Example 7 Let the roots of the odd-degree real-coefficient equation $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}=0$ all lie on the unit circle, and $-a_{n}=a_{0} \neq 0$, find $a_{0}+a_{1}+\cdots+a_{n}$.
Let $f(c)=a_{0} c^{n}+a_{1} c^{n-1}+\cdots+a_{n-1} c+a_{n}=0$, i.e., the complex number $c$ is a root of $f(x)=0$, then the conjugate complex number $\bar{c}$ is also a root of $f(x)=0$. However, $f(x)=0$ is an equation of odd degree, so apart from the conjugate pairs of imaginary roots, it must have an odd number of real roots.
Since all roots of $f(x)=0$ lie on the unit circle, the real roots of $f(x)=0$ can only be 1 or -1.
Furthermore, by the relationship between roots and coefficients, the product of all roots of $f(x)=0$ equals $(-1)^{n} a_{n} / a_{0}=(-1)^{n} \cdot(-1)=1$ (for $n$ being odd). However, the product of conjugate imaginary numbers must be positive, so -1 is either not a root of $f(x)=0$, or it is a root of even multiplicity. Therefore, 1 must be a root of $f(x)=0$. Hence, $a_{0}+a_{1}+\cdots+a_{n}=f(1)=0$.