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} $

Question

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?

Answer 1

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 $1+z+z^2+z^3+z^4+z^5=\frac{z^6-1}{z-1}=\frac {-2}{z-1} \not \in \Bbb R$

Answer 2

Assume $z$ is not equal to $1$. By using the formula for the geometric series we have

$$1+z+...+z^5=\frac{1-z^6}{1-z}=\frac{1-e^{\frac{2\pi i k 6}{12}}}{1-e^\frac{2\pi i k}{12}}=\frac{1-(-1)^k}{1-e^\frac{\pi i k}{6}} $$ Since $1\leq k \leq 11$ we see that $e^{\frac{\pi i k}{6}}$ is only a real number if $k=6$(it is a real number iff the exponent is $i$ times a multiple of $\pi$ because then the sine vanishes). Thus the above number is real iff it is zero. So the above number is real iff $k$ is even.

Your argument is wrong because you can't cancel anything in the equation

$$ 1+e^{(2\pi i k)/12}+...+e^{(10\pi i k)/12}=1+e^{-(2\pi i k)/12}+...+e^{-(10\pi i k)/12} $$

at least i don't see any obvious way to do this.

Answer 3

I misread the question at first, so here comes the edit:

From $z^{12}-1=0$ we get $(z^6-1)(z^6+1)=0$, thus, we get $z^6=\pm 1$.

In the case that $z^6 = 1$ we get that $1+z+\ldots + z^5 = 6$ if $z=1$ and $1+z+\ldots + z^5 = \frac{z^6-1}{z-1} = 0$ if $z\neq 1$.

In the case that $z^6 = -1$, we have $1+z+\ldots + z^5 = \frac{z^6-1}{z-1} = \frac{-2}{z-1}$ which is real if and only if $z$ is real. But, no real satisfies $z^6 = -1$.

Finally concluding that $z$ is sixth root of unity.

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About your argument. If I understand correctly, you want to show that $$\overline{1+z+\ldots + z^5} = 1+z+\ldots + z^5 \iff 1+\bar z+\ldots + \bar z^5 = 1+z+\ldots + z^5$$ This is obvious for sixth roots of unity, but not otherwise. Why for sixth roots of unity? Well, if $\omega$ is sixth root of unity, then $\{1,\omega,\ldots,\omega^5\}$ are sixth roots of unity and that set is invariant under conjugation, i.e. $\{1,\bar\omega,\ldots,\bar\omega^5\} = \{1,\omega,\ldots,\omega^5\}$.

This kind of argument can probably be made precise if we study dihedral group $D_{12}$ and it's action on twelfth roots of unity.