8. (Question 2 of the 5th Test, *The American Mathematical Monthly*, pages 119 to 121, 1953)

Given a positive integer $n \geqslant 3$, for $n$ complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ with modulus 1, find
$$
\min _{x_{1}, z_{2} \cdots, s_{n}}\left[\max _{\omega \in C,|\omega|=1} \prod_{j=1}^{n}\left|\omega-z_{j}\right|\right] \text {, }
$$

and discuss the conditions that the complex numbers $z_{1}$, $z_{2}, \cdots, z_{n}$ satisfy when the minimum of the maximum value is achieved.
$$
\begin{array}{l}
f(u)=\prod_{j=1}^{n}\left(u-x_{j}\right) \\
=u^{n}+C_{n-1} u^{n-1}+\cdots+C_{1} u+C_{0} .
\end{array}
$$

Here, $\left|C_{0}\right|=\left|z_{1} z_{2} \cdots z_{n}\right|=1$. Let $C_{0}=e^{i\theta}, 0 \leqslant \theta<2 \pi$. Take $n$ complex numbers of modulus 1, $\varepsilon_{k}=e^{i \frac{2 x+\theta}{n}}, k=1,2, \cdots, n$ (Note: for any angle $\psi, e^{i \phi}=\cos \psi+i \sin \psi$). Then, $\varepsilon_{i}^{i}=e^{i \theta}=C_{0}$, and when $1 \leqslant m \leqslant n-1$, since $\left(e^{i \frac{2 \pi}{n}}\right)^{m} \neq 1$, we have
Thus, we have
$$
\begin{array}{l}
\sum_{k=1}^{n}\left|f\left(\varepsilon_{k}\right)\right| \geqslant\left|\sum_{k=1}^{n} f\left(\varepsilon_{k}\right)\right| \\
=\left|\sum_{k=1}^{n} \varepsilon_{i}^{n}+\sum_{i=1}^{n-1} C_{i} \sum_{k=1}^{n} \varepsilon_{k}^{k}+n C_{0}\right| \\
=\left|2 n C_{0}\right|=2 n .
\end{array}
$$

From (2), there is at least one $\varepsilon_{k}$ such that $\left|f\left(\varepsilon_{k}\right)\right| \geqslant 2$, i.e.,
$$
\max _{\{u \in \mathbb{C} \mid |u|=1\}} \prod_{j=1}^{n}\left|u-z_{j}\right| \geqslant 2.
$$

If (3) holds with equality, then from (2) we know that for $k=1,2, \cdots, n, f\left(\varepsilon_{k}\right)$ not only have the same modulus 2, but also the same argument. Using (2) again, we have $\sum_{k=1}^{n} f\left(\varepsilon_{k}\right)=2 n C_{0}$, so $f\left(\varepsilon_{k}\right)=2 C_{0}$, for $k=1, 2, \cdots, n$. Using (1), we can see that the $(n-1)$-degree polynomial $C_{n-1} u^{n-1} + C_{n-2} u^{n-2} + \cdots + C_{1} u$ is zero when $u=\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n}$, thus this polynomial is identically zero. Therefore,
$$
f(u)=u^{n}+C_{0}.
$$

Since $f\left(z_{k}\right)=0, k=1,2, \cdots, n$, we can conclude that $x_{1}, z_{1}, \cdots, z_{n}$ are the $n$-th roots of $-C_{0}$. When the minimum of the maximum value is reached, in the complex plane, the points corresponding to the complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ are the $n$ vertices of a regular $n$-sided polygon inscribed in the unit circle.