2. Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ with its right focus at $F$, and $B$ as a moving point on the ellipse, $\triangle F A B$ is an equilateral triangle, and $F$, $A$, $B$ are in counterclockwise order, find the locus of point $A$.
2. According to the problem, we know $2a=6, F(2,0)$, so $z_{B}$ satisfies the equation $\left|z_{B}+2\right|+\left|z_{B}-2\right|=6$. Let point $A$ correspond to the complex number $z$, then $z_{B}-2=(z-2)\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)=(z-2)\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right), z_{B}+2=\frac{1+\sqrt{3} \mathrm{i}}{2}(z-2 \sqrt{3} \mathrm{i})$, so $|z_{B}-2|+|z_{B}+2|=|z-2|+|z-2 \sqrt{3} \mathrm{i}|$, hence the trajectory equation of $A$ is $|z-2|+|z-2 \sqrt{3} \mathrm{i}|=6$, and the trajectory of $A$ is an ellipse with foci at $(2,0),(0,2 \sqrt{3})$ and a major axis length of 6.