Finding the value of $|z|$

Question

Let $z$ is a non-real number such that,
$$\frac{1+z+z^2}{1-z+z^2}$$
is purely real. Find the value of $|z|$.

Hello everyone!
For this question, when I set $\displaystyle{a}$=$\dfrac{1+z+z^2}{1-z+z^2}$ and used $a=\bar{a}$ but it got very messy. Is there any alternate, elegant approach?
Thanks to take your time to read my question.

Answer 1

Given: $\dfrac{1+z+z^2}{1-z+z^2}\in\mathbb{R}$

$\implies 1+\dfrac{2z}{1-z+z^2}\in\mathbb{R}$

$\implies \dfrac{1-z+z^2}{z}\in\mathbb{R}$

$\implies z+\dfrac{1}{z}-1 \in\mathbb{R}$

$\implies z+\dfrac{1}{z}\in\mathbb{R}$

$\implies z+\dfrac{1}{z}=\bar{z}+{\dfrac{1}{\bar{z}}}$

$\implies z-\bar{z}=\dfrac{z-\bar{z}}{z\bar{z}}$

$\implies z\bar{z}=1$ [ Since $z\in\mathbb{C/R}$ ]

$\implies |z|^{2}=1$

$\implies |z|=\boxed{1}$

Answer 2

$$\begin{align} \frac{1+z+z^2}{1-z+z^2}\in\Bbb R&\iff1+\frac{2z}{z^2-z+1}\in\Bbb R\\ &\iff\frac z{z^2-z+1}\in\Bbb R\\ &\iff \frac{z^2-z+1}z\in\Bbb R\\ &\iff z+\frac1z\in\Bbb R \end{align}$$

Now, the equation $z+\frac 1z=a$, for $a\in \Bbb R$, has this solution:

$$z=\frac{a\pm\sqrt{a^2-4}}2$$

For $|a|<2$ this solution is not real, and $$|z|^2=\frac{a^2}4+\frac{4-a^2}4=\frac44=1$$

But for $|a|\ge 2$, $z$ spans the whole set of real numbers.