8. [6] Suppose that $\omega$ is a primitive $2007^{\text {th }}$ root of unity. Find $\left(2^{2007}-1\right) \sum_{j=1}^{2000} \frac{1}{2-\omega^{j}}$.
For this problem only, you may express your answer in the form $m \cdot n^{k}+p$, where $m, n, k$, and $p$ are positive integers. Note that a number $z$ is a primitive $n^{\text {th }}$ root of unity if $z^{n}=1$ and $n$ is the smallest number amongst $k=1,2, \ldots, n$ such that $z^{k}=1$.
Answer: $2005 \cdot 2^{2006}+1$. Note that
$$
\begin{array}{l}
\frac{1}{z-\omega}+\cdots+\frac{1}{z-\omega^{2006}}=\frac{\sum_{j=1}^{2006} \prod_{i \neq j}\left(z-\omega^{i}\right)}{(z-\omega) \cdots\left(z-\omega^{2006}\right)} \\
\quad=\frac{\frac{\mathrm{d}}{\mathrm{d} z}\left[z^{2006}+z^{2005}+\cdots+1\right]}{z^{2006}+z^{2005}+\cdots+1}=\frac{2006 z^{2005}+2005 z^{2004}+\cdots+1}{z^{2006}+z^{2005}+\cdots+1} \cdot \frac{z-1}{z-1} \\
=\frac{2006 z^{2006}-z^{2005}-z^{2004}-\cdots-1}{z^{2007}-1} \cdot \frac{z-1}{z-1}=\frac{2006 z^{2007}-2007 z^{2006}+1}{\left(z^{2007}-1\right)(z-1)} .
\end{array}
$$

Plugging in $z=2$ gives $\frac{2005 \cdot 2^{2006}+1}{2^{2007}-1}$; whence the answer.