Example 1 Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=\left|z_{2}\right|=1$.
(1) If $z_{1}-z_{2}=\frac{\sqrt{6}}{3}+\frac{\sqrt{3}}{3} i$, find the values of $z_{1}, z_{2}$;
(2) If $z_{1}+z_{2}=\frac{12}{13}-\frac{5}{13} \mathrm{i}$, find the value of $z_{1} z_{2}$.
Analysis Notice that $\left|z_{1}+z_{2}\right|,\left|z_{1}-z_{2}\right|$ are the diagonals of a parallelogram with sides $\left|z_{1}\right|,\left|z_{2}\right|$. Therefore, we can solve the problem by combining the geometric meanings of complex number multiplication and division.

Solution (1) Since $\left|z_{1}-z_{2}\right|=1=\left|z_{1}\right|=\left|z_{2}\right|$, $\left|z_{1}\right|,\left|z_{2}\right|,\left|z_{1}-z_{2}\right|$ are the side lengths of an equilateral triangle. The complex numbers $z_{1}, z_{2}$ can be seen as $z_{1}-z_{2}$ rotated by $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$ in a clockwise or counterclockwise direction. Thus, 
$$
z_{1}=\left(\frac{\sqrt{6}}{3}+\frac{\sqrt{3}}{3} \mathrm{i}\right)\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)=\frac{1}{6}[(\sqrt{6}+3)+(\sqrt{3}-3 \sqrt{2}) \mathrm{i}],
$$
$$
z_{2}=\left(\frac{\sqrt{6}}{3}+\frac{\sqrt{3}}{3} \mathrm{i}\right)\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)=\frac{1}{6}[(-\sqrt{6}+3)+(-\sqrt{3}-3 \sqrt{2}) \mathrm{i}] ;
$$

or 
$$
z_{1}=\left(\frac{\sqrt{6}}{3}+\frac{\sqrt{3}}{3} \mathrm{i}\right)\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=\frac{1}{6}[(\sqrt{6}-3)+(\sqrt{3}+3 \sqrt{2}) \mathrm{i}],
$$
$$
z_{2}=\left(\frac{\sqrt{6}}{3}+\frac{\sqrt{3}}{3} \mathrm{i}\right)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=\frac{1}{6}[(-\sqrt{6}-3)+(-\sqrt{3}+3 \sqrt{2}) \mathrm{i}] .
$$
(2) Since $\left|z_{1}+z_{2}\right|=1=\left|z_{1}\right|=\left|z_{2}\right|$, $\left|z_{1}\right|,\left|z_{2}\right|$, $\left|z_{1}-z_{2}\right|$ form an isosceles triangle with a vertex angle of $\frac{2 \pi}{3}$ and legs $\left|z_{1}\right|,\left|z_{2}\right|$. Rotating $z_{1}+z_{2}$ by $\frac{\pi}{3}$ in a counterclockwise (or clockwise) direction yields $z_{1}$ or $z_{2}$ (see Figure 2-1).

Let $z_{0}=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}$. Then, when $z_{1}=\left(z_{1}+z_{2}\right) z_{0}$, $z_{2}=\frac{z_{1}+z_{2}}{z_{0}}$; when $z_{1}=\frac{z_{1}+z_{2}}{z_{0}}$, $z_{2}=\left(z_{1}+z_{2}\right) z_{0}$. Therefore, $z_{1} z_{2}=\left(z_{1}+z_{2}\right)^{2}=\left(\frac{12}{13}-\frac{5}{13} \mathrm{i}\right)^{2}$ $=\frac{119}{169}-\frac{120}{169} \mathrm{i}$.