Example 8 (Olympiad Training Problem from "Intermediate Mathematics" Issue 1, 2004) The equation $x^{3}+p x+q=0(p, q \in \mathbf{C})$ 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 $p+q=$ $\qquad$
Solve: Fill in $-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$. Reason: The circumradius of an equilateral triangle with side length $\sqrt{3}$ is 1, so the complex numbers corresponding to the 3 vertices satisfy the equation $(x+a)^{3}=b(a, b \in \mathbf{C},|b|=1)$, i.e., $x^{3}+3 a x^{2}+3 a^{2} x+a^{2}-b=0$, comparing with the given equation, we have $3 a=0,3 a^{2}=9, a^{3}-b=q$, hence $p=0, q=-b$.
Since $|b|=1$, then $|q|=1$, and $\arg q=\frac{2 \pi}{3}, q=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$, thus $p+q=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$.