Let $p$ be an odd prime and let $\zeta$ be a $p$-th root of unity. Is it true that $$\Bigg\lvert\frac{1 - \zeta^s }{ 1 - \zeta}\Bigg\rvert \in\mathbb{Q}(\zeta)$$ for $s = 1, \ldots, p-1$?
Remark: By $\lvert \cdot \rvert$ I mean the standard norm over $\mathbb{C}$.
I do not know how to prove this. I was trying to use the definition of the norm, so
\begin{align*} \lvert 1-\zeta \rvert &= \sqrt{(1-\Re(\zeta))^2+\Im(\zeta)^2}, \\ \lvert 1-\zeta^s \rvert &= \sqrt{(1-\Re(\zeta^s))^2+\Im(\zeta^s)^2}, \end{align*}
but I do not see what to do now. I know Kummer's Lemma, but I am not sure whether it could be of use here.
Let $z$ be the complex number $$ z=\frac{1-\zeta^s}{1-\zeta}\ . $$ Then the complex conjugate of $z$ is $$ \bar z=\frac{\overline{1-\zeta^s}}{\overline{1-\zeta}} =\frac{1-\zeta^{-s}}{1-\zeta^{-1}} =\frac{\zeta^{-s}}{\zeta^{-1}}\cdot\frac{1-\zeta^s}{1-\zeta} =\zeta^{1-s}\cdot z=\zeta^{2t}\cdot z \ , $$ where $t$ is chosen so that $1-s\equiv 2t$ modulo $p$. (If $s$ is odd, then set $t=(1-s)/2\in\Bbb Z$, if $s$ is even, then use $p+s$ instead of $s$, which is odd and moves to the other case.) Then: $$ |z|^2=z\cdot \bar z=\zeta^{2t}\cdot z^2=(\zeta^t\cdot z)^2\ , $$ so $|z|=\pm \zeta^t\cdot z\in\Bbb Q(\zeta)$.
Note: Not needed, but to be more convincing, let us show explicitly that $\zeta^t\cdot z$ is real. Its conjugate is equal to itself: $$ \overline{\zeta^t\cdot z}=\zeta^{-t}\cdot\bar z=\zeta^{-t}\cdot\zeta^{2t}\cdot z=\zeta^t\cdot z \ . $$
A further note for the downvoters. The problem wants the following from us. Fix some primitive root $\zeta$ of unity of odd order $N$. (We do not need a prime.) We consider the special $z$, let $z$ be the number from $K=\Bbb Q(\zeta)$: $$ z = \frac{1-\zeta^6}{1-\zeta}=1+\zeta+\zeta^2+\zeta^3+\zeta^4+\zeta^5\ . $$ The conjugate value is: $$ \bar z = \frac{1-\zeta^{-5}}{1-\zeta}=1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}+\zeta^{-4}+\zeta^{-5}\ . $$ The question is now if $|z|= \sqrt{z\bar z}$ is an element of the number field $K$. Yes, $z\bar z$ is an element of $K$, because $z,\bar z$ are in $K$. But now we take the square root, this may live $K$! In our case we rewrite $\bar z$ as $$ \bar z = 1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}+\zeta^{-4}+\zeta^{-5} \\ =\zeta^{-5}(1+\zeta+\zeta^2+\zeta^3+\zeta^4+\zeta^5)\ . $$ Now we have a quick chance to see $|z|^2 =z\cdot \bar z$ as a square element of $K$: $$ |z|^2=z\cdot \bar z =z\cdot \zeta^{-5}z =\zeta^{-5}\cdot z^2=\zeta^{N-5}\cdot z^2\ , $$ and $N$ is odd, we need this fact, so that $\zeta^{N-4}$ is zeta to an even power. Finally we can take the square from the R.H.S. - which is $\pm \zeta^{(N-5)/2}\cdot z$, an element of the field $K$.
This later note is a pedestrian way to proceed, the first given argument is the same as this one.