3. Let complex numbers $a, b, c$ satisfy: for any complex number $z$ with modulus not exceeding 1, we have $\left|a z^{2}+b z+c\right| \leqslant 1$. Find the maximum value of $|b c|$.
(Li Weiguo, problem contributor)
3. Let $f(z)=a z^{2}+b z+c$,
$$
\begin{array}{l}
g(z)=z^{-2} f(z)=a+b z^{-1}+c z^{-2}, \\
h(z)=\mathrm{e}^{\mathrm{i} \alpha} g\left(\mathrm{e}^{\mathrm{i} \beta} z\right)=c^{\prime} z^{-2}+b^{\prime} z^{-1}+a^{\prime} .
\end{array}
$$

Choose appropriate real numbers $\alpha, \beta$ such that $c^{\prime}, b^{\prime} \geqslant 0$. For $r \leqslant 1$, we have
$$
\begin{array}{l}
\frac{1}{r^{2}} \geqslant\left|h\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right| \geqslant\left|\operatorname{Im} h\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right| \\
=\left|r^{-2} c^{\prime} \sin 2 \theta+r^{-1} b^{\prime} \sin \theta+\operatorname{Im} a^{\prime}\right| .
\end{array}
$$

Assume $\operatorname{Im} a^{\prime} \geqslant 0$, otherwise we can make the transformation $\theta \rightarrow -\theta$. Thus, for any $\theta\left(0<\theta<\frac{\pi}{2}\right)$, we have
$$
\begin{array}{l} 
\frac{1}{r^{2}} \geqslant r^{-2} c^{\prime} \sin 2 \theta+r^{-1} b^{\prime} \sin \theta \\
\geqslant 2 r^{-\frac{3}{2}} \sqrt{b^{\prime} c^{\prime} \sin 2 \theta \cdot \sin \theta} \\
\Rightarrow|b c|=b^{\prime} c^{\prime} \leqslant \frac{1}{4 r \sin 2 \theta \cdot \sin \theta} \\
\left(\text { for any } r \leqslant 1, \theta \in\left(0, \frac{\pi}{2}\right)\right) \\
\Rightarrow|b c| \leqslant \min _{r \leqslant 1, \theta \in\left(0, \frac{\pi}{2}\right)} \frac{1}{4 r \sin 2 \theta \cdot \sin \theta} \\
=\min _{\theta \in\left(0, \frac{\pi}{2}\right)} \frac{1}{4 \sin 2 \theta \cdot \sin \theta} \\
=\frac{1}{4 \max _{\theta \in\left(0, \frac{\pi}{2}\right)} \sin 2 \theta \cdot \sin \theta}=\frac{3 \sqrt{3}}{16} .
\end{array}
$$
(for any $r \leqslant 1, \theta \in\left(0, \frac{\pi}{2}\right)$ )
An example where $|b c|=\frac{3 \sqrt{3}}{16}$:
$$
\begin{array}{l}
f(z)=\frac{\sqrt{2}}{8} z^{2}-\frac{\sqrt{6}}{4} z-\frac{3 \sqrt{2}}{8} . \\
\text { For } z=r \mathrm{e}^{\mathrm{i} \theta}(r \leqslant 1), \text { we have } \\
\left|f\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right|^{2} \\
=\frac{1}{32}\left[\left(r^{2} \cos 2 \theta-2 \sqrt{3} r \cos \theta-3\right)^{2}+\right. \\
\left.\quad\left(r^{2} \sin 2 \theta-2 \sqrt{3} r \sin \theta\right)^{2}\right] \\
=\frac{1}{32}\left[2 r^{4}+12 r^{2}+18-\left(2 \sqrt{3} r \cos \theta+r^{2}-3\right)^{2}\right] \\
\leqslant \frac{1}{32}\left(2 r^{4}+12 r^{2}+18\right) \leqslant 1 .
\end{array}
$$