6. Let $\omega=\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}$. Then the equation with roots $\omega, \omega^{3}, \omega^{7}, \omega^{9}$ is ( ).
(A) $x^{4}+x^{3}+x^{2}+x+1=0$
(B) $x^{4}-x^{3}+x^{2}-x+1=0$
(C) $x^{4}-x^{3}-x^{2}+x+1=0$
(D) $x^{4}+x^{3}+x^{2}-x-1=0$
6. (B).

From $\omega=\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}$, we know that $\omega, \omega^{2}, \omega^{3}, \cdots, \omega^{10}$. $(=1)$ are the 10 10th roots of 1:
$$
\begin{array}{l}
(x-\omega)\left(x-\omega^{2}\right)\left(x-\omega^{3}\right) \cdots\left(x-\omega^{10}\right) \\
=x^{10}-1 .
\end{array}
$$

Also, $\omega^{2}, \omega^{4}, \omega^{6}, \omega^{8}, \omega^{10}$ are the 5 5th roots of 1, hence
$$
\begin{array}{l}
\left(x-\omega^{2}\right)\left(x-\omega^{4}\right)\left(x-\omega^{6}\right)\left(x-\omega^{8}\right)\left(x-\omega^{10}\right) \\
=x^{5}-1 \text {. }
\end{array}
$$
(1) $\div$ (2) gives
$$
\begin{array}{l}
(x-\omega)\left(x-\omega^{3}\right)\left(x-\omega^{5}\right)\left(x-\omega^{7}\right)\left(x-\omega^{9}\right) \\
=x^{5}+1 .
\end{array}
$$
(3) Dividing both sides by $x-\omega^{5}(=x+1)$, we get
$$
\begin{array}{l}
(x-\omega)\left(x-\omega^{3}\right)\left(x-\omega^{7}\right)\left(x-\omega^{9}\right) \\
=x^{4}-x^{3}+x^{2}-x+1 .
\end{array}
$$