Example 7 Let three complex numbers
\[
\begin{array}{l}
z_{1}=\cos A+\mathrm{i} \sin A, \\
z_{2}=\cos B+\mathrm{i} \sin B, \\
z_{3}=\cos C+\mathrm{i} \sin C,
\end{array}
\]

and \( z_{1}+z_{2}+z_{3}=0 \).
Find \( \cos (A-B) \).
Given $z_{1}+z_{2}+z_{3}=0$, we have
$$
\begin{array}{l}
\cos A+\cos B+\cos C=0, \\
\sin A+\sin B+\sin C=0 . \\
\text { Therefore, } \cos A+\cos B=-\cos C, \\
\sin A+\sin B=-\sin C .
\end{array}
$$

Squaring both equations and then adding them, we get
$$
2+2 \cos (A-B)=1 \text { . }
$$

Thus, $\cos (A-B)=-\frac{1}{2}$.