$\mid 1 + w + w \mid = \mid 1 + w^2 + w^2 \mid$ where $w=e_p(1)$

Question

I was trying to digest something related to exponential sum however there was obstacle for me. My question is the following:

$\mid \sum _{i=1} ^3 e_p(a_i) \mid$ where $e_p (a_i)=e^{\frac{2\pi.i}{p}a_i}$. Hence, by definition, $e_p(x)=\cos \frac {2\pi}{p}x+i\sin\frac{2\pi}{p}x$

$e_p(1)=\cos \frac {2\pi}{p}+i\sin\frac{2\pi}{p}$, $e_p(-1)=\cos \frac {2\pi}{p}+i\sin\frac{2\pi}{p}$ and finally $e_p(0)= 1$.

Let $a_1=0,a_2=1,a_3=-1$ and let's denote $w=e_p(1)$. Before the calculations, I observed that $w^p=1$ so $1+w+\dots+w^{p-1}=0$. Now, I am going to evaluate following:

$\mid \sum _{i=1} ^3 e_p(a_i) \mid= \mid e_p(0)+e_p(1)+e_p(-1) \mid= \mid 1 + w + \frac{1}{w} \mid =(??)\mid 1+w+w^2 \mid $.

Alright, I see that if $w=e_p(1)$ then $\frac{1}{w}=e_p(-1)$ by just using definition but I couldn't see that why $\frac{1}{w}=w^2$

My second question is in the paper they are saying that $\mid 1 + w + w \mid = \mid 1 + w^2 + w^2 \mid$. Why? I can't see the transition there.

Answer 1

Since $w$ is a root of unity, we have $|w| = 1$, and therefore $$|1+w+w^{-1}| = |w||1+w+w^{-1}| = |w+w^2+1|.$$

The second identity doesn't hold for, say, $p = 5$.

If $p = 3$, then $1+w+w^2 = 0$ and $$|1+w+w| = |1+w+w^2+w-w^2|=|w-w^2|$$ $$|1+w^2+w^2| = |1+w+w^2+w^2-w| = |w^2-w|.$$

Edit: Even cleaner, when $p = 3$, we have $\overline w = w^2$, so $|1+2w| = |\overline{1+2w}| = |1+2w^2|$, where $\overline{\,\cdot\,}$ stands for complex conjugation. This, in fact, shows $|a+bw| = |a+bw^2|$ for all real $a,b$.

Edit2: In fact, let $|z| = 1$ be any unit complex number. Then, $|1+2z| = |1+2z^2|$ if and only if $z^3 = 1$.

Proof. Let $z = a+bi$ with $a,b$ real and $|z| = 1$, i.e. $a^2+b^2 = 1$. Then, $$|1+2z|^2 = (1+2z)(1+2\overline z) = 1+2z+2\overline z+4|z|^2 = 5+4\operatorname{Re}z,$$ and the same holds for $z^2$ since $|z^2|=1$ as well, so $|1+2z| = |1+2z^2|$ if and only if $\operatorname{Re}z=\operatorname{Re}z^2$, i.e. $a = a^2-b^2 = 2a^2 -1$ which has two solutions $a=1$ and $a = -\frac 12$. Solving for $b$ gives us three solutions in total $z = 1,-\frac 12\pm i\frac{\sqrt3}2$.

Answer 2

By definition $\omega^3 = 1$, so by dividing both sides by $\omega$ we have that $$\omega^2 = \frac{1}{\omega}$$

Also, from definition we have that $$1- \omega^3 = 0,$$ i.e., since $\omega \neq 1$ $$1+\omega+\omega^2 = 0.$$

Now, add/subtract $\omega^2$ on one side and $\omega$ on the other.