Example 3 Given real numbers $a, b, c$. It is known that complex numbers $z_{1}, z_{2}, z_{3}$ satisfy $\left\{\begin{array}{l}\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1 \\ \frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{3}}+\frac{z_{3}}{z_{1}}=1\end{array}\right.$ (1), find the value of $\left|a z_{1}+b z_{2}+c z_{3}\right|$.
Analysis 1 From (1) we get $\left|\frac{z_{1}}{z_{2}}\right|=\left|\frac{z_{2}}{z_{3}}\right|=\left|\frac{z_{3}}{z_{1}}\right|=1$. Using the trigonometric or exponential form of complex numbers, we can set $\frac{z_{1}}{z_{2}}=e^{i \alpha}, \frac{z_{2}}{z_{3}}=e^{i \beta}$, then $\frac{z_{3}}{z_{1}}=e^{-i(\alpha+\beta)}$. Substituting into (2) and using the necessary and sufficient conditions for the equality of complex numbers to solve.

Solution 1 From condition (1), we can set $\frac{z_{1}}{z_{2}}=e^{i \alpha}, \frac{z_{2}}{z_{3}}=e^{i \beta}$, then $\frac{z_{3}}{z_{1}}=e^{-i(\alpha+\beta)}$. Substituting into condition (2), we have $e^{i \alpha}+e^{i \beta}+e^{-i(\alpha+\beta)}=1$.
By Euler's formula $e^{i \theta}=\cos \theta+i \sin \theta$, taking the imaginary part of both sides of the above equation, we get
$$
\begin{aligned}
0 & =\sin \alpha+\sin \beta-\sin (\alpha+\beta)=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}-2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2} \\
& =2 \sin \frac{\alpha+\beta}{2}\left(\cos \frac{\alpha-\beta}{2}-\cos \frac{\alpha+\beta}{2}\right)=4 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha}{2} \sin \frac{\beta}{2} .
\end{aligned}
$$

Therefore, $\alpha=2 k \pi$ or $\beta=2 k \pi$ or $\alpha+\beta=2 k \pi$, so $z_{1}=z_{2}$ or $z_{2}=z_{3}$ or $z_{3}=z_{1}$.
If $z_{1}=z_{2}$, substituting into condition (2), we get $1+\frac{z_{1}}{z_{3}}+\frac{z_{3}}{z_{1}}=1$, hence $\left(\frac{z_{3}}{z_{1}}\right)^{2}=-1, z_{3}= \pm i z_{1}$,
at this time $\quad\left|a z_{1}+b z_{2}+c z_{3}\right|=\left|z_{1}\right| \cdot|a+b \pm c i|=\sqrt{(a+b)^{2}+c^{2}}$.
Similarly, if $z_{2}=z_{3}$, then $\left|a z_{1}+b z_{2}+c z_{3}\right|=\sqrt{(b+c)^{2}+a^{2}}$; if $z_{3}=z_{1}$, then $\mid a z_{1}+b z_{2}+c z_{3} \mid=\sqrt{(c+a)^{2}+b^{2}}$. Therefore, $\left|a z_{1}+b z_{2}+c z_{3}\right|$ is $\sqrt{(a+b)^{2}+c^{2}}$ or $\sqrt{(b+c)^{2}+a^{2}}$ or $\sqrt{(c+a)^{2}+b^{2}}$.

Analysis 2 From condition (1) and the properties of conjugate complex numbers, we get $\overline{z_{1}}=\frac{1}{z_{1}}, \overline{z_{2}}=\frac{1}{z_{2}}, \overline{z_{3}}=\frac{1}{z_{3}}$, then use algebraic constant transformation (factorization) to solve.
Solution 2 From $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{3}}+\frac{z_{3}}{z_{1}}=1$ we get $z_{1}^{2} z_{3}+z_{2}^{2} z_{1}+z_{3}^{2} z_{2}=z_{1} z_{2} z_{3}$
$$
\text { and } \frac{\overline{z_{1}}}{\overline{z_{2}}}+\frac{\overline{z_{2}}}{\overline{z_{3}}}+\frac{\overline{z_{3}}}{\overline{z_{1}}}=1
$$

Since $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1$, we have $\overline{z_{1}}=\frac{1}{z_{1}}, \overline{z_{2}}=\frac{1}{z_{2}}, \overline{z_{3}}=\frac{1}{z_{3}}$, so equation (4) can be transformed into $\frac{z_{2}}{z_{1}}+\frac{z_{3}}{z_{2}}+\frac{z_{1}}{z_{3}}=1$,
i.e., $z_{2}^{2} z_{3}+z_{3}^{2} z_{1}+z_{1}^{2} z_{2}=z_{1} z_{2} z_{3}$
From (3) and (5), 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.