Problem source

By considering the imaginary part of the equation $z^{7}=1,$ or otherwise, find all the roots of the equation 
	\[
	t^{6}-21t^{4}+35t^{2}-7=0.
	\]
	You should justify each step carefully. 
	Hence, or otherwise, prove that 
	\[
	\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7}=\sqrt{7}.
	\]
	Find the corresponding result for 
	\[
	\tan\frac{2\pi}{n}\tan\frac{4\pi}{n}\cdots\tan\frac{(n-1)\pi}{n}
	\]
	in the two cases $n=9$ and $n=11.$

Solution source

Suppose $z^7 = 1$, then we can write $z = \cos \theta + i \sin \theta$ and we must have that:

\begin{align*}
0 &= \textrm{Im}((\cos \theta + i \sin \theta)^7) \\
&= \binom{7}{6}\cos^6 \theta \sin \theta - \binom{7}{4} \cos^4 \theta \sin^3 \theta + \binom{7}{2} \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\
&= 7 \cos^6 \sin \theta - 35 \cos^4 \theta \sin ^3 \theta + 21 \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\
&= -\cos^7 \theta \l \tan^7 \theta - 21 \tan^5 \theta + 35 \tan^3 \theta - 7 \tan \theta\r \\
&= \cos^7 \theta \cdot  t (t^7-21t^4+35t^2-7)
\end{align*}
Where $t = \tan \theta$. So if $z$ is a root of $z^7 = 1$ and $\cos \theta \neq 0, \tan \theta \neq 0$ then $t$ is a root of the equation. Thererefore the roots are:

$\tan \frac{2\pi k}{7}$ where $k = 1, 2, \ldots 6$.

Noting that $\tan \frac{\pi}7 = -\tan \frac{6\pi}{7}, \tan \frac{3\pi}{7} = -\tan \frac{4 \pi}{7}, \tan \frac{5\pi}{7} = -\tan \frac{2 \pi}{7}$ we can conclude that:

\begin{align*}
&& 7 &= \prod_{k=1}^k \tan \frac{k \pi}{6} \\
&&&= \l \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \r^2 \\
\Rightarrow&& \pm \sqrt{7} &= \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7}
\end{align*}

However, we know that $\tan \frac{2\pi}{7}$ is positive, $\tan \frac{4\pi}{7},\tan \frac{6\pi}{7}$ are negative, therefore the result must be positive, ie $+\sqrt{7}$

Using a similar method, we notice that:

\begin{align*}
0 &= \textrm{Im} \l (\cos \theta + i \sin \theta)^n \r \\
&= \cos^n \theta \cdot t (t^{n-1} + \cdots - \binom{n}{n-1})
\end{align*}

Therefore $\prod_{k=0}^{n-1} \tan \frac{k \pi}{n} = n$ and since $\tan \frac{(2k+1) \pi}{n} = \tan \frac{(n-2k-1)\pi}{n}$ is a map of all the odd numbers to the even numbers (and vice versa) when $n$ is odd. We also know that the terms less where $\tan \theta$ has $\theta < \frac{\pi}{2}$ are positive, and the others even, we can determine the signs:

\begin{align*}
\tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} \tan \frac{6 \pi}{9} \tan \frac{8 \pi}{9} & = 3 \\
 \tan \frac{2 \pi}{11} \tan \frac{4 \pi}{11} \tan \frac{6 \pi}{11} \tan \frac{8 \pi}{11} \tan \frac{10 \pi}{11} &= -\sqrt{11}
\end{align*}