6. Let the complex number $z \neq 1, z^{13}=1$. Then the value of $z^{1}+z^{3}+z^{4}+z^{9}+z^{10}+z^{12}$ is ( ).
(A) Rational number
(B) Irrational number
(C) Imaginary number
(D) Pure imaginary number
6. B.

From $z^{13}-1=0$, and $z \neq 1$, we have $z^{1}+z^{2}+\cdots+z^{12}=-1$. Let $x=z^{1}+z^{3}+z^{4}+z^{0}+z^{10}+z^{12}$, then
$$
\begin{aligned}
x^{2}= & z^{2}+z^{6}+z^{8}+z^{5}+z^{7}+z^{11}+2\left(z^{4}+z^{5}+z^{10}\right. \\
& +z^{11}+1+z^{7}+z^{12}+1+z^{2}+1+z+z^{3} \\
& \left.+z^{6}+z^{8}+z^{0}\right) \\
= & -1-x+2(3-1)=3-x,
\end{aligned}
$$

then $x^{2}+x-3=0$,
so $x=\frac{-1 \pm \sqrt{13}}{2}$ is an irrational number.