The sixty-fifth question: Find all complex parameters $\mathrm{a}, \mathrm{~b}(\mathrm{a} \neq 0)$, such that for any complex root $w$ of the equation $\mathrm{z}^{4}-\mathrm{az}^{3}-\mathrm{bz}-1=0$, the inequality $|a-w| \geq|w|$ is satisfied.
The 65th problem,
Solution: According to the conditions, for any complex root \( w \) of \( z^4 - az^3 - bz - 1 = 0 \), we have
\[
|a - w| \geq |w| \Leftrightarrow \left|1 - \frac{w}{a}\right| \geq \left|\frac{w}{a}\right|
\]

This indicates that the distance from \(\frac{w}{a}\) to 1 is not less than its distance to the origin, thus \(\frac{w}{a}\) lies to the left of the line \(\operatorname{Re}(z) = \frac{1}{2}\), i.e., \(\operatorname{Re}\left(\frac{w}{a}\right) \leq \frac{1}{2}\).
Let the four roots of the equation \( z^4 - az^3 - bz - 1 = 0 \) be \( z_1, z_2, z_3, z_4 \). According to Vieta's formulas, we have:
\[
\left\{\begin{array}{c}
\sum_{k=1}^{4} z_k = a \\
\sum_{1 \leq i < j \leq 4} z_i z_j = 0 \\
\sum_{1 \leq i < j < k \leq 4} z_i z_j z_k \\
z_1 z_2 z_3 z_4 = -1
\end{array} \Leftrightarrow \left\{\begin{array}{c}
\sum_{k=1}^{4} \frac{z_k}{a} = 1 \\
\sum_{1 \leq i < j \leq 4} \frac{z_i}{a} \cdot \frac{z_j}{a} = 1 \\
\sum_{1 \leq i < j < k \leq 4} z_i z_j z_k = b \\
z_1 z_2 z_3 z_4 = -1
\end{array}\right.\right.
\]

Let \(\frac{z_k}{a} = x_k + i y_k (1 \leq k \leq 4)\), where \( x_k, y_k \in \mathbb{R} \), then \( x_k \leq \frac{1}{2} \). From \(\left\{\begin{array}{c}\sum_{k=1}^{4} \frac{z_k}{a} = 1 \\ \sum_{1 \leq i < j \leq 4} \frac{z_i}{a} \cdot \frac{z_j}{a} = 1\end{array}\right.\),
we know \(\left\{\begin{array}{c}\sum_{k=1}^{4} x_k = 1 \\ \sum_{k=1}^{4} (x_k^2 - y_k^2) = 1 \\ \sum_{k=1}^{4} x_k = \sum_{k=1}^{4} x_k y_k = 0\end{array}\right.\).
Since \( x_k \leq \frac{1}{2} (1 \leq k \leq 4) \), we have \( x_1 = 1 - (x_2 + x_3 + x_4) \geq -\frac{1}{2} \), thus \( x_1 \in \left[-\frac{1}{2}, \frac{1}{2}\right] \). Similarly, \( x_1, x_2, x_3, x_4 \in \left[-\frac{1}{2}, \frac{1}{2}\right] \), so \( x_k^2 \leq \frac{1}{4} (1 \leq k \leq 4) \). Therefore,
\[
\sum_{k=1}^{4} x_k^2 \leq 4 \times \frac{1}{4} = 1 = \sum_{k=1}^{4} x_k^2 - \sum_{k=1}^{4} y_k^2
\]

Thus, \( x_k^2 = \frac{1}{4}, y_k = 0 (1 \leq k \leq 4) \). Combining \(\sum_{k=1}^{4} x_k = 1\), we know that exactly three of \( x_k \) are \(\frac{1}{2}\) and one is \(-\frac{1}{2}\).
Assume \( x_1 = x_2 = x_3 = \frac{1}{2}, x_4 = -\frac{1}{2} \), then \( z_1 = z_2 = z_3 = \frac{a}{2}, z_4 = -\frac{a}{2} \). Therefore,
\[
z_1 z_2 z_3 z_4 = -\frac{a^4}{16} = -1 \Rightarrow a = \pm 2, \pm 2i
\]

Combining \(\sum_{1 \leq i < j < k \leq 4} z_i z_j z_k = b\), we have \((a, b) = (2, -2), (-2, 2), (2i, 2i), (-2i, -2i)\).
On the other hand, when \((a, b) = (2, -2), (-2, 2), (2i, 2i), (-2i, -2i)\),
\[
\begin{array}{l}
z^4 - 2z^3 + 2z - 1 = (z + 1)(z - 1)^2 = 0 \Rightarrow z_1 = z_2 = z_3 = 1, z_4 = -1; \\
z^4 + 2z^3 - 2z - 1 = (z - 1)(z + 1)^2 = 0 \Rightarrow z_1 = z_2 = z_3 = -1, z_4 = 1; \\
z^4 - 2iz^3 - 2iz - 1 = (z + i)(z - i)^2 = 0 \Rightarrow z_1 = z_2 = z_3 = i, z_4 = -i; \\
z^4 + 2iz^3 + 2iz - 1 = (z - i)(z + i)^2 = 0 \Rightarrow z_1 = z_2 = z_3 = -i, z_4 = i;
\end{array}
\]

All satisfy the conditions.
In summary, the pairs \((a, b)\) that satisfy the conditions are \((2, -2), (-2, 2), (2i, 2i), (-2i, -2i)\).
Note: We can also explain \(\operatorname{Re}\left(\frac{w}{a}\right) \leq \frac{1}{2}\) from an algebraic perspective. In fact,
\[
\begin{array}{l}
|a - w| \geq |w| \Leftrightarrow \left|1 - \frac{w}{a}\right| \geq \left|\frac{w}{a}\right| \Leftrightarrow \left(1 - \frac{w}{a}\right)^2 \left(1 - \frac{\bar{w}}{\bar{a}}\right) \geq \frac{w}{a} \cdot \frac{\bar{w}}{\bar{a}} \\
\Leftrightarrow 1 - \left(\frac{w}{a} + \frac{\bar{w}}{\bar{a}}\right) + \frac{w}{a} \cdot \frac{\bar{w}}{\bar{a}} \geq \frac{w}{a} \cdot \frac{\bar{w}}{\bar{a}} \\
\Leftrightarrow \frac{w}{a} + \frac{\bar{w}}{\bar{a}} = 2 \operatorname{Re}\left(\frac{w}{a}\right) \leq 1
\end{array}
\]

Thus, \(\operatorname{Re}\left(\frac{w}{a}\right) \leq \frac{1}{2}\).