Example 22 Simplify 
$$
\frac{\left(z-z_{1}\right)\left(z^{\prime}-z_{1}\right)}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)}+
$$
$$
\frac{\left(z-z_{2}\right)\left(z^{\prime}-z_{2}\right)}{\left(z_{2}-z_{3}\right)\left(z_{2}-z_{1}\right)}+\frac{\left(z-z_{3}\right)\left(z^{\prime}-z_{3}\right)}{\left(z_{3}-z_{1}\right)\left(z_{3}-z_{2}\right)} .
$$
The original expression is a polynomial in $z$ with a degree less than or equal to 1.
When $z=z_{1}$, the value of the polynomial is $\frac{\left(z^{\prime}-z_{2}\right)}{\left(z_{3}-z_{2}\right)}- \frac{\left(z^{\prime}-z_{3}\right)}{\left(z_{3}-z_{2}\right)}=1$. Similarly, when $z=z_{2}$, the value of the polynomial is also 1. Therefore, the original polynomial is always equal to 1.
(1) is a symmetric expression in $z_{1}, z_{2}, z_{3}$, meaning that swapping any two of $z_{1}, z_{2}, z_{3}$ does not change (1).
For $n$ variables $x_{1}, x_{2}, \cdots, x_{n}$, the symmetric polynomials
$$
\begin{array}{l}
\sigma_{1}=x_{1}+x_{2}+\cdots+x_{n}=\sum x_{i}, \\
\sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+\cdots+x_{n-1} x_{n}=\sum_{i \neq j} x_{i} x_{j}, \\
\cdots \cdots \\
\sigma_{k}=\sum_{\left(i_{i}, i_{i}, \cdots, i_{i}\right)} x_{i_{1}} x_{i_{i}} \cdots x_{i_{k}}, \\
\cdots \cdots \cdots \\
\sigma_{n}=x_{1} x_{2} \cdots x_{n},
\end{array}
$$

are called the elementary symmetric polynomials of $x_{1}, x_{2}, \cdots, x_{n}$ (where $\sigma_{k}$ is the sum over all $k$-element combinations of $1,2, \cdots, n$), and they have many applications.
According to Vieta's formulas,
$$
\begin{array}{l}
\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)=x^{n}-\sigma_{1} x^{n-1} \\
+\sigma_{2} x^{n-2}+\cdots+(-1)^{n} \sigma_{n},
\end{array}
$$

so $x_{i}^{n}-\sigma_{1} x_{i}^{n-1}+\sigma_{2} x_{i}^{n-2}+\cdots+(-1)^{n} \sigma_{n}$ $=0 \quad(i=1,2, \cdots, n)$.