Proof roots of unity being in $\mathbb R$

Question

Let $n \in \mathbb N$ even, and be $w,z \in \mathbb G_n$ primitives.

Proof that $(w+z)^{n/2} \in \mathbb R$.

Ok, as I didn't really know how to start, I tried several things, such using the Binomial theorem, letting $w,z$ be $e^{i\pi k/n}$ with $(n,k)= 1$ and operating, but I can't seem to find the way to proof this.

Any hint on how should I start?

Answer 1

Hint: It suffices to show:

$$ \left(zw\right)^{\frac{n}{2}} = 1.$$

If that is the case, then (using the fact that conjugate of unity root is its inverse): $$\overline{\left(z+w\right)^{\frac{n}{2}}} = \left(z+w\right)^{\frac{n}{2}}.$$

Edit: We have: $\alpha \in \mathbb{R}$ iff $\overline{\alpha} = \alpha$. Also, for a root of unity $u$ we have: $\overline{u} = u^{-1}.$ So: $$ \overline{\left(z+w\right)^{\frac{n}{2}}} = {\left(\overline{z}+\overline{w}\right)^{\frac{n}{2}}} = \left(z^{-1}+w^{-1}\right)^{\frac{n}{2}} = \left(z+w\right)^{\frac{n}{2}}(zw)^{-\frac{n}{2}}.$$

For $k,l$ such that $k+l$ is even, we have indeed: $$ \left(\mathrm e^{\frac{2ik\pi}{n}} \mathrm e^{\frac{2il\pi}{n}} \right)^{n/2} = \mathrm e^{i(k+l)\pi} = 1.$$

Answer 2

For every term $\binom{n/2}{k}w^kz^{n/2-k}$ in the binomial expansion there is also the term $\binom{n/2}{k}w^{n/2-k}z^{k}$.

It hence suffices to show that $w^{n/2-k}z^{k}$ and $w^kz^{n/2-k}$ are conjugates.

Indeed, if $w=\exp(2ij\pi/n)$ and $z=\exp(2i\ell\pi/n)$, then $$ w^{n/2-k}z^{k}=\exp(2i((n/2-k)j+k\ell)\pi/n)=\exp(2in\pi j/2)\exp(2ik(\ell-j)/n)=-\exp(2ik(\ell-j)/n), $$ since $\exp(2in\pi j/2)=-1$, as $(j,n)=1$, while $$ w^kz^{n/2-k}=\exp(2in\pi k/2)\exp(2ik(j-\ell)/n)=-\exp(2ik(j-\ell)/n). $$ Clearly the two terms are conjugate and their contribution is real.

Answer 3

Let $\zeta_1=re^{2\pi i\alpha}$ and $\zeta_2=re^{2\pi i\beta}$ two complex numbers of the same norm $r$. Summing $\zeta_1+\zeta_2$ with the parallelogram rule in $\Bbb C\simeq\Bbb R^2$ makes obvious that $$ \zeta_1+\zeta_2=\rho e^{2\pi i\frac{\alpha+\beta}2} $$ for some (irrelevant for the following) $\rho\in\Bbb R$.

Applying this to $w$ and $z$ primitive $2m$-th roots of $1$, we get $$ w+z=\rho e^{2\pi i\frac{h+k}{4m}} $$ where $h$ and $k$ are prime to $2m$, thus odd. Now $$ (w+z)^m=\rho^me^{\pi i\frac{h+k}{2}}=\pm\rho^m\in\Bbb R $$ because $\frac{h+k}2\in\Bbb Z$.

Answer 4

Let $n=2l$ for some $l\in\mathbb{Z}_+$, since both $w$ and $z$ are primitive $n$-th roots, then there exists some odd number $m\in\mathbb{Z}_+$ such that $z=w^m$. By the binomial formula, we have \begin{eqnarray*} (w+z)^{\frac{n}{2}}&=&(w+w^m)^l=\sum_{k=0}^{l}\binom{l}{k}w^kw^{m(l-k)}\\ &=&\sum_{k=0}^{l}\binom{l}{k}w^{ml+(1-m)k}. \end{eqnarray*}

Case I: $l$ is odd. Then \begin{eqnarray*} (w+z)^{\frac{n}{2}}&=&\sum_{k=0}^{\frac{l-1}{2}}\binom{l}{k}w^{ml+(1-m)k}+\sum_{k=\frac{l+1}{2}}^{l}\binom{l}{k}w^{ml+(1-m)k}\\ &=&\sum_{k=0}^{\frac{l-1}{2}}\binom{l}{k}w^{ml+(1-m)k}+\sum_{k=0}^{\frac{l-1}{2}}\binom{l}{l-k}w^{ml+(1-m)(l-k)}\\ &=&\sum_{k=0}^{\frac{l-1}{2}}\binom{l}{k}\left[w^{ml+(1-m)k}+w^{ml+(1-m)(l-k)}\right]. \end{eqnarray*}

Notice that \begin{eqnarray*} w^{ml+(1-m)k}\cdot w^{ml+(1-m)(l-k)}=w^{ml+(1-m)k+ml+(1-m)(l-k)}=w^{2ml+(1-m)l}. \end{eqnarray*}

Since $m$ is odd and $n=2l$, then $n\mid 2ml+(1-m)l$, which implies that \begin{eqnarray*} \overline{w^{ml+(1-m)(l-k)}}=w^{ml+(1-m)k}. \end{eqnarray*}

Hence $\displaystyle w^{ml+(1-m)k}+w^{ml+(1-m)(l-k)}\in\mathbb{R}$ for all $k=0,\cdots,\frac{l-1}{2}$, in particular, $(w+z)^{\frac{n}{2}}\in\mathbb{R}$.

Case II: $l$ is even. Then \begin{eqnarray*} (w+z)^{\frac{n}{2}}&=&\sum_{k=0}^{\frac{l}{2}-1}\binom{l}{k}w^{ml+(1-m)k}+\binom{l}{\frac{l}{2}}w^{ml+(1-m)\frac{l}{2}}+\sum_{k=\frac{l}{2}+1}^{l}\binom{l}{k}w^{ml+(1-m)k}\\ &=&\sum_{k=0}^{\frac{l}{2}-1}\binom{l}{k}w^{ml+(1-m)k}+\sum_{k=0}^{\frac{l}{2}-1}\binom{l}{l-k}w^{ml+(1-m)(l-k)}+\binom{l}{\frac{l}{2}}w^{ml+(1-m)\frac{l}{2}}\\ &=&\sum_{k=0}^{\frac{l}{2}-1}\binom{l}{k}\left[w^{ml+(1-m)k}+w^{ml+(1-m)(l-k)}\right]+\binom{l}{\frac{l}{2}}w^{ml+(1-m)\frac{l}{2}}. \end{eqnarray*}

Since $m$ is odd and $n=2l$, then $n\mid 2ml+(1-m)l$, which implies that \begin{eqnarray*} \overline{w^{ml+(1-m)(l-k)}}=w^{ml+(1-m)k}. \end{eqnarray*}

Hence $\displaystyle w^{ml+(1-m)k}+w^{ml+(1-m)(l-k)}\in\mathbb{R}$ for all $k=0,\cdots,\frac{l}{2}-1$. Notice that \begin{eqnarray*} w^{ml+(1-m)\frac{l}{2}}\cdot w^{ml+(1-m)\frac{l}{2}}&=&w^{2ml+(1-m)l} \end{eqnarray*}

Since $m$ is odd and $n=2l$, then $n\mid 2ml+(1-m)l$, which implies that \begin{eqnarray*} \overline{w^{ml+(1-m)\frac{l}{2}}}=w^{ml+(1-m)\frac{l}{2}}, \end{eqnarray*}

That is, $w^{ml+(1-m)\frac{l}{2}}\in\mathbb{R}$. Hence $(w+z)^{\frac{n}{2}}\in\mathbb{R}$.

In summary, we have $(w+z)^{\frac{n}{2}}\in\mathbb{R}$.