Example 3 (2000 National High School Competition Question) Let $\omega=\cos \frac{\pi}{5}+\mathrm{i} \sin \frac{\pi}{5}$, 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. $x^{4}+x^{3}+x^{2}-x-1=0$
Solution: Choose B. Reason: From $\omega=\cos \frac{2 \pi}{10}+\mathrm{i} \sin \frac{2 \pi}{10}$, we know that $\omega, \omega^{2}, \omega^{3}, \cdots, \omega^{10}(=1)$ are the 10 10th roots of unity. And
$$
(x-\omega)\left(x-\omega^{2}\right)\left(x-\omega^{3}\right) \cdots\left(x-\omega^{10}\right)=x^{10}-1 .
$$

Also, $\omega^{2}, \omega^{4}, \omega^{6}, \omega^{8}, \omega^{10}$ are the 5 5th roots of unity, so
$$
\left(x-\omega^{2}\right)\left(x-\omega^{4}\right)\left(x-\omega^{6}\right)\left(x-\omega^{8}\right)\left(x-\omega^{10}\right)=x^{5}-1 \text {. }
$$

Dividing (1) by (2) gives $(x-\omega)\left(x-\omega^{3}\right)\left(x-\omega^{5}\right)\left(x-\omega^{7}\right)\left(x-\omega^{9}\right)=x^{5}+1$.
Dividing both sides of (3) by $x-\omega^{5}=x+1$ yields
$$
(x-\omega)\left(x-\omega^{4}\right)\left(x-\omega^{7}\right)\left(x-\omega^{9}\right)=x^{4}-x^{3}+x^{2}-x+1 .
$$