If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$   is purely  real then find the value of $|z|$ .
I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck.
The answer given is |z|=1

Let's come up with a more interesting question (which might be what you intended to ask). 

Find an $\alpha \in \mathbb{R}$ such that $\forall z \in \mathbb{C}: |z| = \alpha \implies \dfrac{z^2+z+1}{z^2-z+1} \in \mathbb{R} \cup \{\infty\}$. 

I'll show that $\alpha = 1$ works. If $|z| = 1$, then $z = e^{i\theta}$ for some $\theta \in [0,2\pi)$. 
Thus, $\dfrac{z^2+z+1}{z^2-z+1} = \dfrac{z+1+z^{-1}}{z-1+z^{-1}} = \dfrac{e^{i\theta}+e^{-i\theta}+1}{e^{i\theta}+e^{-i\theta}-1} = \dfrac{2\cos\theta+1}{2\cos\theta-1} \in \mathbb{R} \cup \{\infty\}$, as desired. 
Therefore, if $|z| = 1$, then $\dfrac{z^2+z+1}{z^2-z+1}$ is either purely real or undefined. 
