The seventy-fourth question: Let complex numbers $a$, $b$, $c$ satisfy: for any complex number $\mathrm{z}$ with modulus not exceeding 1, we have $\left|\mathrm{az}^{2}+\mathrm{bz}+\mathrm{c}\right| \leq 1$. Find the maximum value of $|bc|$.
The 74th question:
Solution: Let $\omega_{1}=\mathrm{e}^{-\mathrm{i} \cdot \arg c}, \omega_{2}=\mathrm{e}^{\mathrm{i}(\text { argc-arg b) }}$, and
$$
f(z)=\left(a\left(\omega_{2} z\right)^{2}+b\left(\omega_{2} z\right)+c\right) \omega_{1}=a^{\prime} z^{2}+b^{\prime} z+c^{\prime}
$$

Then $\mathrm{b}^{\prime}=|b|, c^{\prime}=|c|$. For any non-negative real number $r \leq 1$ and $\theta \in \mathbb{R}$, we have
$$
\begin{array}{l}
\left|f\left(r^{i \theta}\right)\right| \leq 1 \\
\Leftrightarrow\left|a^{2} r^{2} e^{2 \theta i}+b^{\prime} r e^{i \theta}+c^{\prime}\right| \leq 1 \\
\Leftrightarrow\left|a^{2}+b^{\prime} r e^{-i \theta}+c^{\prime} e^{-2 \theta i}\right| \leq 1 \\
\Leftrightarrow\left|\left(\operatorname{Re}\left(a^{\prime}\right) r^{2}+b^{\prime} \cdot \cos \theta+c^{\prime} \cdot \cos 2 \theta\right)+\left(\operatorname{Im}\left(a^{\prime}\right) r^{2}-b^{\prime} r \cdot \sin \theta-c^{\prime} \cdot \sin 2 \theta\right) i\right| \leq 1
\end{array}
$$

Thus, for any non-negative real number $r \leq 1$ and $\theta \in \mathbb{R}$, we have
$$
\left|\operatorname{Im}\left(a^{\prime}\right) r^{2}-b^{\prime} r \cdot \sin \theta-c^{\prime} \cdot \sin 2 \theta\right| \leq 1
$$

Without loss of generality, we can assume $\operatorname{Im}\left(\mathrm{a}^{\prime}\right) \leq 0$ (otherwise, we can make the transformation $\theta \rightarrow -\theta$). We take $r=1$ in (2), then for any $\theta \in \left(0, \frac{\pi}{2}\right)$ we have
$$
\begin{array}{l}
1 \geq\left|\operatorname{Im}\left(\mathrm{a}^{\prime}\right)-\mathrm{b}^{\prime} \cdot \sin \theta-\mathrm{c}^{\prime} \cdot \sin 2 \theta\right| \geq \mathrm{b}^{\prime} \cdot \sin \theta+\mathrm{c}^{\prime} \cdot \sin 2 \theta \geq 2 \sqrt{\mathrm{b}^{\prime} \mathrm{c}^{\prime} \sin 2 \theta \sin \theta} \\
\Rightarrow|\mathrm{bc}|=\mathrm{b}^{\prime} \mathrm{c}^{\prime} \leq \frac{1}{4 \sin 2 \theta \sin \theta} \\
|\mathrm{bc}| \leq \min _{\theta \in\left(0, \frac{\pi}{2}\right)} \frac{1}{4 \sin 2 \theta \sin \theta} \\
\text { According to the AM-GM inequality, we know: }
\end{array}
$$

Thus, $|b c| \leq \min _{\theta \in\left(0, \frac{\pi}{2}\right)} \frac{1}{4 \sin 2 \theta \sin \theta}$
According to the AM-GM inequality, we know:
$$
\frac{1}{4 \sin 2 \theta \sin \theta}=\frac{1}{8 \cos \theta \sin ^{2} \theta}=\frac{1}{16 \sqrt{\cos ^{2} \theta \cdot \frac{\sin ^{2} \theta \sin ^{2} \theta}{2}}} \geq \frac{1}{16 \sqrt{\left(\frac{\cos ^{2} \theta+\frac{\sin ^{2} \theta, \sin ^{2} \theta}{2}}{3}\right)^{3}}}=\frac{3 \sqrt{3}}{16}
$$

When $\theta=\arcsin \frac{\sqrt{6}}{3}$, the equality holds, so $\min _{\theta \in\left(0, \frac{\pi}{2}\right.} \frac{1}{4 \sin 2 \theta \sin \theta}=\frac{3 \sqrt{3}}{16}$. Therefore, $|b c| \leq \frac{3 \sqrt{3}}{16}$. According to the condition for the equality in the AM-GM inequality, we must have $\left\{\begin{array}{c}b^{\prime} \cdot \sin \left(\arcsin \frac{\sqrt{6}}{3}\right)=c^{\prime} \cdot \sin \left(2 \arcsin \frac{\sqrt{6}}{3}\right) \\ b^{\prime} c^{\prime}=\frac{3 \sqrt{3}}{16}\end{array} \Rightarrow\left\{\begin{array}{l}b^{\prime}=\frac{\sqrt{6}}{4} \\ c^{\prime}=\frac{3 \sqrt{2}}{8}\end{array}\right.\right.$
Furthermore, according to the condition for equality in (1), we must have $\operatorname{Im}\left(a^{\prime}\right)=0$, and
$$
\begin{array}{l}
\operatorname{Re}\left(\mathrm{a}^{\prime}\right)+b^{\prime} \cdot \cos \left(\arcsin \frac{\sqrt{6}}{3}\right)+c^{\prime} \cdot \cos \left(2 \arcsin \frac{\sqrt{6}}{3}\right)=0 \\
\Rightarrow \operatorname{Re}\left(a^{\prime}\right)=-\frac{\sqrt{2}}{8}
\end{array}
$$

Thus, we take $f(z)=-\frac{\sqrt{2}}{8} z^{2}+\frac{\sqrt{6}}{4} z+\frac{3 \sqrt{2}}{8}$, then $|b c|=\frac{3 \sqrt{3}}{16}$. At this point, for $z=r e^{i \theta}(r \leq 1)$, we have
$$
\begin{array}{l}
\left|f\left(r e^{i \theta}\right)\right|^{2}=\frac{1}{32}\left[\left(r^{2} \cos 2 \theta-2 \sqrt{3} r \cos \theta-3\right)^{2}+\left(r^{2} \sin 2 \theta-2 \sqrt{3} r \sin \theta\right)^{2}\right] \\
=\frac{1}{32}\left[2 r^{4}+12 r^{2}+18-\left(2 \sqrt{3} r \cos \theta+r^{2}-3\right)^{2}\right] \\
\leq \frac{1}{32}\left(2 r^{4}+12 r^{2}+18\right) \leq 1
\end{array}
$$

In summary, the maximum value of $|\mathrm{bc}|$ is $\frac{3 \sqrt{3}}{16}$.