6. On the three sides of $\triangle A B C$, take points $P_{1}, P_{2}, \cdots, P_{6}, \cdots$, such that $P_{1}, P_{4}, P_{7}, \cdots$ are on $A C$, $P_{2}, P_{5}, P_{8}, \cdots$ are on $A B$, and $P_{3}, P_{6}, P_{9}, \cdots$ are on $B C$, and $A P_{1}=A P_{2}, B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_{4}=A P_{5}, B P_{5}=B P_{6}, \cdots$, then $\left|P_{5} P_{2003}\right|=(\quad)$.
A. 0
B. 1
C. 2
D. Uncertain
6. A

Establish the complex plane, and let the letter at each point represent the complex number of that point. We have $P_{2}-A=e^{A_{1}}\left(P_{1}-\right.$ A), $P_{5}-A=e^{A i}\left(P_{4}-A\right)$. Subtracting these, we get $P_{5}-P_{2}=e^{A 4}\left(P_{4}-P_{1}\right)$. Similarly, $P_{6}-P_{3}=e^{R_{1}}$ $\left(P_{5}-P_{2}\right)=e^{(A+B)!}\left(P_{4}-P_{1}\right), P_{7}-P_{4}=e^{G_{i}}\left(P_{6}-P_{3}\right)=e^{(A+B+C) i}\left(P_{4}-\right.$ $\left.P_{1}\right)=e^{x i}\left(P_{4}-P_{1}\right)=-\left(P_{4}^{2}-P_{1}\right)$, which gives $P_{7}=P_{1}$. Similarly, $P_{n+6}=P_{n}$. Since $2003=$ $6 \times 335+5$, it follows that $P_{2003}=P_{5}$, hence $\left|P_{5} P_{2003}\right|=0$.