{\theta_2}{2}\cos^2\frac{\theta_3}{2}}{\cos^2\frac{\theta_1+\theta_2+\theta_3}{2}} \end{align*} which is obviously real.