Example 4 Given real numbers $a, b, c$. It is known that complex numbers $z_{1}, z_{2}, z_{3}$ satisfy:
$$
\left\{\begin{array}{l}
\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1, \\
\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{3}}+\frac{z_{3}}{z_{1}}=1 .
\end{array}\right.
$$

Find $|a z_{1}+b z_{2}+c z_{3}|$.
Solution 1: Let $\mathrm{e}^{\mathrm{i} \theta}=\cos \theta+\mathrm{i} \sin \theta$. Suppose $\frac{z_{1}}{z_{2}}=\mathrm{e}^{\mathrm{i} \theta}, \frac{z_{2}}{z_{3}}=\mathrm{e}^{\mathrm{i} \varphi}$, then $\frac{z_{3}}{z_{1}}=\mathrm{e}^{-\mathrm{i}(\theta+\varphi)}$.
From the given condition, we have $\mathrm{e}^{\mathrm{i} \theta}+\mathrm{e}^{\mathrm{i} \varphi}+\mathrm{e}^{-\mathrm{i}(\theta+\varphi)}=1$.
Taking the imaginary part on both sides, we get
$$
\begin{array}{l}
\sin \theta+\sin \varphi-\sin (\theta+\varphi)=0, \\
2 \sin \frac{\theta+\varphi}{2} \cos \frac{\theta-\varphi}{2}-2 \sin \frac{\theta+\varphi}{2} \cos \frac{\theta+\varphi}{2}=0, \\
\sin \frac{\theta+\varphi}{2} \cdot \sin \frac{\theta}{2} \sin \frac{\varphi}{2}=0 .
\end{array}
$$

Therefore, we have $\theta=2 k \pi$ or $\varphi=2 k \pi$ or $\theta+\varphi=2 k \pi, k \in \mathbf{Z}$.
Thus, $z_{1}=z_{2}$ or $z_{2}=z_{3}$ or $z_{3}=z_{1}$.
If $z_{1}=z_{2}$, substituting into the given equation, we get
$$
1+\frac{z_{1}}{z_{3}}=\frac{z_{3}}{z_{1}}=1,\left(\frac{z_{1}}{z_{3}}\right)^{2}=-1, \frac{z_{1}}{z_{3}}= \pm \mathrm{i} .
$$

Thus, we have
$$
\left|a z_{1}+b z_{2}+c z_{3}\right|=\left|z_{1}\right| \cdot|a+b \pm c \mathrm{i}|=\sqrt{(a+b)^{2}+c^{2}} .
$$

Similarly, if $z_{2}=z_{3}$, then
$$
\left|a z_{1}+b z_{2}+c z_{3}\right|=\sqrt{a^{2}+(b+c)^{2}} ;
$$

If $z_{3}=z_{1}$, then
$$
\left|a z_{1}+b z_{2}+c z_{3}\right|=\sqrt{(a+c)^{2}+b^{2}} .
$$

The required value is $\sqrt{(a+b)^{2}+c^{2}}$ or $\sqrt{a^{2}+(b+c)^{2}}$ or $\sqrt{(c+a)^{2}+b^{2}}$.
Solution 2: From $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{3}}+\frac{z_{3}}{z_{1}}=1$, we get $z_{1}^{2} z_{3}+z_{2}^{2} z_{1}+z_{3}^{2} z_{2}=z_{1} z_{2} z_{3}$,
$$
\text { and } \frac{\overline{z_{1}}}{\overline{z_{2}}}+\frac{\overline{z_{2}}}{\overline{z_{3}}}+\frac{\overline{z_{3}}}{\overline{z_{1}}}=1 \text {. }
$$

Since $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1$, we have $\overline{z_{1}}=\frac{1}{z_{1}}, \overline{z_{2}}=\frac{1}{z_{2}}, \overline{z_{3}}=\frac{1}{z_{3}}$,
Thus, (2) can be rewritten as $\frac{z_{2}}{z_{1}}+\frac{z_{3}}{z_{2}}+\frac{z_{1}}{z_{3}}=1$,

i.e., $z_{2}^{2} z_{3}+z_{3}^{2} z_{1}+z_{1}^{2} z_{2}=z_{1} z_{2} z_{3}$.
From (1) and (3), we get $z_{1}^{2} z_{3}+z_{2}^{2} z_{1}+z_{3}^{2} z_{2}=z_{2}^{2} z_{3}+z_{3}^{2} z_{1}+z_{1}^{2} z_{2}$,
i.e., $\left(z_{1}-z_{2}\right)\left(z_{2}-z_{3}\right)\left(z_{3}-z_{1}\right)=0$, so $z_{1}=z_{2}$ or $z_{2}=z_{3}$ or $z_{3}=z_{1}$. The rest is the same as Solution 1.