几何转代数：复数单位根

For $w^k=1$, $w$ is the $k$-th root of unity. Because $x=w,w^2,\cdots,w^k$ are roots of $x^k-1=0$, $x^k-1=(x-w)(x-w^2)\cdots(x-w^k)$.

Consider a circle of radius 1 divided into $k$ congruent arcs. Let $w=\cos\frac{2\pi}{k}+i\sin\frac{2\pi}{k}$, so that $w^k=1$. Thus, $x=1,w,\cdots w^{k-1}$ are roots of the polynomial $x^k-1=0$. Thus, $(x-w)\cdots(x-w^{k-1})=\frac{x^k-1}{x-1}=x^{k-1}+\cdots+x+1$.

The length of the smallest chord is $|1-w|=|1-w^{k-1}|$, the length of the second-smallest chord is $|1-w^2|$, and so on.

\begin{aligned}
&|1-w|\cdot|1-w^2|\cdots|1-w^{k-1}|=\big|(1-w)(1-w^2)\cdots(1-w^{k-1})\big|=k\\
&|1+w|\cdot|1+w^2|\cdots|1+w^{k-1}|=\big|(-1-w)(-1-w^2)\cdots(-1-w^{k-1})\big|=\begin{cases}0\text{, if }2\mid k\\1\text{, if }2\nmid k\end{cases}\\
\end{aligned}

So, 分情况.

Case 1: $k=2t$

\begin{aligned}
&|1-w|\cdots|1-w^{t-1}|=|1-w^{t+1}|\cdots|1-w^{2t-1}|=\sqrt{\frac{|1-w|\cdots|1-w^{k-1}|}{|1-w^t|}}=\sqrt{\frac{k}{2}}=\sqrt{t}\\
&|1+w|\cdots|1+w^{t-1}|=|1+w^{t+1}|\cdots|1+w^{2t-1}|=\sqrt{t}\\
&w^{2t}-1=0\Longrightarrow 1-w+w^2-\cdots+w^{2t-1}=0
\end{aligned}

Case 2: $k=2t+1$

\begin{aligned}
&|1-w|\cdots|1-w^t|=|1-w^{t+1}|\cdots|1-w^{2t}|=\sqrt{|1-w|\cdots|1-w^{k-1}|}=\sqrt{k}=\sqrt{2t+1}\\
&|1+w|\cdots|1+w^t|=|1+w^{t+1}|\cdots|1+w^{2t}|=1
\end{aligned}

然后，就能算出来很多东西了。