Example 13 Let $a, b$ be non-zero complex numbers, and $\frac{a}{b}$ is not a real number. Define:
$$
\begin{array}{l}
L_{a, b}=\{r a+s b \mid r, s \in \mathbf{Z}\}, \\
R_{a, b}=\left\{z \mid z \text { is a non-zero complex number, and } L_{a, b}=L_{z a, z}\right\} .
\end{array}
$$

Try to find the maximum number of elements in the set $R_{a, b}$ as $a, b$ vary.
【Analysis】Notice that $L_{a, b}$ represents the set of all integer points $(r, s)$ in the oblique coordinate system with complex vectors $a, b$ as basis vectors, while $L_{z a, z b}$ is the result of multiplying each element in $L_{a, b}$ by $z$ (a combination of scaling and rotation transformations). From $L_{a a, b b}=L_{a, b}$, we can determine the scaling factor $r$ and the smallest positive angle $\theta$ of rotation. Consequently, we can estimate the upper bound of $\mid R_{a, b} \backslash$.

Solution: In $L_{a, b}$, there exist two points $A, B$ such that $|A B|$ achieves its minimum value. By the invariance of modulus under translation, we can assume $A$ is the origin $O$ on the complex plane.

Let $z=r \mathrm{e}^{\mathrm{i} \theta}$, such that $L_{z a, z b}=L_{a, b}$, and $L_{z a, z b}$ is the set of points obtained by scaling $L_{a, b}$ by $r$ with the origin $O$ as the center, and then rotating by $\theta$. Thus,
$|z B|=|B| \Rightarrow|z|=1 \Rightarrow r=1$.
Clearly, $1 \in R_{a, b}$.
If non-zero complex numbers $z_{1}, z_{2} \in R_{a, b}$, then
$$
z_{1} B, z_{2} B \in L_{a, b} .
$$

Since $|B|$ is the minimum distance between any two points in $L_{a, b}$, we have
$$
\begin{array}{l}
\left|z_{1} B-z_{2} B\right| \geqslant|B| \\
\Rightarrow\left|z_{1}-z_{2}\right| \geqslant 1 . \\
\text { Also, }\left|z_{1}\right|=\left|z_{2}\right|=1, \\
\angle Z_{1} o Z_{1} o z_{2} \geqslant \frac{\pi}{3} .
\end{array}
$$

From equation (1) and the pigeonhole principle, we know that $R_{a, b}$ contains at most $2 \pi \div \frac{\pi}{3}=6$ elements, i.e., $\left|R_{a, b}\right| \leqslant 6$. When $a=1$, $b=\mathrm{e}^{\mathrm{i} \frac{\pi}{3}}$, and $z=\mathrm{e}^{\mathrm{i} \frac{\hbar}{3}}(k \in\{0,1, \cdots, 5\})$, then $z \in R_{a, b}$. In this case, $\left|R_{a, b}\right|=6$.
In summary, $\left|R_{a, b}\right|_{\text {max }}=6$.