The 71st problem,
Solution: First, we prove that for any $\{z_1, z_2, \ldots, z_n\} \in \mathcal{A}$, there exists a unit complex number $u$ such that $\prod_{j=1}^{n} |u - z_j| \geq 2$. In fact, let
$$f(z) = \prod_{j=1}^{n} (z - z_j) = z^n + c_{n-1} z^{n-1} + c_{n-2} z^{n-2} + \ldots + c_1 z + c_0$$

where $c_0 = (-1)^n z_1 z_2 \ldots z_n$, and $|c_0| = 1$.
Let $\omega = e^{\frac{2 \pi i}{n}}$, $\varepsilon = e^{\frac{\arg c_0}{n} i}$, then
$$\sum_{k=1}^{n} f(\varepsilon \cdot \omega^k) = n \cdot \varepsilon^n + n \cdot c_0 = 2n c_0$$
$$\Rightarrow \sum_{k=1}^{n} |f(\varepsilon \cdot \omega^k)| \geq \left| \sum_{k=1}^{n} f(\varepsilon \cdot \omega^k) \right| = |2nc_0| = 2n$$

By the mean value principle, there exists a $k \in \{1, 2, \ldots, n\}$ such that $|f(\varepsilon \cdot \omega^k)| \geq 2$. Let $u = \varepsilon \cdot \omega^k$, then $\Pi_{j=1}^{n} |u - z_j| = |f(u)| \geq 2$. Therefore, for any $A = \{z_1, z_2, \ldots, z_n\} \in \mathcal{A}$, we have
$$\max_{\substack{u \in C \\ |u| = 1}} \left\{ \Pi_{z \in A} |u - z| \right\} \geq 2$$

Next, we find all $\{z_1, z_2, \ldots, z_n\} \in \mathcal{A}$ such that $\max_{\substack{u \in C \\ |u| = 1}} \left\{ \prod_{z \in A} |u - z| \right\} = 2$. According to the above analysis, in this case, we must have:
$$f(\varepsilon \cdot \omega^1) = f(\varepsilon \cdot \omega^2) = \ldots = f(\varepsilon \cdot \omega^n) = 2c_0$$

Let $g(z) = f(z) - z^n - c_0$, then $\deg g(z) \leq n-1$, and for any $k \in \{1, 2, \ldots, n\}$,
$$g(\varepsilon \cdot \omega^k) = f(\varepsilon \cdot \omega^k) - (\varepsilon \cdot \omega^k)^n - c_0 = 2c_0 - c_0 - c_0 = 0$$

This indicates that $g(z) \equiv 0$. Therefore,
$$f(z) = \prod_{j=1}^{n} (z - z_j) = z^n + c_0 = z^n + (-1)^n z_1 z_2 \ldots z_n$$

This shows that $z_1, z_2, \ldots, z_n$ are $n$ points uniformly distributed on the unit circle in the complex plane, i.e., $z_1, z_2, \ldots, z_n$ are the $n$ vertices of a regular $n$-gon.
Conversely, if $z_1, z_2, \ldots, z_n$ are the $n$ vertices of a regular $n$-gon, then
$$\prod_{j=1}^{n} (z - z_j) = z^n + (-1)^n z_1 z_2 \ldots z_n$$

Thus, for any unit complex number $u$,
$$\Pi_{z \in A} |u - z| = \left| u^n + (-1)^n z_1 z_2 \ldots z_n \right| \leq \left| u^n \right| + \left| (-1)^n z_1 z_2 \ldots z_n \right| = 2$$

Therefore, such $\{z_1, z_2, \ldots, z_n\}$ satisfy the given conditions.

Note: Generally, if $f(z) = c_n z^n + c_{n-1} z^{n-1} + \ldots + c_1 z + c_0$ is an $n$-degree polynomial with complex coefficients, then there exists a unit complex number $z_0$ such that $|f(z_0)| \geq |c_0| + |c_n|$.

More generally, we have the following proposition: Given $f(z) = c_n z^n + c_{n-1} z^{n-1} + \ldots + c_1 z + c_0$ as an $n$-degree polynomial with complex coefficients, prove that there exists a complex number $z_0 (|z_0| \leq 1)$ such that $|f(z_0)| \geq |c_0| + \max_{1 \leq k \leq n} \frac{|c_k|}{\left[ \frac{n}{k} \right]}$.

Proof: We only need to prove that for $k \in \{1, 2, \ldots, n\}$, there exists $z \in C, |z| \leq 1$, such that $|f(z)| \geq |c_0| + \frac{|c_k|}{\left[ \frac{n}{k} \right]}$.
When $k = 1$, we first prove that there exists a complex number $z_1 (|z_1| \leq 1)$ such that $|f(z_1)| \geq |c_0| + \frac{1}{n} \cdot |c_1|$.
Take the unit complex number $s_0 = e^{i \alpha}, \alpha = -\arg(c_0)$, then $s_0 c_0 \in \mathbb{R}$, and $s_0 c_0 \geq 0$; take the unit complex number $s_1 = e^{i \beta}$, $\beta = -\arg(c_1 s_0)$, then $s_1 s_0 c_1 \in \mathbb{R}$, and $s_1 s_0 c_1 \geq 0$. Let
$$\begin{array}{l} g(z) = s_0 f(s_1 z) \\ = s_0 c_n s_1^n z^n + s_0 c_{n-1} s_1^{n-1} z^{n-1} + \ldots + s_0 c_1 s_1 z + s_0 c_0 \\ = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_1 z + a_0 \end{array}$$

Then $|f(s_1 z)| = |s_0| \cdot |f(s_1 z)| = |g(z)|$, $|a_0| = |s_0 c_0| = |c_0|$, $|a_1| = |s_0 c_1 s_1| = |c_1|$. Next, we only need to prove that there exists $|z| \leq 1$ such that $|g(z)| \geq |a_0| + \frac{|a_1|}{n} = a_0 + \frac{a_1}{n}$.
Otherwise, for any $|z| \leq 1$, $|g(z)| < a_0 + \frac{a_1}{n}$, hence $|g(z)| < a_0 + \frac{a_1}{n}$. Let $g(z) = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_1 z + a_0$, and let \(g(z) = a_n (z - b_1)(z - b_2) \