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

raj
Since the fraction is real, so it is equal to its conjugate. That is, $\frac{z^2+z+1}{z^2-z+1}={\bar{(\frac{z^2+z+1}{z^2-z+1})}}$. Then by simple calculations one can obtain
$(z\bar{z}-1)(z-\bar{z})=0$. Since $z$ is complex so $z\bar{z}=\mid{{z}\mid^2}=1$
