5. $\frac{\pi}{6}$.

As shown in Figure 2, construct the point $D$ which is the central symmetric point of $A$ with respect to $C$. Then the point $D$ corresponds to the complex number $2 z_{3}-z_{1}$.
Notice that,
$$
\left(k+\frac{1}{2}\right)^{2}=-\frac{3}{4} \Rightarrow k^{2}+k+1=0,
$$

and $k=\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}$.
Thus, the condition is
$$
\begin{array}{l}
z_{1}+k z_{2}=(1+k)\left(2 z_{3}-z_{1}\right) \\
\Rightarrow\left(2 z_{3}-z_{1}\right)-z_{1} \\
\quad=k\left(z_{2}-\left(2 z_{3}-z_{1}\right)\right)
\end{array}
$$
$\Rightarrow \overrightarrow{D B}$ after a counterclockwise rotation of $\frac{2 \pi}{3}$ equals $\overrightarrow{A D}$
$\Rightarrow \triangle A B D$ is an equilateral triangle.
Since $C$ is the midpoint of $A D$, therefore,
$$
\angle A C B=\frac{\pi}{2}, \angle A B C=\frac{\pi}{6} \text {. }
$$
5. Let $k=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$, and the vertices of $\triangle A B C$ in the complex plane correspond to the complex numbers $z_{1}, z_{2}, z_{3}$ which satisfy
$$
z_{1}+k z_{2}+k^{2}\left(2 z_{3}-z_{1}\right)=0 \text {. }
$$

Then the smallest interior angle of the triangle in radians is
$\qquad$