5. The equation $x^{3}+p x+q=0(p, q \in$ C) has three complex roots, and the points corresponding to these roots in the complex plane are the three vertices of an equilateral triangle with side length $\sqrt{3}$. If the principal value of the argument of the complex number $q$ is $\frac{2 \pi}{3}$, then $p+q=$ $\qquad$.
5. $-\frac{1}{2}+\frac{\sqrt{3}}{2}$ i.

The circumradius of an equilateral triangle with side length $\sqrt{3}$ is 1, so the complex numbers corresponding to the three vertices satisfy the equation
$$
(x+a)^{3}=b \quad(a, b \in \mathbf{C},|b|=1),
$$

which is $x^{3}+3 a x^{2}+3 a^{2} x+a^{3}-b=0$.
Comparing with the given equation, we have
$$
3 a=0,3 a^{2}=p, a^{3}-b=q \text {, }
$$

so $p=0, q=-b$.
Since $|b|=1$, then $|q|=1$.
Also, $\arg q=\frac{2 \pi}{3}, q=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$, hence
$$
p+q=-\frac{1}{2}+\frac{\sqrt{3}}{2} \text { i. }
$$