2. Let the complex number $z \neq 1, z^{11}=1$. Then $z+z^{3}+z^{4}+z^{5}+z^{9}=$ $\qquad$
2. $\frac{-1 \pm \sqrt{11} \mathrm{i}}{2}$.

Given $z \neq 1, z^{11}=1$
$$
\begin{array}{l}
\Rightarrow 0=\frac{z^{11}-1}{z-1}=1+z+z^{2}+\cdots+z^{10} \\
\Rightarrow z+z^{2}+\cdots+z^{10}=-1 .
\end{array}
$$

Let $x=z+z^{3}+z^{4}+z^{5}+z^{9}$.
$$
\begin{array}{c}
\text { Then } x^{2}=z^{2}+z^{6}+z^{8}+z^{10}+z^{18}+ \\
2\left(z^{4}+z^{5}+z^{6}+z^{10}+z^{7}+\right. \\
\left.z^{8}+z+z^{9}+z^{2}+z^{3}\right) \\
=z^{2}+z^{6}+z^{8}+z^{10}+z^{7}-2 \\
=-x-1-2=-3-x .
\end{array}
$$

Solving the equation $x^{2}+x+3=0$, we get
$$
x=\frac{-1 \pm \sqrt{11} \mathrm{i}}{2} \text {. }
$$