Find all $ z \in {\mathbb C} $ such that $z^{12}=1 $ and $ 1+z+z^2+z^3+z^4+z^5 \in {\mathbb R} $ So, my first thought. If $z^{12} = 1, z \in $ is a twelfth root of unity. Knowing this, I can write $ z = e^{i {2k \pi} \over {12}} $, with k $\in \{0,1,2,3,4,5,6,7,8,9,10,11\} $.
Then if I just replace z with these new form in the second equation and using the known property that $ z = \overline z $ iff $ z \in {\mathbb R}$, I get that this is valid for every z that satisfies the first equation, because I can cancel the left side with the right side and I get $ 0 = 0 $.
I don't see the flaw with my argument, but a teacher has told me it's wrong, and that in fact the two equations are valid if $z$ is a sixth root of unity.
What's my mistake? How can I fix it?

Your $12^{\text{th}}$ roots of unity have a typo, it should be $z = e^{i {12k \pi} \over {12}}$.  Then for $k=1, z^6=-1$, not $1$.  We have $1+z+z^2+z^3+z^4+z^5=\frac{z^6-1}{z-1}=\frac {-2}{z-1} \not \in \Bbb R$
