2. Let $k$ be a complex number, and the quadratic equation in $x$, $x^{2}+k x-\frac{1}{2}=0$, has two complex roots $x_{1}$ and $x_{2}$. If $x_{1}+2 x_{2}^{3}=k$, then $k=$ $\qquad$ .
2.0 or $\frac{3}{2} \mathrm{i}$ or $-\frac{3}{2} \mathrm{i}$.

Since $x_{2}^{2}+k x_{2}-\frac{1}{2}=0$, we have
$$
x_{2}^{3}+k x_{2}^{2}-\frac{1}{2} x_{2}=0 \text {. }
$$

Thus, $x_{2}^{3}=-k x_{2}^{2}+\frac{1}{2} x_{2}$
$=-k\left(-k x_{2}+\frac{1}{2}\right)+\frac{1}{2} x_{2}$
$$
=\left(k^{2}+\frac{1}{2}\right) x_{2}-\frac{k}{2} \text {. }
$$

Substituting into $x_{1}+2 x_{2}^{3}=k$, we get
$$
\begin{array}{l}
x_{1}+\left(2 k^{2}+1\right) x_{2}-k=k \\
\Rightarrow-k+2 k^{2} x_{2}=2 k \Rightarrow\left(2 k x_{2}-3\right) k=0
\end{array}
$$
$\Rightarrow k=0$ or $x_{2}=\frac{3}{2 k}$ (when $k \neq 0$).
When $k \neq 0$, substituting $x_{2}=\frac{3}{2 k}$ into $x_{2}^{2}+k x_{2}-\frac{1}{2}$ $=0$, we get $\frac{9}{4 k^{2}}+\frac{3}{2}-\frac{1}{2}=0$.
Solving for $k$ yields $k= \pm \frac{3}{2} \mathrm{i}$.
In summary, $k=0$ or $\pm \frac{3}{2} \mathrm{i}$.