Example 5 Given that the complex number $z$ satisfies $\arg (z+3)=\frac{3}{4} \pi$, find the value of $z$ for which $u=\frac{1}{|z+6|+|z-3 \mathrm{i}|}$ is maximized, and determine this maximum value.
Analyze: In the complex plane, the equation $\arg (z+3)=\frac{3}{4} \pi$ represents a ray $A E$ with endpoint $A(-3,0)$ and an inclination angle of $\frac{3}{4} \pi$. The expression $|z+6|+|z-3 \mathrm{i}|$ represents the sum of the distances from a point on the ray $A E$ to the points $B(-6,0)$ and $C(0,3)$. Therefore, the problem can be solved using the intuitive properties of the graph and knowledge of analytic geometry.

Solution: In the complex plane, the equation $\arg (z+3)=\frac{3}{4} \pi$ represents a ray $A E$ with endpoint $A(-3,0)$ and an inclination angle of $\frac{3}{4} \pi$ (as shown in Figure 2-3). Therefore, the problem is reduced to finding a point $Z$ (corresponding to the complex number $z$) on the ray $A E$ such that $|z+6|+|z-3 \mathrm{i}|$ is minimized. The expression $|z+6|+|z-3 \mathrm{i}|$ represents the sum of the distances from a point on the ray $A E$ to the points $B(-6,0)$ and $C(0,3)$.

As shown in Figure 2-3, it is clear that the line segment $B C$ intersects the ray $A E$. Therefore, the minimum value of $|z+6|+|z-3 \mathrm{i}|$ is $|B C|=\sqrt{(-6)^{2}+3^{2}}=3 \sqrt{5}$. Thus, $u_{\text {max }}=\frac{1}{3 \sqrt{5}}=\frac{\sqrt{5}}{15}$. When $u$ reaches its maximum value, the coordinates of the intersection point $Z$ of the line segment $B C$ and the ray $A E$ can be easily found to be $(-4,1)$, i.e., $z=-4+\mathrm{i}$.

In summary, when $z=-4+\mathrm{i}$, $u_{\max }=\frac{\sqrt{5}}{15}$.